A THEORY OF TIME AND SPACE CAMBRIDGE UNIVERSITY PRESS C. F. CLAY, MANAGEB 3LontJon: FETTER LANE, E.G. lEUmburgf) : 100 PRINCES STREET $cto Sorft: G. P. PUTNAM'S SONS anto Calcutta: MACMILLAN AND CO., LTD. Toronto: J. M. DENT AND SONS, LTD. Kofcgo: THE MARUZEN-KABUSHIKI-KAISHA All rights reserved A THEORY OF TIME AND SPACE BY ALFRED A. ROBB, M.A Cambridge : at the University Press 1914 Cum PREFACE introduction to the present volume was first published in 1913*, in substantially the form here reproduced. The plan there outlined is here carried out in detail, and, although much remains to be done along similar lines, yet the present work is fairly complete in itself. The special object here aimed at has been to show that spacial relations may be analyzed in terms of the time relations of before and after, and the demonstration of this thesis has been carried out up to the stage of introducing coordinates, leaving further developments for a future volume. The present work is the outcome of an endeavour to get rid of certain obscurities in connection with some of the fundamental parts of Physical Science. Thus the meaning ordinarily attached to the equality of time intervals and of lengths, although sufficiently precise for the ordinary purposes of daily life, will yet be found to be somewhat vague when examined more closely. Again, although the idea of acceleration plays such an important part in Dynamics, yet, so far as the writer is aware, no satisfactory definition has hitherto been given of what we mean when we say that a particle is " unaccelerated." All attempts in this direction tacitly assumed that we had already at our disposal some unaccelerated body or system to which the motions of other bodies or systems might be referred. * A Theory of Time and Space, Heffer and Sons, Cambridge, 1913. VI . PREFACE The present work aims at giving precision to ideas of this kind by reducing them to more fundamental concepts. The method here pursued leads to the construction of a system of Geometry having many curious characteristics, and of which ordinary Euclidean Geometry forms a part. Although the use of figures is of great assistance in this as in other types of Geometry, yet it is to be remembered that they are here merely aids to the imagination, and it has been deemed advisable, in the majority of cases, to leave the construction of them to the reader. In conclusion I desire to express my best thanks to Professor Sir Joseph Larmor, whose kindly encouragement, given from time to time, has done much to counteract those occasional feelings of dis- couragement which are perhaps inseparable from work of this sort. I further desire to express my thanks to the officials of the University Press for their never-failing courtesy and for the care and trouble which they have taken in the production of this book. ALFRED A. ROBB. CAMBRIDGE, September 19, 1914. A THEORY OF TIME. AND SPACE INTEODUCTION In the following pages the writer proposes to give an account of an investigation of the relations of Time and Space in connection with the physical phenomena of Optics. The subject is thus in part philosophical, in part mathematical, and in part physical. Under the name of "The Theory of Relativity" this subject has been much under discussion, but it is still in a condition of consider- able obscurity. Although generally associated with the names of Einstein and Minkowski, the really essential physical considerations underlying the theories are due to Larmor and Lorentz. According to the Newtonian mechanics there is no perceptible dis- tinction between a system of bodies " at rest " and one moving with uniform velocity in a straight line. The velocity is only apparent when two distinct bodies are compared, and, so far as the mathematics is concerned, it is a matter of indifference which of the two be re- garded as "at rest." It remained to be considered whether the phenomena of Optics and Electricity might not afford some means of distinguishing between the two bodies in this respect, but experiment failed to show any. It was then shown by Larmor that the electromagnetic equations could, by a linear substitution, be made to assume the same form when taken with respect to a system moving with uniform velocity as they had when taken with respect to a system " at rest," and similar results were arrived at by Lorentz. This seemed to indicate that, even if such a thing as " absolute rest " did exist, we should not be able in this way to distinguish it from motion with a uniform velocity. The question thus naturally arose as to whether any real dis- tinction existed; since philosophers had long contended that all motion was relative. OF TIME AND SPACE ie- sul^ject wgHj rendered more complex and difficult to grasp by the circumstance that* in "passing from a system "at rest" to one moving with uniform velocity, a "local time" had to be introduced, and further, bodies appeared to contract in the direction of their motion. The question of " local time " appeared the greatest obstacle to an acceptance of the view that velocity is merely a relative phenomenon. We are all familiar with 4}he use of " local time " at different places on the earth's surface, but the two cases are not analogous. Although noon in Greenwich and noon in New York are both represented as twelve o'clock local time, yet no one would contend that noon in Greenwich is at the same instant as noon in New York. In the case, however, of two material systems which were moving with uniform velocity relative to one another, events which were regarded as " simultaneous " in the one system according to the one "local time" could not, in general, be regarded as "simultaneous" according to the " local time " of the other system. If one of these "local times" were taken as the true time and the other regarded as a mathematical fiction, the two systems could not be considered as exactly on a par with one another, and the motion could not be a purely relative phenomenon. Thus on the one hand, the mathematics seemed to suggest that either of the systems might be regarded as "at rest," while con- siderations as to the " simultaneousness " of events, on the other hand, appeared to introduce dissymmetry. It was in order to preserve symmetry that Einstein made the suggestion that events might be simultaneous to one observer, but not simultaneous to another. This remarkable suggestion was at once seized upon, without it apparently being noticed that it struck at the very foundations of Logic. That "a thing cannot both be and not be at the same time" has long been accepted as one of the first principles of reasoning, but here it appeared for the first time in science to be definitely laid aside, and although many of those who accepted Einstein's view saw that there was something which was psychologically very strange about it, yet this was allowed to pass in view of the beauty and symmetry which seemed, in this way, to be brought about in the mutual relations of material systems. To others, however, this view of Einstein's appeared too difficult to grasp or analyse, and to this group the writer must confess to belong. Much of the subsequent development, such as that of Minkowski, has been of a purely analytical character, while the philosophical A THEORY OF TIME AND SPACE . 3 difficulty seems to remain much in the same state as it was left by Einstein. In 1911, the writer published a short tract entitled "Optical Geometry of Motion, a New View of the Theory of Relativity*," in which was put forward an outline of a method of treatment in which he avoided any attempt to identify instants of time at different places. The view was advanced that the axioms of Geometry might be regarded mostly as the formal expression of certain optical facts. As was re- marked in the preface to this tract, there are more aspects of things than one, and the fact that we give a physical signification to axioms of geometry, by no means implies that we do not regard their logical aspect as of equal importance. As regards the physical significance, there is, of course, also more than one ; but it appears desirable that we should give the axioms, so far as is possible, an optical significance primarily, rather than have the significance of some of them an optical one, while that of others is made to depend upon the properties of so-called rigid bodies, &c. This is particularly desirable because we have to consider what appears to be a contraction in the line of motion of a body which moves relatively to another, and the question naturally arises how is such contraction to be measured ? How, too, is velocity to be measured, since velocity depends upon distance as well as time ? The whole subject is one which can bristle with circular definitions, which, in fact, are very difficult to escape, since ideas which we have been in the habit of regarding as funda- mental are here called in question; and the only way out of the difficulty appears to be that we should make a careful re-examination of these apparently fundamental ideas so as to analyse those which are really complex into their separate components. It was with this object in view that the work here briefly sketched was undertaken. In the above-mentioned tract this analysis was made to a certain extent; but a number of logical details were omitted, and, since its publication, the author has devised a new method of approaching the problem, which illuminates certain points that were formerly obscure. This method involves an idea which is believed to be new, and which may be shortly described as the idea of Conical Order. This idea we shall now proceed to present in a general way and point out its connection with our problem. The systematic working out of the idea by means of a series of * Heffer and Sons, Cambridge, 1911. 12 4 A THEORY OF TIME AND SPACE postulates and definitions will form the main subject of a detailed discussion, and we are at present only concerned with an attempt to convey the general point of view. An element of time is called an instant and is to be regarded as a fundamental concept. Of any two elements of time of which I am directly conscious one is after the other. The relation of two instants, one of which is after the other, is an asymmetrical relation, and the converse asymmetrical relation is denoted by the term before, so that if an instant A be after an instant B, the instant B is before the instant A. The set of instants of which I am directly conscious form a series in linear order. Thus they satisfy the following conditions : (1) If an instant A be after an instant B, the instant B is not after the instant A, and is said to be before it. (2) If A be any instant, I can conceive of an instant which is after A and also of one which is before A. (3) If an instant A be after an instant B, I can conceive of an instant which is both after B and before A. (4) If an instant B be after an instant A and an instant G be after the instant B, the instant G is after the instant A. (5) If an instant A be neither before nor after an instant B, the instants A and B are identical. If, now, we examine the fifth of these conditions, it might perhaps be thought that it was a necessary consequence of our conceptions of before and after. That it is in reality no logical consequence of the other conditions may be shown by the help of a geometrical illustration. This illus- tration is suggestive, but the development of our theory is in no logical sense dependent upon it. Let us consider a system of cones having their axes parallel and having equal vertical angles. Let us regard any cone of the set as terminating in the vertex and as having the opening pointed upwards, let us say. We may call such a cone an a cone, and one with the opening pointed downwards a ft cone. The vertex in either case is to be regarded as belonging to the respective cones. Thus, corresponding to any point of space, there is an a cone of the A THEORY OF TIME AND SPACE set having that point as vertex, and similarly there is also a /3 cone of the set having the point as vertex. Now it is possible, by using such cones and by making a certain convention with respect to the use of the words before and after, to set up a type of order of the points of space. If J.! be any point and x the corresponding a cone, then we shall say that any point A 2 is after A 1} provided that the points are distinct and A 2 lies either on or inside the cone o^. Similarly, if A l be any point and ft the corresponding /3 cone, then we shall say that any point A 2 is before A ly provided that the two points are distinct and A 2 lies either on or inside the cone ft. It is easy to see that : (1) If a point A be after a point B, the point B is not after the point A, and is said to be before it. (2) If A be any point, there is a point which is after A and also one which is before A. (3) If a point A be after a point B, there is a point which is both after B and before A. (4) If a point B be after a point A and a point C be after the point B, the point C is after the point A. We cannot, however, say in this case that, if a point A be neither before nor after a point B, the points A and B must be identical. We have here, in fact, that not only is A neither before nor after A, but also any element B lying in the region outside both the a and ft cones of A is neither before nor after A. (See figure.) Thus the fact that an element is neither before nor after an element is not sufficient in this case to enable us to say that the elements are identical. It is to be observed, however, that if an element A be neither before nor after another element B, there are elements which are after both, and also elements which are before both ; since the a cones of A and B intersect in this case, as do also the ft cones of A and B. We shall speak of the type of order thus obtained as Conical Order, but shall not confine ourselves in the use of the name to the geometrical Fig. .1. 6 A THEORY OF TIME AND SPACE example considered above, or to the number of dimensions in which the example is presented. The main point to be considered at present is that we may have two elements, of -which one is neither before nor after the other, but which yet are not identical, without our being involved in any logical absurdity. Let us now return to the consideration of elements of time. We recognise events as taking place in time and as having a time order. We further recognise certain events as of an instantaneous character ; such, for example, as two particles striking one another. In speaking of events hereafter, unless otherwise stated, we shall refer to instantaneous events. Two events may occur at the same instant ; thus a particle P may strike a particle R at the same instant as another particle Q strikes it. Events which occur at the same instant will be said to be simul- taneous. According to the view here put forward, the only events which are really simultaneous are events which occur at the same place. The standpoint is practically this : that although the set of instants of which any one individual is directly conscious, or the set of instants which a single particle of matter occupies, are sets in linear orcler, yet the aggregate of all instants forms a set in conical order. Thus, of any two events A and B, we may have A before B or A after B, or A neither before nor after B. According to the view generally held, A being neither before nor after B is taken as equivalent to A and B being simultaneous. Accord- ing to the view here adopted, this is only so when the events A and B occur at the same place. If such events occur at different places, we are only entitled to say that the one is neither before nor after the other. The standpoint is rendered fairly clear by a consideration of the geometrical type of conical order which we have explained. If we consider, for instance, the case of a straight line which makes an angle with the axes of the cones not greater than their semi- vertical angle, it is evident that of any two distinct elements of such a line one must be after the other ; and so, if an element A be neither before nor after an element B, the elements A and B must be identical. Thus, although the whole aggregate of elements are in conical order, it is possible to select certain sets of elements of the aggregate A THEORY OF TIME AND SPACE . 7 which have a linear order analogous to that possessed by the instants of time, of which any one individual is directly conscious. Having thus given a brief outline of the idea of conical order as applied to instants of time, it is desirable to give some further justi- fication for our adopting it. Let us consider more closely what we practically mean when we say that one instant is after another, or one event after another. One fact is clear : If an instant B be distinct from an instant A, and if I, at the instant A, can produce any effect, however slight, at the instant J9, then this is sufficient to imply that B is after A. Now our contention is that we have here not merely a sufficient but also a necessary condition that B is after A. Thus we have the following definition : If an instant B be distinct from an instant A, then B will be said to be after A, if, and only if, it be abstractly possible for a person, at the instant A, to produce an effect at the instant B. If this be granted, then it follows that if A and B be distinct instants, and if I, at the instant A, cannot produce any effect at the instant B, then B is not after A. It does not, however, follow that B is before A, unless a person, at the instant B, can produce an effect at the instant A, since before and after are converse relations. Thus the significance of an instant A being neither before nor after a distinct instant B is that I, at the instant A, should be unable to produce any effect at the instant B, and that another person, at the instant J5, should be unable to produce any effect at the instant A. The question arises : are there any grounds for thinking that instants can stand in , such a relation ? The answer appears to be in the affirmative. Let us consider a flash of light or other electromagnetic disturbance to be sent out from a particle P at an instant A to a separate particle Q not in contact with P, and let the flash or disturbance be reflected back again directly to P. Now, Fizeau's apparatus for determining what we usually speak of as the " Velocity of light " is an arrangement in which this is practically carried out, and it indicates that if the flash returns to P at an instant, say C, then G is after A. If it be granted that the flash sent out from P arrives at Q at a 8 A THEORY OF TIME AND SPACE definite instant, say B, which is distinct from A and C, it follows from the meaning of after that B is after A and C is after B. Now, there are strong reasons for thinking that no influence or material particle could be sent out from P at the instant A so as to arrive at Q at any instant before B] and, similarly, there are strong reasons for thinking that no influence or material particle could be sent out from P at any instant after A so as to arrive at Q at the instant B. If these be granted, we see that any instant at the particle P, which is after A, and before C, will, according to our view of the matter, be neither before nor after B. Thus, if we consider Fizeaus arrangement, any instant at the sending apparatus, which is after the instant of departure of a flash of light and before the instant of its return, is neither before nor after the instant of arrival of the flash at the distant reflector. As a matter of fact, we have no means of identifying any particular instant at the sending apparatus after the instant of departure and before the instant of return with the instant of arrival at the reflector. , Einstein attempts to identify the time of arrival with that midway between the times of departure and return, but we have already pointed out the logical difficulty in which this involves us. Further, Einstein, in order to determine the instant midway between the instants of departure and return, postulates the existence of a clock. It does not appear a satisfactory mode of procedure to found a philosophical theory upon a complicated mechanism like a clock without any precise definition of what constitutes equal intervals of time. According to the theory here put forward, we avoid both these difficulties and base the logical superstructure upon the ideas of before and after, giving to them the philosophical and physical meanings above described. Thus, instead of starting from ordinary geometric cones with a definite angle and giving thereby an interpretation to before and after, it is proposed to reverse this process, and, starting from the ideas of before and after, to formulate in terms of them a system of postulates and definitions, and thereby build up a system of geometry. Instead of speaking of a and /5 cones, we shall speak of a and ft sub-sets, but shall find it often exceedingly suggestive to picture these to our minds by cones. The physical interpretation is this : If a flash of light or other instantaneous electromagnetic disturbance be sent out, let us say from a particle P at the instant A, so as to arrive directly at another particle Q at the instant B, then the instant B A THEORY OF TIME AND SPACE 9 lies in the a sub-set of the instant A, while the instant A lies in the ft sub-set of the instant B. The system of geometry thus built up will ultimately assume a sort of four dimensional character, or rather, we should say, any element of it is determined by four coordinates. It thus appears that the theory of space becomes absorbed in the theory of time, spacial relations being regarded as the manifestation of the fact that the elements of time form a system in conical order : a conception which may be analysed in terms of the relations of after and before. If I am directly conscious of an instant A, and if .B be a distinct instant which is neither before nor after A, then B is an instant of which I am only indirectly aware, and so it assumes an external character. We say that it is an instant elsewhere, and we can thus see in a general way how time relations and space relations can yet be both relations of one continuum. We shall now proceed with the formal development of the theory briefly sketched above. It will be observed that the postulates given generally consist of two parts marked (a) and (6) in which the relations of after and before are interchanged. In some however, such as those numbered I, III and IV, the one part follows as a direct consequence of the mutual relations of after and before, while in others, such as V, these relations are involved symmetrically. CONICAL ORDER We shall suppose that we have a set of elements and that certain of these elements stand in a relation to certain other elements of the set which we denote by saying that one element is after another. We shall further assume the following conditions : POSTULATE I. If an element B be after an element A, then the element A is not after the element B. Definition. If an element B be after an element A, then the element A will be said to be before the element B. POSTULATE II. (a) If A be any element, there is at least one element which is after A. (b) If A be any element there is at least one element which is before A. POSTULATE III. If an element B be after an element A, and if an element C be after the element B, the element C is after the element A. POSTULATE IV. If an element B be after an element A, there is at least one element which is both after A and before B. o POSTULATE V. If A be any element, there is at least one other element distinct from A which is neither before nor after A. *. POSTULATE VI. (a) If A and B be two distinct elements, one of which is neither before nor after the other, there is at least one element which is after both A and B, but is not after any other element which is after both A and B. (b) If A and B be two distinct elements, one of which is neither after nor before the other, there is at least one element which is before both A and B, but is not before any other element which is before both A and B. A THEORY OF TIME AND SPACE 11 Definition, (a) If A be any element of the set, then an element X will be said to be a member of the a sub-set of A provided X is either identical with A, or else provided there exists at least one element Y distinct from A and neither before nor after A and such that X is after both A and Y but is not after any other element which is after both A and Y. (b) If A be any element of the set, then an element X will be said to be a member of the (3 sub-set of A provided X is either identical with A, or else provided there exists at least one element Y distinct from A and neither after nor before A and such that X is before both A and Y but is not before any other element which is before both A and Y. If A be any element, then, by Post. V, there is at least one other element distinct from A which is neither before nor after A and so it follows directly by Post. VI (a) that there is at least one other element besides A which is a member of the a sub-set of A. Similarly, by Post. VI (b), there is at least one other element besides A which is a member of the y3 sub-set of A. Notation. We shall denote by a, and & the sub-sets corresponding to an element A lf and by 03 and /3 2 those corresponding to an element A, &c. POSTULATE VII. (a) If A l and A 2 be elements and if A 2 be a member of a lf then Aj is a member of j& 2 . (b) If A] and A 2 be elements and if A 2 be a member of jS lf then AJ is a member of a 2 . POSTULATE VIII. (a) If A l be any element and A 2 be any other element in a lf there is at least one other element distinct from A 2 which is a member both of a l and of a 2 . (b) If A x be any element and A 2 be any other element in /? 1? there is at least one other element distinct from A, which is a member both of fl 1 and of J& 2 . THEOREM 1. If A l be any element and A. 2 be any other element in a lt then any element A 3 which is both after A^ and before A 2 must be a member both of ! and #>. By the definition of a member of the sub-set ! there exists at least one element, say A 4 , distinct from A l and neither before nor after A l 12 A THEORY OF TIME AND SPACE and such that A z is after both A t and A 4 but is not after any other element which is after both A^ and A 4 . Fig. 2. Then A 4 cannot be after A 2 , for if it were then, by Post. Ill, A 4 would be after A t contrary to hypothesis. Further A 4 cannot be identical with A 3 , for then again we should have A 4 after A contrary to hypothesis. Again A 4 cannot be before A s , for then we should have A 2 after the element A 3 which would be after both A-^ and A 4 contrary to the hypothesis that A 2 is after both A and A 4 but not after any other element which is after both A l and A 4 . Thus At is distinct from A 3 and is neither before nor after A. Now A 2 cannot be after any other element which is after both A 3 and A 4 , for if A 5 were such an element it would follow by Post. Ill that since A s is after A-^ we should have A 5 after A^. Thus we should have A 2 after A 5 which would be after both A l and A 4 contrary to hypothesis. Thus no such element as A 5 can exist and so A 2 satisfies the defini- tion of being a member of 3 . Thus by Post. VII (a) it follows that A 3 is a member of ft. Again by Post. VII (a) since A 2 is a member of j it follows that A l is a member of ft, and so by a similar method we may prove that A s is a member of a 1 . Thus the theorem is proved. We may state the results of this theorem as follows : (a) If A l be any element and A 2 be any other distinct element in ! then A l is before A 2 , but is not before any other element outside the sub-set ^ and before A 2 . (b) If A l be any element and A 2 be any other distinct element in ft then A l is after A> 2 , but is not after any other element outside the sub-set and o3ter A 2 . A THEORY OF TIME AND SPACE 13 THEOREM 2. (a) If A 1 be any element and A z be any other element in ct lt there is at least one other element in ! distinct from A 2 which is neither before nor after A 2 . Since A z is a member of x it follows by Post. VII (a) that A^ is a member of /3 2 . Thus there exists at least one other element, say A s , distinct from A 2 and neither before nor after A 2 and such that A l is before A 2 and A 3) but is not before any other element which is before both A 2 and A s . Thus A l satisfies the definition of being a member both of j3 2 and /3 3 and so, by Post. VII (6), A 3 is also a member of a 1 . Thus since A 3 is distinct from A 2 and neither before nor after A 2) the theorem is proved. (b) If A! be any element and A 2 be any other element in ft, there is at least one other element in ft distinct from A 2 which is neither after nor before A 2 . Definition. If A l be any element and A 2 be any other element in !, the optical line A^A 2 is defined as the aggregate of all elements which lie either (1) both in ttj and 2 , or (2) both in j and j3 2) or (3) both in ft and ft. THEOREM 3. (a) If a be any optical line, there exists at least one element which is not an element of the optical line, but is before some element of it. If A^ be any element and A 2 be any other element in a x then, by Post. VII (a), A l is a member of ft. Thus by Theorem 2 (b) there is at least one other element in ft distinct from A l which is neither after nor before A lf Call such an element A 3 . Then since A s is in /3 2 and distinct from A 2 it is before A 2 . But A 3 cannot lie in the optical line A 1 A 2 , for by the definition of the optical line A l A 2 , in order to lie in it A 3 would require to lie also either in a x or '^. But if A s lay in j it would be either after A T or identical with A lt while if A 3 lay in & it would be either before A 1 or identical with A lt 14 A THEORY OF TIME AND SPACE But A 3 is distinct from A l and is neither after nor before A-^ and therefore does not lie in the optical line A^A^ although it is before A 2 an element of it. (6) // a be any optical line, there exists at least one element which is not an element of the optical line, but is after some element of it. POSTULATE IX. (a) If a be an optical line and if AJ be any element which is not in the optical line but before some element of it, there is one single element which is an element both of the optical line a and the sub-set a 1 . (6) If a be an optical line and if A x be any element which is not in the optical line but after some element of it, there is one single element which is an element both of the optical line a and the sub-set jS l . THEOREM 4. (a) If A! be any element there is at least one other element which is after A l but is not a member of the sub-set ^. Let A 2 be any other member of the sub-set ! distinct from A 1 . Then A 2 is after A and so by Post. IV there is at least one element, say A 3 , which is both after AI and before A 2 . By Theorem 1 A 3 is a member both of e^ and of /3 2 and is therefore an element of the optical line A t A 2 . But since A 9 is a member of -/8 a it follows that A 2 is a member of a s and so by Theorem 2 there is at least one other element in Qf 3 distinct from A z which is neither before nor after A 2 . Let A 4 be such an element. Then since A 4 is neither before nor after A 2 it cannot be a member either of /3 2 or 2 and so A is not an element of the optical line A 1 A 2 although it is after A 3 an element of it. A THEORY OF TIME AND SPACE 15 But since A is a member of 3 it follows by Post. VII (a) that A 3 is a member of the sub-set /3 4 . Thus A 3 is the one single element which by Post. IX (6) is an element both of the optical line and the sub-set /3 4 . But A cannot be a member of 15 for then AI would be a member of /3 4 and so A v would be a second element common to the optical line A 1 A 2 and the sub-set /3 4 , which is impossible by Post. IX (6). Further A 4 is after A 3 and A- 3 is after A l and therefore A 4 is after A lf Thus A 4 is after A l but is not a member of the sub-set a^. (b) If A l be any element there is at least one other element which is before A l bat is not a member of the sub-set {3 l . THEOREM 5. If A l be any element and A z be any other element which is after AI, there is at least one other distinct element which is a member of both ! and j3 z . Two cases arise : (1) A 2 may be a member of a x or (2) A z may not be a member of cr lt If A 2 is a member of j then by Post. IV there is at least one element which is both after A 1 and before A 2) and by Theorem 1 such an element is a member both of j and /3 2 . Thus case (1) is proved. Suppose next that A 2 is not a member of ^ and let A s be any element of er a distinct from A 2 . Then the optical line A 2 A 3) which for brevity we may call a, consists of the aggregate of all elements which lie either (1) both in oig and Cf 3 , or (2) both in or 2 an d @ 3 , or (3) both in /3 2 and j3 s . Since A 2 is not a member of otj it follows that A! is not a member of ft p. 4 . and so, since A l is before A 2 it follows that A! is not an element of the optical line a. Then by Post. IX (a) since A 1 is not an element of the optical line a but is before an element of it, it follows that there is one single element which is an element both of the optical line a and the sub-set a l . Let A 4 be this element. 16 A THEORY OF TIME AND SPACE Then since we have supposed that A 2 is not a member of ^ it follows that A 4 is not identical with A 2 . Further A 4 cannot be after A 2 for then we should have A 2 after A and before A 4 and so by Theorem 1 we should have A z a member of ^ contrary to hypothesis. Thus A cannot be a member of a 2 and therefore since it is an element of the optical line a it must be a member of /3 2 and f3 s . Thus the element A 4 is a member of both ofj and Q 2 and so the theorem is proved. THEOREM 6. (a) If A l be any element and A 2 be any other element in a,, while A 3 is an element distinct from A 2 , which is a member both of a r and of or 2 , then there is at least one other element which is a member of a ly of c* 2 and of ct 3 . By Post. VIII (a) since A 3 is an element of a 2 distinct from A 2 there is at least one other element distinct from A 3 which is a member both of 2 and of 3 . Call such an element A. Then since A 4 is in 3 and distinct from A 3 it is after A 3 . Thus A 4 is after an element of the optical line A 1 A 2 . But At is a member of a 2 and also of a s and so by Post. VII (a) A 2 and A 8 are each members of fa. Now if A 4 were not in the optical line A 1 A 2 it would follow by Post. IX (6) that there was one single element which was an element both of the optical line and the sub-set fa. There are however at least two elements A 2 and A s with this property and so A must be in the optical line A 1 A 2 . Also since A 4 is in 2 it must also be in otj from the definition of the optical line. Thus A is a member of a^ of 2 and of 3 . (6) If A l be any element and A^ be any other element in fa, while A s is an element distinct from A 2 , which is a member both of fa and of fa, then there is at least one other element which is a member of fa, of fa and of @ s . THEOREM 7. (a) If X be any element of an optical line there is at least one element of the optical line which is after X. Let the optical line be denned by any element A l and another element A 2 in a lP Then X may lie either (1) both in j and a a , A THEORY OF TIME AND SPACE 17 or (2) both in ctj and /3 2 , or (3) both in & and {3 2 . If X be not identical with A z then in cases (2) and (3) since X lies in /8 2 , the element A 2 is after X. If X be identical with A 2 then by Post. VIII (a) there is at least one other element distinct from A 2 which is a member both of ct l and of a, and is therefore an element of the optical line. Since such an element is not identical with A 2 it must be after A 2 ; that is to say it must be after X. Next suppose X is in both a^ and 2 and is distinct from A 2 . It follows by Theorem 6 (a) that there is at least- one other element which is a member of T and ot 2 and of the a sub-set of X. Since such an element is not identical with X and lies in the a sub- set of X it must be after X. Further since it is an element both of j and of a. 2 it lies in the optical line. Thus in all cases there is at least one element of the optical line which is after X. (b) If X be any element of an optical line there is at least one element of the optical line which is before X. THEOREM 8. (a) If A l be any element and A 2 be any other element in 15 and if A s and A 4 be other distinct elements which are members of both ^ and a 2 , one of the two elements A% and A 4 is in the a sub-set of the other. Since A s is in or 2 and distinct from A 2 therefore A 2 and A 9 define an optical line. Further since A 2 and A 9 both lie in a^ therefore A l lies in both /3 2 and /&,. Thus A! is an element of the optical line A 2 A 3 . But A 4 , since it is a member of a x and not identical with A l , is after A l . That is to say, it is after an element of the optical line A 2 A 3 . If then A 4 were not an element of the optical line A 2 A 3 there would, by Post. IX (b), be one single element which would be an element both of the optical line A 2 A 3 and the sub-set /3 4 . But A 4 is a member both of a x and of a 2 an d so both A l and A 2 are members of /3 4 . Thus since A and A 2 are two distinct elements of the optical line A 2 A 3 it follows that A 4 must be an element of the same optical line. R. 2 18 A THEORY OF TIME AND SPACE But A 4 is a member of 2 and therefore by the definition of the optical line A 4 must be either a member of a 3 or of ft. If A 4 be a member of ft then we should have A 3 a member of cr 4 . Thus one of the two elements A 3 and A 4 lies in the a sub-set of the other. It also follows, since A 3 and A 4 are supposed to be distinct, that the one is after the other. (b) If A! be any element and A* be any other element in ft, and if A 3 and A 4 be other distinct elements which are members of both ft and ft, one of the two elements A 3 and A 4 is in the @ sub-set of the other. It also follows, since A 3 and A 4 are supposed to be distinct, that the one is before the other. THEOREM 9. If a pair of elements be in an optical line defined by another pair of elements, then one of the first pair is in the a sub-set of the other. Consider the optical line defined by the element A l and another element A 2 in j. Suppose now in the first place that we have an element A 3 distinct from A l and A 2 and lying in the optical line. Then by the definition of an optical line A 3 may be (1) both in ! and a 2) or (2) both in j and ft, or (3) both in ft and ft. Thus if A and A 3 be taken as a pair of elements in the optical line defined by A l and A 2 , we have in the first and second cases A 3 is in a lt while in the third we have A 3 in ft and consequently A 1 in a 3 . Thus one of the pair A l , A 3 is in the a sub-set of the other. Again if A 2 and A 3 be taken as a pair of elements in the optical line defined by A and A 3t we have in the first case A 3 is in 2 , while in the second and third we have A 3 in ft and consequently A 2 in 3 . Thus one of the pair A 2) A 3 is in the a s,ub-set of the other. Next suppose that we have another element A 4 lying in the optical line and distinct from A 1} A 2 and A 3 . Then there are the following possibilities : I A 4, both in j and a 2 (1), or A+ both in ^ and ft (2), or J. 4 both in ft and ft (3). A THEORY OF TIME AND SPACE 19 IA 4 both in j and a z (4), or A 4 both in otj and ft (5), or A 4 both in ft and ft (6). ( A 4 both in j and 2 (7), J. 3 both in ft and ft with ^^d being distinct from A% it must be o/i^er A 2 an element of the optical line determined by A 4 and A z . 20 A THEORY OF TIME AND SPACE Also since A s is in both c^ and Og it follows that both A l and A. 2 lie in /3 3 . But by Post. IX (b), if A 3 were not in the optical line determined by A 4 and A 2 there would be one single element which would be an element both of the optical line and the sub-set ft. Thus since there are at least two distinct elements A l and A 2 common to the optical line and the sub-set /3 3 it follows that A 3 must be an element of the optical line A 4 A 2 . Further, since A s lies in 2 it must, by the definition of the optical line, lie also in a 4 . We may in a similar manner show in case (7) that A must lie in 3 . We are thus left with only case (5) to prove. Now since A 2 is an element distinct from A l and lying in ct lt therefore, by Post. VIII (a), there is at least one other element distinct from A. 2 which is a member both of j and of 2 . Call such an element A 5 . Now since A s is in /3 2 and distinct from A 2 it follows that A 2 is after A 3 , and since A 5 is in or 2 an d distinct from A 2 , therefore A 5 is after A 2 . . Thus by Post. Ill, A, is after A 3 . Further, since A 5 is an element of a lt therefore A l is an element of $5 and similarly A% is an element of /3 5 . But since A^ is an element of /3 5 it follows by Theorem 1 (6) that A 5 is after A l but is not after any element outside /3 5 which is after A^ But we have seen that A 6 is after A 3) which, being an element of ^ and not identical with A lt is after A lt Thus A 3 must be an element of the sub-set /3 5 . Similarly A 4 must be an element of the sub-set /3 5 . Also both A 3 and A 4 are elements of @ 2 and so by Theorem 8 (b) one of the two elements A 3 and A 4 is in the sub-set of the other, and therefore by Post. VII (b) the one is in the a sub-set of the other. Thus the theorem is true in all cases. It follows directly from this theorem that of any two distinct elements in an optical line one is after the other. THEOREM 10. Any two elements of an optical line determine that optical line. Let A! be any element and A 2 any other element in ct lt then the optical line A^A^ is defined as the aggregate of all elements which lie either (1) both in ttj and 2 , or (2) both in j and /3 2 , or (3) both in ft and /3 2 . A THEORY OF TIME AND SPACE 21 Suppose A 3 and A 4 to be any pair of elements in the optical line A^Ac,; then by Theorem 9 one of the pair A 3 , A 4 is in the a sub-set of the other. We may suppose without loss of generality that it is A 4 which is in the sub-set 3 . Consider now any element A 5 of the optical line A^A^ such that A s is distinct from A 3 and A 4 . Then by Theorem 9 there are the following possibilities : A 4 in 5 and also A 3 in a 5 (1), A in 5 and also A 5 in 3 (2), A 6 in 4 and also A 5 in a :5 (3), A 6 in a 4 and also A 3 in a 5 (4). Case (4) must however be excluded, for since A 3 , A* and A s are supposed distinct we should have A 5 after A 4 and A 3 after A 5 and therefore, by Post. Ill, A 3 after A 4 . We however supposed A 4 to be after A s and by Post. I we cannot have also A 3 after A 4 . Thus case (4) is impossible. The three permissible cases may be expressed thus : A 5 both in & and ft (1), AS both in oto and ft (2), A 5 both in 3 and 4 (3). Thus in all cases A s lies in the optical line denned by A 3 and A 4 . Similarly it may be shown that every element in the optical line denned by A 3 and A 4 lies in the optical line defined by A^ and A z . Thus the optical lines A-^A^ and A 3 A 4 are identical. THEOREM 11. If AX and A 4 be any two elements of an optical line A 1 A 2 there is at least one element of the optical line which is after the one and before the other. Since A 3 and A 4 are both elements of the same optical line the one must be in the a sub-set of the other by Theorem 9. We shall suppose that A 4 lies in 3 . Then since A 3 and A 4 are distinct, A 4 will be after A 3 , and so by Theorem 5 there is at least one other distinct element which is a member both of 3 and of ft. Call such an element A 5 . Then A 5 is in the optical line AA 4 , and therefore by Theorem 10 in the optical line A-iA*. 22 A THEORY OF TIME AND SPACE Further since A 5 is distinct from A 3 and A 4 it must be after A 3 and before A^. From the preceding results it follows that an optical line contains an infinite number of elements. THEOREM 12. If an element A l be before an element of an optical line a, and be also after an element of a, then A-^ must be itself an element of the optical line a. Suppose that A l is before the element A 2 of a and also after the element A 3 of a. Then by Post. I A s cannot be identical with A 2 , and by Theorem 9 one of the elements A 2 and A 3 must be in the a sub-set of the other. Since A l is after A 3 and A z is after A l it follows that A 2 is after A 3 and so it must be A 2 which is in the a sub-set of A 3 . But by Theorem 1 it follows that A l must lie in a, and also in /3 2 , and accordingly A t lies in the optical line A 3 A 2 . Thus since, by Theorem 10, any two elements of an optical line determine that optical line, it follows that A l lies in the optical line a. THEOREM 13. (a) If A l be any element and A 2 be any other element in j and if A s be any element in j which is either before or after A 2 , then A s lies in the optical line A^A*. (1) Suppose A 3 is before A 2 . Then since A 3 lies in otj it must be either identical with A l : in which case it lies in the optical line A^A^\ or else A 3 is after A^. in which case by Theorem 1 A s must lie both in a a and {3 2 and therefore must lie in the optical line A^A 2 . (2) Suppose A 3 is after A 2 . Then A 3 lies in ^ and A 2 is after A 1 and before A 3 and therefore, by Theorem 1, A 2 must lie both in j and j3 3 . But if A 2 lies in /3 3 , it follows by Post. VII (6) that A 3 lies in 2 . Thus A 3 lies both in a x and 2 and therefore lies in the optical line A A ^1 1 ^1 2 . (b) If A l be any element and A 2 be any other element in /3 1 and if A 3 be any element in & which is either after or before A 2 , then A 3 lies in the optical line A 1 A 2 . A THEORY OF TIME AND SPACE 23 THEOREM 14. (a) If A! be any element and A 2 be any other element in ^ and A 3 be any element in ^ distinct from A 2 which is neither before nor after A 2 , then A 3 is neither before nor after any element of the optical line A 1 A 2 which is after A^. The element A s cannot lie in the optical line A^A 2 , for then since it is distinct from A z it would be either before or after it, contrary to hypothesis. Now any element of the optical line A 1 A Z which is after A l must lie in fltj. Let A 4 by any such element. Then if A 3 were either before or after A 4 it would by Theorem 13 lie in the optical line A 1 A 4f which by Theorem 10 is identical with the optical line A 1 A 2 , and this we have shown to be impossible. Thus AS cannot be either before or after any element of the optical line A! A 2 which is after A lt (6) If AI be any element and A 2 be any other element in ft and A 3 be any element in ft distinct from A 2 which is neither after nor before A a , then A s is neither after nor before any element of the optical line A 2 A l which is before A lf POSTULATE X. (a) If a be an optical line and if A be any element not in the optical line but before some element of it, there is one single optical line containing A and such that each element of it is before an element of a. (b) If a be an optical line and if A be any element not in the optical line but after some element of it, there is one single optical line containing A and such that each element of it is after an element of a. THEOREM 15. (a) If each element of one optical line be before an element of another optical line the two optical lines cannot have an element in common. It is evident from Theorem 10 that two distinct optical lines cannot have more than one element in common. Consider now any two optical lines a and b having the element A l in common and let A^ be another element of a in cc l . Then by Theorem 13 if A 3 be any other distinct element in a l which is either before or after A z , then A s lies in the optical line A 1 A 2 : that is in a. 24 A THEOKY OF TIME AND SPACE Further if A s be after A- it cannot lie in b, for then the optical lines a and b would have the two elements A and A 3 in common and therefore would be identical. Thus A 2 cannot be before any element of b which is after A^ But A 2 is after A l and is therefore after any element of b which is before A^. Thus by Post. I A 2 cannot be before A l or any element of b which is before A l . It follows that A 2 is not before any element of the optical line b and therefore, if two optical lines a and b have an element in common, we cannot have each element of a before an element of 6. Thus conversely if each element of an optical line a be before an element of another optical line b the two optical lines cannot have an element in common. (b) If each element of one optical line be after an element of another optical line the two optical lines cannot have an element in common. THEOREM 16. (a) If each element of an optical line a be before an element of another optical line b, then through each element of a there is one single optical line which contains also an element of b. By Theorem 15 an element of a cannot also be an element of b. Suppose then that A l be any element of a. Then A l is not an element of b, but is before an element of 6 and therefore by Post. IX (a) there is one single element, say A 2 , which is an element both of the optical line b and the sub-set a^. Since A 2 cannot be identical with A- it follows that A l and A 2 determine an optical line which contains an element of a and also an element of b. Further there cannot be more than one optical line through A l which contains also an element of b ; for such an element of b must, by Theorem 9, lie either in j or ^. But by Post. IX (a) there is only one single element common to b and the sub-set a,, and so if such an element of b existed it would have to lie in ,. Call such a hypothetical element A s . Then since A s is supposed to lie in &, we should have A l in 3 . But A 2 lies in ^ and so A l lies in /3 2 , and thus if such an element as A s existed, A^ would lie in the optical line A S A 2 : that is, in the optical line 6, which is impossible, and so there cannot be any such element as A 3 . A THEORY OF TIME AND SPACE 25 Thus there is only one single optical line through A l which contains also an element of b. (b) If each element of an optical line a be after an element of another optical line b, then through each element of a there is one single optical line which contains also an element of b. Definition. If two distinct optical lines have an element in common they will be said to intersect one another in that element. If A l and A 2 be two distinct elements one of which is neither before nor after the other, then we know by Post. VI that there is at least one element, say X, which is after both A^ and A 2 , but is not after any other element which is after both A v and A 2 . From the definition of a sub-sets it follows that X lies both in ^ and 2, so that there is at least one element which is a member both of o^ and 2 . Similarly there is at least one element which is a member both of /9j and /3 2 . These remarks prepare the way for Postulate XI (a) and (b). POSTULATE XI. (a) If A 1 and A 2 be two distinct elements one of which is neither before nor after the other and X be an element which is a member both of a 1 and 2 , then there is at least one other element distinct from X which is a member both of OL and a 2 . (6) If AJ and A 2 be two distinct elements one of which is neither after nor before the other and X be an element which is a member both of jS l and fi 2 , then there is at least one other element distinct from X which is a member both of fi l and /? 2 . The above is the first of our postulates which requires more than two dimensions for its representation. It is to be noted that it may easily be combined with Postulate VI as follows : (a) If A and B be two distinct elements one of which is neither before nor after the other, there are at least two distinct elements either of which is after both A and B but is not after any other element which is after both A and B. (b) If A and B be two distinct elements one of which is neither after nor before the other, there are at least two distinct elements either of which is before both A and B but is not before any other element which is before both A and B. 26 A THEORY OF TIME AND SPACE THEOREM 17. (a) If A 1 and A 2 be any two distinct elements, one of which is neither before nor after the other, and if A s and A 4 be distinct elements which lie both in a z and o^, then one of these latter two elements is neither before nor after the other. By the definition of a sub-sets A 3 is after both A l and A 2 but is not after any other element which is after both A l and A 2 . Similarly A 4 is after both A l and A 2 but is not after any other element which is after both A-^ and A 2 . Thus A s is not after A 4 , and A 4 is not after A 2 . Thus A 3 is neither before nor after A 4 . (b) If A! and A 2 be any two distinct elements, one of which is neither after nor before the other, and if A 3 and A 4 be distinct elements which lie both in fa and fa, then one of these latter two elements is neither after nor before the other. THEOREM 18. (a) If A! be any element and A 2 and A s be two other distinct elements of a l} one of which is neither before nor after the other, there is at least one other distinct element in a x which is neither before nor after A 2 and neither before nor after J. 3 . Since A% is a member of a l , therefore A l is a member of 2 . Thus by Post. VIII (6) there is at least one other element distinct from A l which is a member both of fa and of fa. Call such an element A 5 . Then A l and A 2 are both members of 5 . Thus by Theorem 2 (a) there is at least one other element in a 5 distinct from AI which is neither before nor after A lf A THEORY OF TIME AND SPACE 27 Call such an element A e . Now A s cannot lie in 6 for then, as it is an element of a l5 it would lie in the optical line A 5 A^ along with A 2 and so A 2 and A 3 would either be identical or else A 2 would be either before or after A s , contrary to hypothesis. Now A 3 is after A l and A^ is after A 5 and so by Post. Ill A s is after A 5 , and since A 3 is not an element of 5 it cannot lie in the optical line A 5 A 6 . Thus by Post. IX (6) there is one single element (say A 7 } which is an element both of the optical line A 5 A 6 and the sub-set ft. Now A 5 cannot be after A 7) for A 7 lies in ft, and so by Theorem 1 (6) A s is after A 7 but is not after any element outside the sub-set ft which is after A 7 . But A 2 is after A 5 , and since A s does not lie in a 5 , therefore A 5 does not lie in ft. Thus A 5 is not after A 7 . Also A 5 cannot coincide with A 7 for then it would be in ft}. Thus A 7 must be after A 5 , and so by Theorem 14 A^ is neither before nor after A 7 . Now AS lies both in j and in cr 7 , and so by Post. XI (a) there is at least one other distinct element, say A 4 , which lies both in ^ and in 7 . Then by Theorem 17 A 4 is neither before nor after A s . Further A 4 cannot be either before or after A u , for since A 2 and A 4 are both members of ^ it would follow by Theorem 13 that A 4 lay in the optical line Aj^A^. This would also be the case if A 4 coincided with A z . But then (since A 4 is after A 1 and therefore after A 5 ) we should have A 4 in 5 and A l and A 7 both in a 5 and ft, and thus A^ and ^4 7 would lie in one optical line. Thus A l and A 7 would either coincide or else the one would be after the other, which is impossible. Thus A 4 is neither before nor after A 2 and is neither before nor after A s and is distinct from either. (b) If A! be any element and A z and A 3 be two other distinct elements f ft> ne f which is neither after nor before the other, there is at least one other distinct element in ft which is neither after nor before A 2 and neither after nor before A 3 . THEOREM 19. If A ! be any element there are at least three distinct optical lines containing A-. 28 A THEORY OF TIME AND SPACE Let A 2 be any element in a, distinct from A lm Then by Theorem 2 (a) there is at least one other element in otj distinct from A 2 which is neither before nor after A 2 . Call such an element A 3 . Further by Theorem 18 there is at least one other distinct element in ! which is neither before nor after A% and neither before nor after A 3 . Call such an element A 4 . Then A l and A 2 determine one optical line; A and A$ determine a second optical line ; A l and A 4 determine a third optical line. These are all distinct and all contain A l . If a be an optical line and if A be any element not in the optical line but before some element of it we have by Post. X (a) one single optical line containing A and such that each element of it is before an element of A. Further we have seen in Theorem 16 that there is one single optical line containing A and also intersecting a. Also by Theorem 19 there are at least three optical lines containing A and so there must be at least one optical line containing A in addition to the two particular ones which we have already mentioned. Similarly if a be an optical line and if A be any element not in the optical line but after some element of it, there is one single optical line containing A and such that each element of it is after an element of a and there is one single optical line containing A and intersecting a. In addition to these two particular optical lines Theorem 19 shows that there is at least one other optical line containing A. These considerations prepare the way for Postulate XII (a) and (6). POSTULATE XII. (a) If a be an optical line and if A be any element not in the optical line but before some element of it, then each optical line through A, except the one which inter- sects a and the one of which each element is before an element of a, has one single element which is neither before nor after any element of a. (//) If a be an optical line and if A be any element not in the optical line but after some element of it, then each optical line through A, except the one which intersects a and the one of which each element is after an element of a, has one single element which is neither after nor before any element of a. A THEORY OF TIME AND SPACE 29 THEOREM 20. (a) If each element of an optical line a be after an element of a distinct optical line b, tlien each element of b is before an element of a. Let A-L be any element of a ; then since A l is not in b but after an element of b, there is one single element (say J. 2 ) common to the optical line b and the sub-set ft (Post. IX (b)). Then A z is not an element of a but is before the element A^ of a and so by Post. X (a) there is one single optical line (say c) containing A 2 and such that each element of it is before an element of a. Now b cannot be identical with the optical line A 2 A 1} for then a and b would have the element A-^ in common, which is impossible by Theorem 15 (b). Suppose now, if possible, that b is not identical with c; then by Post. XII (a) there will be one single element in 6 (say A 3 ) which will be neither before nor after any element of a. Now A 3 cannot be before A 2 , for if so we should have A 2 after A 3 and A! after A 2 and therefore by Post. Ill A l after A 3 , contrary to the hypothesis that A 3 is neither before nor after any element of a. Also since A 2 is before A l we cannot have A s identical with A 2 . Suppose now that A 3 is after A. z and consider an element A 4 in b and after A 3 . Since there can only be one element in b which is neither before nor after any element of a, it follows that A must be either before or after some element of a. Since A 3 is before A^ it would follow, if A 4 were before an element of a, that A 3 was also before an element of a, contrary to hypothesis. We must therefore suppose that A 4 is after an element of a. Then since A 4 cannot be an element of a it would follow by Post. IX (6) that there was one single element which was an element both of the optical line a and the sub-set ft. Call such an element A 5 . Now since A 5 is an element of a it must be after some element of 6, and since by Theorem 15 (b) A 5 cannot be an element of b, there must by Post. IX (6) be one single element (say A 6 ) which is an element both of the optical line b and the sub-set ft. Then A 6 must be in a 6 . But we were led to the conclusion that A s lay in ft and since A 5 could not be both before and after the same element, A 6 would have to be distinct from A and both A 6 and A^ are supposed to be elements of the optical line b. 30 -A THEORY OF TIME AND SPACE Thus A 5 would have to lie in the optical line 6, which is impossible by Theorem 15 (b). Thus the supposition that b is distinct from c leads to a contradiction and therefore is not true. Thus b must be identical with c and so each element of b is before an element of a. (b) If each element of cm optical line a be before an element of a distinct optical line b, then each element of b is after an element of a. THEOREM 21. If a be an optical line and if A l be any element which is neither before nor after any element of a, there is one single optical line containing A l and such that no element of it is either before or after any element of a. Let A 2 be any selected element of a; then A l is neither before nor after A 2 , and so by Post. VI (b) an element exists which is a member both of & and of {3 2 . Call such an element A s . Now A 3 is before A 2 , an element of a, and does not lie in a and therefore by Post. X (a) there is one single optical line (say c) containing A 3 and such that each element of c is before an element of a. Further A l is after A 3 but is not before any element of a and so does not lie in c. Thus by Post. X (b) there is one single optical line (say 6) containing A l and such that each element of 6 is after an element of c. Consider now any element A 4 other than A l in the optical line 6 ; then J. 4 cannot be an element of a since otherwise A l would be either before or after an element of a, contrary to hypothesis. Suppose now if possible that A 4 is after some element of a. Then by Post. X (6) there is one single optical line (say d) containing A 4 and such that each element of d is after an element of a. But since each element of a is after an element of c therefore by Post. Ill each element of d is after an element of c. But by Post. X (b) there is only one single optical line containing A 4 which has this property and the optical line b is such a one. Thus the optical line d must be identical with the optical line b. Thus each element of b would be after an element of a, contrary to the hypothesis that A was neither before nor after any element of a. Thus A 4 is not after any element of a. Next suppose if possible that A 4 is before some element (say A 5 ) of a. A THEORY OF TIME AND SPACE 31 * Then A 5 is not an element of b, but is after an element of 6, and so by Post. X (6) there is one single optical line (say e) containing A s and such that each element of e is after an element of 6. But each element of 6 is after an element of c and so by Post. Ill each element of e is after an element of c. There is however by Post. X (6) only one single optical line con- taining A 5 and having this property, and a is such an optical line. Thus e must be identical with a and so each element of a must be after an element of 6. But if this were so then by Theorem 20 (a) each element of b must be before an element of a, contrary to the hypothesis that A^ is neither before nor after any element of a. Thus A 4 is not before any element of a, and so no element of b is either before or after any element of a. We have thus shown that there is one optical line containing A 1 and having this property. We have now to show that there is only one. Consider any optical line containing A l other than the optical lines b and A 8 Aj_. Call such an optical line/! Then by Post. XII (6) there is one single element in / (say A 6 ) such that A s is neither before nor after any element in c. If then we take any element A 7 in/ and before A 6 , such an element cannot be after any element in c, for then A 6 being after A 7 would be after an element of c, contrary to hypothesis. Also since there is only one element having the property of A 6 and lying in/, therefore A 7 must be before some element of c. But this element is before some element of a and so A 7 is before some element of a. Thus there is only one optical line containing A l and such that no element of it is either before or after any element of a. THEOREM 22. If a be an optical line and A^ be any element which is neither before nor after any element of a while b is the one single optical line containing A l and such that no element of it is either before or after any element of a, then every optical line through A ly ivith the exception of b, is divided by A! into elements which are before an element of a and elements which are after an element of a. We proved in Theorem 21 that there is only one optical line through A t having the property of 6. 32 "A THEORY OF TIME AND SPACE * Thus if we take any other optical line d through A 1 there must be at least one element of d which is either before or after some element of a. Suppose first that there is an element A 3 which is before some element of a. Then A 3 cannot be after A l} for since there is an element of a after AS there would by Post. Ill be an element of a after A l} contrary to hypothesis. Thus A 3 must be before A l . Further A s cannot be an element of a, for then A l would be after an element of a, contrary to hypothesis. Thus A s is not an element of a but before an element of it, and so by Post. IX (a) there is one single element (say A a ) which is an element both of the optical line a and the sub-set 3 . Further by Post. X (a) there is one single optical line (say c) con- taining A s and such that each element of it is before an element of a. Then by Post. XII (a) since the optical line d contains A s and is not identical with either of the optical lines A 3 A 2 or c it follows that there is one single element of d which is neither before nor after any element of a. But by hypothesis AI has this property and so every other element of d is either before or after an element of a. However, as we have already seen, an element which is after A l in d cannot be before an element of a and so it must be after an element of a. Similarly an element which is before AI in d cannot be after an element of a for then A l would be after an element of a contrary to hypothesis, and so an element which is before AI in d must be before an element of a. We arrive at the same conclusion if we start off by supposing the existence in d of an element A s ' which is after some element of a. Thus the theorem is proved. THEOREM 23. (a) If each element of each of two distinct optical lines a and b be after elements of a third optical line c, and if one element A} of the optical line b be after some element of the optical line a, then each element of b is after an element of a. Let &' be the one single optical line containing A l and such that each element of b' is after an element of a. A THEORY OF TIME AND SPACE 33 Then since each element of a is after an element of c therefore by Post. Ill each element of b' is after an element of c. But by hypothesis each element of b is after an element of c, and b contains A 1 an element not in the optical line c but after some element of it. Thus by Post. X (b), since there is only one single optical line containing A l and having this property, it follows that b' must be identical with b. Thus each element of b is after an element of a. (b) If each element of each of two distinct optical lines a and b be before elements of a third optical line c, and if one element A l of the optical line b be before some element of the optical line a, then each element of b is before an element of a. THEOREM 24. (a) If each element of each of two distinct optical lines a and b be after elements of a third optical line c, and if one element A l of the optical line b be neither before nor after any element of the optical line a, then no element of the optical line b is either before or after any element of the optical line a. Since A l is not an element of c but is after some element of it, therefore by Post. IX (6) there is one single element (say A s ) which is common to the optical line c and the sub-set &. Then since A 3 is not an element of a, but is before an element of a (Theorem 20 (a)), therefore by Post. IX (a) there is one single element (say A 2 ) which is common to the optical line a and the sub- set 3 . The demonstration then follows as in Theorem 21. (b) If each element of each of two distinct optical lines a and b be before elements of a third optical line c, and if one element A 1 of the optical line b be neither after nor before any element of the optical line a, then no element of the optical line b is either after or before any element of the optical line a. This may be demonstrated in an analogous manner. THEOREM 25. (a) If an optical line a be such that no element of it is either before or after any element of the optical line c, and if another optical line b be such that each element of it is before an element of c, then each element of b is before an element of a. R. 3 34 " A THEOKY OF TIME AND SPACE Since each element of b is before an element of c, it follows by Theorem 20 (b) that each element of c is after an element of b. Let A l be any element of c. Then since A l is not an element of b but is after an element of 6, there is one single element common to the optical line b and the sub-set A (Post. IX (6)). Let A 2 be this element. Then A 2 and A t determine an optical line. But by Theorem 22 every optical line containing A l except c is divided by A^ into elements which are before an element of a and elements which are after an element of a, and since A 2 is before A l and lies in the optical line AiA 2 , it follows that A 2 is also before an element of a and is not an element of a. Thus by Post. IX (a) there is one single element (say A s ) common to the optical line a and the sub-set a 2 . Now A 3 is neither before nor after any element of c and therefore if an optical line a be taken through A s such that each element of it is after an element of b, then by Theorem 24 (a) no element of a' is either before or after any element of c. But by Theorem 21 there is only one optical line through A 3 having this property and a is such an optical line. Thus a' is identical with a and so each element of a is after an element of b and thus by Theorem 20 (a) each element of b is before an element of a. (b) If an optical line a be such that no element of it is either after or before any element of the optical line c, and if another optical line b be such that each element of it is after an element of c, then each element of b is after an element of a. THEOREM 26. (a) If each element of an optical line a be after an element of a distinct optical line c, and each element of another optical line b be before an element of c, then each element of a is after an element of b. By Theorem 20 (b) each element of c is after an element of b and since each element of a is after an element of c, therefore by Post. Ill each element of a is after an element of b. (b) If each element of an optical line a be before an element of a distinct optical line c, and each element of another optical line b be after an element of c, then each element of a is before an element of b. A THEORY OF TIME AND SPACE 35 THEOREM 27. If two distinct optical lines a and b be such that no element of either of them is either before or after any element of a third optical line c, then no element of a is either before or after any element of b. For suppose, if possible, that some element AI of a is after an element of b ; then A l cannot lie in b and by Post. IX (b) there is one single element (say A 2 ) common to the optical line b and the sub-set ft. But by Theorem 22 every optical line through A l except a is divided by A l into elements which are before an element of c and elements which are after an element of c. Thus since A 2 and A l determine an optical line through A l} and since A 2 is before A 1} therefore A 2 must be before an element of c, contrary to the hypothesis that no element of b is either before or after any element of c. Similarly if we suppose A l to be before an element of b we are led to a conclusion contrary to hypothesis. Thus no element of a is either before or after any element of b. Definitions. An optical line a will be said to be parallel to a second distinct optical line b when either : (1) each element of a is after an element of 6, or (2) each element of a is before an element of b, or (3) no element of a is either before or after any element of b. In case (I) a will be said to be an after-parallel of b. In case (2) a will be said to be a before-parallel of b. In case (3) a will be said to be a neutral-parallel of b. It follows from these definitions in conjunction with Theorem 20 that: If an optical line a be parallel to an optical line b, then the optical line b is parallel to the optical line a. Again if a be any optical line and A be any element not in the optical line, A may be before an element of a, or may be after an element of a, but by Theorem 12 A cannot be before one element of a and after another element of a. By Post. XII (a) and (6) it follows that A may be neither before nor after any element of a. If A be before an element of a, then by Post. X (a) there is one single parallel to a containing A. 32 36 A THEORY OF TIME AND SPACE If A be after an element of a, then by Post. X (b) there is one single parallel to a containing A. If A be neither before nor after any element of a, then by Theorem 21 there is one single parallel to a containing A. Thus we can say in general : If a be any optical line and A be any element which is not iti the optical line, then there is one single optical line parallel to a and con- taining A. Further, combining Theorems 23 (a), 23(6), 24 (a), 24(6), 25 (a), 25 (6), 26 (a), 26 (6), 27, we have the general result that : If two distinct optical lines a and b are each parallel to a third optical line c, then the optical lines a and b are parallel one to another. Definition. If a and b be any pair, of distinct optical lines one of which is an after-parallel of the other, then the aggregate of all elements of all optical lines which intersect both a and 6 will be called an acceleration plane* . THEOREM 28. If a be an optical line there are an infinite number of distinct acceleration planes which all contain a. From Post. XII (a) and (b) it follows that there is at least one element, say A lt which is neither before nor after any element of a. If b be the one optical line through A l such that no element of it is either before or after any element of a, then by Theorem 22 every optical line through A 1 except b is divided by A l into elements which are before an element of a and elements which are after an element of a. Let / be one particular optical line containing A 1 and distinct from b. Let A 2 be any element in /other than A l ; then A 2 must be either before or after some element of a but is not itself an element of a. Thus if an optical line c be taken through A 2 parallel to a then c is either a before or after- parallel of a and therefore along with a serves to define an acceleration plane. Let A 3 be another element of/ distinct from A 2 . Then in order that A s should lie in the acceleration plane defined by a and c it would have to lie in an optical line intersecting both a and c. * The reason for adopting this name is that, as will be seen hereafter, any acceleration of a particle determines an acceleration plane. A THEORY OF TIME AND SPACE 37 But since A 9 is distinct from A 2 and lies in the optical line f which also contains A 2 it must be either before or after A 2 , and so by Post. IX (a) or Post. IX (b) there must be one single element which is an element both of the optical line c and the sub-set a s or /3 3 as the case may be. But the element A 2 is such an element and therefore the optical line / containing A 3 and A 2 is the only optical line which intersects c and contains A 3 . Thus in order that A 3 should lie in the acceleration plane defined by a and c it would be necessary for / to intersect a and this we know it does not do since if it did A l would be either before or after an element of a, contrary to hypothesis. If then A s be distinct from A^ it is either before or after an element of a and so if we take the optical line through A 3 parallel to a, it will be either a before or after-parallel of a. Call such an optical line d. Then d and a define another acceleration plane which is distinct from that defined by c and a, since the latter does not contain A 3 . If any other element A n in the optical line f be selected other than A 2 or A s and an optical line be taken through it parallel to a, then provided A n is distinct from A lt the parallel to a through A n will, along with a, define an acceleration plane distinct from the others. Thus each element of /except A l corresponds to a distinct accelera- tion plane and the number of elements in / is infinite, while all the acceleration planes contain a. Thus there are an infinite number of distinct acceleration planes all containing the optical line a. From the last theorem it follows directly that it is permissible to speak of three or more acceleration planes which have two elements in common. This prepares the way for Postulate XIII. POSTULATE XIII. If two distinct acceleration planes have two elements in common, then any other acceleration plane containing these two elements contains all elements common to the two first-mentioned acceleration planes. THEOREM $9. If a and b be two distinct optical lines and if a be an after-parallel of b, then if c and d be two other distinct optical lines intersecting both a and b, one of these latter two optical lines is an after-parallel of the other. 38 A THEORY OF TIME AND SPACE Let the optical line c intersect b in A 1 and a in A 2 and let the other optical line d intersect 6 in A 3 and a in A 4 . \e f f \ iJ Then, by Theorem 16 (a), it is not possible for A l and A s to be coincident while A 2 and A 4 are distinct; while, by Theorem 16 (b), it is not possible for A 2 and A 4 to be coincident while A l and A 3 are distinct. We may suppose without loss of generality that A s is after A l . Then since a is an after-parallel of b we must have A^ after A s and therefore by Post. Ill A 4 is after A l} or A l before A 4 . Further, since a is an after-parallel of 6, and since A l and A 2 lie in the optical line c, we must have A 2 after A and therefore A. 2 must lie in a l . Thus by Theorem 1 (a) AI is before A 2 but is not before any element outside the sub-set a a which is before A 2 and so, since J., is before A 4 , and there is only one element common to the optical line a and the sub-set !, it follows that J. 4 is not before A 2 . Thus since A 4 cannot coincide with A 2 and since J. 4 and A 2 both lie in the optical line a we must have A after A z and so A 4 lies in 2 . Now let e be the optical line through A s parallel to c ; then e is an after-parallel of c, since A 3 is a/ter ^L^ Again, there is one single optical line (say/") through A 2 intersecting e in some element, say A 5 , which lies in 2 . A THEORY OF TIME AND SPACE 39 Now since A a and A 3 are distinct elements both lying in a l} and since A 2 does not lie in the optical line A^^ it follows by Theorem 13 that A 2 is neither before nor after A 3 and therefore A 5 lies in 3 . Suppose now, if possible, that A 5 is distinct from J. 4 ; then by Theorem 17 (a) since A 4 and A s lie both in 0% and cr 3 , the one is neither before nor after the other. Thus 2 would be a member of 3 and so we should have the two elements A z and A s both common to the inertia line a and the sub-set a s , contrary to Post. XIV (a). Now since A lt A z and A 3 are distinct we must have A 2 after A l and A! after A 3 and therefore A 2 after A 3 . Further, since A 2 does not lie in 3 it follows that A 2 and A s are not in the same optical line. But since A 2 is in a x and A 3 in & it follows that A l and A 2 lie in an optical line through A l} and also A 3 and AI lie in an optical line through AI, and these optical lines are distinct. Now by Theorem 32 there are only two distinct optical lines in the acceleration plane which pass through A lt and so one of them must be A^A?. and the other A 3 A lt and since b must be identical with one of these optical lines, it follows that a and b must have one single element in common. THEOREM 37. Of any two distinct elements of an inertia line one is after the other. Let AI and A. 2 be any two distinct elements of the inertia line a, and let b be one of the two optical lines in an acceleration plane containing a which pass through A^ Now of the two optical lines in this acceleration plane which pass through A 2 , the one is parallel to b and the other intersects it in some element, say A 3 . Now A! and A 2 being distinct cannot both lie in a 3 by Post. XIV (a) and they Cannot both lie in ft s by Post. XIV (b). Thus one of the two elements A 1 and A 2 must lie in a 3 and the other in /3 3 , and so one of them must be after A 3 and the other before A s . Thus by Post. Ill one of the two elements A and A. 2 must be after the other. A THEORY OF TIME AND SPACE 49 From the definition of a separation line it contains a pair of elements one of which is neither before nor after the other. Thus it follows from the above theorem that no inertia line can be a separation line and no separation line can be an inertia line. THEOREM 38. If A i be any element in an inertia line a, there is at least one other element in the inertia line which is after AI and also at least one other element in it which is before AI. Let b be one of the two optical lines through A^ in any acceleration plane which contains a and let A 2 be any element in b which is after A lt Then by Post. XIV (a) there is one single element, say A 3 , common to the inertia line a and the sub-set or 2 . Then A 3 cannot be identical with A 2 since then we should have two elements common to the inertia line a and the optical line 6, contrary to Theorem 36. Thus A s is after A 2 and A 2 is after A 1 and therefore A 3 is after A l and is an element of the inertia line a. Similarly if we take any element A 4, in the optical line b and before A l there will by Post. XIV (b) be one single element, say A 5 , common to the inertia line a and the sub-set &. Then A l will be after A 4 and A 4 after A 5 and therefore A 1 after A 5 . Thus A s is before AI and is an element of the inertia line a. THEOREM 39. If A l and A 2 be any two distinct elements of an inertia line a, there is at least one other distinct element of a which is after one of the two elements and before the other. By Theorem 37 one of the two elements A l and A 2 is after the other. We shall suppose that A 2 is after A^. Let b and c be the two optical lines through A 1 in any acceleration plane containing a. Then the after-parallel to c through A 2 will intersect b in some element A 3 which must be after A 1 while the after-parallel to b through A z will intersect c in some element A 4 after A lf Now take any element A s in b after A r and before A 3 and take the optical line through A 5 parallel to c. This parallel through A 5 will intersect the optical line A 4 A 2 in some element A 6 . 4 50 A THEORY OF TIME AND SPACE Then A 5 A 6 is an after-parallel of A^A but a before-parallel of A S A 2 , and so A 6 is before A 2 . But by Theorem 36 there is one single element (say A 7 ) common to the inertia line a and the optical line A r) A s and this element must be distinct from A^ and J, 2 . Also by Post. XIV there is one single element common to the inertia line a and the sub-set a e and one single element common to the inertia line a and the sub-set # 5 and since A l is before A 5 and so lies in /3 5 , it follows that A 7 must lie in 5 . Similarly A 7 must lie in /3 6 . As A 7 cannot be identical with A 5 or A 6 and since it lies in a 5 and in /3 6 , it follows that A 7 is after A 5 and A 8 is after A 7 . Thus since A 5 is after A l} therefore by Post. Ill A 7 is after A 1} and since A 2 is after A 6 , therefore A 2 is after A 7 . Thus A 7 is after A 1 and before A 2 and lies in the inertia line a. It follows from the above results that there are an infinite number of elements in any inertia line. POSTULATE XV. If two general lines, one of which is a separation line and the other is not, lie in the same acceleration plane, then they have an element in common. Since there are an infinite number of optical lines in an acceleration plane, and since only two of them pass through any given element, and since by Post. XV each of them has an element in common with any separation line lying in the acceleration plane, it follows that there are an infinite number of elements in any separation line. Further, since as we have remarked in connection with Theorem 37 no inertia line can be a separation line, it follows that no element of a separation line is either before or after another element of it. THEOREM 40. If J.j and A 2 be two distinct elements one of which is neither before nor after the other, and if a and b be the two optical lines through A 1 in an acceleration plane containing the two elements, then A z is before an element of one of these optical lines and after an element of the other. By Theorem 33 A 2 must be either before or after an element of a and also must be either before or after an element of b ; but A 2 cannot lie either in a or b since it is distinct from A l and is neither before nor after it. Suppose first that A 2 is before an element of a. A THEORY OF TIME AND SPACE 51 Then one of the two optical lines through A 2 in the acceleration plane will intersect a in some element, say A 3 , while the other optical line through A 2 in the acceleration plane will intersect 6 in some element, say A 4 . Then A 2 must be before A s since A 2 cannot either lie in a or be after any element of it. But A 3 cannot either coincide with A 1 or be before A l} for then we should have A 2 before A ly contrary to hypothesis. Thus A s must be after A^ But A l is an element of 6 and so the optical line A 2 A 3 (which since it intersects a must be parallel to 6) must be an after-parallel of 6. Thus A 2 must be after an element of 6, and since A 2 must be either before or after A 4 , it follows that A 2 is after A 4 . In a similar manner we may prove that if A 2 be before an element of b it must be after an element of a. Also in an analogous manner we may show that if A 2 be after an element of b it must be before an element of a, and if A 2 be after an element of a it must be before an element of- 6. Thus A 2 must be before an element of one of the optical lines a and b and after an element of the other. Definition. An element in, an acceleration plane will be said to be between a pair of parallel optical lines in the acceleration plane if it be after an element of the one optical line and before an element of the other and does not lie in either optical line. THEOREM 41. If AT. and A 2 be any two distinct elements of a separation line, there is at least one other element of the separation line which lies between a pair, of parallel optical lines through A 1 and A 2 respectively in an acceleration plane containing the separation line. Let a l and 6j be the two optical lines passing through A 1 in any acceleration plane containing the separation line. Then, since A 2 is neither before nor after A ly it follows that A 2 is before an element of one of the two optical lines a a and 6 : and is after an element of the other. (Theorem 40.) Suppose that A 2 is before an element of a^. Then it is after an element of 6 lt Let a 2 and b 2 be the two optical lines through A 2 parallel respectively to aj and &!. Then a 2 and b 2 lie in the acceleration plane and since A 2 is before an element of a 1 therefore a 2 is a before-parallel of a^ 42 52 A THEORY OF TIME AND SPACE Similarly since A 2 is after an element of 6j it follows that b 2 is an after-parallel of 6j. Further b 2 must intersect a l in some element, say A 3 , which must be after A 2 since a 1 is an after-parallel of a 2 . Let A be any element of 6 2 which is after A 2 and before A 3 and consider the optical line through A parallel to c^. We shall call this optical line of. Then since A 4 is before A 3 it follows that a' is a before-parallel of a x and since A is a/fer .A 2 therefore a' is an after-parallel of a 2 . Also a/ lies in the acceleration plane. Thus by Post. XV a' must have an element in common with the separation line A 1 A 2 . Call this element A 5 . Then since a' is a before-parallel of a l therefore A 5 is before an element of a l and since a' is an after-parallel of a 2 therefore A 5 is after an element of a 2 . Thus AS is between the parallel optical lines a 1 and a 2 . THEOREM 42. //" ^Ij, A 2 a?ic A s be three elements in a separation line and if A 3 lies between a pair of parallel optical lines through A 1 and A 2 in an acceleration plane containing the separation line, then A 3 also lies between a second pair of parallel optical lines through A l and A 2 in the accelera- tion plane. Let a l and a 2 be a pair of parallel optical lines through A and A 2 respectively in the acceleration plane and suppose that A 3 lies between them. We may without loss of generality suppose that A s is after an element of a 2 and before an element of c^. Let 6 X be the second optical line which passes through A l in the acceleration plane and let 6 2 be the second optical line which passes through A 2 in the acceleration plane. Then, since a 1 and a 2 are parallel, 6j and b 2 are also parallel. But since A 3 and A l lie in a separation line, A 3 is neither before nor after A^ and since A 3 is before an element of c^ therefore, by Theorem 40, A s is after an element of 6^ Similarly A 3 is neither before nor after A 2 and, since A 3 is after an element of a 2t therefore, by Theorem 40, A 3 is before an element of 6 2 . Thus A 3 is between the parallel optical lines 6j and b 2 passing through A l and A 2 respectively in the acceleration plane. A THEORY OF TIME AND SPACE 53 Since there are only two optical lines in an acceleration plane which pass through a given element of it, it follows directly from the above theorem that if A lt A 2 and A 3 be three elements in a separation line and if A 3 lies between a pair of parallel optical lines through A l and A 2 in an acceleration plane containing the separation line, then A 2 does not lie between a pair of parallel optical lines through A v and A 3 in the acceleration plane. Similarly A 1 does not lie between a pair of parallel optical lines through A 2 and A 3 in the acceleration plane. THEOREM 43. If A l and A 2 be any two elements of a separation line, there is at least one other element of the separation line such that A 2 lies between a pair of parallel optical lines through A l and that element is an accelera- tion plane containing the separation line. Using the notation employed in Theorem 41 let us take any element, say A 6 , in the optical line b 2 and before A 2 and consider the optical line through A 6 parallel to a 2 . Call this optical line a". Then since A 6 is before A 2 therefore a" is a before-parallel of a 2 , and since a, 2 is a before-parallel of a therefore a" is also a before-parallel Of ttj. Further a" lies in the acceleration plane and so by Post. XV it has an element in common with the separation line. Call this element A 7 . Then A 2 is before A 3 an element of a^ and is after A 6 an element of a". Thus A 2 is between the parallel optical lines a l and a" passing through A l and the element A 7 respectively and lying in the acceleration plane. THEOREM 44. Of any three distinct elements of a separation line in a given accelera- tion plane there is one which lies between a pair of parallel optical lines through the other two and in the acceleration plane. Let A lt A 2 and A 3 be any three distinct elements in the separation line. Then since there are two optical lines in an acceleration plane passing through any element of it, let us select one of those passing through one of these elements, say A lt and the parallel optical lines through A 2 and A 3 . 54 A THEORY OF TIME AND SPACE Call these optical lines a l} a 2 and a s respectively. Then a lf a 2 and A 2 and A 3 being all elements of the optical line 6, and being all distinct, it follows that of any two of them one must be after the other. Thus remembering that Post. Ill must be satisfied it follows that either A 2 is after A^ and A 3 f after A 2 f (1) or A 2 is after A 3 and A,' after A 2 (2) or A 3 ' is after A-t and A 2 after A 3 (3)1 or A s ' is after A 2 and A-[ after A s ' (4)J or AS is after A 2 and A 3 after A-[ (5)1 or A i is after A 3 and A 2 after A^ (6)J In case (1) a 2 is an after-parallel of a l and a before-parallel of a s and so each element of o 2 is between the parallel optical lines a^ and a s . Thus A 2 is between a pair of parallel optical lines through A^ and A 3 in the acceleration plane. Similarly in case (2) a 2 is an after-parallel of a 3 and a before-parallel of a l and therefore again A 2 is between a pair of parallel optical lines through A l and A 3 in the acceleration plane. In a similar manner in cases (3) and (4) A 3 is between a pair of parallel optical lines through A and A 2 in the acceleration plane ; while in cases (5) and (6) A l is between a pair of parallel optical lines through A 2 and A 3 in the acceleration plane. Thus in all cases one of the three elements is between a pair of parallel optical lines through the other two and in the acceleration plane. THEOREM 45. If A be an element of an optical line a and if B be an element which is neither before nor after any element of a, then no element of the general line AB, with the exception of A, is either before or after any element of a. Let C be any element of the general line AB other than A, and let c be an optical line through C parallel to a. Suppose, if possible, that C is either before or after some element of a. A THEORY OF TIME AND SPACE 55 Then c would be either a before or after-parallel of a and accordingly c and a would be generators of an acceleration plane which would contain the two elements A and G of the general line AB and would therefore contain every element of AB. Thus the element B would lie in an acceleration plane containing the optical line a, and therefore, by Theorem 33, B would be either before or after an element of a, contrary to hypothesis. Thus the assumption that any element of the general line A B, other than A, is either before or after any element of a leads to a contradiction and therefore is not true and so no element of AB with the exception of A is either before or after any element of a. SETS OF THREE ELEMENTS WHICH DETERMINE ACCELERATION PLANES. Let A lt A 2 and A 3 be three distinct elements which do not all lie in one general line, then A^ and A 2 must lie in one general line, A 2 and A% in a second and A 3 and A l in a third. These three general lines need not however lie in one acceleration plane, although they do in certain cases. In these latter cases the three elements determine the acceleration plane containing them, since if they lay in two distinct acceleration planes they would lie in one general line, contrary to hypothesis. It is important to have criteria by which we can say that a set of three elements does lie in one acceleration plane. CASE I. Three elements A ly A 2 , A s lie in one acceleration plane if A-i and A 2 lie in an optical line while A 3 is an element not in the optical line but before some element of it, or after some element of it. This is clearly true, since, if A t and A 2 lie in the optical line a, while A 3 does not lie in a but is before some element of it, then there is a before-parallel optical line, say b, containing A 3 and so a and b are a pair of parallel generators of an acceleration plane, containing A lf A 2 and A s and which is determined by them. Similarly if A ? be after some element of a there is a definite after- parallel optical line b containing A 3 and so the two optical lines a and b are a pair of parallel generators of an acceleration plane containing A lt A 2 and A 3 and which is determined by them. CASE II. Three elements A lt A 2 , A s lie in one acceleration plane if A l and A 2 lie in an inertia line and A 3 be any element outside the inertia line. This can also be readily seen to hold since if a denote the inertia 56 A THEOKY OF TIME AND SPACE line containing A^ and A 2 then by Post. XIV (a) there is one single element, say A 4 , common to the inertia line a and the sub-set 3 , and by Post. XIV (b) there is one single element, say A B , common to the inertia line a and the sub-set /9 3 . Thus AS and A 4 lie in one optical line while A 3 and A 5 lie in another optical line. These two optical lines may be taken as generators of opposite sets of an acceleration plane containing A 3 , A 4 and A 5 . But since this acceleration plane contains the two elements A 4 and A 5 of the inertia line a, it must contain every element of a and therefore contains A l and A 2 . Thus the three elements A lf A 2 and A 3 lie in one acceleration plane which is determined by them. CASE III. Three elements A l9 A 2t A 3 lie in one acceleration plane if A l and A 2 lie in a separation line and if A 3 be an element not in the separation line but before at least two elements of it or after at least two elements of it. In order to show this let a be the separation line containing A^ and A z and suppose A 3 is before the elements A 4 and A 5 of a which are supposed distinct. Then A 3 and A 4 must lie either in an optical line or an inertia line since A 4 is after A 3) and similarly A 3 and A 5 must lie either in an optical line or an inertia line and the two general lines A 3 A 4 and A 3 A 5 are distinct. If A 3 A 4 and A 3 A 5 be both optical lines, then they may be taken as generators . of opposite sets of an acceleration plane containing A s , A 4 and A 5 . But this acceleration plane, since it contains the two distinct elements A 4 and A 5 of the separation line a, must contain every element of it and so must contain A l and A 2 . Thus A l} A 2 and A 3 lie in one acceleration plane which is determined by them. We shall suppose next that at least one of the general lines A 3 A 4 and A 3 A 5 is an inertia line. Let us say that A 3 A 4 is an inertia line. Then by Case II the three elements A 3 , A* and A 5 lie in one acceleration plane which is determined by them. But since this acceleration plane contains the two elements A 4 and A 5 of the separation line a, therefore it contains every element of a and so must contain A^ and A 2 . A THEORY OF TIME AND SPACE 57 Thus A lt A 2 and A 3 lie in one acceleration plane which is determined by them. The case when A 3 is after two distinct elements of a is quite analogous. If A! and A 2 lie in an optical line a while A 3 is an element which is neither before nor after any element of a, then the three elements do not lie in one acceleration plane, for by Theorem 45 no element of the general line A t A s with the exception of A l is either before or after any element of a. But if A lt A 2 and A a lay in an acceleration plane there would be two optical lines through A 2 in the acceleration plane and both of these would have an element in common with the separation line A^A*. Thus there would be at least two elements of A 1 A 3 which would be before or after A 2 , contrary to Theorem 45. Thus A lt A 2 and A s do not lie in one acceleration plane. If A 1 and A 2 lie in a separation line a, while A 3 is before one single element of a or after one single element of a, then the three elements A lt A 2 , AS cannot lie in one acceleration plane. This is easily seen, for if we suppose that they do all lie in one acceleration plane, there are two optical lines through A 9 in the acceleration plane which have each an element in common with a. If these elements be called A 4 and A 5 then, since a is a separation line, A 4 is neither before nor after A 5 and so A 4 and A 5 must be either both before or both after A 3 , contrary to the hypothesis that there is only one single element of a which A s is after or before. If A l and A 2 lie in a separation line a, while A 9 does not lie in a and is neither before nor after any element of a, it is also evident from the above considerations that the three elements A l} A 2 , A 3 cannot lie in one acceleration plane. We have not however as yet proved the possibility of this last case, but shall do so hereafter (Theorem 99). Till then no use will be made of it, and it is merely mentioned here for the sake of completeness. Definition. If an acceleration plane have its two sets of generators respectively parallel to the two sets of generators of another distinct acceleration plane, then the two acceleration planes will be said to be parallel to one another. It is clear that if P be an acceleration plane and A be any element outside it, then there is one single acceleration plane containing A and parallel to P ; for there is one single optical line through A parallel to 58 A THEORY OF TIME AND SPACE the one set of generators of P and one single optical line through A parallel to the other set of generators of P. These are generators of opposite sets of an acceleration plane containing A and determine that acceleration plane, which is therefore unique. It is further clear that two parallel acceleration planes can have no element in common, for if the element A lies outside the acceleration plane P and if a be an optical line passing through A and parallel to a generator of P, then a can have no element in common with P since otherwise it would require to lie entirely in P, contrary to the hypothesis that A is outside P. Similarly any optical line b which intersects a and is parallel to a generator of P of the opposite set can have no element in common with P. But if Q be the acceleration plane passing through A and parallel to P, every element of Q must lie in an optical line such as b and so P and Q can have no element in common. It is. also clear from the definition that two distinct acceleration planes which are parallel to the same acceleration plane are parallel to one another ; since distinct optical lines which are parallel to the same optical line are parallel to one another. THEOREM 46. If an acceleration plane P have one element in common with each of a pair of parallel acceleration planes Q and R then, if P have a second element in common with Q, it has also a second element in common with R. If P and Q have two elements in common they must have a general line in common which we may call a. Let B l be the element which by hypothesis P and R have in common. Then if a be an inertia or separation line it follows by Theorem 36 and Post. XV that both the optical lines through B l in the acceleration plane P have an element in common with a, while if a be an optical line one of the optical lines through B l in P has an element in common with a. Thus in all cases at least one of the optical lines through B l in the acceleration plane P has an element in common with a. Let A l be such an element. Suppose first that a is an optical line. Then a is one of the generators of Q and since the acceleration A THEORY OF TIME AND SPACE 59 plane R is parallel to Q and since B l lies in R there will be one of the generators of R passing through B^ and parallel to a. Since A l and B 1 lie in an optical line and are distinct, the one must be after the other and so this parallel to a through B 1 must be either a before or after-parallel of a. Let us denote it by b. Then a and b determine an acceleration plane which contains three distinct elements of P which are not all in one general line and so this acceleration plane must be identical with P. Thus since it contains the optical line 6 it follows that P has a second element in common with R. Suppose next that a is an inertia or separation line and let c be one of the generators of Q which pass through AI. Then since R is parallel to Q and since B lies in R there will be one of the generators of R passing through B 1 and parallel to c. Since A l and B l lie in an optical line and are distinct, the one must be after the other and so this parallel to c through B 1 must be a before or after-parallel. Let G be any element of c distinct from A l and let an optical line through G intersect the optical line through B l parallel to c in the element D. Then by Theorem 29 the optical line GD must be a before or after- parallel of the optical line A^B^ Let the second optical line through G in the acceleration plane Q meet a in the element A 2 . The element A 2 must exist since a is an inertia or separation line. Since the optical line GA 2 must be a generator of Q of the opposite set to c, there must be an optical line through D in the acceleration plane R which is parallel to CA 2 and is a generator of R of the opposite set to DB. Since C and D lie in an optical line and are distinct, the one must be after the other and so this parallel to CA 2 through D must be a before or after-parallel. Let an optical line through A 2 intersect the optical line through D parallel to CA 2 in the element B 2 . Then by Theorem 29 the optical line A 2 B 2 must be a before or after- parallel of GD and GD is a before or after-parallel of AiBi and so if A l B l and A 2 B 2 be distinct they must be parallel to one another. Now the optical lines GA l and GA 2 are distinct from the inertia or separation line a and are also distinct from one another. 60 A THEORY OF TIME AND SPACE Also the element G cannot lie in a since then CA j would have to be an inertia or separation line. Thus the elements A l and A 2 are distinct and since they lie in an inertia or separation line they cannot lie in one optical line. Thus A 2 B 2 is distinct from A^ and is therefore parallel to it. Also since the general line a and the optical line A l B l lie in the acceleration plane P and since the element A 2 does not lie in A l B l it follows by Theorem 33 that A 2 is either before or after some element of A&. Thus A 2 B 2 must be either a before or after-parallel of A l B l and so the optical lines A^ and A 2 B 2 lie in an acceleration plane containing the general line a and the element B lf This acceleration plane must therefore be identical with P and it contains the element B 2 in common with R where B 2 is distinct from B lf Thus the theorem holds in all cases. REMARKS. It follows directly from this theorem that if two distinct acceleration planes P and Q have a general line in common and, if further, P has one element in common with an acceleration plane R which is parallel to Q, then P and R have a general line in common. Further, since Q and R can have no element in common, it follows that these two general lines have no element in common. Again if Q and R be two parallel acceleration planes and if a be any general line in Q, then there is at least one acceleration plane containing a and another general line in R. This may be shown in the following way : Let A be any element of a and let /be any inertia line in R. Then by Post. XIV (a) there is one single element common to the inertia line /and the sub- set cfj. Let B be this element and let A 2 be any element of / which is after B. Then A 2 is after A^ but does not lie in a : and so A l and A 2 lie in an inertia line. Thus A 2 and a lie in an acceleration plane, say P, which by Theorem 46 must contain a second element in common with R. Thus P contains a and another general line in R. It is easy to see that there are really an infinite number of accelera- tion planes which have this property of P. We have seen that if two distinct optical lines intersect a pair of optical lines one of which is an after-parallel of the other, then of the A THEORY OF TIME AND SPACE 61 two first- mentioned optical lines one is an after-parallel of the other (Theorem 29). We have also seen that it is impossible for an optical line to intersect a pair of neutral-parallel optical lines. Thus we may state the following definition. Definition. If two distinct optical lines intersect a pair of optical lines one of which is an after-parallel of the other, then the four optical lines will be said to form an optical parallelogram. It is evident that an optical parallelogram must lie in an acceleration plane. The elements of intersection will be spoken of as the corners of the optical parallelogram. A pair of corners which lie in one optical line will be spoken of as adjacent. A pair of corners which do not lie in one optical line will be spoken of as opposite. A general line passing through a pair of opposite corners of an optical parallelogram will be spoken of as a diagonal line of the optical parallelogram. We make a distinction between two optical parallelograms having a diagonal line in common and having a diagonal in common. When we speak of two optical parallelograms having a diagonal line in common we shall mean that a pair of opposite corners of each of the optical parallelograms lie in the same general line. When, on the other hamj, we speak of two optical parallelograms having a diagonal in common, we mean that they have a pair of opposite corners in common. It is obvious that an optical parallelogram has two diagonal lines and it is easy to see that one of these must be an inertia line, and the other a separation line. For if we call the four optical lines a, b, c and d, and if a be an after- parallel of b while c is an after-parallel of d, then the intersection element of a and c must be after the intersection element of d and b so that these two intersection elements lie in an inertia line. Further if we denote the intersection element of a and c by A lt that of a and d by A 2 , that of c and b by A 3 and that of d and b by A, it follows by Theorem 13 (6) that if A 3 were either before or after A 2 then A s would have to lie in the optical line A 2 A l or a, contrary to hypothesis. 62 A THEORY OF TIME AND SPACE Thus A s is neither before nor after A 2 and so A 2 and A 3 lie in a separation line. Definition. If a general line a have orae single element in common with a general line b } then a will be said to intersect b. Since a general line does not intersect itself and since we may have two optical parallelograms in the same acceleration plane having a diagonal line in common, it is permissible to speak of two optical parallelograms in the same acceleration plane whose diagonal lines of one kind or the other do not intersect. This prepares the way for Postulate XVI. POSTULATE XVI. If two optical parallelograms lie in the same acceleration plane, then if their diagonal lines of one kind do not intersect, their diagonal lines of the other kind do not intersect. THEOREM 47. If a be any general line in an acceleration plane P and A l be any element of the acceleration plane which is not in the general line, then there is one single general line through A l in the acceleration plane which does not intersect a. Let Q be any other acceleration plane distinct from P and containing the general line a, and let R be an acceleration plane passing through J. x and parallel to Q. Then by Theorem 46 P and R will have a general line in common which can have no element in common with a, and so there is at least one general line through A in the acceleration plane P which does not intersect a. We must next show that there is only one such general line. Consider first the case where a is an optical line. Then of the two optical lines through AT. in the acceleration plane P we know that one is parallel to a while the other intersects it. Further by Theorem 36 any inertia line through A-^ in the accelera- tion plane P must intersect a. Also by Post. XV any separation line through A l in the acceleration plane P must intersect a. Thus if a be an optical line there is one single general line through A l in the acceleration plane P which does not intersect a. Consider next the cases where a is an inertia or a separation line. If a be an inertia line, then by Theorem 36 both the optical lines through A l in the acceleration plane P intersect a, while by Post. XV every separation line in P intersects A t . A THEORY OF TIME AND SPACE 63 Thus when a is an inertia line any general line through A l in the acceleration plane P which does not intersect a can only be an inertia line. Also from Post. XV it follows that when a is a separation line any general line through A l in the acceleration plane P which does not intersect a can only be a separation line. With these provisos the demonstration of the unique character of the non-intersecting general line is similar in the two cases. m Fig. 8. Suppose, if possible, that there are two general lines through A l in the acceleration plane, say i and j, which do not intersect a. Then i and j must intersect in A lt Let 6 and c be the two optical lines through A l in the acceleration plane and let them intersect a in A z and A 3 respectively. Let d be the second optical line through A 2 in the acceleration plane and let e be the second optical line through A s in the acceleration plane and let d and e intersect in A. Then the optical lines b, c, d and e form an optical parallelogram. 64 A THEORY OF TIME AND SPACE Let m be the diagonal line through A l and A 4 . Let the optical line d intersect i in A 5 and let the optical line e intersect i in A 6 . Let /be the second optical line through A 5 in the acceleration plane and let g be the second optical line through A 6 in the acceleration plane and let /and g intersect in A 7 . Then the optical lines / g, d and e form an optical parallelogram and the diagonal line i is of the same kind as the diagonal line a of the optical parallelogram formed by b, c, d and e. Thus since the diagonal lines a and i do not intersect it follows by Post. XVI that the diagonal lines of the other kind of the two optical parallelograms do not intersect. But the two optical parallelograms have the corner A^ in common and so they must have the diagonal line through A 4 in common. Thus A 7 must lie in m. Now suppose that the optical line d intersects j in A 8 and that the optical line e intersects j in A 9 . Let k be the second optical line through A s in the acceleration plane and let I be the second optical line through A 9 in the acceleration plane and let k and I intersect in A w . Then the optical lines k, I, d arid e form an optical parallelogram and since j is supposed not to intersect a it follows as before that A lo must lie in m. But now we have the optical parallelograms formed by/ g, d and e, and by k, I, d and e having the diagonal line m in common, and so, by Post. XVI, their other diagonal lines do not intersect, which is contrary to the hypothesis that i a,ndj intersected in A lf Thus the hypothesis that there are two general lines through A 1 in the acceleration plane which do not intersect a leads to a contradiction and therefore is not true. Thus there is in all cases one. single general line through A t in the acceleration plane which does not intersect a. THEOREM 48. If two acceleration planes P and Q have a general line a in common, and if A l be any element which does not lie either in P or Q, then the acceleration planes through A 1 parallel to P and Q respectively have a general line in common. Let R and 8 be the acceleration planes through Aj_ parallel to P and Q respectively. A THEORY OF TIME AND SPACE 65 Two possibilities are open : either (1) Q has one element at least in common with R, or (2) Q has no element in common with R. Consider first the case where Q has one element at least in common with R. Here since Q has two elements in common with P and since P and R are parallel, it follows by Theorem 46 that Q has a second element in common with R. Further, since Q and S are parallel and R has two elements in common with Q and has the element A l in common with S, it follows that R has a second element in common with S and therefore R and S have a general line, say c, in common. Next consider the case where Q has no element in common with R. This case has no analogue in ordinary three-dimensional Euclidean geometry, but must be considered in our system. We have seen that there is at least one acceleration plane con- taining a and another general line, say 6, in R since P and R are parallel. Let T be such an acceleration plane, let A 2 be any element in b and let U be the acceleration plane through A 2 parallel to Q. Then since Q and U are parallel and since T contains the general line a and also the element A 2 of U, it follows that T contains a general line, say b', in U. But the general lines b and b' both contain the element A* and neither of them can intersect a. Thus since b and b' both lie in one acceleration plane T, it follows by Theorem 47 that they must be identical, and so b must be common to U and R. Now the acceleration planes S and U are both parallel to Q and therefore must be either parallel to one another or else identical. If they are not identical, the acceleration plane R has the general line b in common with U and has the element A l in common with 8. Thus in either case R and 8 have a general line in common. If we consider case (2) of the last theorem it is clear that, if the general line a be an optical line, then since the general line 6 lies in the same acceleration plane T and has no element in common with a, it follows by Theorem 47 that b must also be an optical line and be parallel to a. If c be the general line common to R and 8 then, provided c and b are distinct, it follows in a similar manner that c is an optical line parallel to b and therefore also parallel- to a. R. 5 66 A THEORY OF TIME AND SPACE A similar result follows in case (1), and so we always have c parallel to a provided a be an optical line. Now we have as yet given no definition of the parallelism of any type of general lines except optical lines, but are now in a position to do so. Definition. If a be a general line and A be any element which does not lie in it and if two acceleration planes R and 8 through A are parallel respectively to two others P and Q containing a, then the general line which R and S have in common is said to be parallel to a. THEOREM 49. If a be a general line and A l be any element which does not lie in it, then there is one single general line containing AI and parallel to a. Two cases have to be considered. (1) The element A^ lies in an acceleration plane containing a. (2) The element AI does not lie in an acceleration plane con- taining a. Consider first case (1) and let T be the acceleration plane containing A 1 and a. Let P!, P 2 , P 3 , P 4 be any other acceleration planes containing a, and let Q lt Q 2 , Q 3 , ft b e acceleration planes through A l parallel to P 1; P 2 , P 3 , P 4 respectively. Then since the acceleration plane T has the general line a in common with P l and has the element A l in common with Q lf it follows that it has a general line, say b, in common with Q x and b does not intersect a. But, by Theorem 47, there is only one general line through A l in the acceleration plane T which does not intersect a and so 6 must be this general line. Similarly Q 2 , Q s , Q 4 must all contain the general line b in common with T and so any pair of the acceleration planes Q lt Q 2 , Q 3 , Q 4 have the same general line b in common. Thus b is independent of the particular pair of acceleration planes P!, P 2 , P 3 , P 4 which we may select and so there is only one general line through A! parallel to a. Suppose next that A 1 does not lie in an acceleration plane containing a and suppose that P lt P 2 , P 3 , P 4 are any acceleration planes which are distinct from one another and all contain a. Let Q lt Q 2 , Q 3 , Q 4 be acceleration planes through A l and parallel to P 1} P 2 , P 3 , P 4 respectively. A THEORY OF TIME AND SPACE 67 Let P n be an acceleration plane containing a and a general line b in ft. Then b is parallel to a and lies in the same acceleration plane P n with it. If then we take acceleration planes ft', ft', ft' through any element of b and parallel to P 2 , P 3 , P 4 respectively, these will all contain b and will also be respectively parallel to ft, ft, ft which contain the element A lf But the general line b and the element A 1 lie in the acceleration plane ft and so, by case (1), ft, ft, ft all have the same general line, say c, in common with ft. Thus any pair of the acceleration planes ft, ft, ft, ft have the same general line c in common. It follows that c is independent of the particular pair of the accelera- tion planes P lt P 2 , P 3 , P 4 which we may select and so there is only one general line through A l parallel to a. Thus the theorem holds in general. THEOREM 50. If two distinct general lines are each parallel to a third, then they are parallel to one another. Let a and b be two distinct general lines which are each parallel to the general line c. Let R! and R z be two acceleration planes each containing c but not containing a or b. Let Pj and P 2 be two acceleration planes parallel respectively to R 1 and R 2 and through any element of a. Then P l and P 2 each contain a. Similarly let ft and ft be two acceleration planes parallel respectively to .Rj and R 2 and containing b. Then ft is either parallel to P l or identical with it, while ft is either parallel to P 2 or identical with it. In either case we must have a parallel to b. REMARKS. If a and b be any pair of parallel general lines, it is easy to see that they must be general lines of the same kind, for we know already that two parallel general lines in one acceleration plane must be of the same kind and by two applications of this result it follows that if a and b do not lie in one acceleration plane they must also be of the same kind. 52 68 A THEORY OF TIME AND SPACE THEOREM 51. If two parallel general lines a and b lie in one acceleration plane R and if two other distinct acceleration planes P and Q containing a and b respectively have an element A l in common, then P and Q have a general line in common which is parallel to a and b. Let any element in b be selected and let $ be the 'acceleration plane through this element and parallel to P. . Then the general line b must lie in 8 and so since Q contains the general line b and the element A l it follows that P and Q contain a general line in common which is parallel to 6 and therefore also parallel to a. THEOREM 52. If a pair of non-parallel general lines a and b lie in one acceleration plane P and if through an element A 1 not lying in the acceleration plane there are two other general lines c and d respectively parallel to a and b, then c and d lie in an acceleration plane parallel to P. Let R be any acceleration plane distinct from' P which contains a but not Ai 9 and let S be any acceleration plane distinct from P which contains b but not A lf Let P be the acceleration plane through A l parallel to P, while R' and S' are the acceleration planes through A l parallel to R and 8 respectively. Then P' and R' have a general line in common which is parallel to a and since it passes through A 1 must be identical with c ; while P' and S' have a general line in common which is parallel to b and since it passes through AI must be identical with d. Thus c and d lie in the acceleration plane P' which is parallel to P. THEOREM 53. If three distinct acceleration planes P, Q and R and three parallel general lines a, b and c be such that a lies in P and R, b in Q and P and c in R and Q, then if Q' be an acceleration plane parallel to Q through some element of P which does not lie in b the acceleration planes R and Q' have a general line in common which is parallel to c. Since the acceleration plane P contains two elements in common with Q and one element in common with the parallel acceleration plane Q', it follows by Theorem 46 that P and Q' have two elements in common and therefore have a general line in common which is parallel to b. Call this general line d. A THEORY OF TIME AND SPACE 69 If this general line should happen to coincide with a, the result follows directly. We shall therefore consider the case where it does not coincide with a. Let A be any element in a. Then, in case a be an optical line, the other optical line through A in the acceleration plane P will intersect b, while, if a be an inertia or separation line, both the optical lines through A in the acceleration plane P will intersect b. Thus in all cases there is at least one optical line through A in the acceleration plane P which intersects b. Let such an optical line intersect b in B and let an optical line through B in the acceleration plane Q intersect c in G. Then BA and BG may be taken as generators of opposite sets of an acceleration plane, say S, which contains A, B and G. Now the general line a is parallel to b and therefore also parallel to d, and, since BA passes through A, is distinct from a, and lies in the acceleration plane P, it follows that BA intersects d in some element, say Z), which accordingly lies in the acceleration plane Q / . But since D lies in BA it lies in the acceleration plane S and thus S contains two elements (B and G) in common with Q and an element D in common with the parallel acceleration plane Q'. It follows by Theorem 46 that S contains a second element in common with Q' and so S and Q' contain a general line in common which must be parallel to GB. If we call this general line in S and Q' g, then any general line through G in the acceleration plane S, with the exception of CB, must intersect g. But the element A does not lie in b and so does not lie in the acceleration plane Q and therefore does not lie in GB. Thus since the general line GA is distinct from GB, and since GA must lie in S, it follows that GA must intersect g in some element, say F. But C and A both lie in the acceleration plane R which accordingly must contain the general line CA and therefore the element F. Thus since the acceleration plane R contains the general line c in common with Q and contains the element F in the parallel accelera- tion plane Q', it follows that R must have a general line in common with Q' and this general line must be parallel to c. 70 A THEORY OF TIME AND SPACE THEOREM 54. (a) If a and b be two parallel separation lines in the same accelera- tion plane and if one element of b be before an element of a, then each element of b is before an element of a. Let A be the element of b which by hypothesis is before an element of a. Let the two optical lines through A in the acceleration plane be called c and d. Let B be any other element of b. Then by Theorem 40 B must be before an element of one of the optical lines c and d and after an element of the other. It will be sufficient to consider the case when B is before an element of c and after an element of d, since the proof in the other case is similar. Let e and / be the two optical lines through B in the acceleration plane and let e be the one which is parallel to c. Then / intersects c in some element C. Also c intersects a in some element D (Post. XV) and D must be after A ; for since A is before an element of a, we should otherwise have one element of a after another, contrary to the hypothesis that a is a separation line. Now since B is before an element of c and cannot also be after an element of c, and since G lies in the optical line / through B, it follows that C is after B. Now C cannot be before A for then A would be after B, contrary to the hypothesis that A and B lie in a separation line. If C be either before D or coincident with D, then B is before D an element of a. Suppose next that C is after D and let E be the element in which / intersects a. Let h be the second optical line through D in the acceleration plane and let g be the second optical line through E in the acceleration plane and let g and h intersect in F. Then the optical lines c, /, h and g form an optical parallelogram whose diagonal line through D and E is a. Let j be the other diagonal line through c and/; then j is an inertia line. Let the optical lines d and e intersect in G. Then the optical lines c, f, d and e form an optical parallelogram whose diagonal line through A and B is b. A THEORY OF TIME AND SPACE 71 Thus in the two optical parallelograms since the diagonal lines a and b do not intersect, it follows that the diagonal lines of the other kind do not intersect (Post. XVI). But the two optical parallelograms have the corner C in common and so they must have a diagonal line in common and so must lie in j. Also D is after A and so h must be an after-parallel of d. But since F and are elements of j which is an inertia line, it follows that the one is after the other ; and since no element of d can be after an element of h, it follows that F must be after G. Thus since F is an element of g and G is an element of e, it follows that g is an after-parallel of e. But since E and B lie in the optical line f, one of them must be after the other, and since B lies in e it cannot be after E which is an element of g. Thus E is after B and so B is before an element of a. Thus in all cases B is before an element of a. (b) If a and b be two parallel separation lines in the same accelera- tion plane and if one element of b be after an element of a, then each element of b is after an element of a. THEOREM 55. (a) If a and b be a pair of parallel separation lines in the same acceleration plane and if an optical line c intersects a in A l and b in B l while a parallel optical line d intersects a in A 2 and b in B 2 , then if B l is before A 1 we have also B 2 before A 2 . By Theorem 54, since B l is before A ly therefore B 2 is before an element of a. But since A 2 and B 2 are distinct elements in the optical line d, therefore one of them is after the other. Further B z could not be after A z for then since B 2 is before an element of a we should have A 2 before this element of a, contrary to the hypothesis that a is a separation line. Thus B 2 must be before A 2 . (b) If a and b be a pair of parallel separation lines in the same acceleration plane and if an optical line c intersects a in A l and b in B l while a parallel optical line d intersects a in A 2 and b in B 2 , then if B v is after A l we have also B 2 after A 2 . 72 A THEORY OF TIME AND SPACE THEOREM 56. (a) If a and b be a pair of parallel inertia lines in the same acceleration plane and if an optical line c intersect a in A l and b in B l} while a parallel optical line d intersects a in A 2 and b in B 2 ; then if B l is before A l we have also B 2 before A 2 . Since B^ and B 2 are elements of an inertia line 6, one of them must be after the other. We shall first consider the case when B 2 is after B lf Let e be the- second optical line through B 2 in the acceleration plane. Then since by hypothesis d is parallel to c, it follows that e must intersect c in some element C. Then G must be after B l , for if C were before B l} then B l would lie in the a sub-set of C and by Post. XIV (a) B l would be the only element common to the inertia line b and the a sub-set of C. Also by Post. XIV (b) there would be one single element common to the inertia line b and the (3 sub-set of C and since there are only two optical lines through G in the acceleration plane this element would have to be identical with B 2 . Thus we should have B 2 l>efwe G and C before B t and therefore B 2 before B l} contrary to the hypothesis that B 2 is after B^ Thus we see that C must be after B l and since thus B l must be in the ft sub-set of G, it follows by Post. XIV (a) that there is one single element common to the inertia line b and the a sub-set of C. Since there are only two optical lines through C in the acceleration plane, it follows that this element must be identical with B 2 . Let the optical line e intersect a in D. If then G is before A 1 we shall have A l in the a sub-set of G and by Post. XIV (b) there is one single element common to the inertia line a and the ft sub-set of G, and since there are only two optical lines through G in the acceleration plane, it follows that this element must be identical with D. Thus D is before C and C is before B 2 and consequently D is before B 2 and since D and B 2 lie in one optical line it follows that D lies in the ft sub-set of B 2 . If G were identical with A lt it would also be identical with D and again D would lie in the ft sub-set of B 2 . But by Post. XIV (a) there is one single element common to the inertia line a and the a sub-set of B 2 and since there are only two optical lines through B 2 in the acceleration plane this element must lie in d and must therefore be identical with A 2 . A THEORY OF TIME AND SPACE 73 Thus since A 2 lies in the a sub-set of B 2 'a,ud is not identical with B 2 , therefore B 2 must be before A 2 . Thus in case C is either before A l or identical with A l we have B 2 before A 2 . Next suppose G is after A^ and let / be the second optical line through B l in the acceleration plane. Let / and d intersect in the element F. Then the optical lines e, d, c and f form an optical parallelogram whose diagonal line through B l and B 2 is b. Let j be the other diagonal line through C and F. Then since b is an inertia line, j must be a separation line. Again let g be the second optical line through D in the acceleration plane and let h be the second optical line through A l in the acceleration plane and let g and h intersect in E. Then the optical lines e, g, c and h form an optical parallelogram whose diagonal line through A l and D is a. Thus the two optical parallelograms formed by e } d, c and/ and by e, g, c and h have diagonal lines of one kind, b and a, which do not intersect and so by Post. XVI their diagonal lines of the other kind do not intersect. But the two optical parallelograms have the corner C in common and so they have the diagonal line through G in common. Thus E lies in j and since j is a separation line E is neither before nor after F. But since A l is after B l it follows that h is an after-parallel of f and so E must be after an element of /. But since E is neither before nor after F it follows by Theorem 40 that since E is after an element of / it must be before an element of d. Thus g is a before-parallel of d and since D and B 2 lie in the optical line e which intersects g in D and d in B 2 , it follows that D is before B 2 . Thus D lies in the ft sub-set of B 2 and in the optical line e. But by Post. XIV (a) there is one single element common to the inertia line a and the a. sub-set of B 2 and since there are only two optical lines through B 2 in the acceleration plane it follows that this element must lie in d and is therefore identical with A 2 . Thus since A 2 is in the a sub-set of B 2 and is not identical with J5 2 , therefore B 2 is before A 2 . This proves the theorem provided B 2 is after B^ Suppose now that B 1 is after B 2 . 74 A THEORY OF TIME AND SPACE Then A l must be after A 2 for in the first place it cannot be identical with it since the optical lines c and d are parallel and so cannot have an element in common. Further, since the element Si is after B 2 , it follows that c must be an after-parallel of d and so A l must be after an element of the optical line d. But since A l and A 2 are distinct elements of the inertia line a, the one must be after the other, and since, by Theorem 12, A l cannot be before one element of the optical line d and after another element of it without lying in the optical line, it follows that the element A l must be after the element A 2 . Suppose now, if possible, that A 2 is before B 2 , then reversing the rdles of the inertia lines a and b it would follow from what we have already proved that, c and d being parallel, A l would have to be before B l) contrary to hypothesis. Thus since B 2 must be either after or before A 2 and cannot be after, it follows that B 2 is before A 2 . (b) If a and b be a pair of parallel inertia lines in the same acceleration plane and if an optical line c intersect a in A l and b in B l} while a parallel optical line d intersects a in A 2 and b in B 2 ; then if B l is after A t we have also B 2 after J. 2 . Since a pair of parallel inertia lines always lie in an acceleration plane, the words " in the same acceleration plane " may be omitted in the enunciation of this theorem. THEOREM 57. If two elements A and B lie in one optical line and if two other elements G and D lie in a parallel optical line in the same acceleration plane, then if A be after B and G after D the general lines AD and BG intersect. Let a be the optical line containing A and B, and let b be the parallel optical line containing C and D. Then the general lines AD and BC cannot be parallel optical lines, for since B is before A an optical line through B which intersected b would be a before-parallel of an optical line through A which inter- sected b and so the element in which the former optical line intersected b would be before the element in which the latter optical line inter- sected b. Further, Theorems 55 and 56 show that AD and BG cannot be either parallel separation lines or parallel inertia lines. A THEORY OF TIME AND SPACE 75 Again AD and EG cannot both be optical lines for we know that two optical lines which intersect a pair of parallel optical lines are themselves parallel. Thus we are left with the following possibilities as to the general lines AD and EG. (1) One is an optical line and the other an inertia line. (2) One is an optical line and the other a separation line. (3) One is a separation line and the other an inertia line. (4) Both are inertia lines. (5) Both are separation lines. In case (1) Theorem 36 shows that the general lines intersect. In cases (2) and (3) it follows from Post. XV that the general lines intersect. In cases (4) and (5), since we have shown that the two general lines cannot be parallel, it follows by Theorem 47 that they must intersect. Thus in all cases the general lines AD and BG intersect. Definitions. If four optical lines form an optical parallelogram, they will be spoken of as the side lines of the optical parallelogram. A pair of side lines which do not intersect will be called opposite. The element of intersection of the diagonal lines will be spoken of as the centre of the optical parallelogram. THEOREM 58. If any two distinct elements A and be taken in an inertia or separation line i in a given acceleration plane, then there is one single optical parallelogram in the acceleration plane having as the centre and A as one of its corners. Let a and b be the two optical lines through A in the acceleration plane while c and d are the ones through ; the optical line c being parallel to a and the optical line d parallel to b. Let j be the second diagonal line of the optical .parallelogram formed by a, b, c and d. Then by Theorem 47 there is one single general line through and parallel to^'. Call this general line k and let a intersect & in D while b intersects I? in a The elements of intersection must exist since k, being parallel to j, must be an inertia or separation line according as i is a separation or inertia line ; while a and b are both optical lines. 76 A THEORY OF TIME AND SPACE Let e be the second optical line through C in the acceleration plane, while / is the second optical line through D in the acceleration plane, and let e and / intersect in B. Then a, b, e and / form an optical parallelogram in the same acceleration plane with that formed by a, b, c and d and their diagonal lines of one kind k and j do not intersect and so by Post. XVI their diagonal lines of the other kind do not intersect. But the corner A is common to both optical parallelograms and so the diagonal line i which passes through that corner must be a diagonal line of both optical parallelograms. Thus B must lie in i and so is the centre of the optical parallelo- gram formed by a, b, e and/, while A is one of its corners. Again, if there were a second optical parallelogram in the acceleration plane having as centre and A one of its corners, then such an optical parallelogram would have i as one of its diagonal lines and so the other diagonal lines of the two optical parallelograms would not intersect. Further . since the two optical parallelograms have the element common to these other diagonal lines, the latter must be identical. But there are only two optical lines, a and b, through A in the acceleration plane and these intersect k m D and C respectively, which must accordingly be a pair of opposite corners of the second optical parallelogram. But then the second optical parallelogram would have e and/" as its remaining side lines and so could not be distinct from the first optical parallelogram. Thus there is no second optical parallelogram in the acceleration plane having as centre and A as one of its corners. THEOREM 59. If two optical parallelograms have two opposite corners in common, then they have a common centre. Two cases are possible : (1) The common opposite corners may lie in an inertia line. (2) The common opposite corners may lie in a separation line. We shall consider first the case where they lie in an inertia line. Let A and B be the two common opposite corners of the optical parallelograms: B being after A. Let G and D be the other pair of opposite corners of the one optical parallelogram which we shall suppose to lie in an acceleration plane P, while C' and D' are the other pair of opposite corners of the other A THEORY OF TIME AND SPACE 77 optical parallelogram which we shall suppose to lie in an acceleration plane P'. Fig. 9. Then P and P f must be distinct if the optical parallelograms are distinct. Let be the centre of the optical parallelogram whose corners are A, B, C, D and let OE and OF be optical lines through parallel to GB and AC respectively and intersecting AC and GB in E and F re- spectively. Then E, G, F and form the corners of an optical parallelogram in the acceleration plane P, and this optical parallelogram and the one whose corners are A, G, B and D have the common diagonal line CD and so their diagonal lines of the other kind do not intersect. Thus AB and EF are parallel and EF is an inertia line. Now let OE' and OF' be optical lines through parallel to G'B and AC' respectively and intersecting AC' and C'B in E' and F' respectively. Then AC and AC' may be taken as generators of opposite sets of an acceleration plane Q lt while OP and OF' will be generators of opposite sets of a parallel acceleration plane Q z . Similarly BG and BG' may be taken as generators of opposite sets of an acceleration plane R lt while OE and OE' will be generators of opposite sets of a parallel acceleration plane R z . 78 A THEORY OF TIME AND SPACE But Q 1 and R 1 have the general line GO' in common while ft and R 2 have the general line EE' in common and so since R 1 and R 2 are parallel it follows that CC' and EE' are parallel. Again since R^ and ft have the general line FF' in common and since ft and ft ar e parallel, it follows that FF' and (7(7 ' are parallel. Thus FF' is parallel to EE'. But since EF is an inertia line there exists an acceleration plane containing E, F and F'. Let S be this acceleration plane. Then there exists in S a general line through E which is parallel to FF' and, since there can be only one parallel to FF' through E, this must be identical with the general line EE'. Thus E' must lie in the acceleration plane S. But since AB and EF are parallel and lie in P while P' and 8 are two other distinct acceleration planes containing AB and EF respectively and since P' and S have an element F' in common, it follows by Theorem 51 that the general line E'F' which is common to P and S is parallel to AB. But now E', G', F' and form the corners of an optical parallelogram in the acceleration plane P', and this optical parallelogram and the one whose corners are A, G', B and D' have one pair of diagonal lines, namely E'F' and AB, which do not intersect and so their diagonal lines of the other kind do not intersect. But these latter diagonal lines are C'O and G'D' respectively and so since they have the element G' in common it follows that they are identical. Thus the element must lie in G'D' and since it also lies in AB it follows that is the centre of the optical parallelogram whose corners are A, G', B, D'. Thus the optical parallelograms having A and B as opposite corners have a common centre 0. We have next to consider the case where the common opposite corners lie in a separation line. Let A and B be the two common opposite corners of the optical parallelograms: B being neither before nor after A. Let G and D be the other pair of opposite corners of the one optical parallelogram, which we shall suppose to lie in an acceleration plane P, while G' and D' are the other pair of opposite corners of the other optical parallelogram, which we shall suppose to lie in an acceleration plane P. A THEORY OF TIME AND SPACE 79 Then P and P' must be distinct if the optical parallelograms are distinct. Fig. 10. We shall further suppose D to be after G and D' after G r . Now the following pairs of intersecting optical lines may be taken as generators of opposite sets of certain acceleration planes which we shall denote by the following symbols opposite each pair. Optical lines GA and C'A BD tmdBD' CB and C'B AD and AD' AC' and AD BD' and BO BC' and BD AD' and AC Acceleration plane . . - <2, . . r, T, Of these acceleration planes we evidently have those pairs parallel which are represented by the same letters. 80 A THEORY OF TIME AND SPACE Thus the general line C'D, since it lies in S^ and T lt must be parallel to the general line CD', since the latter lies in S 2 and T%. Similarly the general line DD\ since it lies in Q 2 and R 2 , must be parallel to the general line C'C, since the latter lies in Q l and R lt But CD is an inertia line and so there is an acceleration plane containing C, D and D', and if we call this acceleration plane U then U contains the general lines CD' and DD' and so U must also contain the general lines through D parallel to CD' and through C parallel to DD'. That is : the acceleration plane U must contain G'D and C'C. Thus U must contain C' and therefore contains C'D'. Thus the centres of the two optical parallelograms must lie in the acceleration plane U and in the separation line AB. The acceleration plane U cannot however have more than one element in common with AB, for otherwise it would contain both A and B, and since U contains D we should have U identical with P ; but U contains D' which does not lie in P and so this is impossible. Thus the element in which CD intersects AB must be identical with the element in which C'D' intersects AB, or in other words the two optical parallelograms have a common centre. Thus the theorem is proved. THEOREM 60. If two optical parallelograms have two adjacent corners in common, then optical lines through the centres of the optical parallelograms and intersecting their common side line intersect it in the same element. Let A and B be the two common adjacent corners of two optical parallelograms which we shall suppose to lie in separate acceleration planes P and P'. We shall suppose C and D to be the other corners of the optical parallelogram in P and shall suppose C to be opposite to B and D opposite to A. We may further, without limitation of generality, take the diagonal line CB as the inertia diagonal line. We shall suppose C' and D' to be the remaining corners of the optical parallelogram in P' and we shall take C' opposite to B and D' opposite to A. Let be the centre of the optical parallelogram in P and let the one optical line through in the acceleration plane P intersect AB in M, while the other optical line in P through intersects AC in E. Then A, E, and M form the corners of an optical parallelogram also in the acceleration plane P. A THEORY OF TIME AND SPACE 81 The optical parallelograms whose corners are A, E, 0, M and A, C, D, B have the diagonal line AD in common and so, by Post. XVI, their diagonal lines of the other kind do not intersect. Fig. 11. Thus EM and GB are parallel. Now let MN be the optical line through M parallel to AC' and let MN intersect the diagonal line G'B in 0'. Let O'E' be the optical line through 0' parallel to MA and inter- secting AC' in E'. Then O'E' is parallel to OE and unless it be a neutral- parallel we have O'E' and OE in one acceleration plane. Now since MN is an optical line through M which neither intersects OE nor is parallel to it, it follows by Post. XII that there is one single element in MN which is neither before nor after any element of OE. If Q be this element, we shall suppose first that 0' is distinct from and thereby ensure that O'E' and OE lie in one acceleration plane. Call this acceleration plane Q. Now since MO and MO' are respectively parallel to AE and AE' and all four are optical lines, it follows that M, and 0' lie in one acceleration plane, say R li while A, E and E' lie in a parallel acceleration plane, say R 2 . But Q has the elements and 0' in common with R^ and has the K. 6 82 A THEORY OF TIME AND SPACE elements E and E' in common with R 2 and so the general lines 00' and EE' are parallel. We have however further seen that OB and EM are parallel and are both inertia lines. Thus 0, 0' and B lie in one acceleration plane, say S lt while E, E' and M lie in a parallel acceleration plane, say S 2 . But the acceleration plane P f has the elements 0' and B in common with $! and has the elements E' and M in common with $ 2 . Thus 50' and .#' are parallel. But BO' is the same general line as BC', which is a diagonal line of the optical parallelogram whose corners are A, C', D' } B, while ME' is a diagonal line of the optical parallelogram whose corners are A, E', 0', M and these diagonal lines do not intersect. It follows by Post. XVI that their other diagonal lines AD' and AO' do not intersect and so since they have the element A in common they must be identical. Thus 0' must lie in AD' and since it also lies in BC', it follows th,at 0' is the centre of the optical parallelogram whose corners are A, C", D', B. Thus the optical lines through the centres and 0' and intersecting AB, intersect it in the same element M. Now this same method of proof holds for the case of any optical parallelogram in the acceleration plane P' which has A and B as adjacent corners, provided that the diagonal line through B does not intersect MN in , and so all such optical parallelograms have their centres in the optical line MN. Again, if we select a second optical parallelogram in the acceleration plane P having A and B as adjacent corners but not having as centre, we may use a similar method of proof and show that all optical parallelograms in the acceleration plane P' having A and B as adjacent corners have, with one possible exception, got their centres in one optical line. This one possible exception is however different from the one possible exception which we found before and so it follows that no exception exists. Similar considerations show that all optical parallelograms in the acceleration plane P having A and B as adjacent corners, have their centres in one optical line MO. Thus the theorem holds for optical parallelograms in the acceleration planes P and P' and will therefore also hold for optical parallelograms in any other acceleration planes which contain A and B. A THEORY OF TIME AND SPACE 83 Definition. If A and B be two distinct elements lying in an inertia line or in a separation line, then the centre of an optical parallelogram of which A and B are a pair of opposite corners will be spoken of as the mean of the elements A and B. Theorem 59 shows that if two elements A and B lie in an inertia or separation line their mean is independent of the particular optical parallelogram used to define it. Since a diagonal line of an optical parallelogram is either an inertia or a separation line, the above definition fails for the case of two distinct elements lying in an optical line. In this case we adopt the following definition. Definition. If A and B be two distinct elements lying in an optical line, then an optical line through the centre of an optical parallelogram of which A and B are a pair of adjacent corners and intersecting the optical line AB, intersects it in an element which will be spoken of as the mean of the elements A and B. Theorem 60 shows that if two elements A and B lie in an optical line, their mean is independent of the particular optical parallelogram used to define it. THEOREM 61. If two or more optical parallelograms have a pair of opposite side lines in common, their centres lie in a parallel optical line in the same acceleration plane. We have already seen in the course of proving the last theorem that this result must hold if the two optical parallelograms have a third side in common. In case this is not so, let A lt B lt C 1 , D l be four distinct elements in an optical line a and let b be a parallel optical line in an acceleration plane containing a. Let the second optical lines through A lt B l} G 1} D 1 respectively in the acceleration plane intersect b in A 2 , B 2 , C 2) D 2 respectively and let Ai t Ai A 2) B z be the corners of one of the optical parallelograms under consideration and C l} D l} C 2 , D 2 the corners of another. Then A lt D lt A 2) D 2 is a third optical parallelogram. Call these optical parallelograms (1), (2) and (3) and let their centres be 0, 0', 0" respectively. Then by the first case and 0" lie in an optical line parallel to a and b since (1) and (3) have the pair of adjacent corners J. x and A 2 in common. 62 84 A THEORY OF TIME AND SPACE Similarly 0' and 0" lie in an optical line parallel to a and 6 since (2) and (3) have the pair of adjacent corners D l and D z in common. But there is only one optical line through 0" parallel to a and 6 and so 0, 0' and 0" lie in one optical line parallel to a and b. Thus all optical parallelograms having a and b as a pair of opposite side lines must have their centres in the optical line 00'. THEOREM 62. If two optical parallelograms have a pair of opposite side lines in common and if one diagonal line of the one optical parallelogram passes through the centre of the other, then the two optical parallelograms have a common centre. Since the centre of an optical parallelogram is the element of intersection of its diagonal lines, and since by hypothesis one diagonal line of the one optical parallelogram passes through the centre of the other, it follows that both centres must lie in that diagonal line. Now we know that in any optical parallelogram the one diagonal line is an inertia line, while the other is a separation line. Thus the centres of the two optical parallelograms must lie in an inertia line or a separation line. But we have already seen by Theorem 61 that they lie in an optical line, and since any two distinct elements determine a general line, it follows that the centres cannot be distinct. Thus the two optical parallelograms have a common centre. THEOREM 63. If two optical parallelograms P and Q in the same acceleration plane have a common centre, then the elements in which a pair of opposite side lines of P intersect the diagonal lines of Qform the corners of an optical parallelogram with the same centre. Let be the common centre of the two optical parallelograms P and Q and let i and j be the two diagonal lines of Q while a and b are a pair of opposite side lines of P. Let a intersect i in E and j in F, while b intersects i in G and j in H. Denote the second optical line through E in the acceleration plane by c and suppose it intersects b in H'. Denote the second optical line through G in the acceleration plane by d, and suppose it intersects a in F'. Then the optical lines a, c, b and d form an optical parallelogram one of whose diagonal lines, namely i } passes through the centre A THEORY OF TIME AND SPACE 85 of the optical parallelogram P of which a and b are opposite side lines, and so by Theorem 62 these two optical parallelograms have a common centre 0. Fig. 12. Thus if j' be the second diagonal line of the optical parallelogram formed by a, c, b and d, it has the element in common with j. The two optical parallelograms Q and that formed by a, c, b and d have however the diagonal line i in common and thus their diagonal lines of one kind do not intersect and so by Post. XVI their diagonal lines of the other kind do not intersect. But these diagonal lines are j and / which as we have seen have the element in common and therefore must be identical. Thus F' must be identical with F and H' must be identical with H and so the elements E, F, G and H must form the corners of an optical parallelogram having the same centre as the two original optical parallelograms, as was to be proved. REMARKS AND DEFINITIONS. If a and b be any two distinct inertia lines and A be any element in a which is not an element of intersection with b, then from Post. XIV (a) it follows that there is one single element common to the inertia line b and the a sub-set of A Q . Call this element B Q . Then B is distinct from A and cannot be an element of intersection 86 A THEORY OF TIME AND SPACE of the two inertia lines, for if it were A and B would lie both in an inertia line and an optical line, which is impossible. Further there cannot be an element of intersection of the two inertia lines lying after A Q and before B for by Theorem 1 (a) A Q is before B but is not before any element outside the a sub-set of A which is before B , and an element of intersection of the two inertia lines lying after A Q and before B Q would have this character. Thus any element of intersection of the two inertia lines, if such an element exists, must lie either before A or after B Q . Again from Post. XIV (a) it follows that there is one single element, say A lt common to the inertia line a and the a sub-set of B , and again A! cannot be an element of intersection of the inertia lines. Further any such element, if it exists, must lie either before A or after A^. Proceeding again in the same way there is one single element, say B lt common to the inertia line b and the a sub-set of A l and one single element A z common to the inertia line a and the a sub-set of B l and so on. Thus we get an infinite series of elements A , A l} A 2 , A S) ... in the inertia line a and another infinite series of elements B , B l> B 2 , B 3 , ... in the inertia line b. An element of intersection of the two inertia lines if such an element exists must lie either before A Q or after A n , where n is any finite integer whatever. This process will be spoken of as taking steps along the inertia line a with respect to the inertia line b. The passing from A to A l is the first step, the passing from A l to AZ the second, and so on. If X be an element which is after A in the inertia line a and before A n but not before A n - 1} then the element X will be said to be surpassed from A Q in n steps taken with respect to b. If G be an element of intersection of the two inertia lines and if G be after A , it is evident from what we have said that G cannot be surpassed from A in any finite number of steps. These remarks and definitions prepare the way for Postulate XVII. POSTULATE XVII. If A and A x be two elements of an inertia line a such that A x is after A , and if b be a second inertia line which does not intersect a either in A , A x or any element both after A and before A x , then A x may be surpassed in a finite number of steps taken from A along a with respect to b. A THEORY OF TIME AND SPACE 87 This postulate will be found to take the place of the well-known aodom of Archimedes, to which it will be seen to bear a certain re- semblance. It, however, unlike the axiom of Archimedes, contains no reference to congruence. It follows directly from Post. XVII that if the two inertia lines a and b do not intersect at all then A x may always be surpassed in a finite number of steps. If A x lies before A ti a similar result holds, but this may be proved without further assumptions and forms the subject of our next theorem. THEOREM 64. If A and A x be two elements of an inertia line a such that. A x is before A , and if b be a second inertia line which does not intersect a either in A , A x , or any element both before A and after A x , then A may be reached in a finite number of steps taken along a from an element before A x in a and with respect to b. By Post. XVII since A is after A x it follows that A Q may be surpassed in a finite number of steps, say n, taken from A x along a with respect to b. Let the elements marking these steps in a be denoted by A x+l , A x+2 , A x+3 , ... A x+n and let the elements in b lying in the ft sub-sets of these be denoted by B x , B x+l , B x+2 , ... B x+n _ l respectively. Then A may either coincide with A x+n _ 1 or be after it. If A coincides with A x+n _ l} then it is reached in n 1 steps taken along a from A x . Now there is one single element, say -B^-i? common to the inertia line b and the j3 sub-set of A x and also one single element, say A x _ l9 common to the inertia line a and the (3 sub-set of B x _^. Then Ay^ is before A x and A is reached in n steps taken along a from A x _^ with respect to b. This proves the theorem if A Q coincides with J. a;+w _ 1 . Suppose next that A Q does not coincide with A x + n ^. Then A is after A x + n _^ and before A x+n . Let J5_j be the one single element common to the inertia line b and the @ sub-set of A and let A^ be the one single element common to the inertia line a and the /? sub-set of J5_ x . Let 5_ 2 be the one single element common to the inertia line b and the {3 sub-set of A_ l and let A_ 2 be the one single element common to the inertia line a and the ft sub-set of B- 2 , and so on, till we get to an element A. n . 88 A THEORY OF TIME AND SPACE Now jB_j cannot coincide with B^n^ for then A Q and A x+n would be two distinct elements of the inertia line a both lying in the a sub-set of _!, contrary to Post. XIV (a). Further B^ cannot be after B x+n - ly for A x+n is after A Q and A is after B_^ and therefore A x+n is after B^. But B x+n _ l lies in the ft sub-set of A x+n and is distinct from it, and therefore A x+n is after B x+n -^ but is not after any element outside the sub-set which is after B x+n ^. Thus #_! cannot be after B x+n ^. It follows that B_ l must be before B x+n ^. Similarly B x+n _ 2 must be before B^. Also J._! must be before A x+n _ l} A x+n _ 2 must be before A- lf B_ 2 must be before B x+n _ 2 , B x+n _s must be before B_ 2 , A_ 2 must be before A x+n _ 2 , A x+n _ 3 must be before A_ 2 , A x must be before A_ n+l , A_ n must be before A x . Thus A_ n is an element in a which is before A X) and A Q may be reached in a finite number of steps taken from A_ n with respect to b along a. Thus the theorem holds in general. THEOREM 65. (a) If A and A x be two elements in an inertia line a which lies in the same acceleration plane with another inertia line b which does not intersect a in A , A X) or any element after the one and before the other, and if an optical line through A intersects b in B so that B is after A , then a parallel optical line through A x will intersect b in an element which is after A x . We shall first suppose that A x is after A Q , Let the optical line through A x parallel to A B intersect b in B x . Then by Post. XVII A x may be surpassed in a finite number of steps, say n, taken from A along a with respect to b. Let the elements (including A Q ) marking these steps in a be A , A l} A 2 , ... A n and let the elements in b lying in the a sub-sets of these be B , B l} B 2 , ... B n respectively. A THEORY OF TIME AND SPACE 89 Then A x may either coincide with A n ^ or be after it. Now the optical line B Q A l intersects the two optical lines A B and A^ and so these latter two optical lines belong to one set and are therefore parallel. Similarly A^ intersects the two optical lines B^ and B l A z and so these two are also parallel but belong to the other set. Proceeding thus we see that the optical lines A B , A l B l> A 2 B 2) ... A n B n belong to one set and are all parallel, while B A l) B 1 A 2) B 2 A 3 , ... B n ^A n belong to the other set and are all parallel. But A 1 lies in the a sub-set of B , H\ A l} -Oj, --n -On 1> -L*n v )) " Thus if A x coincides with A n _ lt then B x must coincide with B n ^ and therefore B x must lie in the a sub-set of A x and since B x and A x are distinct it follows that B x is after A x and the optical lines A B and A X B X are parallel. This proves the theorem in this case. If A x does not coincide with A n -^, then it must be after A n _^ and before A n . Also since A X B X is parallel to A B it must be parallel to A n - l B nr . l and to A n B n . But since A x is after An^ and before A n it follows that A X B X is an after-parallel of An^Bn^ and a before-parallel of A n B n . Further A X B X must intersect the optical line B n ^A n in some element, say (7, since E n _^A n is an optical line of the opposite set to A X B X and so G must be after B n ^ and before A n . Thus B n ^ must lie in the ft sub-set of (7, while A n lies in the a sub- set of C. But by Post. XIV (a) tfyere is one single element common to the inertia line b and the a sub-set of C and this must lie in the other optical line through G in the acceleration plane ; that is to say in the optical line A X B X and must therefore be identical with B x . Similarly by Post. XIV (b) there is one single element common to the inertia line a and the /3 sub-set of C and this must be identical with A x . Thus C is after A x and before B x and therefore B x is after A x . Thus the theorem is proved for all cases in which A x is after A Q . 90 A THEORY OF TIME AND SPACE A similar method shows that the theorem is true when A x is before AQ except that Theorem 64 takes the place of Post. XVII. Thus the theorem holds in general. (b) If A and A x be two elements in an inertia line a which lies in the same acceleration plane with another inertia line b which does not intersect a in A Q , A x , or any element before the one and after the other, and if an optical line through A Q intersects b in B so that B is before AQ, then a parallel optical line through A x will intersect b in an element which is before A x . THEOREM 66. (a) If A and A x be two elements in a separation line a which lies in the same acceleration plane with another separation line b which does not intersect a in A Q , A x or any element lying between a pair of parallel optical lines through A Q and A x in the acceleration plane, and if an optical line through A intersects b in B so that B "is after A , then a parallel optical line through A x will intersect b in an element which is after A x . Fig. 13. In case the separation lines a and b do not intersect at all, then since they lie in one acceleration plane they are parallel and the result follows directly from Theorem 55 (b). We shall therefore consider the case in which an element of A THEORY OF TIME AND SPACE 91 intersection < of a and b does exist and we shall denote this element by 0. We shall suppose first that A x is between a pair of parallel optical lines through A Q and in the acceleration plane. There are however two sub-cases of this according as is before an element of the given optical line through A or after an element of it. It is sufficient however to consider the case where is before an element of this optical line, since the case of its being after is quite analogous. Now by Theorem 58 there exists a definite optical parallelogram in the acceleration plane having the separation line a as one of its diagonal lines and having as its centre and A as one of its corners. Let C be the opposite corner to A and let the optical line A B intersect the other diagonal line in E while the second optical line through A Q in the acceleration plane intersects the same diagonal line in 6r . Then A , E , G and G Q are the corners of the optical parallelogram. Let the separation line b intersect the optical line G G in D and let the optical line through D parallel to G E Q intersect A Q E in F , while the optical line through B parallel to G E intersects G G in H Q . Then B Q , F , D and H Q are the corners of an optical parallelogram having a pair of side lines in common with the optical parallelogram whose corners are A , E , (7 and G and having its diagonal line b passing through the centre of this optical parallelogram, and therefore by Theorem 62 the two optical parallelograms have a common centre 0. Denote the optical parallelogram whose corners are A , E Q , G and G by P Q , and the one whose corners are B , F , D Q and H by Q . Suppose now that the optical line H Q B intersects the diagonal line A Q G in A l and the diagonal line G E Q in G 1 and that the optical line D F intersects the diagonal line G E Q in E l and the diagonal line A G Q m-Ai Then by Theorem 63, A lt E l} C^ and G l form the corners of an optical parallelogram having also the centre 0. Call it P x . Suppose now that the optical line A 1 E l intersects the diagonal line B Q D in B l and the diagonal line H F in F ly while the optical line GiCi intersects the diagonal line H F in H and the diagonal line B Q D Q in A- Then as before by Theorem 63, B l} F l} D l and H l form the corners of an optical parallelogram Q l which bears the same relation to the optical parallelogram P l whose corners are A lt E l} G l and G l as the optical parallelogram Q to the optical parallelogram P . 92 A THEORY OF TIME AND SPACE This construction may be repeated indefinitely and we obtain a series of parallel optical lines A E Q , A^E^ A 2 E 2 , A 3 E 3 , etc. intersecting the separation line a in the elements A , A lt A 2 , A 3) etc. and the other diagonal line of the optical parallelogram P in the elements E , E lt E 2 , E 3 , etc. Further, these same optical lines intersect the separation line b in the elements B , B l} B 2 , B 3 , etc. and the other diagonal line of the optical parallelogram Q in the elements F 0) F l} F 2) F 3) etc. Again we have another set of parallel optical lines A 1 B , A 2 B l} A 3 B 2 , A,B 3) etc. and a further set E,F , E 2 F^ E 3 F 2 , E,F 3 , etc. Now by hypothesis is before an element of the optical line A E Q but is not an element of it, and since OE is an inertia line we must have E after 0. Similarly we must have F Q after 0. But B being after A must lie in the a sub-set of A , while A l must lie either in the a or ft sub-set of J5 . A! however cannot lie in the a sub-set of B since then it would be after A 0) contrary to' the hypothesis that A Q and A-^ are elements of the same separation line. Further, A l and B cannot coincide, for then .B could not be after A Q . 'Thus A l must lie in the @ sub-set of B and must be before B . It follows that AiE l is a before -parallel of A E and E l must be before E since E l and E are elements of an inertia line. Again F must lie either in the a or ft sub-set of E l and since E l cannot be after F or coincide with it. it follows that F must be in the a sub-set of E^ and after E l . Thus by Theorem 1 (a) E l is before F Q but is not before any element outside the sub-set which is before F . But is before F and is outside the sub-set and therefore E r is not before 0. Further, E l cannot coincide with and therefore must be after it since and E lie in an inertia line. Thus is before an element of the optical line A 1 E 1 and it may similarly be proved that is before an .element of each of the parallel optical lines A 2 E 2 , A S E 3 , A 4 E 4 , etc. Further A 2 E 2 is a before-parallel of A l E l , A 3 E 3 A 2 E 2 , A THEORY OF TIME AND SPACE 93 Now since B and B 1 are elements of a separation line and since B Q lies in the a sub-set of A lt we must have B l in the a sub-set of A l and after A^ Thus B l is after A lt B 2 A 2 , If therefore A x should coincide with any element of the series A lt A 2 , A s , ..., an optical line through A x parallel to A B Q would intersect b in an element which would be after A x . If however A x does not coincide with any of these, we may suppose that the optical line through it and parallel to A Q B intersects b in B x , the inertia line OF Q in F x and the inertia line OE in E x . Now we have supposed A x to be between a pair of parallel optical lines through A and in the acceleration plane and so if an optical line be taken through parallel to A E Q , such optical line must be a before-parallel of A Q E Q since is before an element of it. Thus the optical line through A x parallel to A E must be a before- parallel of A E and an after-parallel of the optical line through 0. Thus F x and E x must be after 0. We thus see by Theorem 64 that there can only be a finite number of the elements E 1} E 2) E 3 , E 4 until we reach one E n which is before E x . Let E n be the one immediately before E x , then E n ^ l is after E x . Thus A X E X is an after-parallel of A n E n and a before-parallel of A n ^ E n -\ or using J5's instead of E's we have A X B X is an after-parallel of A n B n and a before-parallel of An^Bn^. Now the optical line A n E n ^ being of the opposite set to A Q B and so also of the opposite set to A X B X must inter- sect A x B x m some element, say K. Then K must be after A n and before A-I. Fig ' 14 ' Also since A n and A x are elements of a separation line, K being after A n must also be after A x , while similarly since B^i and B x are elements of a separation line and K is before B n _ lt therefore K must also be before B x . Thus K is after A x and B x is after K from which it follows that B x is after A x , which was to be proved. 94 A THEORY OF TIME AND SPACE Now we started out by considering the case where A x is between a pair of parallel optical lines through A and in the acceleration plane ; if instead we had taken the case where A is between a pair of parallel optical lines through A x and in the acceleration plane, then the supposition that A x was after B x would, in a similar manner, lead to the conclusion that A Q was after B , contrary to the hypothesis that BQ is after A . Also since A x and B x could not coincide without the separation lines being identical, it follows that we must also in this case have B x after A x . Thus the theorem holds in general. (b) If A and A x be two elements in a separation line a which lies in the same acceleration plane with another separation line b which does not intersect a in A , A x or any element lying between a pair of parallel optical lines through A and A x in the acceleration plane, and if an optical line through A intersects b in B so that B is before A , then a parallel optical line through A x will intersect b in an element which is before A x . THEOREM 67. If two elements A and B lie in one optical line and if two other elements C and D lie in a parallel optical line in the same acceleration plane, then if A be after B and C after D the element of intersection of the general lines AD and BC (which was proved in Theorem 57 to exist) lies between the two given optical lines. Let a be the optical line containing A and B, and let b be the parallel optical line containing G and D. Since one of the optical lines must be an after-parallel of the other and since it is immaterial which of them is, we shall suppose that a is an after-parallel of b. Now the general lines AD and BC cannot both be optical lines since two optical lines which intersect a pair of parallel optical lines are themselves parallel and have no element of intersection. One of them however may be an optical line. Suppose first that BC is an optical line and that E is the element of intersection of A D and BC. Then since a is an after-parallel of b and CB is an optical line, there- fore B is after C. But C is after D and therefore B is after D, and since A is after B it follows that A is after D. A THEORY OF TIME AND SPACE 95 Thus since AD cannot be an optical line and has one element which is after another, it must be an inertia line. Now since G is after D and lies in an optical line containing D, it follows that D is in the (B sub-set of C ; and since E lies in the second optical line through C in the acceleration plane, it follows by Post. XIV (a) that E must be in the a sub-set of G. Thus since E cannot be identical with (7, it follows that E is after C. Similarly since A is after B and A and B lie in an optical line, it follows that A is in the a sub-set of B ; and since E lies in the second optical line through B in the acceleration plane, it follows by Post. XIV (6) that E must be in the fi sub-set of B. Thus since E cannot be identical with B, it follows that E is before B. This proves that E lies between a and b. Suppose secondly that AD is an optical line and again let E be the element of intersection of AD and BC. Let the optical line through G parallel to D A intersect a in F. Then G being after D it follows that F must be after A and since A is after B therefore F must be after B. Now F must be after G and therefore lies in the a sub-set of G and is distinct from G, and so by Theorem 1 (a) G is not before any element outside the sub-set which is before F. Thus since B is outside the sub-set and before F, it follows that C is not before B. But since a is an after-parallel of 6, it follows that G is not after B. Thus GB must be a separation line. Now D cannot be after E, for since G is after D we should then have G after E which is impossible since G and E lie in a separation line. But since D and E are distinct elements of an optical line, the one must be after the other and thus E must be after D. Again E cannot be after A, for since A is after B we should then have E after B which is impossible since E and B are elements of a separation line. But E must be either before or after A since E and A are distinct elements of an optical line, and since E cannot be after, it must be before A. Thus again in this case E lies between a and 6. Next take the case where one of the two general lines AD and BG is an inertia line and the other a separation line. If BG is an inertia line, we must have B after G since a is an after- parallel of 6. 96 A THEORY OF TIME AND SPACE Since then G is after D we have B after D, and since A is after B we must have A after D. This shows that AD could not be a separation line in this case, and so we must instead take BG as a separation line and AD as an inertia line. If BG be a separation line and E be the element of intersection with AD, then E is neither before nor after G and also neither before nor after B. But E cannot be before D, for since D is before G we should then have G after E which is impossible. Thus since D and E are distinct elements of an inertia line, we must have E after D. Again E cannot be after A, for since A is after B we should then have E after B which is impossible. Thus since A and E are distinct elements of an inertia line, we must have E before A. Thus again in this case E lies between a and b. We shall next take the case where both the general lines AD and BG are inertia lines and E is their element of intersection. By Theorem '65, if A were after D and before E then G being after D would imply that B was after A, contrary to hypothesis ; while if D were after E and before A, then A being after B would imply that D was after G, contrary again to hypothesis. Thus since E cannot be identical with either A or Z), it follows that E must be after D and before A and so E lies between a and b. Finally we have the case where AD and BG are both separation lines and E their element of intersection. Let c be an optical line through E parallel to a and b. First suppose, if possible, that c is an after-parallel of a ; then c would also be an after-parallel of b since a is an after-parallel of 6. Thus AD and BG would intersect in an element which was not between a and b and did not lie either in a or 6, and so by Theorem 66, A being after B would imply that D was after (7, contrary to hypothesis. The same would hold if we supposed c to be a before-parallel of b. Thus c cannot be an after-parallel of a and cannot be identical with a and therefore must be a before-parallel of a. Also c cannot be a before-parallel of b and cannot be identical with 6, and thus c must be an after-parallel of b. Thus the element E must be after an element of b and before an element of a and so E lies between a and b. A THEORY OF TIME AND SPACE 97 This exhausts all the possibilities and so we see that the theorem holds in general. THEOREM 68. If two elements A and B lie in one optical line and if two other elements C and D lie in a parallel optical line in the same acceleration plane, then if A be after B and if the general lines AD and BC intersect in an element E lying between the parallel optical lines, we must also have C after D. Let a be the optical line containing A and B, and let b be the parallel optical line containing G and D. Then one of the optical lines a and b is an after-parallel of the other, but as the demonstration is quite analogous in the two cases we shall only consider that in which a is an after-parallel of b. We must therefore have E after an element of b and before an element of a. Now AD and BG cannot both be optical lines since two optical lines which both intersect a pair of parallel optical lines are themselves parallel and so the element E could not exist. We may however have one of them an optical line and shall first consider the case in which AD is such. In this case E is before A and A is therefore in the a sub-set of E. It follows by Theorem 1 (a) that E is not before any element outside the sub-set which is before A. But B is not in the a sub-set of E but is before A and so E is not before B. Also E being before A cannot be after any element of the optical line a and thus, E being neither before nor after B, the general line BE must be a separation line. Thus G can be neither before nor after E. But D is before E and so if G were before D we should have G before E, which is impossible. Further C cannot coincide with D and therefore G must be after D. We shall next consider the case where BG is an optical line. Then we have B after E and since A is after B, it follows that A is after E and so AE is an inertia line. Again E is after G and so E lies in the a sub-set of G and therefore by Post. XIV (6) D must lie in the /3 sub-set of G. Thus since G and D cannot be identical, we must have G after D. We shall next consider the case where one of the general lines BG and AD is an inertia line and the other a separation line. R. 7 98 A THEORY OF TIME AND SPACE Now if BG were an inertia line we should have B after E and so, since A is after B, we should have also A after E. Thus in this case both general lines would be inertia lines and so we must suppose instead that BC is a separation line and AD an inertia line. Then since E cannot be before any element of b and since it must be either before or after D, it follows that E must be after D. But D cannot be after G, for then we should have E after C, which is impossible since C and E lie in a separation line. Thus since C and D cannot be identical, we must have C after D. We have next to consider the cases where the general lines BC and AD are both separation lines and where they are both inertia lines. The constructions and demonstrations are analogous in both cases up to a certain point. By Theorem 58 there is an optical parallelogram in the acceleration plane having E as centre and B as one of its corners. Let C' be the corner opposite to B and let the optical line through C' in the acceleration plane and of the opposite set to AB intersect AB in the element G. Then GE is the other diagonal line of the optical parallelogram. Let the second optical line through B in the acceleration plane intersect GE in F. Then B, F, G' and G are the corners of the optical parallelogram. Let AE intersect the optical line FC f in D' ; let an optical line through A parallel to BF intersect FC' in H, and let an optical line through D' parallel to G'G intersect BG in /. Then A, H, D' and / are the corners of an optical parallelogram having a pair of opposite side lines in common with the optical parallelogram whose corners are B, F, G' and G and having one of its diagonal lines . AD' passing through E the centre of this optical parallelogram. It follows from Theorem 62 that these two optical parallelograms have a common centre. Let AH intersect BG' in' A l and FG in F l and let ID' intersect BG' in Ci and FG in G^ Then by Theorem 63 the elements A lt F lt C, and Q 1 form the corners of another optical parallelogram with the same centre. Suppose now first that AE and BE are both separation lines, then EG and El are both inertia lines, and by hypothesis E is before an element of BG and so E must be before G and also before I. A THEORY OF TIME AND SPACE 99 Also since B and A^ lie in a separation line and since A is after B, it follows that A must also be after A lt Thus A l G l must be a before-parallel of BG and so G l must be before G. Thus G! D' must be a before-parallel of GG', and since 0' and D' lie in an optical line we must have C' after D'. Fig. 15. Now E being the centre of the optical parallelogram whose corners are B, G, C' and F and being before an element of BG must be after an element of FC f . Thus E is between the parallel optical lines BG and FG' . Now the optical line b containing C and D may either coincide with FC' in which case G is after D or else 6 may be a before-parallel of FG' or an after-parallel of FG', but in any case b is a before-parallel of a parallel optical line through E. Thus if D does not coincide with D' we must have either D' between a pair of parallel optical lines through E and D in the acceleration plane or else D between a pair of parallel optical lines through E and D' in the acceleration plane. Thus by Theorem 66 since G' is after D' we must have G after D. Suppose next that AE and BE are both inertia lines, then EG and El are both separation lines, and by hypothesis E is before an element of BG, so E is before A and also before B. Also since B and J^ lie in an inertia line and since B is in the /9 sub-set of A and distinct from it, therefore A^ must be in the a sub-set of A, and since B and J. are distinct, A and J.j must also be distinct and therefore A l is after A. 72 100 A THEORY OF TIME AND SPACE Thus A l G l must be an after-parallel of AI, and since G l and / lie in an optical line we must have G after I. But since G l and G lie in a separation line, the one is neither before nor after the other and so G must also be after I. Fig. 16. Thus GO' must be an after-parallel of ID', and since C' and D' lie in an optical line we must have C' after D'. From this point the demonstration is similar to that of the case where AE and BE are both separation lines, except that the reference is to Theorem 65 instead of Theorem 66. This exhausts all the possibilities, and so the theorem holds in general. THEOREM 69. If A, B and G be three elements in a separation line and if B be between a pair of parallel optical lines through A and C in an accelera- tion plane containing the separation line, then B is also between a pair of parallel optical lines through A and C in any other acceleration plane containing the separation line. A THEORY OF TIME S'PACFj ''/ V K j^i i /- l ,101 Let a be an optical line through A, and c a parallel optical line through G ; both lying in the given acceleration plane, say P, and such that B lies between a and c. We may suppose that B is before an element of a and a/fer an element of c without any essential loss of generality. Let an optical line through B in the acceleration plane, and of the opposite set to a and c, intersect a in D and c in E. Then D must be o/iter B } and ^ must be before B. Further, since A t B and C lie in a separation line, we must have D after A and E before C. Now let Q be any other acceleration plane containing the separation line, and let a', b' and c' be three parallel optical lines through A, B and G respectively in the acceleration plane Q. Now the element D is after B, an element of the optical line &', while the optical line a passes through D but does not intersect b', since then it would have to be identical with the optical line DB which belongs to the opposite set. Further the optical line a cannot be parallel to b' for since a passes through A it would in that case have to be identical with a' and the acceleration planes P and Q could not be distinct. Thus each element of a is not after an element of b', and so by Post. XII (b) there is one single element of a, say F, which is neither after nor before any element of b'. Thus by Theorem 21 there is one single optical line containing F and such that no element of it is either before or after any element of 6'. If / be this optical line, then / is a neutral-parallel of b'. But since a' and b' lie in the acceleration plane Q and are parallel, the one must be an after-parallel of the other and so a' cannot be identical with /. Thus F must be either after or before A and cannot be identical with it. Now the general line FB lies in the acceleration plane P and is clearly a separation line since F is neither before nor after B. 102^ J ^L THEORY OF TIME AND SPACE Let FB intersect the optical line c in G. Then, by Theorem 45, G is neither before nor after any element of b' and so if an optical line g be taken through G parallel to b' it will be a neutral-parallel. Now by Theorem 68, since B lies between the parallel optical lines a and c passing through A and C respectively and lying in the accelera- tion plane P, it follows that if F be after A then C is after G ; while if A be after F then G is after C. If however F be o/ifer J., then a' must be a before-parallel of /, and therefore by Theorem 25 (a) a must be a before-parallel of 6'. Then we shall have also c' an after-parallel of g and therefore by Theorem 25 (b) c must be an after-parallel of b'. Thus J5 will be after an element of a' and before an element of c' : that is, B will be between the parallel optical lines .a' and c' passing through A and G respectively in the acceleration plane Q. Similarly if F be before A, then a' must be an after-parallel of/ and therefore, by Theorem 25 (b), a' must be an after-parallel of &'. We shall in that case have also c' a before-parallel of g and therefore, by Theorem 25 (a), c 7 must be a before-parallel of &'. Thus again we shall have B between the parallel optical lines a' and c' passing through A and G respectively in the acceleration plane Q. Thus the theorem is proved. REMARKS. If A, B and C be three elements in an optical or inertia line I, and if B be between a pair of parallel optical lines through A and in an acceleration plane containing I, then it is easy to see that B is also between a pair of parallel optical lines through A and C in any other acceleration plane containing I. This follows directly from the consideration that, in this case, of any two of the three elements A, B, C, one is after the other. We accordingly introduce the following definition. Definition. If three distinct elements lie in a general line and if one of them lies between a pair of parallel optical lines through the other two in an acceleration plane containing the general line, then the element which is between the parallel optical lines will be said to be linearly between the other two elements. r - The above definition is so framed as to apply to all three types of general line and for this reason is more complicated than it need be if we were dealing only with optical or inertia lines. A THEORY OF TIME AND SPACE 103 For the case of elements lying in either of these types of general line, one element is linearly between two other elements if it be after the one and before the other. In the case of elements lying in a separation line, however, no one is either before or after another and so we have to fall back on our definition involving parallel optical lines. The distinction between the three cases is interesting. Thus if the three elements A, B and G lie in a general line a, and if B be linearly between A and C, then, in case a be an inertia line, we must have either B after A and C after B or else B after G and A after B, and similarly when a is an optical line. If a be an inertia line and B be after A and G after B, then B will be before elements of both optical lines through G and after elements of both optical lines through A in any acceleration plane containing a. If a be an optical line and B be after A and G after B, then, apart from a itself, there is only one optical line through any element of a in any acceleration plane containing a, and so we should have B before an element of the optical line through G and after an element of the parallel optical line through A. If a be a separation line however, we should have B before an element of one of the optical lines through G and after an element of the parallel optical line through A and also after an element of the second optical line through G and before an element of the parallel optical line through A. The distinctions are perhaps exhibited more clearly by the following figures : Inertia Line Optical Line Separation Line Fig. 18. From Theorem 69 it follows that the property of one element being linearly between two others is independent of the particular acceleration 104 A THEORY OF TJME AND SPACE plane in which the elements are considered as lying and so may be regarded as a relation of the one element to the other two. This relation has been denned in terms of the relations before and after, not only for the cases where the three elements considered are such that of any two of them one is after the other ; but also for the case of elements in a separation line when this is no longer so. It is thus possible to state certain general results which hold for all three types of general line involving the conception linearly between. Peano has given some eleven axioms of the straight line which are as follows : (1) There is at least one point. (2) If A is any point, there is a point distinct from A. (3) If A is a point, there is no point lying between A and A. (4) If A and B are distinct points, there is at least one point lying between A and B. (5) If the point C lies between A and B, it also lies between B and A. (6) The point A does not lie between the points A and B. Definition. If A and B are points, the symbol AB represents the class of points such as C with the property that G lies between A and B. . Definition. If A and B are points, the symbol A'B represents the class of points such as C with the property that B lies between A and C. Thus A'B is the prolongation of the line beyond B, and B'A its prolongation beyond A. (7) If A and B are distinct points, there exists at least one member of A'B. (8) If A and D are distinct points and C is a member of AD and B of AC, then B is a member of AD. (9) If A and D are distinct points and B and C are members of AD, then either B is a member of AC, or B is identical with C, or B is a member of CD. (10) If A and B are distinct points and C and D are members of A'B, then either C is identical with D, or C is a member of BD, or D is a member of BC. (11) If A, B, C, D are points and B is a member of AC and C of BD, then C is a member of AD. A THEORY OF TIME AND SPACE 105 Definition. The straight line possessing A and B> symbolized by str. (A, B), is composed of the three classes A'B, AB, B'A together with the points A and B themselves. Of these axioms the writer has succeeded in proving nos. (6) and (9) from the others, so that they are really redundant*. It is easy to see, with our definition of linearly between, that corresponding results hold for all three types of "general line." As regards axioms (1) and (2) which we shall express thus : (1) There is at least one element, and (2) If A be any element there is an element distinct from A, the first follows from our preliminary statement on page 10, while the second follows directly from Posts. II and I and also from Post. V. As regards axiom (3) we shall put it in the form : (3) If A is an element, there is no element lying linearly between A and A. This follows from the definition of linearly between. (4) If A and B are distinct elements, there is at least one element lying linearly between A and B. From our remarks at the end of Theorem 35 it appears that there are an infinite number of acceleration planes containing any two distinct elements and accordingly any two distinct elements lie in a general line. If A and B lie in an optical line, then Theorem 11 shows that there is at least one element which is after the one and before the other and is therefore linearly between them. If A and B lie in an inertia line, the same result follows from Theorem 39 ; while if they lie in a separation line, it follows from Theorem 41. (5) If the element C lies linearly between A and B, it also lies linearly between B and A. This follows from the definition of linearly between. (6) The element A does not lie linearly between the elements A and B. This follows from the definition of what is meant by an element lying between a pair of parallel optical lines in an acceleration plane. According to this definition the element must not lie in either optical line. * Messenger of Mathematics, vol. xui, pp. T21-123 and 134. 106 A THEORY OF TIME AND SPACE (7) If A and B are distinct elements, there is at least one element such that B lies linearly between it and A. If A and B lie in an optical line or an inertia line, one of them must be after the other. If it be the element A which is after B, then Theorems 7 and 38 show that there is at least one element of the general line which is before B, and so B lies linearly between it and A. Similarly if A be before B there is an element of the general line which is after B, and so B is linearly between it and A. If A and B lie in a separation line, the result follows from Theorem 43. (8) If A and D are distinct elements and C is linearly between A and D, and B linearly between A and C, then B is linearly between A and D. This is readily seen to be true if we take a set of parallel optical lines a, b, c and d through A, B, C and D respectively in any acceleration plane containing the four elements. Let these optical lines intersect an optical line/ of the opposite set in A', B', C' and D' respectively. Remembering that Post. Ill must be satisfied, it is clear that we must have either : (i) C' after D' and A after C' together with B' after C' and A' after B ; or (ii) C' before D' and A' before C' together with B f before C' and A' before B'. In the first case it follows by Post. Ill that B' is after D' and consequently since B' is before A we have B linearly between A and D. Similarly in the second case we have B' before D' and after A, and therefore again B linearly between A and D. (9) If A and D are distinct elements and B and G are each linearly between A and D, then either B is linearly between A and C or B is identical with G or B is linearly between G and D. This result may be deduced in a similar manner to the last. We must have either (i) B' after D' and A after B together with G' after D' and A' after G' '; or (ii) B before D and A' before B' together with G' before D' and A before G'. A THEORY OF TIME AND SPACE 107 Then the elements B' and G' must either be identical or else the one is after the other. In the first case if B' be after C', since also B' is before A', we have B linearly between A and C. If B' is identical with C', then B is identical with C. If C' be after B', then since also D' is before B' we have B linearly between G and D. Similarly in the second case we must either have B Hnearly between A and C or B identical with G or B linearly between G and D. (10) If A and B are distinct elements and if B is linearly between A and G and also linearly between A and D, then either G is identical with D, or G is linearly between B and D, or D is linearly between B and G. This result may also be deduced in a similar way. We must have either : (i) B' after C' and A' after B' together with B' after D' ; or (ii) B' before C' and A' before B' together with B f before H. Then the elements C' and D' must either be identical or else the one is after the other. In the first case if C' is after D r , then since G' is before B' we have G linearly between B and D. If G' is identical with D', then G is identical with D. If D' is after G', then since D' is before B' we have D linearly between B and G. Similarly in the second case we must either have G linearly between B and D or G identical with D, or D linearly between B and G. (11) If A, B, G, D are elements and B is linearly between A and G, and G is linearly between B and D, then G is linearly between A and D. This result may also be deduced in a similar way. We must have either : (i) B' after G' and A' after B' together with C' after D' ; or (ii) B before C' and A' before B' together with G' before D'. In the first case, since B' is after G' and A' after B', it follows by Post. Ill that A' is after G', and so G must be linearly between A and D. Similarly in the second case we must also have G linearly between A and D. Thus all these axioms of Peano hold for the general line. 108 A THEORY OF TIME AND SPACE THEOREM 70. (a) If AQ and A x be two elements in a general line a which lies in the same acceleration plane with another general line b which intersects a in the element C such that either A is linearly between C and A X) or A x is linearly between C and A , and if an optical line through A intersects b in B so that B is after A , then a parallel optical line through A x will intersect b in an element which is after A x . We have already proved special cases of this in Theorems 65 and 66, and have now to prove the general theorem. The optical line through A x parallel to A B Q must intersect b since b intersects A B Q in B . Let the element of intersection of this optical line through A x with b be B.. Then B x cannot be identical with A X) for then the general lines a and b would have two distinct elements C and A x in common and would therefore be identical, which is impossible since a and b intersect by hypothesis. Further, if A x were after B x the general lines a and b would inter- sect in some element between the parallel optical lines (Theorem 67). That is to say in some element linearly between A Q and A x . But a and b have only one element C in common, so that if A x were after B x we should require C to be linearly between A Q and A x , contrary to the hypothesis that either A is linearly between C and A x or A x is linearly between C and A . Thus B x must be after A x . (b) If A and A x be two elements in a general line a which lies in the same acceleration plane with another general line b which intersects a in the element G such that either A is linearly between C and A X) or A x is linearly between C and A , and if an optical line through A intersects b in B so that B is before A 0) then a parallel optical line through A x will intersect b in an element which is before A x . The following five theorems (71 to 75 inclusive) are special cases of Theorems 76 and 77, but as the proofs of the general theorems are reduced to depend on these special cases the latter are treated separately. THEOREM 71. If A, B and C be three distinct elements in an acceleration plane, of which A and B lie in an optical line, but C does not lie in it, and if D be an element linearly between A and B while E is an element linearly A THEOKY OF TIME AND SPACE 109 between B and G, then there exists an element which lies both linearly between G and D and also linearly between A and E. It will be sufficient to consider the case where A is after B, since the case where B is after A is quite analogous. Since E is linearly between B and (7, therefore E is between the optical line AB and a parallel optical line through (7. Thus since A is after B this optical line through G intersects the general line AE in some element, say G, such that G is after G (Theorem 68). But since D is linearly between A and B and A is after B, therefore A is after D and D is after B. But A being after D and G after G it follows by Theorem 67 that the general lines A G and DG intersect in some element, say F, which is between the optical lines AD and GG. That is, F lies linearly between G and D. But now D is after B and E is linearly between G and B, and so by Theorem 70 an optical line through E parallel to BD will intersect GD in some element, say H, such that H is after E. But since A is after D and H is after E, therefore by Theorem 67 the element F of intersection of AE and DH is between the parallel optical lines AD and HE. That is, F lies linearly between A and E, and we have already shown that F is linearly between G and D. This proves the theorem. THEOREM 72. If A, B and G be three distinct elements in an acceleration plane, of which A and B lie in an optical line, but G does not lie in it, and if E be an element linearly between B and G while I is an element linearly between G and A, then there exists an element which lies both linearly between A and E and also linearly between B and I. It will again be sufficient to consider only the case where A is after B. As in Theorem 71, since A is after B and E is linearly between B and G, therefore the general line AE intersects an optical line through G parallel to A B in some element, say G, such that G is after G. Again, since A is after B and / is linearly between G and A, it follows in a similar manner that the general line BI intersects the optical line GG in some element, say J, such that J is after G. Thus since J is after G and G is after G, we have J after G. 110 A THEORY OF TIME AND SPACE But since A is after B and J after 0, therefore by Theorem 67 the general lines AG and BJ intersect in some element, say F, which is between the parallel optical lines AB and JG. But since also G is after G, the general line CF must intersect AB in some element, say D, such that A is after D. Also since J is after G and F between the optical lines JG and AB, therefore D is after B. Thus D is linearly between A and B, and so by Theorem 71, F is linearly between A and E and in a similar manner F is linearly between B and J. This proves the theorem. THEOREM 73. If A, B and G be three distinct elements in an acceleration plane, of which A and B lie in an optical line, but G does not lie in it, and if D be an element linearly between A and B, while F is an element linearly between G and D, then there exists an element, say E, which lies linearly between B and G and such that F lies linearly between A and E. As in the previous two theorems it will be sufficient to consider only the case where A is after B. Then A will be after D and D after B. Now since F is linearly between G and D, therefore F is between the optical line AD and a parallel optical line through G and so since A is after D it follows by Theorem 68 that this optical line through G intersects the general line AF in some element, say G, such that G is after G. But since G is after G and A is after B, therefore the general lines AG and BG intersect in some element, say E, such that E is between AB and CG. That is, E is linearly between B and C. But since D is after B, it follows by Theorem 70 that an optical line through E parallel to BD will intersect GD in some element, say H, such that H is after E. But since A is after D, it 'follows that the general lines AE and HD intersect in an element which is between AD and HE. Thus F must be between AD and HE ; that is, F must be linearly between A and E. ( We have however already shown that E is linearly between B and C, so that the theorem is proved. A THEORY OF TIME AND SPACE 111 THEOREM 74. If A, B and G be three distinct elements in an acceleration plane, of which A and B lie in an optical line, but G does not lie in it, and if E be an element linearly between B and C, while F is an element linearly between A and E, then there exists an element, say D, which lies linearly between A and B and such that F lies linearly between G and D. It will again be sufficient to consider only the case where A is after B. As in Theorem 71, since A is after B and E is linearly between B and G, therefore the general line AE intersects an optical line through G parallel to A B in some element, say G, such that G is after G. Also E is linearly between A and G. But F is linearly between A and E and so, by the analogue of Peano's eighth axiom, F is linearly between A and G. Thus F is between the parallel optical lines AB and GG, and since G is after G it follows that GF intersects AB in an element, say D, such that A is after D. Further F is linearly between G and D. Now F being linearly between A and E is between AD and a parallel optical line through E, and since A is after D the optical line through E parallel to AD will intersect DF in some element, say H, such that H is after E. But since E is linearly between B and G, therefore, by Theorem 70, D must be after B. Thus D is linearly between A and B while F is linearly between G and D. Thus the theorem is proved. THEOREM 75. If A, B and G be three distinct elements in an acceleration plane, of which A and B lie in an optical line but G does not lie in it, and if E be an element linearly between B and G, while F is an element linearly between A and E, then there exists an element, say I, which lies linearly between G and A and such that F lies linearly between B and I. It will again be sufficient to consider the case where A is after B. Since F is linearly between A and E, therefore F is between AB and a parallel optical line through E, and since A is after B this optical line through E will intersect the general line BF in some element K, such that K is after E. But since E is linearly between B and G, therefore an optical line 112 A THEORY OF TIME AND SPACE through G parallel to EK will intersect BK in some element J, such that J is after C. But since J is after G and A after B, therefore the general lines AC and BJ will intersect in some element, say /, such that / is between AB and JC. Thus / is linearly between G and A and we have also E linearly between B and (7, and so by Theorem 72, F must be linearly between B and /. Thus the theorem is proved. THEOREM 76. If A, B and C be three elements in an acceleration plane which do not all lie in one general line and if D be an element linearly between A and B, while E is an element linearly between B and C, there exists an element which lies both linearly between A and E and linearly between C and D. By combining Theorems 71 and 72 we see that the above holds for the special case where one of the general lines AB, BC or GA is an optical line. In considering the remaining cases it is evident that an optical line through A in the acceleration plane will intersect the general line BC in some element M which cannot coincide with either B or C. It will be sufficient to consider the cases where A is after M, as the cases where A is before M are quite analogous. Three special cases have to be considered. We may have either (i) M linearly between B and C, or (ii) B linearly between M and C, or (iii) C linearly between M and B. We shall take these in order. CASE (i). Since the general lines AB, BC and CA are none of them optical lines, they must be inertia or separation lines and so it is evident that if we take optical lines through B, C and D parallel to AM then the general line BA must intersect the optical line through C in some element, say G, while the general line CA must intersect the optical line through B in some element H and the optical line through D in some element, say J. Then since M is linearly between B and C and since A is after M, therefore, by Theorem 70, G is after C. A THEORY OF TIME AND SPACE 113 Similarly H is after B, and so by Theorem 67 the element A is between CG and EH. That is, A is linearly between E and 0. But D is linearly between B and A, and so by the analogue of Peano's axiom (8) we have D linearly between B and G. That is, D is between the parallel optical lines GG and BH, and so by Theorem 68, since G is after C and CD is not parallel to HE, we have CD intersecting BH in some element / such that / is after B. But since / is after B and M is linearly between C and B, therefore by Theorem 70 the optical line AM must intersect CI in some element, say F lt such that F l is after M. Similarly, since H is after B and D is linearly between A and 5 we must have J after D. But since G is a/for (7 and J after D, therefore by Theorem 67 we must have A between CG and DJ. That is, A linearly between G and D. Thus the optical line F^A must be either an after-parallel of CG and a before-parallel of DJ, or else F 1 A must be a before-parallel of (7(7 and an after-parallel of DJ. In either case F^ is linearly between C and D, and since G is a/ter C we must have ^1 after F^. Thus ^\ is linearly between A and . Now the general line CD must intersect the optical line through B parallel to MA in some element / since it cannot be parallel to it, and since G is after G and D is linearly between G and B it follows by Theorem 68 that / must be after B. Then since D is linearly between A and B and therefore between the parallel optical lines MA and BI, and since / is after B, it follows that the element of intersection, say F lt of AM and GI must be such that A is after F. Again, since / is after B and M is linearly between B and G there- fore by Theorem 70 F l must be after M. Thus F l is linearly between A and M. But since A is linearly between Gr and D we must have AF l either an after-parallel of GG and a before-parallel of a parallel optical line through D, or else ^4^ is a before-parallel of (76r and an after-parallel of a parallel optical line through D. In either case FI is linearly between C and D. Consider now the element F. Since F is linearly between (7 and D, and ^ is also linearly between G and D, therefore either A THEORY OF TIME AND SPACE 117 (1) F coincides with F lt or (2) F is linearly between C and F l} or (3) F is linearly between F 1 and D. (1) If F coincides with F l then the element M is identical with the element E which is linearly between B and C while F is linearly between A and E. (2) If F is linearly between G and F lt then since F l is linearly between A and .M", and since AM is an optical line, it follows by Theorem 73 that there exists an element, say E, which lies linearly between C and M and such that F is linearly between A and E. But since E is linearly between G and M, while M is linearly between i? and (7, therefore i? is linearly between B and (7. (3) If F is linearly between F l and D, then since F^ is linearly between A and Jf and since AM is an optical line it follows by Theorem 73 that there exists an element, say K, such that K is linearly between M and D while F is linearly between A and K. But since D is linearly between A and .B, while JT is linearly between M and D, and since .4^ is an optical line, it follows by Theorem 75 that there exists an element, say E, such that E is linearly between M and B while K is linearly between J. and E. But since .F is linearly between A and K while ^T is linearly between A and ^ therefore F is linearly between J. and E. Also since j57 is linearly between M and B while M is linearly between B and (7 therefore E is linearly between .B and G. CASE (ii). Here we have B linearly between M and G and since also D is linearly between A and 5 and since AM is an optical line, it follows by Theorem 74 that there exists an element, say F lt such that F l is linearly between A and M while D is linearly between G and .FL Consider now the element F. Since F is linearly between and D while D is linearly between G and .F!, therefore F is linearly between G and .F^ But since also F l is linearly between A and If and since AM is an optical line, it follows by Theorem 73 that there exists an element, say E, which is linearly between G and M and such that F is linearly between A and E. Now since F is linearly between G and F l while D is also linearly between G and F lt therefore either D is linearly between .Fand F lt or D coincides with F, or D is linearly between G and .F. 118 A THEORY OF TIME AND SPACE But since F is linearly between C and D, we cannot have either D linearly between C and F or D coincident with F. Thus D is linearly between F and F^. Since further F is linearly between A and E and since AF l is an optical line, it follows by Theorem 75 that there exists an element, say K, which is linearly between E and F l and such that D is linearly between A and K. Also since K is linearly between E and F lt while F l is linearly between A and M, and since AM is an optical line, it follows by Theorem 73 that there is an element, say B', which is linearly between E and M and such that K is linearly between A and B'. But the general line AK is identical with the general line AD and so B' must be identical with B. Thus B is linearly between E and M. But J is linearly between M and C while jB also is linearly between M and (7 and so either E is linearly between C and 5, or E is identical with B, or ^ is linearly between B and Jlf. Since however B is linearly between E and Jf we cannot have either E identical with B or E linearly between B and M. Thus ^ must be linearly between B and C and we have already shown that F is linearly between A and ^. CASE (iii). Here we have C linearly between M and B and so, by Theorem 70, since A is after M an optical line through C parallel to MA will intersect AB in some element, say M', such that M.' is after C. Further, the optical line CM' must be either a before-parallel of MA and an after-parallel of a parallel optical line through B, or else CM' is an after-parallel of MA and a before-parallel of a parallel optical line through B. In either case M' must be linearly between A and 5. But D is also linearly between A and 5 and so we must have either D linearly between A and M', or D identical with M', or D linearly between M' and B. First suppose D linearly between A and Jlf '. Then since F is linearly between C and D and since CM' is an optical line, therefore by Theorem 74 there exists an element, say F lt which is linearly between C and M' and such that F is linearly between A and J^. Secondly, suppose D to be identical with M'. Then the general line CD is identical with the optical line CM ' and A THEORY OF TIME AND SPACE 119 F may then be taken as identical with F l and so F l is linearly between C and M'. Thirdly, suppose D linearly between M ' and B. Then since M ' is linearly between A and B it follows that M' is linearly between A and D. Also since F is linearly between and D it follows by Theorem 76 that there exists an element, say F lt which is both linearly between A and F and also linearly between G and M '. Thus in all three cases we have an element F l linearly between C and M' and lying in the general line AF. But since M' is after G, therefore F l is after G and before M'. Since however M' is linearly between A and B and F l is before M', therefore by Theorem 70 an optical line through B parallel to M'G will intersect AF t in some element, say G, such that G is before B. But since F 1 is after G and B after G, therefore by Theorem 67 the general lines BG and GFi intersect in some element, say E, which is between the parallel optical lines CM' and GB. Thus E is linearly between B and C. But since D is linearly between A and B, therefore by Theorem 76 there is an element which is both linearly between A and E and also linearly between G and D. But since the general lines AE and CD have only the element F in common, it follows that F is linearly between A and E. Thus the theorem holds in all cases. REMARKS. Peano has given the following three axioms of the plane : (12) If r is a straight line, there exists a point which does not lie on r. (13) If A, B, G are three non-collinear points and D lies on the segment BG, and E on the segment AD, there exists a point F on both the segment A G and the prolongation B'E. (14) If A, B, G are three non-collinear points and D lies on the segment BG, and F on the segment AC, there exists a point E lying on both the segments AD and BF. Now since there is always an element outside any general line it follows that the analogue of Peano's axiom (12) holds for our geometry. Further, provided the three elements A, B, G lie in an acceleration plane, Theorem 76 corresponds to Peano's axiom (14) while Theorem 77 corresponds to his axiom (13). 120 A THEORY OF TIME AND SPACE Also Theorem 47 corresponds to the axiom of parallels in Euclidean geometry so far as an acceleration plane is concerned. An acceleration plane however differs from a Euclidean plane, since there are three types of general line in the former and only one type of straight line in the latter. Further, although closed figures exist in an acceleration plane, there is no closed figure which corresponds to a circle. How this comes about will be seen hereafter. THEOREM 78. If A, B and G be three elements in an acceleration plane which do not all lie in one general line and if D be an element linearly between A and B while DE is a general line through D parallel to BC and inter- secting AC in the element E, then E is linearly between A and G. In the first place E cannot be identical with A for then the general line DE would be identical with the general line BA and would there- fore intersect BC. Again E cannot be identical with C for once more BC and DE would intersect. Thus we must either have C linearly between A and E, or A linearly between C and E, or E linearly between A and C. If C were linearly between A and E then since D is linearly between A and B it would follow by Theorem 76 that there existed an element which was both linearly between B and C and linearly between E and D. Thus in this case also BC and DE would intersect. Next if A were linearly between C and E, then since D is linearly between A and B it would follow similarly by Theorem 77 that BC and DE must intersect. Thus the only possibility is that E is linearly between A and G. THEOREM 79. If three parallel general lines a, b and c in one acceleration plane intersect a general line d l in A l , B 1 and C l respectively and intersect a second general line d 2 in A 2) B 2 and C 2 respectively, then if B^ is linearly between A l and C 1 we shall also have B 2 linearly between A 2 and C 2 . If AI should be identical with A 2 the result follows directly from Theorem 78. Similarly it follows directly if Ci should be identical with C 2 . A THEORY OF TIME AND SPACE 121 If B^ should be identical with B 2 the following method is still valid. The general line CiA z cannot be identical with the general line c and therefore G 1 A 2 must intersect the general line b (which is parallel to c) in some element, say B'. Then since B^ is linearly between AI and C l , it follows by Theorem 78 that B is linearly between Ci and A 2 . Similarly, since B' is linearly between A 2 and G 1} it follows that B z is linearly between A z and (7 2 . Definition. If two parallel general lines in an acceleration plane be both intersected by another pair of parallel general lines then the four general lines will be said to form a general parallelogram in the accelera- tion plane. It will be seen hereafter that it is necessary to extend the meaning of the phrase general parallelogram to the case of figures which do not lie in an acceleration plane and so the words " in an acceleration plane " are important. The general lines which form a general parallelogram in an accelera- tion plane will be called the side lines of the general parallelogram. A pair of parallel side lines will be said to be opposite. The elements of intersection of pairs of side lines which are not parallel will be called the corners of the general parallelogram. A pair of corners which do not lie in the same side line will be said to be opposite. A general line passing through a pair of opposite corners will be called a diagonal line of the general parallelogram. It is clear that a general parallelogram in an acceleration plane has two diagonal lines. Further it is clear that an optical parallelogram is a particular case of a general parallelogram in an acceleration plane. THEOREM 80. If we have a general parallelogram in an acceleration plane, then : (1) The two diagonal lines intersect in an element which is the mean of either pair of opposite corners. (2) A general line through the element of intersection of the diagonal lines and parallel to either pair of side lines, intersects either of the other side lines in an element which is the mean of the pair of corners through which that side line passes. We shall first consider the cases in which one pair of opposite side lines are optical lines. 122 A THEORY OF TIME AND SPACE Let A, B, C, D be the corners of the general parallelogram and let AC and BD be the one pair of opposite side lines which are optical, while AB and CD are the other pair of opposite side lines. We shall suppose that AB and CD are not optical lines, since, in case they are, the result follows directly from the definitions of the mean of a pair of elements. Let the second optical line through B in the acceleration plane intersect the optical line AC in G, and let the second optical line through D in the acceleration plane intersect AC in F. Fig. 19. Further let AE be an optical line through A parallel to GB and intersecting BD in E, and let CH be an optical line through C parallel to FD and intersecting BD in H. Then A, E, D and F form the corners of an optical parallelogram as do also F, D, C and H and also G, B, A and E. Further, if C and G do not coincide then C, H, G and B also form the corners of an optical parallelogram. Let the diagonal lines AD and EF intersect in /, the diagonal lines AB and EG in J and the diagonal lines CD and HF in K. Then by Theorem 61, 7, / and K lie in an optical line parallel to AC and BD, and if C and G are distinct the centre of the optical parallelogram whose corners are 0, G, B and H also lies in the same optical line. Now since AB and CD are parallel it follows that AB and FH must intersect in some element, say L. Also, since AE is an optical line while FH is not, it follows that AE and FH intersect in some element, say Q. Similarly GB and FH intersect in some element, say M. Again, since FD is an optical line and AB is not, it follows that FD and AB intersect in some element, say R, and similarly CH and AB intersect in .some element, say N. If now the second optical line through Q in the acceleration plane A THEORY OF TIME AND SPACE 123 intersects FR in R', then F, A, Q and R' are the corners of an optical parallelogram. But if we consider the optical parallelogram whose corners are F, C, H, D, this has the diagonal line FH which is identical with the diagonal line FQ and so the diagonal lines of the other kind, namely CD and AR', do not intersect. Now by hypothesis CD and AB are parallel, and so AB and AR' must be the same general line. It follows that R must be identical with R. Thus A, Q, R, F are the corners of an optical parallelogram whose centre is L. If we consider the case where C and G do not coincide we may prove in a similar manner that H, B, M, N are also the corners of an optical parallelogram having L as centre. Confining our attention for the present to the case where C and G are distinct, let CB and GH intersect in J'. Then we have seen that /' lies in the optical line JK as does also /. Further, since the optical parallelograms whose corners are A, E, D, F and A, Q, R, F have a pair of opposite side lines in common, there- fore their centres J and L lie in an optical line parallel to FR. Also since the optical parallelograms whose corners are G, B, H, C and B, M, N, H have a pair of opposite side lines in common, therefore their centres /' and L lie in an optical line parallel to HN. That is, parallel to FR. Thus / and /' both lie in the optical line through L parallel to FR and as we have seen they also both lie in the optical line JK. Thus /' must be identical with /. But since CB and GH intersect in /, therefore / is the mean of B and C. Also since AD and EF intersect also in / therefore / is the mean of A and D. Thus the element of intersection of the two diagonal lines BC and AD is the mean both of B and C and also of A and D. Again, since CD and FH intersect in K, therefore K is the mean of C and D, while similarly J is the mean of A and B. Further K and J both lie in a general line through the element / parallel to A G and BD. Thus the first part of the theorem is proved provided C and G are not identical as is also the second part if the general line through the element of intersection of the diagonal lines be taken parallel to the optical side lines. 124 A THEORY OF TIME AND SPACE As regards the exceptional case in which G and G coincide we also have H and B coincident with one another and also with L, N and M. The diagonal line CB becomes an optical line and so the mean of and B is the element in which CB is intersected by JK. Also since B coincides with H and with L the optical line CB passes through L. But as before the mean of A and D lies both in the optical line JK and an optical line through L parallel to FR. Thus as before the mean of C and B coincides with the mean of A and D in the element /, while K the mean of G and D and J the mean of A and B both lie in the optical line through / parallel to AG and BD. Thus the theorem holds in this exceptional case as regards part (1) and also as regards part (2), when the general line through the element of intersection of the diagonal lines is taken parallel to the optical side lines. It remains to prove the theorem for the case of a general line taken through / parallel to AB and CD. Taking first the case in which C and G do not coincide, let GB intersect QR in S. Then Z), B, S, R are the corners of an optical parallelogram while F, A, Q, R are the corners of another optical parallelogram having the diagonal line AR in common with it. Thus the other diagonal lines DS and FQ do not intersect. Let DS and BR intersect in U and let an optical line through U parallel to FR intersect the optical line BD in 6 and FQ in W. Then 6 is the mean of B and D by definition. Let an optical line through W parallel to AG intersect AR in 0, GN in Z and FR in X. A THEORY OF TIME AND SPACE 125 Let an optical line through U parallel to AC intersect FQ in P, CN in Y and FR in V. Let an optical line through parallel to WU intersect UV in P'. Then W t 0, P', U are the corners of an optical parallelogram having the diagonal line OU in common with the optical parallelogram whose corners are F, A, Q, R. Thus the other diagonal lines cannot intersect. But the corner W lies in the diagonal line FQ and so P' must also lie in FQ. Thus P' is identical with P, and WP and OU intersect in L the centre of the optical parallelogram whose corners are F, A, Q, R. Now D, 6, U, V form the corners of an optical parallelogram whose diagonal line D Udoes not intersect the diagonal line FQ of the optical parallelogram whose corners are F, A, Q, R. Thus the diagonal lines 6V and AR do not intersect and therefore are parallel. Now provided U does not coincide with N the elements 6 and W will both be distinct from H and from one another. DX Fig. 21. If however U coincides with N the elements 0, H, W will all coincide with one another and also with the element Z. Further, in this case X coincides with D, with B y Y with U and * N and finally P with M. If we suppose U is distinct from N then H, 0, W, Z form the corners of an optical parallelogram whose diagonal line H W does not intersect the diagonal line DU of the optical parallelogram whose corners are A 6, U, V. Thus their other diagonal lines 6Z and 6V do not intersect and since they have the element 6 in common they must be identical. Thus whether U coincides with N or not, the general lines ZV and 6 V are identical. Now X, Z, F, V form the corners of an optical parallelogram. Let the diagonal lines XY and ZV intersect in T, then T is the 126 A THEORY OF TIME AND SPACE centre of the optical parallelogram and lies in the general line ZV or But now the optical parallelograms whose corners are X, Z, Y, V and W, 0, P, U have a pair of opposite side lines in common and so their centres T and L lie in an optical line parallel to OX and P V. Also the optical parallelograms whose corners are X, Z, Y, V and F, C, H, D have a pair of opposite side lines in common and so their centres T and K lie in an optical line parallel to FV. But K, /, L, T form the corners of an optical parallelogram having the diagonal line KL which does not intersect the diagonal line D U of the optical parallelogram whose corners are D, 6, U, V. Thus their other diagonal lines IT and 6 V do not intersect. But the general lines IT and 6V have the element T in common and so must be identical. Thus IT is parallel to AR ': that is, to AB and CD ; and passes through 6 the mean of B and D. Similarly it must pass through the mean of A and G. Thus the second part of the theorem is proved also in this case, provided C and G are not identical. There is no special difficulty in proving it for this case also. As we have already seen CB and AD intersect in the element / which is the mean both of G and B and of A and D. Also we have seen that an optical line through / parallel to AC and BD intersects CD in the element K which is the centre of the optical parallelogram whose corners are F, G, H, D. Now CB is in this case an optical line and so, if we take an optical line through K parallel to CB and intersecting BD in 0, we shall have K, I, B, 6 the corners of an optical parallelogram having the diagonal line KB in common with the optical parallelogram whose corners are F, G, H, D. Thus the other diagonal lines 16 and GD do not intersect and therefore are parallel. But by definition, 6 is the mean of B and D and so in this case also a general line through the element J of intersection of the diagonal lines CB and AD parallel to CD intersects BD in an element which is the mean of B and D. Similarly it intersects A C in an element which is the mean of A and C. Thus the theorem holds in this case also. We have thus proved the theorem to hold for the case of a general parallelogram in an acceleration plane having a pair of opposite side lines optical lines. A THEORY OF TIME AND SPACE 127 It remains to prove it without this latter restriction. Let A, B t C, D be the corners of a general parallelogram in an acceleration plane and let A C and BD be one pair of opposite side lines while AB and CD are the other pair of opposite side lines. Fig. 22. We shall suppose that neither pair of opposite side lines are optical lines, since we have already proved the theorem for that case. Let AE and CG be parallel optical lines in the acceleration plane through the elements A and C and intersecting the general line BD in E and G respectively. Let other optical lines parallel to AE and CG through the elements B and D intersect the general line AC in F and H respectively. Then by the case already proved a general line through the mean of B and C parallel to BD will intersect BF in the mean of B and F, say P, and will intersect GC in the mean of G and C, say Q. Similarly a general line through the mean of D and C parallel to BD will intersect GC in the mean of G and C: that is, in Q; and will intersect DH in the mean of D and H, say M. Since there is only one parallel to BD through Q, therefore M must lie in PQ. In a similar manner a general line through the mean of B and A parallel to BD will intersect BF in the mean of B and F : that is, in P : and will intersect EA in the mean of E and A, say L. Then L must also lie in PQ. Let J be the mean of B and A, while K is the mean of D and C. Again a general line through the mean of D and A parallel to BD will intersect EA in the mean of E and A : that is, in L ; and will intersect DH in the mean of D and H: that is, in M. 128 A THEORY OF TIME AND SPACE Thus the mean of B and C and also the mean of D and A both lie in the general line JK which passes through the mean of B and A and the mean of D and C and is parallel to BD. Similarly it may be proved that the mean of B and G and also the mean of D and A both lie in a general line which passes through the mean of A and C and also through the mean of B and D and is parallel to AB. Thus since the mean of B and C and the mean of D and A both lie in two distinct general lines these means must be identical and accordingly the two diagonal lines BC and DA must intersect in an element, say /, such that / is the mean of B and G and also the mean of D and A. Further a general line through / parallel to either pair of side lines intersects either of the other side lines in an element which is the mean ' of the pair of corners through which that side line passes. Thus the theorem holds in general. THEOREM 81. If A y B and G be three elements in an acceleration plane which do not all lie in one general line and if D be the mean of A and B, then : (1) A general line through D parallel to BG intersects AC in an element which is the mean of A and G. . (2) If E be the mean of A and G the general line DE is parallel to BG. Let a general line be taken through A parallel to BG and let a general line be taken through G parallel to BA and let the two general lines intersect in F. Then A, B, C, F are the corners of a general parallelogram in the acceleration plane and A and C are a pair of opposite corners. Then, by Theorem 80, a general line through the mean of A and C parallel to BC passes through the mean of A and B. That is, through D. Thus, since there is only one general line through D parallel to BG, the parallel to BG through D-must intersect A G in an element which is the mean of A and G. Again, if E be the mean of A and G there is only one general line passing through both D and E and accordingly this must be identical with the parallel to BG through either of these elements. Thus both parts of the theorem are proved. A THEORY OF TIME AND SPACE 129 THEOREM 82. If three parallel general lines a, b and c in one acceleration plane intersect a general line d l in A l , B l and C- respectively and intersect a second general line d 2 in A 2 , B 2 and G 2 respectively, and if B l be the mean of A l and G l} then B 2 will be the mean of A 2 and C 2 . If A 2 should happen to coincide with A lt or if G 2 should happen to coincide with G lt the result follows directly from Theorem 81. If d 2 should happen to be parallel to d lt then the result follows from Theorem 80 (2). In any other case let a general line through A 1 parallel to d 2 intersect b in B and c in G. Then, by Theorem 81, B is the mean of A l and G and so, by Theorem 80 (2), B 2 will be the mean of A 2 and G 2 . REMARKS. If AQ and A n be two distinct elements in a general line a, we can always find n 1 elements A lt A z , ... A n ^ in a (where n 1 is any integer) such that: A l is the mean of A Q and A 2) A 2 is the mean of A 1 and A 3) An-\ is the mean of A n _ 2 and A n . For let P be any acceleration plane containing a and let b be any general line in P which passes through A and is distinct from a. Let AI be any element in b distinct from A and let A 2> A s ', ... A' n ^ lt An be other elements in b such that : AI is the mean of A and A 2 , A 2 is the mean of AI and A 3 ', -A'n-i is the mean of A' n -s and A n '. Let general lines through AI , A 2 , ... J/n-i parallel to A n 'A n inter- sect a in the elements A lt A 2 , ... Am-i. Then, by Theorem 82, it follows that : AI is the mean of A and A 2 , A 2 is the mean of J., and A 3) An-! is the mean of A n - 2 and A n , and so the n1 elements A lt A 2) ... A n -i can be found as stated. B. 9 130 A THEORY OF TIME AND SPACE THEOREM 83. (a) If A be any element in an optical line a and A' be any element in a neutral-parallel optical line a', then, if B be any element in a which is after A, the general line through B parallel to A A' intersects a' in an element which is after A'. Since A and A lie in the neutral-parallel optical lines a and a' respectively, it follows that A is neither before nor after A and so there is at least one element which is common to the a sub-sets of A and A. Fig. 23. Let C be such an element and let b be the optical line through C parallel to a or a'. Then since C is after both A and A', it follows that b is an after- parallel of both a and a' and accordingly b and a lie in one acceleration plane while b and a' lie in another. Let the optical line through B parallel to AC intersect b in the element D and let the optical line through D parallel to CA intersect a' in the element H. Then A, C, D, B form the corners of an optical parallelogram in an acceleration plane which we shall call P, while A, C, D, B' form the corners of another optical parallelogram in another acceleration plane which we shall call P. NoW since B is after A and C is also after A, while AC and AB are A THEORY OF TIME AND SPACE 131 both optical lines, it follows that the diagonal line CB is a separation line and accordingly the diagonal line AD is an inertia line having D after A. Further D must be after C and, since G is after A', it follows that the diagonal line AD is an inertia line having D after A, and ac- cordingly the diagonal line CB is a separation line. Thus since C is after A, we must also have B after A. Let the general line through B parallel to AD intersect b in E and CA in F, and let the optical lines through E and F respectively parallel to OF and CE intersect one another in G. Then F t (7, E, G are the corners of an optical parallelogram in the same acceleration plane as the optical parallelogram whose corners are A, C, D y B and the diagonal lines FE and AD do not intersect and so the diagonal lines CG and CB do not intersect. Thus B must lie in CG and since it also lies in FE it follows that B is the centre of the optical parallelogram whose corners are F, C, E, G. Now let the general line through E parallel to DA' intersect CA' in F' and let the optical lines through E arid F r respectively parallel to CF' and CE intersect one another in G'. Then F', (7, E, G f are the corners of an optical parallelogram in the same acceleration plane as the optical parallelogram whose corners are A, (7, D, B and the diagonal lines F'E and AD do not intersect and so the diagonal lines CG' and CB' do not intersect. Thus B' lies in CG'. But the optical parallelograms whose corners are F, (7, E, G and F' t C, E, G' have the pair of adjacent corners C and E in common and the optical line BD through the centre of the first of these intersects CE in D and so it follows by Theorem 60 that the centre of the second optical parallelogram lies in the optical line through D parallel to CF' and EG'. Thus the centre of the optical parallelogram whose corners are F', C, E, G' lies in DB'. But this centre also lies in CG' and therefore it must be the element B'. Thus B' must lie in F'E. But we saw that AD and AD were both inertia lines and so they lie in an acceleration plane, say Q lt while BE and B'E which are respectively parallel to these must lie in a parallel acceleration plane, say Q 2 . Further AC and A'C are both optical lines and so they lie in an 92 132 A THEORY OF TIME AND SPACE acceleration plane, say R lt while BD and B'D which are respectively parallel to these must lie in a parallel acceleration plane, say R z . But the general lines A A' and BB' lie in the parallel acceleration planes Qi and Q 2 respectively and also in the parallel acceleration planes R l and R z respectively, and since these acceleration planes are distinct it follows that BB' is parallel to A A'. Thus the parallel to A A' through B intersects a' in the element B r which is after A'. (b) If A be any element in an optical line a and A be any element in a neutral-parallel optical line a , then, if B be any element in a which is before A t the general line through B parallel to A A intersects a' in an element which is before A. THEOREM 84. If A and B be two elements lying respectively in the two neutral- parallel optical lines a and 6, and if A be a second and distinct element in a, there is only one general line through A' and intersecting b which does not intersect the general line AB. We have seen by Theorem 83 (a) and (b) that the general line through A' parallel to AB must intersect b. Fig. 24. Let B' be the element of intersection. Then the general lines AB and A'B', being parallel, cannot inter- sect. A THEORY OF TIME AND SPACE 133 Let any other general line through A' and intersecting b intersect it in the element C. Then if C should coincide with B the general lines A'G and AB have the element B in common and therefore intersect. Suppose next that C does not coincide with B. Let P l be any acceleration plane containing a and let P 2 be the parallel acceleration plane containing 6. Let a l be any inertia line through A in the acceleration plane P l and let Q be the acceleration plane containing a l and the element B. Then Q must contain a general line, say b lt in common with P 2 and the general lines a x and 6 X must be parallel. Again let a/ be a general line through A' parallel to c^. Then a/ must lie in the acceleration plane P l and must be an inertia line. Thus the general line a/ and the element B' must lie in an accelera- tion plane, say Q', and since a/ is parallel to a l and A' B' is parallel to AB, it follows by Theorem 52 that Q' is parallel to Q. But the acceleration plane Q' contains the general line a/ in P 1 and the element B' in P 2 and therefore since P x and P 2 are ^parallel it follows that Q' and P 2 contain a general line, say &/, in common, which will be parallel to a/. Again, since a/ is an inertia line, there is an acceleration plane con- taining a/ and the element G. If we call this acceleration plane R, then by Theorem 51 the acceleration planes P 2 and R have a general line, say c lt in common and Cj is parallel to a/ and &/. Thus since GI lies in P 2 and 12, &/ in Q' and P 2 , and a/ in .R and Q', and since Q is an acceleration plane parallel to Q' through the element B of P 2 which does not lie in &/, it follows by Theorem 53 that the acceleration planes R and Q have a general line in common, say /, which is parallel to a/. Now since G is neither before nor after A', it follows that A'G is a separation line and therefore must intersect the inertia line / since both lie in one acceleration plane R. Similarly AB is a separation line and must intersect the inertia line / since both lie in the acceleration plane Q. Let AB intersect/! in F and let A'G intersect /i in F'. We have to show that F' is identical with F. Let /be the optical line through F parallel to a and let/' be the optical line through F' parallel to a. Then since B is neither before nor after any element of a, it follows 134 A THEORY OF TIME AND SPACE by Theorem 45 that no element of the general line AB with the exception of A is either before or after any element of a ; and similarly no element of the general line A'C with the exception of A' is either before or after any element of a. But F cannot be identical with A, for this would require C to lie in P 1} which is impossible, and F' cannot be identical with A' since F' and A lie in parallel acceleration planes Q and Q'. Thus F is neither before nor after any element of a and F' is neither before nor a/te?* any element of a. It follows that / is a neutral-parallel of a and also /' is a neutral- parallel of a. Suppose now, if possible, that F' is distinct from F ; then since F and P' lie. in the inertia line/i, it would follow that the one was after the other. Also if they were distinct, since they both lie in the same inertia line they could not also lie in one optical line and so the optical lines / and /' would be distinct and the one would be an after-parallel of the other. But we have seen that / and /' are each neutral -parallels of a and so it would follow by Theorem 27 that they were neutrally parallel to one another. But one optical line cannot be both a neutral-parallel and an after- parallel of another optical line and so the supposition that F' is distinct from F leads to a contradiction and therefore is not true. Thus F' is identical with F and therefore the general line A'G intersects the general line AB. Thus there is no general line through A' and intersecting b which does not also intersect AB, except the parallel general line A'B'. THEOREM 85. If a and b be two neutral-parallel optical lines and if one general line intersects a in A and b in B, while a second general line intersects a in A and b in B', then an optical line through any element of AB and parallel to a or b intersects A'B'. Let D be any element of AB and let d be an optical line through D parallel to a or 6. We have to show that d intersects A'B'. If D should coincide with either A or B, no proof is required. If A'B' be parallel to AB, then the result follows directly by Theorem 83 (a) and (6). A THEORY OF TIME AND SPACE 135 If A'B' be not parallel to AB, then by Theorem 84 the general lines AB and A'B' must intersect in some element, say C. Now, the general lines AB and A'B' being supposed distinct, G must be distinct from at least one of the elements A and B and without limitation of generality we may suppose that C is distinct from B. Fig. 25. Let Q be any acceleration plane containing the optical line b and let &! be any inertia line through B in Q. Let bi be the parallel inertia line through B' which will also lie in Q. Let P be the acceleration plane containing b^ and G, while R is the acceleration plane containing &/ and G. Then by Theorem 51 P and R have a general line, say c lf in common, which is parallel to 6 X and &/. Suppose that D is not identical with B and let Q 7 be the acceleration plane through D and parallel to Q. Then we have the three distinct acceleration planes P, Q and R and the three parallel general lines c l} ^ and &/, such that Cj lies in P and .R, &! in Q and P, and &/ in J? and Q, while Q' is an acceleration plane parallel to Q through an element of P which does not lie in &,, and so by Theorem 53 the acceleration planes R and Q' have a general line in common which is parallel to &/. 136 A THEORY OF TIME AND SPACE Call this general line d/. Then di is an inertia line. Now the optical line d must lie in Q' and must therefore intersect di' in some element, say D'. Also A'B being a separation line in the acceleration plane R must intersect the inertia line di in some element, say D". We have to show that D" is identical with D'. Suppose if possible that D" is distinct from D' and let d" be the optical line through D" parallel to b. Then since, by Theorem 45, D is neither before nor after any element of b, it follows that d is a neutral-parallel of b. Similarly d" is a neutral-parallel of b and so if D' and D" were distinct and did not lie in the same optical line, it would follow by Theorem 27 that d" was a neutral-parallel of d. But D' and D" lie in d[ which is an inertia line and so if D' and D" were distinct one of them would have to be after the other and so d and d" could not be neutral-parallels. Thus the supposition that D" is distinct from D' leads to a con- tradiction and so D" must be identical with D'. Thus the optical line d intersects A'B' in D' which proves the theorem. THEOREM 86. If a and b be two neutral-parallel optical lines and E be any element in a separation line AB which intersects a in A and b in B, and if A'B' be any other separation line intersecting a in A' and b in B' but not parallel to AB, then E either lies in AB' or in a separation line parallel to A'B' which intersects both a and b. If E does not lie in A'B', then by Theorem 85 an optical line through E parallel to a or b intersects A'B' in some element, say E', which is either before or after E. , Thus by Theorem 83 the general line through E parallel to A'B' intersects a and similarly it intersects b. Thus E must lie in a separation line parallel to A'B' and intersecting both a and b when it does not lie in A'B' itself. REMARKS. If a and b be two neutral-parallel optical lines and if c and d be any two non-parallel separation lines intersecting both a and 6, then it is evident from Theorem 86 that : the aggregate consisting of all the elements in c and in all separation lines intersecting a and b which are A THEORY OF TIME AND SPACE 137 parallel to c must be identical with the aggregate consisting of all the elements in d and in all separation lines intersecting a and b which are parallel to d. This follows since each element in the one set of separation lines must also lie in the other set. Thus the aggregate which we obtain in this way is independent of the particular set of parallel separation lines intersecting a and b which we may select and so we have the following definition. Definition. The aggregate of all elements of all mutually parallel separation lines which intersect two neutral-parallel optical lines will be called an optical plane*. It is evident that through any element of an optical plane there is one single optical line lying in the optical plane. For if a and b be two neutral-parallel optical lines which are intersected by a separation line d in the elements A and B respectively, and if C be any other element in d, then there is a neutral-parallel to a and b through C which we may call c. But through each element of c other than G there is a separation line parallel to d which, by Theorem 83 (a) and (6), must intersect both a and b, and so every element of the optical line c lies in the optical plane defined by a and b. An optical plane differs in this respect from an acceleration plane, since the latter contains two optical lines passing through any element of it. Definition. In analogy with the case of an acceleration plane, an optical line which lies in any optical plane will be called a generator of the optical plane. THEOREM 87. If two distinct elements of a general line lie in an optical plane, then every element of the general line lies in the optical plane. Let the optical plane be determined by the two neutral-parallel optical lines a and b. If the two elements lie in a general line which is known to intersect both a and 6, no proof is required. Let C be any element in any separation line AB which intersects a in A and b in B, and let D' be any element in any separation line A'E' parallel to AB and intersecting a in A' and b in B'. * The name "optical plane" has been adopted because of certain analogies with an optical line. 138 A THEORY OF TIME AND SPACE We have to show that every element of the general line CD' lies in the optical plane. By Theorem 83 (a) or (6) an optical line through C parallel to a or b will intersect A'B' in some element, say C'. If C' should coincide with D', then CD' would be an optical line which would be neutrally parallel to a or 6 and we already know that each element of it must lie either in a separation line parallel to AB and intersecting both a and b, or in AB itself. Thus if C' should coincide with D', the general line CD' is such that every element of it lies in the optical plane. If C' does not coincide with D', then an optical line through D' parallel to CO' will intersect AB in some element, say D (Theorem 83 () or (6)). Now DD' must be a neutral-parallel of CO' and either of the optical lines a or 6 must be either parallel to (7(7' and DD' or identical with one of them. If a is identical with CC' or DD', then a intersects CD', while if b is identical with CC' or DD' then b intersects CD'. If a is not identical with CC' or DD, then, by Theorem 85, a must intersect CD' and similarly if b is not identical with CC' or DD' then b must intersect CD'. Thus in all these cases CD' intersects both a and b and therefore every element of CD' lies in the optical plane determined by a and b. THEOREM 88. If e be a general line in an optical plane and A be any element of the optical plane which does not lie in e, then there is one single general line through A in the optical plane which does not intersect e. We saw in the course of proving Theorem 87 that if an optical plane be determined by two neutral-parallel optical lines a and 6, then any general line containing two elements in the optical plane and therefore any general line lying in the optical plane, must either be a neutral-parallel of a or b, or else must intersect both a and b. Suppose first that e is a separation line in the optical plane determined by a and b, then e must intersect both a and b. Since A does not lie in e it must lie in a separation line / parallel to e and intersecting both a and b. Now through A there is an optical line, say c, which is a neutral- parallel of a or b and which by Theorem 83 (a) and (6) must intersect e and must lie in the optical plane, while any other general line/' through A and lying in the optical plane must intersect both a and b. A THEORY OF TIME AND SPACE 139 But /' is not parallel to e and therefore by Theorem 84 it must intersect it. Suppose next that e is an optical line. Then -e must either be parallel to a and b or be identical with one of them. Through A there is an optical line parallel to a or b and therefore parallel to e, and this optical line must lie in the optical plane. Any other general line through A in the optical plane intersects both a and b and so by Theorem 85 it must also intersect e. Thus there is in all cases one single general line through A in the optical plane which does not intersect e. THEOREM 89. If A, and C be three elements in an optical plane which do not all lie in one general line and if D be an element linearly between A and B, while E is an element linearly between B and C, there exists an element which lies both linearly between A and E and linearly between G and D. Let a l be any inertia line through A while b and c l are parallel inertia lines through B and C respectively. B Fig. 26. Then 6 a and c l lie in one acceleration plane, say P, c l and -a^ in a second acceleration plane, say Q, and a l and b : in a third acceleration plane, say R. Let one of the optical lines through B in the acceleration plane P intersect c l in C' and let one of the optical lines through B in the acceleration plane R intersect a in A'. Then BC' and BA' may be taken as generators of opposite sets of an acceleration plane, say S, containing B, 0' and A'. 140 A THEORY OF TIME AND SPACE Let di be an inertia line through D parallel to bi and let e l be an inertia line through E parallel to 6j. Then d^ will lie in R and since D is linearly between A and B it follows by Theorem 78 that d t must intersect A 'B in some element, say D', such that D' is linearly between A' and B. Similarly e l will lie in P and since E is linearly between B and C it follows that e t will intersect BC' in some element, say E', such that J2" is linearly between B and C'. But since J.', 5, C' are three elements in the acceleration plane 8 and do not all lie in one general line, it follows by Theorem "76 that there exists an element, say F' t which lies both linearly between A' and E' and linearly between C' and D'. Now since a x and d are parallel inertia lines lying in the acceleration plane Q it follows that there is an acceleration plane, say T, containing ! and the element F', and similarly there is an acceleration plane, say U, containing d and the element F'. Thus by Theorem 51 the acceleration planes T and U have a general line, s&y fi, in common, which is parallel to a x and d and is therefore an inertia line. But the acceleration plane T contains the general line A'F' and must therefore contain E' and the inertia line e l which passes through E / and is parallel to a^ Thus T contains the element E and therefore contains the general line AE. Similarly U contains the general line CD. But since F' is linearly between A' and E' it follows by Theorem 79 that the inertia line /i must intersect AE in some element, say F, such that F is linearly between A and E. Similarly since F' is linearly between C' and D' it follows that / x must intersect CD in some element, say F, such that F is linearly between C and D. But both F and F must lie in the optical plane through A, B and C and, if distinct, could therefore only lie in an optical line or a separation line. But F and F each lie in the inertia line / x and so it follows that they cannot be distinct. Thus the element F is both linearly between A and E and linearly between C and D. It may happen in this and the following theorem that C' coincides with (7, or A' with A, but this does not affect the validity of the method of proof. A THEORY OF TIME AND SPACE 141 THEOREM 90. If A, B and C be three elements in an optical plane which do not all lie in one general line and if D be an element linearly between A and B while F is an element linearly between G and D, there exists an element, say E, which is linearly between B and C and such that F is linearly between A and E. Let a l be any inertia line through A while 6j and c l are parallel inertia lines through B and G respectively. Let P, Q, R, C', A', S, d lt D' have the same significance as in the last theorem, and let U be the acceleration plane containing the parallel inertia lines Cj and d,. Let/j be an inertia line through F parallel to c 1 and d l and which will also lie in U. Since F is linearly between G and D it follows by Theorem 79 that yi will intersect C'D' in some element, say F', such that F' is linearly between G' and D'. But as in the last theorem D' is linearly between A' and B and so, since A', B, G' lie in the acceleration plane S, it follows by Theorem 77 that there exists an element, say E' y which is linearly between B and G' and such that F' is linearly between A' and E f . If now we denote the acceleration plane containing a l and /i by T, then T contains the element E' in common with the acceleration plane P. But since a 1 lies in R and T while the parallel inertia line 6 t lies in R and P, it follows by Theorem 51 that P and T have a general line, say -en in common, which is parallel to a l and 6 X and is therefore an inertia line. Now since c l rmist also be parallel to e l and lies in the same accelera- tion plane P with it and since E' is linearly between B and G', it follows, by Theorem 78, that e l must intersect BG in some element, say E, such that E is linearly between B and G. Again, since a 1 , / x and e all lie in the acceleration plane T and since F' is linearly between A and E', it follows by Theorem 79 that AF must intersect e l in some element E such that F is linearly between A and E. But E and E must each lie in the optical plane through A, B and C and so E and E, if distinct, can only lie in an optical or separation line. But E and E each lie in the inertia line e l and so E and E cannot be distinct. 142 A THEORY OF TIME AND SPACE Thus the element E is linearly between B and G and the element F is linearly between A and E. * REMARKS. It will be observed that Theorem 89 is the analogue of Peano's axiom (14) for the case of elements in an optical plane, while Theorem 90 is the corresponding analogue of his axiom (13). Further, Theorem 88 corresponds to the Euclidean axiom of parallels for the case of general lines in an optical plane. THEOREM 91. If two elements A and B lie in one optical line and if two other elements G and D lie in a neutral-parallel optical line, and if A be after B, then: (1) If G be after D the general lines AD and BC intersect in an element which is both linearly between A and D and linearly between B and C. (2) // the general lines AD and BG intersect in an element E which is either linearly between A and D or linearly between B and C, we shall also have G after D. Let A and B lie in an optical line a and let G and D lie in a neutral- parallel optical line c. Let a 1 be any inertia line through A and let 6 Z be a parallel inertia line through B. Then a l and 6 X lie in an acceleration plane, say P. Let B f be any element in b^ which is after B and let a' be an optical line through B' parallel to a. Then a' will intersect a l in some element, say A', and, by Theorem 56, since A is after B ; we must have A' after B'. But since B' is not an element of a but is after B an element of a, it follows that a' is an after-parallel of a. Since further a and c are neutral -parallels, it follows by Theorem 25 (6) that a' is an after-parallel of c. Thus a' and c lie in an acceleration plane, say Q. Proceeding now to prove the first part of the theorem, we have A' after B' and G after D and so it follows by Theorem 67 and the definition of " linearly between " that A'D and B'G intersect in an element, say E', which is linearly between A' and D and also linearly between B' and G. A THEORY OF TIME AND SPACE 143 But since ^ is an inertia line there is an acceleration plane con- taining ! and the element E' which we may call R, and similarly there is an acceleration plane containing h and the element E f which we may call S. Now since ^ and 6j are parallel general lines in the acceleration plane P it follows, by Theorem 51, that the acceleration planes R and S have a general line, say e lt in common, which is parallel to a^ and & a and must therefore be an inertia line. Fig. 27. Since e l lies both in R and S it must intersect BC and AD which lie respectively in 8 and R and are separation lines. Suppose BI intersects BC in E and AD in E, then E and E lie in the inertia line ^ and so, if they were distinct, they could not lie in one optical plane. But E and E each lie in the optical plane determined by the neutral-parallel optical lines a and c and so E is identical with E. But since Z), A and A' are elements in the acceleration plane R which do not all lie in one general line, and since E' is linearly between 144 A THEORY OF TIME AND SPACE A' and D and E'E is parallel to A'A, it follows, by Theorem 78, that E is linearly between A and D. Similarly since E' is linearly between B' and G, and C, B and B' lie in the acceleration plane $ and are not all in one general line and since E'E is parallel to B'B it follows that E is linearly between B and G. Thus the first part of the theorem is proved. Proceeding now to prove the second part of the theorem ; since AD and BG intersect in the element E and since a l and 6j are inertia lines it follows that there is an acceleration plane, say R, which contains 2 . Then by Theorem 129 00 is an inertia line and C is after 0. B. H 210 A THEORY OF TIME AND SPACE But c is normal to the inertia line OC and to the separation line a and therefore by case (4) on page 207 c must be normal to every inertia line (and therefore also every general line) which passes through and lies in the acceleration plane containing OC and a. Let R be this acceleration plane. If R should coincide with P the result follows directly and so we shall suppose that R is distinct from P. Let S be the general plane containing a and 6. Then S will be distinct from both P and R, and, as was pointed out in the remarks at the end of the last theorem, S may be an acceleration plane, an optical plane, or a separation plane. Let one of the generators of R which pass through C intersect a in G and let the generator of the opposite set passing through intersect CG in F. Then since does not lie in the optical line CG but is before the element G of it, it follows that F must lie in the a sub-set of and therefore F is after 0. Let b' be the general line through G parallel to b. Then since G lies in S it follows that b' lies in S. Let Q be the general plane containing &' and G G. Then since D l A is parallel to 6 and is distinct from b' it follows that it is parallel to b', and since A A passes through G it must lie in the general plane Q. Thus A> A and F are three distinct elements in Q which do not all lie in one general line. Now any general line in S which passes through and is distinct from b must intersect b' in some element, say G. Then the general line GG lies in Q and is distinct from A A- If then G does not coincide with G it follows, by Theorem 128, that CG must either intersect QFin an element H^ linearly between A and F, or else must intersect D 2 F in an element H z linearly between A and F. But since is before both A and F it follows by Theorem 129 that OH l is an inertia line and similarly since is before both A and F it follows that OH 2 is an inertia line. Now c is normal to every general line in P which passes through and also to every general line in R which passes through and there- fore c is normal to the three optical lines 0A> OD Z and OF. Thus c must be conjugate to every inertia line which passes through and lies either in the acceleration plane containing OA and OF, or the acceleration plane containing OA and OF. A THEORY OF TIME AND SPACE 211 Thus c is conjugate to OH l and also to OH 2 . But c is conjugate to 00 and therefore is conjugate to every inertia line which passes through and lies in the acceleration plane containing 00 and OH l or the acceleration plane containing 00 arid OH 2 . Thus since OG lies in the acceleration plane containing 00 and OH^ or in the acceleration plane containing 00 and OH Z as the case may be, it follows that c must be normal to OG. Thus, including the separation lines a and 6, the separation line c is normal to every general line which passes through and lies in the general plane 8. THEOREM 137. If two distinct separation lines a and b intersect in an element and if an optical line c passing through be normal to both a and 6, then c is normal to every general line which passes through and lies in the general plane containing a and b. f ! be any separation line in P which passes through and let Q be the acceleration plane containing a 1 and b^ 218 " A THEORY OF TIME AND SPACE Then by Theorem 131 there is one and only one separation line in P and passing through which is conjugate to every inertia line in Q which passes through 0. Let b 2 be this separation line. Then as was remarked at the end of Theorem 133 b 2 is conjugate to certain other inertia lines passing through which do not lie in Q. Let a be any such inertia line and let Q' be the acceleration plane containing a' and 6j. Then, by Theorem 134, b 2 is conjugate to every inertia line in Q' which passes through 0. Let a 2 be the one single inertia line in Q' and passing through which is conjugate to & x and let R be the acceleration plane containing a l and a 2 . Then a l and a 2 are each conjugate to both 6 X and b 2 . Thus both a 1 and a 2 are conjugate to every separation line in P which passes through and so every separation line in P which passes through is conjugate to every inertia line in R which passes through 0. Thus every general line in P is normal to every general line in R and so the acceleration plane R is completely normal to P. Thus, since R passes through 0, the theorem is proved. THEOREM 143. If P be an optical plane and be any element in it, there is at least one optical plane passing through and completely normal to P. Let a be the generator of P which passes through and let b be any separation line in P which passes through 0. Then, by Post. XIX, there is at least one element, say A, which is neither before nor after any element of P. The general line OA is thus a separation line and, by Theorem 45, no element of OA with the exception of is either before or after any element of a. Thus a is normal to OA and it is also normal to b and so, since OA and b must lie in a separation plane, say $, it follows that the optical line a is normal to 8. But now we know that there is one single separation line, say c, which passes through 0, lies in S and is normal to b. Then c is normal to both a and b and therefore is normal to P. But c and a lie in an optical plane which is distinct from P and which we shall call R. Further a is an optical line in P and therefore is normal to P. A THEORY OF TIME AND SPACE 219 Thus any general line in P is normal to the two intersecting general lines a and c which lie in R and so every general line in P is normal to every general line in R. It follows that R is completely normal to P and, since R passes through 0, the theorem is proved. REMARKS. By combining Theorems 141, 142 and 143 we get the general result : If P be any general plane and be any element in it, there is at least one general plane passing through and completely normal to P. If R be this general plane which is completely normal to P and if 0' be any element not lying in P, then 0' either may or may not lie in R. If 0' does not lie in R then there is a general plane, say R', passing through 0' and parallel to R. It is evident that since R is completely normal to P we must also have Pf completely normal to P and so we may generalize the above result and say: // P be any general plane and be any element whatever, there is at least one general plane passing through and completely normal to P. Let be any element and let 8 be any separation plane passing through 0, while P is an acceleration plane also passing through and completely normal to 8. Let a be any separation line in 8 which passes through and let b be the one single separation line in 8 and passing through which is normal to a. Let c be any separation line passing through and lying in P and let d be the one single inertia line in P and passing through which is normal to c. Then both c and d are normal to both a and b and so we have the three separation lines a, b and c all passing through and each of them normal to the other two ; while in addition to these we have the inertia line d also passing through and normal to all three. This result marks an important stage in the development of our theory as it suggests the possibility of setting up a system of coordinate axes one of which axes is of a different character from the remaining three. Another important result is the following : If 8 be a separation plane and if P be an acceleration plane passing through any element of S and completely normal to 8, then there are 220 A THEORY OF TIME AND SPACE two generators of P which pass through and each of them is normal to the separation plane 8. Thus there are at least two optical lines which pass through any element of a separation plane and are normal to it. THEOREM 144. If P be an acceleration or separation plane and be any element which does not lie in it, there is one single general line passing through and normal to P which has an element in common with P. We already know that if a- be a separation line and if be any element which does not lie in it, then, in whatever type of general plane and a may lie, there is one single general line passing through and lying in this general plane which is normal to a. Further, if d be this general line normal to a, then d must intersect a in some element, say A. Now suppose that a lies in the acceleration or separation plane P. Then there is one single general line passing through A and lying in P which is normal to a. Let b be this general line. Then since P is an acceleration or separation plane and a is a separation line, b must be distinct from a and must be either an inertia or separation line and cannot be an optical line. Now we know that in whatever type of general plane and b may lie there in one single general line passing through and lying in this general plane which is normal to b. Let c be this general line. Then since b is not an optical line this normal to it through cannot be parallel to b and therefore must intersect b in some element, say B. Now a is normal to the two general lines d and b which intersect in A and accordingly a is normal to every general line in the general plane containing d and b and therefore is normal to c. But c is normal to the two intersecting general lines a and b which lie in P and therefore c is normal to P. Since c has the element B in common with P we have proved that there is at least one general line through and normal to P which has an element in common with P. It remains to show that there is only one general line having this property. Consider first the case where P is a separation plane and let B' be any element in P distinct from B. A THEOKY OF TIME AND SPACE 221 Then BB'. is a separation line and so in whatever type of general plane and BB' may lie there is one single general line passing through lying in this general plane and normal to BB'. But OB passes through and is normal to BB' and therefore OB' cannot be normal to BB' and so cannot be normal to P. This proves that OB is the only general line through and normal to P which has an element in common with P provided P be a separation plane. This method does not serve if P be an acceleration plane since BB' might in this case be an optical line. If P be an acceleration plane, let P' be an acceleration plane passing through and parallel to P. Then and B must be representatives of one another in the parallel acceleration planes P' and P. If B' were any other element in P distinct from B and such that OB' were normal to P, then B' would be also the representative of in P which we know is impossible. Thus again OB is the only general line through and normal to P which has an element irr common with P. The theorem thus holds for both separation and acceleration planes. THEOREM 145. If P be an optical plane and be any element which does not lie in it, then: (1) If be neither before nor after any element of P there is one single generator of P such that every general line which passes through and intersects this generator is normal to P. (2) If be either before or after any element of P there is no general line passing through and having an element in common with P which is normal to P. As regards the first part of this theorem, if we carry out the con- struction of Theorem 144 taking a as a separation line, then since P is an optical plane the general line b must be an optical line since it is normal to a. Since is neither before nor after any element of P it is neither before nor after any element of b. If then OB be any general line passing through and intersecting b in the element B, it follows by Theorem 45 that no element of OB with the exception of B is either before or after any element of b. It follows that OB is normal to 6. 222 A THEORY OF TIME AND SPACE But, as in Theorem 144, OB is normal to a and thus OB is normal to the two intersecting general lines a and b which lie in P and there- fore it is normal to P. - Again if B' be any element in P which does not lie in 6, then BB' is a separation line and so as in Theorem 144 OB' cannot be normal to P. Thus all general lines through normal to P which have an element in common with P intersect b. Thus the first part of the theorem is proved. Suppose next that is before some element, say E, in P. Then through E one single generator of P passes which we may denote by /. Since does not lie in / but is before an element of /, it follows that through there is an optical line which is a before-parallel of / and which we shall denote by c. If f be any other generator of P it will be a neutral-parallel of / and so by Theorem 25 (a) c will be a before-parallel of /'. Thus is before elements of every generator of P. Similarly if be after any element of P it is after elements of every generator of P. Thus in case be either before or after any element of P it will lie in an acceleration plane along with any selected generator of P. Let OB be any general line passing through and having the element B in common with P and let b be the generator of P which passes through B. Then OB and b lie in an acceleration plane and intersect in B and so since b is an optical line OB cannot be normal to it. Thus OB cannot be normal to P and therefore there is in this case no general line passing through and having an element in common with P which is normal to P. THEOREM 146. If a general line d have an element A in common with a general plane P, there is at least one general line passing through A and lying in P which is normal to d. If d lies completely in P we already know that the theorem holds and so we shall suppose that A is the only element common to d and P. We= shall first consider the case where P is an acceleration or separa- tion plane. A THEORY OF TIME AND SPACE 223 In this case, if be any element of d distinct from A, there is, by Theorem 144, one single general line passing through and normal to P which has an element in common with P. Let B be this element. If B should coincide with A then every general line passing through A and lying in P would be normal to d. If B does not coincide with A let a be the one single general line passing through A and lying in P which is normal to AB. Then since OB is normal to P it must be normal to a. Thus a is normal to the two intersecting general lines AB and OB and therefore is normal to the general plane containing them. Thus the general line a must be normal to d and since a passes through A and lies in P the theorem is proved for the case where P is an acceleration or separation plane. Suppose next that P is an optical plane and let b be the generator of P which passes through A. Now since b is an optical line it follows that the intersecting general lines b and d must lie in a general plane, say Q, which must be either an optical plane or an acceleration plane. Suppose first that Q is an optical plane. Then since b is an optical line in Q and d intersects b, it follows that d must be a separation line and b must be normal to d. But b passes through A and lies in P and so the theorem is proved for this case. Next consider the case where Q is an acceleration plane. Let b' be any generator of P distinct from 6. Then since b' is a neutral-parallel of b it follows that an acceleration plane Q' through any element of 1) and parallel to Q will contain b'. If then A' be the representative of A in Q', the general line A A' will be normal to Q and therefore will be normal to d. But, since b' is neutrally parallel to b which contains the element A, the element A' must lie in b' and therefore in the optical plane P. Thus the general line A A' must lie in P and, since it passes through A and is normal to d, the theorem holds also in this case. Thus the theorem holds in general. THEOREM 147. If three general lines a, b, and c have an element in common, there is at least one general line passing through which is normal to all three. If we take any two of the three given general lines, say a and b, it 224 A THEORY OF TIME AND SPACE follows, since they have the element in common, that they lie in a general plane, say P. Then by Theorems 141, 142, and 143 there is at least one general plane passing through and completely normal to P. Let Q be this general plane. Then, since c has the element in common with Q, it follows, by Theorem 146, that there is at least one general line, say d, passing through and lying in Q which is normal to c. But since d lies in Q it is normal to both a and b and thus is normal to all three general lines. Thus the theorem is proved. Definition. If a general line and a general plane have one single element in common, they will be said to intersect in that element. Definition. If a general line a and a general plane P intersect, then the aggregate of all elements of P and of all general planes parallel to P which intersect a will be called a general threefold. It will be found that, just as there are three types of general line and three types of general plane, so there are three types of general threefold. In the case of general threefolds however, unlike that of general lines or of general planes, we are able to give a definition which applies to all three types without first considering any of the special cases. From the definition it is clear that if a general threefold W be deter- mined by a general line a intersecting a general plane P then any other general plane P' parallel to P and intersecting a may take the place of P, so that a and P' will also serve to determine W. Again if a intersects P in the element and if a! be a general line parallel to a and intersecting P in another element 0', then a and a will lie in a general plane, say Q. If through any element O l of a distinct from the general plane P T passes parallel to P, then by Theorem 124 the general plane Q must have a second element in common with P lt Thus P! and Q have a general line in common which must be parallel to 00' and so the general line a must intersect P 1 in some element O/. Thus a' intersects every general plane parallel to P which intersects a and similarly a intersects every general plane parallel to P which intersects a'. It follows that every element of a lies in the general threefold deter- mined by a and P, and also : that a and P determine the same general threefold as a and P. A THEORY OF TIME AND SPACE 225 THEOREM 148. // two distinct elements of a general line lie in a general threefold then every element of the general line lies in the general threefold. Let the general threefold W be determined by a general plane P and a general line a which intersects it. Let X l and X 2 be two distinct elements of a general line b and let them both lie in W. If X 1 and X 2 should both lie in P or in any one of the general planes which intersect a and are parallel to P, then the general line b will lie in that general plane and therefore every element of b must lie in W. We shall next suppose that X l lies in one of the set of parallel general planes, say P lt while X 2 lies in another, say P 2 . Then b either may or may not lie in a general plane containing a. Suppose first that b lies in a general plane Q along with a. Then we may have either : (1) b identical with a, or (2) 6 parallel to a, or (3) b intersecting a. If b be identical with a the result is obvious. If b be parallel to a then, as we have already shown, every element of b lies in W. If b intersects a then at least one of the elements X lt X 2 must be distinct from the element of intersection of b and a. We may suppose that X l is distinct from this element of inter- section. Then the element in which a intersects P l must be distinct from X 1 and so the general plane Q has two distinct elements in common with PL Further since the general line a intersects all the general planes parallel to P! whose elements along with the elements P x make up W, it follows, by Theorem 124, that Q has a general line in common with each of these general planes and all these general lines are parallel to one another. Now since 6 does not lie in P x it follows that b must intersect all these general planes and similarly a general plane through any element of b distinct from X 1 and taken parallel to P l must intersect a. Thus we see that in this case also every element of b lies in W and further that b and Pj determine the same general threefold as a and P l : namely W. Thus the theorem holds provided b and a lie in one general plane. R. 15 226 A THEORY OF TIME AND SPACE Finally suppose as before that X l lies in P a and X 2 in P 2 and that 6 and a do not lie in one general plane. Let a intersect P x in the element Y l and let b' be a general line through YI parallel to b. Then b and b' lie in a general plane, say R, which has the two elements X^ and Y l in common with P l and has the element X 2 in common with the parallel general plane P 2 . Thus R has a general line in common with P 2 which is parallel to JTjFj and so b' must intersect P 2 in some element, say Y 2 . But now, from what we have already proved, every element of b must lie in W and also b' and P 2 determine the general threefold W equally with a and P 2 or a and P. Again since b is parallel to b' it follows from what we have already proved that every element of b lies in the general threefold determined by b and P 2 : that is in W ; and that b and P 2 may also be taken as determining the general threefold W. Thus the theorem holds in general. REMAKES. It is evident from the above that if a general threefold W be deter- mined by a general plane P and a general line a which intersects P, then a general line b which has two distinct elements in common with W, which do not both lie in P or do not both lie in one of the general planes parallel to P and intersecting a, will intersect all these general planes including P. Further 6 and P, or b and any one of these general planes, will also determine W. Again if a general plane Q have two distinct elements X and X 2 in common with W, then Q will have at least one general line in common with W: namely the general line X t X 2 since, by the above theorem, every element of X l X 2 must lie in W, and we already know that every element of it must also lie in Q. It is not however possible from this to prove that Q and W have more than one general line in common. THEOREM 149. If a general plane have three distinct elements in common with a general threefold and if these three elements do not all lie in one general line then every element of the general plane lies in the general threefold. Let the general threefold W be determined by a general plane P and a general line a which intersects P. A THEORY OF TIME AND SPACE 227 Let X-i, X. 2 and X 3 be three distinct elements of a general plane Q which do not all lie in one general line and suppose that X l} X z and X 3 all lie in W. If all these three elements should lie in P or if they should all lie in one of the general planes parallel to P which intersect a, then Q would be identical with the general plane in which they all lie and accordingly every element of Q would lie in W. If X 1} Xj and X 3 do not all lie in one of this set of general planes, suppose that X 1 lies in the general plane P l of the set while X a lies in another distinct general plane of the set, say P 2 . Then X s will lie in some general plane P 3 of the set which may be either identical with P! or with P 2 , or may be distinct from both. Now since X l and X 2 lie in two distinct general planes of the set it follows that the general line X : X Z intersects every general plane of the set and therefore must intersect P 3 in some element, say 0. Further, since X 1 , X 2 and X 3 do not all lie in one general line, it follows that X 3 and must be distinct elements. Thus the general planes P 3 and Q have two distinct elements X 3 and in common and therefore have the general line OX 3 in common which accordingly lies in W. Again the general threefold W, as we have seen, may be determined by the general plane P 3 and the general line X 1 X 2 which intersects P 3 in 0. But now every element of Q lies either in X^X 2 or in a general line parallel to X^X^ and intersecting OX 3 . We have however seen that every element of any such general line must lie in W. It follows that every element of Q must lie in W. Thus the theorem holds in all cases. THEOREM 150. (1) If a general line b lies in a general threefold W and if A be any element lying in W but not in 6, then the general line through A parallel to b also lies in W. (2) If a general plane P lies in a general threefold W and if A be any element lying in W but not in P, then the general plane through A parallel to P also lies in W. The first part of the theorem may be proved as follows : The general line b and the element A determine a general plane, say Q, having three elements in common with W which do not all lie in one general line, and so, by Theorem 149, Q lies in W. 152 228 A THEORY OF TIME AND SPACE But the general line through A parallel to 6 must lie in Q and therefore must lie in W. This proves tthe first part of the theorem. In order to prove the second part let b and c be two intersecting general lines which both lie in P and therefore in W. The element A does not lie in P and therefore cannot lie either in b or c. If then b' and c be general lines through A parallel to b and c respectively, it follows from the first part of the theorem that b' and c both lie in W. If then P' be the general plane containing b' and c' it will contain three distinct elements in common with W which do not all lie in one general line and so, by Theorem 149, P' must lie in W. But P' is parallel to P and passes through A and so the theorem is proved. THEOREM 151. If a general threefold W be determined by a general plane P and a general line a which intersects P, then if Q be any general plane lying in W, and if b be any general line lying in W and intersecting Q, the general plane Q and the general line b also determine the same general three- fold W. It is evident from the remarks at the end of Theorem 148, this above holds in the special case where Q is one of the set of general planes con- sisting of P and all general planes parallel to P which intersect a. We shall therefore consider the case where Q is distinct from any one of this set of general planes which we shall for convenience refer to as the primary set. Let X-L be any element in Q and let c l and c/ be any two distinct general lines lying in Q and passing through X t . Then c l and c/ could not both lie in any general plane of the primary set, for if so Q would require to be identical with that general plane, contrary to hypothesis. Thus at least one of the two general lines c 1} c/ does not lie in any general plane of the primary set. Suppose c l be a general line of this character. Then, since Q lies in W, each element of c must lie in a distinct general plane of the primary set, and c l must intersect every general plane of the primary set. Let X 2 be any element of Q which does not lie in c 1? and let c. z be a general line through X 2 parallel to c x . A THEORY OF TIME AND SPACE 229 Then c 2 must also lie in Q and must also intersect every general plane of the primary set. If then P r be any one general plane of the primary set, it is inter- sected both by Cj and by c 2 and the elements of intersection must be distinct since c and c 2 are parallel. Thus P' has two distinct elements in common with Q and therefore has a general line in common with Q. It follows that Q has a general line in common with each general plane of the primary set. Now let A be any element in b other than its element of intersection with Q. Then A must lie in some general plane of the primary set, say P lt since b lies in W. Now as we have seen P x has a general line in common with Q, and since A does not lie in Q it cannot lie in this general line. If then B and C be any two distinct elements in this general line, the three elements A, B and G are three distinct elements in P 1 which do not all lie in one general line. But it is evident that A, B and C all lie in the general threefold determined by Q and b and so, by Theorem 149, the general plane P l must lie in this general threefold which we may call W. Now, since the general line c x intersects every general plane of the primary set, it follows from the remarks at the end of Theorem 148 that Cj and P l determine the general threefold W equally with a and P. Also since Cj lies in Q it must lie in W' t and so, by Theorem 150, every general plane which passes through an element of Cj and is parallel to P l must lie in W. But the general threefold W is the aggregate of all elements of P l and of all general planes parallel to P! which intersect C L , and so every element of W must lie in W. But, since Q and b both lie in W, it follows by Theorem 150 that every general plane which passes an element of 6 and is parallel to Q must lie in W. Since however the general threefold W is the aggregate of all elements of Q and of all general planes parallel to Q which intersect 6, it follows that every element of W must lie in W. Thus the general threefolds W and W consist of the same set of elements and are therefore identical. Thus Q and b determine W as was to be proved. 230 THEOKY OF TIME AND SPACE REMARKS. It follows directly from the above theorem that any four distinct elements which do not all lie in one general plane determine a general threefold containing them. For let A, B, C, D be four distinct elements which do not all lie in one general plane. Then no three of them can lie in one general line. Let Q be the general plane containing A, B and G and let b be the general line DA. Then b cannot have any other element than A in common with Q, for then D would have to lie in Q along with A, B and C contrary to hypothesis. Thus b intersects Q. Let W be the general threefold determined by Q and b and let W be any general threefold containing A, B, G and D. Then since W contains A, B and C it follows by Theorem 149 that W contains Q. Also by Theorem 148, since W' contains A and D it contains b. Thus by Theorem 151 the general threefold W is identical with W: that is to say is identical with one definite general threefold. Again it is clear that : any three distinct general lines having a common element and not all lying in one general plane determine a general threefold containing them. THEOREM 152. If two distinct general planes P and Q lie in a general threefold W, then if P and Q have one element in common they have a second element in common. Let A be any element in P and let B be any element which lies in W but not in P. Let the general line AB be denoted by a. Then a intersects P and since it has two distinct elements in common with W it follows that a lies in W. Then by Theorem 151, P and a may be taken as determining TFand any element of W lies either in P or in a general plane parallel to P and intersecting a. If now we call this set of mutually parallel general planes the "primary set" we have already seen in proving Theorem 151 that Q must either be identical with some general plane of the primary set or else must have a general line in common with each general plane of the primary set. A THEORY OF TIME AND SPACE 231 But now, since P and Q are supposed to be distinct, Q cannot be identical with P, and since Q is supposed to have an element in common with P, it follows that Q is not parallel to P. Thus Q cannot be identical with any general plane of the primary set and therefore must have a general line in common with each of them including P. Thus P and Q must have a second element in common. REMARKS. It is further evident from the above considerations that if two dis- tinct general planes P and Q both lie in a general threefold W, then if P and Q have no element in common they must be parallel to one another. Now we have already seen that we can have a separation plane S and an acceleration plane P having an element in common and which are completely normal to one another. We have seen that in this case P and S cannot have a second element in common. It follows that P and S cannot lie in one general threefold. Now let a l and a z be any two distinct general lines lying in P and passing through 0. Then S and a l determine a general threefold, say TFj, while S and a 2 determine a general threefold, say W 2 . Now W l and W 2 must be distinct, for if W 2 were identical with W 1} then Wi would contain both a x and a 2 and would therefore contain P. But Wi contains S and so this is impossible. Thus W l and W 2 are distinct general threefolds each of which con- tains the separation plane S. Since there are an infinite number of general lines lying in P and passing through it follows that there are an infinite number of general threefolds which all contain any separation plane S. Similarly there are an infinite number of general threefolds which all contain any acceleration plane P. Without Post. XIX or some equivalent, we cannot from our remaining postulates show that there is more than one general three- fold ; for the proof of the existence of an acceleration plane which is completely normal to a separation plane depends upon Post. XIX. THEOREM 153. If a general plane P and a general line a both lie in a general three- fold W and if a does not lie in P, then either a is parallel to a general line in P or else has one single element in common with P. 232 - A THEORY OF TIME AND SPACE Let B be any element lying in P but not in a. Then a and B determine a general plane, say Q, which must lie in W, since it contains three elements in common with W which do not all lie in one general line. . But since P and Q have the element B in common and both lie in W, therefore by Theorem 152 they have a general line in common which we may denote by b. Since then b must pass through the element B which does not lie in a, it follows that a and b are two distinct general lines lying in Q and must therefore either be parallel to one another, or else have one element in common, which is also an element of P. Thus a is either parallel to a general line in P or has an element in common with P. Further, a cannot have more than one element in common with P, since then it would require to lie in P. THEOREM 154. Ifa,b and c be any three distinct general lines having an element in common, but not all lying in one general plane, and if a general line d, also passing through 0, be normal to a, b and c, then d is normal to every general line in the general threefold containing a. b and c. Let P be the general plane containing b and c. Then a intersects P in and so P and a determine a general three- fold, say W, containing a, b and c. Consider now any general line e in W which passes through but is distinct from a, b and c. Then a and e determine a general plane, say Q, which by Theorem 149 must lie in W. , Further Q cannot be identical with P, since Q contains a but P does not contain it. Again Q and P have the element in common and therefore by Theorem 152 they have a general line, say /, in common which passes through 0. Now since d is normal to the two intersecting general lines b and c, it follows that d is normal to every general line in P and therefore is normal to f. Again, since d is normal to the two intersecting general lines a and f t it follows that d is normal to every general line in Q and therefore is normal to e. But e is any general line in W which passes through but is A THEORY OF TIME AND SPACE 233 distinct from a, 6 and c and so d is normal to every general line in W which passes through 0. Next let e be any general line in W which does not pass through and let e be the general line through parallel to e. Then by Theorem 150 e' must also lie in W and so by the first case d is normal to e' and therefore also normal to e. Thus d is normal to every general line in W as was to be proved. Definition. A general line which is normal to every general line in a general threefold will be said to be normal to the general threefold. Since, by Theorem 147, if three distinct general lines not all lying in one general plane have an element in common there is at least one general line passing through and normal to all three, it follows that through any element of a general threefold there is always at least one general line which is normal to the general threefold. THE THREE TYPES OF GENERAL THREEFOLD. As in the case of general lines and general planes there are three types of each, so too there are three types of general threefold. This may be shown in the following way : If S be any separation plane and be any element in it, there is an acceleration plane, say P, which passes through and is completely normal to S. Now if or be any general line in P which passes through 0, then a must be normal to S and must intersect it. But a may be either : (1) a separation line, or (2) an optical line, or (3) an inertia line, and if a general threefold be determined by $ and a, then these three cases give rise to the three different types. Let W be the general threefold determined by a and S and consider first the case where a is a separation line. If now e be any general line in W which passes through and is distinct from a, then a and e determine a general plane Q, which lies in W and since Q has the element in common with S it must have a general line, say/, in common with S. Now /must pass through and since it lies in S therefore a must be normal to/. But a and / are both separation lines and we already know that if 234 A THEORY OF TIME AND SPACE two intersecting separation lines are normal to one another they must lie in a separation plane. Thus Q must be a separation plane and therefore e must be a separa- tion line. Thus every general line in W which passes through must be a separation line. If e' be any other general line in W which does not pass through 0, then there is a general line through parallel to e' and which by Theorem 150 must also lie in W and therefore must be a separation line. But a general line parallel to a separation line must itself be a separation line and so e' is a separation line. Thus every general line in W is a separation line and so no element of W is either before or after any other element. It also follows from this that every general plane in W must be a separation plane. Consider next the case where a is an optical line. As before let e be any general line in W which passes through and is distinct from a. Then a and e determine a general plane Q which has a general line yin common with S. As before a is normal to /, but in this case a is an optical line while / is a separation line and we know that in these circumstances a and / must lie in an optical plane. Thus Q must be an optical plane and since there is only one optical line in an optical plane which passes through any element of it and all other general lines in it which pass through that element are separation lines, it follows that e must be a separation line. Again let e be any other general line in W which does not pass through 0. Then there is a general line through parallel to e and this general line must either be the optical line a or a separation line. Thus e' must be either an optical line or a separation line. Again if 0' be any element of W distinct from then 0' may or may not lie in a. If 0' does not lie in a then 00' is a separation line and there is an optical line through 0' parallel to a and which by Theorem 150 must lie in W f Thus there is at least one optical line passing through any element of W and lying in W. Let e' be any general line in W which passes through 0' but riot through 0, and which is not parallel to a. A THEORY OF TIME AND SPACE 235 Then the general line through parallel to e cannot be identical with a and therefore must be a separation line. Thus e' must be a separation line. It follows that of all the general lines passing through any given element of W and lying in W one and only one is an optical line and all the others are separation lines. Further all the optical lines in W are parallel to one another. Since there are two optical lines in any acceleration plane which pass through any element of it, it follows that no acceleration plane can lie in W. Thus every general plane in W must be either a separation plane or an optical plane. It follows that all the optical lines in W being parallel to one another must be neutral-parallels. Consider finally the case where a is an inertia line. As before let e be any general line in W which passes through and is distinct from a. Then a and e determine a general plane Q, which lies in W and, since a is an inertia line, Q must be an acceleration plane. Thus e may be either an inertia line, an optical line, or a separation line. If 0' be any element in W which is distinct from and if d be any general line passing through and lying in W, but distinct from 00', then through 0' there is a general line parallel to d, which must lie in W and must be of the same type as d. Thus through any element of W there are general lines of all three types lying in W. Again iff be any general line lying in S and passing through then, since a is an inertia line, a and f must lie in an acceleration plane, say R. Now since there are an infinite number of general lines such as f which lie in S and pass through there must be an infinite number of acceleration planes such as R which are all distinct but have the inertia line a in common. In any one of these acceleration planes such as R there are two and only two optical lines which pass through 0. All these optical lines must be distinct since the acceleration planes have only an inertia line in common, and so there are an infinite number of optical lines passing through and lying in W. Further any optical line which passes through and lies in W must clearly lie in one of this set of acceleration planes. 236 A THEORY OF TIME AND SPACE Again if 0' be any element of W distinct from and if g be any optical line passing through and lying in W, but distinct from 00', then there is an optical line through 0' parallel to g and lying in W. The general line 00' either may or may not itself be an optical line. Thus through any element of W there are an infinite number of optical lines which lie in W. Now we have already seen that W contains the separation plane 8 and also contains acceleration planes and we can easily show that it also contains optical planes. Thus let P be any acceleration plane in W and let A be any element in W but not in P. Then through A there is an acceleration plane parallel to P which we may call P'. Let B be the representative of A in P and let GI and c 2 be the two generators of P which pass through B. Then A is neither before nor after any element of either c x or c 2 and so A and ^ lie in one optical plane, say T l} while A and c 2 He in another optical plane, say T 2 . But Tj and T 2 each contain three elements in common with W which do not all lie in one general line and so, by Theorem 149, both 2\ and T 2 lie in W. Thus W contains all three types of general plane. We thus see that there are at least three types of general threefold and we have investigated a few of their characteristic properties. We have next to show that any general threefold must belong to one of these three types. Since any four distinct elements which do not all lie in one general plane lie in one and only one general threefold it will be sufficient if we examine the nature of any such general threefold. SETS OF FOUR ELEMENTS WHICH DETERMINE THE DIFFERENT TYPES OF GENERAL THREEFOLD. Let A, B, 0, D be any four distinct elements which do not all lie in one general plane. Then no three of them can lie in one general line and A, B and G must determine a general plane which we shall call P. Now P may be either : (1) an acceleration plane, or (2) an optical plane, or (3) a separation plane. A THEORY OF TIME AND SPACE 23*7 Suppose first that P is an acceleration plane and that D is any element outside it. Let W be the general threefold containing A, B, C and D and which must evidently contain P. Then by Theorem 144 there is one single general line passing through D and normal to P which has an element in common with P. Let this element be denoted by and let a be any inertia line in P which passes through 0, while b is the separation line in P and passing through which is normal to a. Then since a is an inertia line the general line DO which is normal to it must be a separation line. But DO is also normal to b and since we know that two intersecting separation lines which are normal to one another must lie in a separation plane, it follows that DO and b lie in a separation plane which we shall call 8. Now S contains DO and b and therefore contains three elements in common with W which do not all lie in one general line. It follows by Theorem 149 that S lies in W. Thus by Theorem 151 the general plane determined by S and 'a is identical with the general plane determined by P and DO. This latter is however identical with W and so S and a determine W. But a is an inertia line which is normal to the two intersecting separation lines DO and b which lie in S and therefore a is normal to S. Thus the general threefold W is of the third type. Further it is evident that if any general threefold contains an acceleration plane it must belong to the third type. Next consider the case where P is an optical plane and D an element outside it. Two sub-cases arise here : we may have D before or after some element of P, or D neither before nor after any element of P. We shall suppose first that D is either before or after some element of P and we shall denote the generator of P which passes through this element by a. If as before W denote the general threefold containing A, B, C and D, then W will contain P and will therefore contain a. But since a is an optical line and D is an element which does not lie in a but is either before or after some element of a, it follows that a and D lie in an acceleration plane, say Q. 238 A THEORY OF TIME AND SPACE But Q contains three elements in common with W which do not all lie in one general line and so Q must lie in W. But Q is an acceleration plane and so it follows that in this case also W is a general threefold of the third type. We shall next take the case where P is an optical plane and the element I) is neither before nor after any element of P. Let b be any separation line in P and a be any optical line in P and let b and a intersect in the element 0. If as before W denote the general threefold containing A, B, C and D, then W will contain P and therefore will contain a and b. Now, since D is neither before nor after any element of P, it is neither before nor after any element of b and so D and b lie in a separa- tion plane which we may call S. Further, since 8 has three elements in common with W which do not all lie in one general line, it follows that S lies in W. Again D and a must lie in an optical plane and, since DO is a separation line while a is an optical line, it follows that a is normal to DO. But a must also be normal to b for a similar reason and so, since DO and b are intersecting separation lines in 8, it follows that a is normal toS. But by Theorem 151 the general threefold determined by 8 and a is identical with that determined by P and DO which again is identical with W. Since however S is a separation plane while a is an optical line normal to it, it follows that W is in this case a general threefold of the second type. Consider next the case where P is a separation plane and as in the previous cases let W denote the general threefold containing A, B, C and D and therefore also containing P. Three sub-cases occur here : thus we may have D neither before nor after any element of P, or D either before or after one single element of P, or D either before or after at least two elements of P. Now by Theorem 144 there is one single general line passing through D and normal to P which has an element in common with P. Let be this element. Then DO may be either a separation line, an optical line, or an inertia line. A THEORY OF TIME AND SPACE 239 Consider first the case where D is neither before nor after any element of P. Then D is neither before nor after and so DO is a separation line and the general threefold W is of the first type. Next consider the case where D is either before or after one single element of P and denote this element by 0'. Let b and c be two distinct separation lines in P and passing through 0' ' . Then DO' and b lie in an optical plane and DO' and c lie in another optical plane. Since D is either before or after 0', it follows that DO' is an optical line and therefore is normal to both b and c. Since b and c intersect one another, it follows that DO' is normal to P and therefore 0' must be identical with 0. Thus in this case the general threefold W is of the second type. Next let D be either before or after at least two distinct elements of P, say E and F. Then .fi'.F is a separation line and Z) does not lie in it, and so the three elements D, E and F lie in an acceleration plane, say Q. But D, E and F are elements in W and therefore Q must lie in W. Thus since Q is an acceleration plane it follows that the general threefold W belongs in this case to the third type. This exhausts all the possibilities which are open and so we see that any general threefold whatever must be of one of the three types which we have considered. We shall accordingly give special names to these three types. Definition. If a separation line a intersects a separation plane S and is normal to it, then the aggregate of all elements of $ and of all separation planes parallel to S which intersect a will be called a separa- tion threefold. Definition. If an optical line a intersects a separation plane S and is normal to it, then the aggregate of all elements of S and of all separation planes parallel to S which intersect a will be called an optical threefold. Definition. If an inertia line a intersects a separation plane S and is normal to it, then the aggregate of all elements of S and of all separation planes parallel to 8 which intersect a will be called a rotation threefold. 240 A THEORY OF TIME AND SPACE We are now in a position to introduce a new postulate which limits the number of dimensions of our set of elements. POSTULATE XX. If W be any optical threefold, then any element of the set must be either before or after some element of W. If W be any optical threefold and A be any element of W, then through A there is one single optical line which lies in W and A is before certain elements of this optical line and is after certain others. Thus in this case A is before certain elements of W and after certain other elements of W. If, on the other hand, A be any element outside W, then, by Post. XX, A must be either before some element of W or after some element of W. If A be before the element B of W, then there is an optical line, say b, passing through B and lying in W. If 6' be the optical line through A parallel to b then b' will be a before-parallel of b. But any element of W which does not lie in b must lie in an optical line c neutrally parallel to b and lying in W and so by Theorem 25 b' must be a before-parallel of c. Thus A must be before certain elements of c and since A is not an element of W and therefore not an element of c, it follows that A cannot be after any element of c. Thus A is before elements of every optical line in W and is not after any element of W. Similarly if A be any element outside W and after some element of W, then A will be after elements of every optical line in W and will not be before any element of W. Definition. An optical line which lies in an optical or rotation threefold will be spoken of as a generator of the optical threefold or rotation threefold, as the case may be. THEOREM 155. If P be an optical plane -and be any element in it, there is only one general plane passing through and completely normal to P. Let a be the generator of P which passes through and let b be any separation line in P and passing through 0. Then we already know that there is at least one optical plane, say Q, which passes through and is completely normal to P. Further this optical plane Q contains a. A THEORY OF TIME AND SPACE 241 Now let c be any separation line passing through and lying in Q. Then c is normal to both a and b. Let d be any other general line which passes through and is normal to P and let X be any element in d distinct from 0. Now, P and c determine an optical threefold since no element of c with the exception of is either before or after any element of P. Let this optical threefold be denoted by W. Then, by Post. XX, the element X is either before or after some element of W. If X were outside W, then, as we have seen, X would be before or after elements of every generator of W and therefore before or after elements of a. Since the general line d could not then be either identical with a or be a separation line normal to a, it follows that d could not be normal to P, contrary to hypothesis. Thus X must lie in W and therefore d must lie in W. But now c and d determine a general plane Q f which has three elements in common with W which are not all in one general line and therefore Q' must lie in W. Further since P and Q' have the element in common, therefore by Theorem 152 they have a general line in common, which we may call a. But now b is normal to both p and d and, since these intersect in 0, it follows that b is normal to Qf and therefore normal to a. But b and a lie in the optical plane P and since 6 is a separation line a' must be an optical line. Thus since a' passes through it must be identical with a and so Q' must be identical with Q. It follows that d lies in Q and accordingly every general line which passes through and is normal to P must lie in Q. Thus any general plane which passes through and is completely normal to P must be identical with Q, or there is only one general plane passing through and completely normal to P. THEOKEM 156. If P be a separation plane and be any element in it, there is only one general plane passing through and completely normal to P. We already know that there is at least one acceleration plane, say Q, passing through and completely normal to P. Suppose, if possible, that there is a general line, say a, passing through and normal to P but not lying in Q. Then Q and a will determine a rotation threefold, say W. 11. 16 242 A THEORY OF TIME AND SPACE If 6 and c be any two distinct general lines in P which both pass through 0, then 6 and c will each be normal to three distinct general lines passing through and lying in W, but not all lying in one general plane. Thus, by Theorem 154, 6 and c must each be normal to every general line in W. But now we have seen that any rotation threefold contains optical planes and if we take any such optical plane it would either pass through or else there would be a parallel optical plane passing through which, by Theorem 150, must also lie in W. Thus there would always be at least one optical plane, say R, passing through and lying in W. But then both b and c would be normal to every general line in R and since b and c are intersecting general lines in P we should have every general line in R normal to every general line in P. Thus P would be completely normal to R and would pass through the element in it. But P is a separation plane and we already know by Theorem 155 that there could be only one general plane passing through and completely normal to R, and that one must itself be an optical plane and could not be a separation plane. Thus the assumption that there is.a general line a passing through and normal to P but not lying in Q, leads to a contradiction and therefore is not true. It follows that every general line passing through and normal to P must lie in Q. Thus Q is the only general plane which passes through and is completely normal to P. Thus the theorem is proved. THEOREM 157. If P be an acceleration plane and be any element in it, there is only one general plane passing through and completely normal to P. We already know that there is at least one separation plane, say Q, passing through and completely normal to P. Let b be any separation line in Q which passes through and let c be the one separation line lying in Q and passing through which is normal to b. Suppose now, if possible, that there is a general line d passing through and normal to P but not lying in Q. Then, since any inertia line in P would be normal to d, it would A THEORY OF TIME AND SPACE 243 follow that d must be a separation line and since then any inertia line in P which passed through would be conjugate to the two intersecting separation lines b and d it would follow, as a consequence of Theorem 99, that b and d must lie in a separation plane, say R. Now R would require to be distinct from Q, since d is supposed not to lie in Q. Since however we should then have two intersecting separation lines in R : namely b and d, normal to P, it would follow that R was com- pletely normal to P. Now suppose e to be the one separation line in R and passing through which would be normal to b. Then c and e would be distinct separation lines since b is the only general line common to Q and R. Further since any inertia line in P would be normal to both c and e it follows that c and e would lie in a separation plane, say S. But now P and b would determine a rotation threefold, say W, and since both c and e would be normal to P and to the separation line b (which does not lie in P) it follows, by Theorem 154, that both c and e would be normal to every general line in W. But, as we have seen, there is at least one optical plane passing through and lying in W and if T be such an optical plane we should have both c and e normal to T. Thus the separation plane S would be completely normal to T and this we know by Theorem 155 is impossible, since only an optical plane can have an element in common with an optical plane and be completely normal to it. It follows that no such general line as d can exist and so every general line which passes through and is normal to P must lie in Q. Thus Q is the only general plane which passes through and is completely normal to P and so the theorem is proved. REMARKS. Combining these last three theorems we get the general result : If P be any general plane and be cany element in it, there is one and only one general plane Q passing through and completely normal to P. Further : If P be an optical plane, Q is an optical plane. If P be a separation plane, Q is an acceleration plane. If P be an acceleration plane, Q is a separation plane. 162 244 A THEORY OF TIME AND SPACE Again we know that if 0' be any element outside P there is at least one general plane through 0' which is completely normal to P. If we call this general plane Q', then Q' is either identical with Q or parallel to Q according as 0' does or does not lie in Q. Now there cannot be any other general plane than Q' which passes through 0' and is completely normal to P. For if Q" were such another general plane it would either pass through or else there would be a general plane parallel to Q" and passing through 0, which would also be completely normal to P. Thus there would be two distinct general planes passing through and completely normal to P ; which is impossible. Thus we can say : If P be any general plane and be any element of the set, there is one and only one general plane passing through and completely normal to P. THEOREM 158. (1) If P be an acceleration or separation plane and be any element outside it, then the general plane through and completely normal to P has one single element in common with P. (2) If P be an optical plane and be any element outside it, then the optical plane through and completely normal to P has an optical line in common with P if be neither before nor after any element of P and has no element in common with P if be either before or after any element of P. Let P be an acceleration or separation plane and any element outside it. Then by Theorem 144 there is one single general line passing through and normal to P which has an element in common with P. Let 0' be this element. Then by Theorem 157 or 156 there is one single separation or acceleration plane, say Q, which passes through 0' and is completely normal to P; and Q has only one element in common with P. Thus Q must contain the general line O'O and therefore it must be identical with the one single 'general plane which passes through and is completely normal to P. Thus the general plane through and completely normal to P has one single element in common with P, and so the first part of the theorem is proved. Next let P be an optical plane and any element outside it. Then, by Theorem 145, if be neither before nor after any element A THEORY OF TIME AND SPACE 245 of P there is one single generator of P such that every general line which passes through and intersects this generator is normal to P. Thus if a be this generator and 0' be any element in a, the general lines a and 00' determine an optical plane, say Q, which passes through 0, is completely normal to P and has the optical line a in common with P. Since there is only one optical plane through and completely normal to P, this must be identical with Q and it has the optical line a in common with P if be neither before nor after any element of P. Next consider the case where is either before or after some J J element of P. Here, by Theorem 145, there is no general line passing through and having an element in common with P which is normal to P. Thus the optical plane through and completely normal to P can, in this case, have no element in common with P. Thus all parts of the theorem are proved. THEOREM 159. // a general line a have an element in common with a general threefold W, then there is at least one general plane lying in W and passing through to which a is normal. Let Q be any general plane in W and passing through 0. Then by Theorem 146 there is at least one general line, say b, passing through and lying in Q which is normal to a. Let c be any other general line distinct from b, lying in Q and passing through and let A be any element lying in W but not in Q. Then c and A determine a general plane, say R, which must lie in W since it contains three elements in common with W which do not all lie in one general line. Further R must be distinct from Q, since R contains the element A which does not lie in Q, and moreover R does not contain 6. But again, by Theorem 146, there is at least one general line, say d, passing through and lying in R which is normal to a. Then d must be distinct from b which it intersects in the element and so d and b determine a general plane, say P,. which must lie in W since it contains three elements in common with W which do not all lie in one general line. But since a is normal to the two intersecting general lines d and b, therefore a is normal to P, and thus there is at least one general plane P lying in W and passing through to which a is normal. 246 A THEORY OF TIME AND SPACE It is to be observed in connection with the above theorem that if a were normal to any other general line passing through and lying in W but not in P, then, by Theorem 154, a would be normal to every general line in W. It is also to be observed that the above theorem holds both when the general line a lies in W and when it has only one element in common with W. THEOREM 160. (1) If W be a general threefold and P be a general plane lying in W, while is any element in P, then there is at least one general line passing through and lying in W which is normal to P. (2) There is only one such general line except in the case where W is an optical threefold and P an optical plane, in which case there are an infinite number. To prove the first part of the theorem consider first the case where P is an optical plane. In this case the generator of P which passes through is normal to P and lies in W. Next let P be an acceleration or separation plane and let A be any element lying in W but not in P. Then by Theorem 144 there is one single general line passing through A and normal to P which has an element in common with P. Let B be this element. Then the general line AB has two distinct elements in common with W and therefore lies in W, but does not lie in P. If B should be identical with 0, then AB passes through 0, lies. in W and is normal to P. If B be not identical with 0, then there is a general line passing through and parallel to AB which must also be normal to P. But by Theorem 150 this general line must also lie in W. Thus in all cases there is at least one general line passing through and lying in W which is normal to P. Proceeding now to the second part of the theorem, let us consider first the case where P is either an acceleration or separation plane. Suppose, if possible, that a and b are two distinct general lines both of which pass through 0, lie in W and are normal to P. Then a and b would determine a general plane, say Q, which would have three elements in common with W not all lying in one general line, and so Q would lie in W. A THEORY OF TIME AND SPACE 247 Thus by Theorem 152, since Q and P have the element in common they would have a general line in common. But, since Q is supposed to contain the two intersecting general lines a and b each of which is normal to P, it would follow that Q must be completely normal to P, and since P is by hypothesis either an acceleration or separation plane, it would follow that Q must be either a separation or acceleration plane. But we already know that if an acceleration plane and a separation plane be completely normal to one another, they cannot have more than one element in common. Thus P and Q could not have a general line in common and so the supposition that more than one general line can pass through 0, lie in W, and be normal to P leads in this case to a contradiction and there- fore is not true. Thus if P be an acceleration or separation plane there cannot be more than one such general line. Suppose next that P is an optical plane and let a be the generator of P which passes through and let b be any general line lying in W but not in P and which passes through 0. Let A be any element in b distinct from 0. Then if A be either before or- after any element of P the general threefold W must be a rotation threefold and a and A must lie in an acceleration plane. Thus since a is an optical line and since b intersects a and lies in an acceleration plane with it, it follows that b cannot be normal to a and therefore cannot be normal to P. Further, since a is the only general line in P which passes through and is normal to P, it follows that in this case there is only one general line in W which passes through and is normal to P. Consider now the case where the element A is neither before nor after any element of P. In this case the general threefold W must be an optical threefold and the general line b must be a separation line. Let c be any general line in P and passing through but distinct from a. Then c is a separation line and b and c determine a separation plane, say S, which must lie in W. Now a must, in this case, be normal to both b and c and therefore normal to S. Let d be the one single separation line in 8 which passes through and is normal to c. 248 A THEORY OF TIME AND SPACE Then d is normal to both a and c and therefore is normal to P. If then Q be the general plane containing a and d, it contains two intersecting general lines each of which is normal to P and therefore it follows that Q is completely normal to P. Thus every general line which passes through and lies in Q must be normal to P. But since a and d are two intersecting general lines which both lie in W, it follows that Q contains three distinct elements in common with W which do not all lie in one general line and therefore Q must lie in W. Thrfs in this case there are an infinite number of general lines which pass through 0, lie in W, and are normal to P. This exhausts all the different cases and so the second part of the theorem is proved. THEOREM 161. If W be a general threefold and be any element which does not lie in it, then : (1) If W be a rotation or separation threefold there is one single general line passing through and normal to W which has an element in common with W. (2) If W be an optical threefold there is no general line passing through and normal to W which has an element in common with W. If W be a rotation threefold it contains inertia lines. Let / be any inertia line in W. Then /and lie in an acceleration plane, say R, and if a be any inertia line in R and passing through but not parallel to/ then a and /will intersect in some element, say A, which is an element of W. If on the other hand W be a separation threefold, let A be any element in W and let a be the general line OA. Now whether W be a rotation or separation threefold, it follows by Theorem 159 that there is at least one general plane, say P, lying in W and passing through A to which a is normal. Now if W be a rotation threefold, a has been selected so as to be an inertia line and, since only separation lines can be normal to an inertia line, it follows that P is a separation plane. If on the other hand W be a separation threefold it can contain no other type of general plane, and so in this case also P must be a separation plane. Now by Theorem 160, whether W be a rotation or a separation threefold there is one general line, say b, passing through A and lying A THEORY OF TIME AND SPACE 249 in W which is normal to P t and, since P is a separation plane, b must intersect it. Now a and b must be distinct, since b lies in W while a can only have the one element A in common with W. Thus a and b lie in a general plane, say Q, and, since Q contains two intersecting general lines each of which is normal to P, it follows that Q must be completely normal to P. Further, since P is a separation plane, it follows that Q is an acceleration plane. Now since b is normal to P and lies in W, the general threefold W might be determined by P and b and we know that if b be a separation line, W must be a separation threefold, while if b be an optical line, W must be an optical threefold and if b be an inertia line W must be a rotation threefold. It follows that if W be a rotation threefold then b must be an inertia line, while if W be a separation threefold, b must be a separation line. But now in either of these cases there is a general line, say c, which passes through 0, lies in Q and is normal to 6, and in both cases c intersects b in some element, say 0', which is an element of W. Further c will be a separation line if 6 be an inertia line : that is, if W be a rotation threefold ; while c will be an inertia line if b be a separation line : that is, if W be a separation threefold. Now since c lies in Q and since Q is completely normal to P, it follows that c is normal to P. If then P' be a general plane passing through 0' and parallel to P it follows by Theorem 150 that P must also lie in W. Thus c will be normal to P' and to the general line b which inter- sects P / in 0'. It is thus evident that c is normal to three distinct general lines in W which have the element 0' in common and which do not all lie in one general plane and therefore, by Theorem 154, c is normal to W. Also c passes through and has the element 0' in common with W. Now there can be no other general line passing through and normal to W', for suppose, if possible, that c is such another general line. Then c and c' would determine a general plane, say T, which would contain two intersecting general lines each of which would be normal to every general line in W and therefore normal to every general plane in W. Thus T would be completely normal to every general plane in W. 250 - A THEORY OF TIME AND SPACE But through any element of W there passes more than one general plane which lies in W and so we should have more than one general plane passing through any element of W and completely normal to T, which, as we have seen, is impossible. Thus the supposition that more than one general line can pass through and be normal to W leads to a contradiction and therefore is not true. Thus there is one and only one general line which passes through and is normal to W when W is a rotation or separation threefold, and this general line has an element in common with W. Suppose next that W is an optical threefold. Then, by Post. XX, must be either before or after some element of W and, as we have seen, if be before any element of W it must be before elements of every generator of W, while if be after any element of W it must be after elements of every generator of W. If then a be any general line which passes through and has an element A in common with W, then A must lie in some generator of W, say /, and / and a will lie in an acceleration plane. But since f is an optical line and a is a general line intersecting f and lying in an acceleration plane with it, it follows that a cannot be normal to / and therefore cannot be normal to W. Thus in this case there is no general line passing through and normal to W which has an element in common with W. Thus both parts of the theorem are proved. REMARKS. If W be a rotation or separation threefold and be any element in W, it is easy to see that there is one and only one general line passing through and normal to W. For if A be any element outside W and a be the one general line passing through A and normal to W, then a will have an element B in common with W. If B should coincide with 0, then a is a general line passing through and normal to W. If B does not coincide with 0, then a general line a' passing through and parallel to a must be normal to every general line in W and must therefore be normal to W. Thus we have shown that there is at least one general line passing through and normal to W and the same considerations employed in the last theorem show that there is only one such general line. Further the general line through normal to W cannot have more A THEORY OF TIME AND SPACE 251 than the one element in common with W\ for if it had a second element in common with W it would lie entirely in W y and by Theorem 150 it would follow that a must lie in W, contrary to the hypothesis that the element A of a lies outside W. In this respect an optical threefold is quite different. Through any element in an optical threefold W there passes one single generator of W, say a. Now a is normal to any separation line in W and is also normal to itself. Thus a is normal to W and passes through 0, but lies entirely in W. If 0' be any element outside W and a be an optical line parallel to a, then a! is also normal to W but can have no element in common with W. We may also show, by similar considerations to those employed in the case of a rotation or separation threefold, that there cannot be more than one general line passing through any element and normal to a given optical threefold. Thus for all three types of general threefold we have the result: If W be any general threefold and be any element of the set, there is one and only one general line passing through and normal to W. THEOREM 162. If a be a general line and be any element in it, there is one and only one general threefold passing through and normal to a. Let P be any acceleration plane containing a and let Q be the separation plane passing through and completely normal to P. Then P and Q have only the one element in common. Now through and lying in P there is one single general line, say b, which is normal to a. But b and Q can have only one element in common and therefore they determine a general threefold, say W. Since however a is normal to every general line in Q and is also normal to the general line b which passes through and does not lie in Q, it follows by Theorem 154 that a is normal to W. Thus there is at least one general threefold passing through and normal to a. We shall next show that every general line which passes through and is normal to a must lie in W. Since every such general line which lies in Q must lie in W, it will be sufficient to consider any general line c passing through 0, normal to a and not lying in Q. 252 A THEORY OF TIME AND SPACE Then c and Q determine a general threefold, say W, and by Theorem 160 there is at least one general line, say d, passing through and lying in W' which is normal to Q. Further, since Q is a separation plane, d must lie in the acceleration plane through which is completely normal to Q, and since there is only one such acceleration plane, it follows that d must lie in P. But since a is normal to c and Q it follows that a is normal to W and therefore is normal to d. But there is only one general line passing through and lying in P which is normal to a, and by hypothesis b is this general line. It follows that d must be identical with b and so, by Theorem 151, since d and Q must determine the same general threefold as do c and Q, it follows that W must be identical with W. Thus c must lie in W. But if there were any other general threefold distinct from W which passed through and was normal to a, such general threefold would require to contain a general line which passed through and was normal to a, but which did not lie in W and this we have shown to be impossible. Thus there is one and only one general threefold which passes through and is normal to a. REMARKS. In the above theorem it is to be observed that : if a be an inertia line, b must be a separation line ; if a be a separation line, b must be an inertia line ; while if a be an optical line, 6 must be the same optical line. Thus it follows that : if a be a general line and be any element in. it, while W is a general threefold passing through and normal to a, then: (1) If a be an inertia line, W is a separation threefold. (2) If a be a separation line, W is a rotation threefold. (3) If a be an optical line, W is an optical threefold containing a. On the other hand we have already seen that if W be a general threefold and be any element in it, there is one and only one general line a passing through and normal to W. Thus it follows that : (1) If W be a separation threefold, a is an inertia line. (2) If W be a rotation threefold, a is a separation line. (3) If W be an optical threefold, a is an optical line lying in W. A THEORY OF TIME AND SPACE 253 Again if a be a general line and be any element which does not lie in a, then, through there is one single general line, say of, which is parallel to a and is accordingly a general line of the same type. Thus through there is a general threefold which is normal to of and therefore also normal to a. Further there cannot be a second general threefold passing through and normal to a, for such general threefold would also be normal to a! and so we should have two general threefolds passing through and normal to a contrary to Theorem 162. Thus we can extend Theorem 162 and say: If a be a general line and be any element of the set, there is one and only one general threefold passing through and normal to a. THEOREM 163. If W be an optical threefold and A be any element outside it, then every optical line through A, except the one parallel to the generators of W, has one single element in common with W. Let a be the optical line through A parallel to the generators of W and let b be any such generator. Then, by Post. XX, A must be either before or after some element of W and we have already seen that if A be before an element of W it must be before elements of every generator of W ; while if A be after an element of W it must be after elements of every generator of W. Thus a must be either a before- or after-parallel of b. It will be sufficient to consider the case where a is a before -parallel of b since the proof in the other case is quite analogous. Then a and 6 lie in an acceleration plane and so there is one single optical line passing through A and intersecting b in some element, say B. If we call this optical line c, then c has the element B in common with W. If then d be any optical line passing through A but distinct from c and a, it follows, by Post. XII, that there is one single element in d, say D, which is neither before nor after any element of b. Now if D were outside W it would be either before or after elements of every generator of W, as we have already seen. Thus, since D is neither before nor after any element of the generator 6, it follows that D must lie in W. It follows that every optical line through A with the exception of a has at least one element in common with W. 254 A THEORY OF TIME AND SPACE But if any optical line has more than one element in common with W it must lie entirely in W, which is not possible for any optical line which passes through the element A. It follows that every optical line through A with the exception of a has one single element in common with W, as was to be proved. THEOREM 164. If W be a general threefold and A be any element outside it, then any general line through A is either parallel to a general line in W or else has one single element in common with W. It will be observed that the last theorem is a special case of this one. Let a be any general line which passes through A. Now a cannot have more than one element in common with W, for then it would require to lie entirely in W and therefore could not pass through A. Let B be a second element in a distinct from A. In case W be a rotation or separation threefold, let the general line through A normal to W meet W in the element A', as we have seen in Theorem 161 that it must. Now in case the general line a should coincide with A A it would have an element in common with W, and so we shall suppose it is distinct from it. Again let the general line through B normal to W meet W in the element B'. Then since B does not lie in A A' we must have BB' parallel to A A'. In case W be an optical threefold then, by Theorem 163, any optical line through A except the one parallel to the generators of W must have an element in common with W. Let any optical line through A which is not parallel to the generators of W meet W in the element A'. In case the general line a should coincide with A A' it would have an element in common with W , and so we shall suppose it is distinct from it. Let the optical line through B parallel to A A' meet W in the element B'. Now both in the cases where W is a rotation or separation threefold and where W is an optical threefold, since BB ' is parallel to A A', it follows that BB' and AA' lie in a general plane which we may call Q. But A'B' and a must also lie in Q, and therefore a is either parallel to A'B' or intersects A'B' in some element, say C. But A'B' has two distinct elements A' and B' in common with W, A THEORY OF TIME AND SPACE 255 and therefore A'B' must lie in W, and if the element C exists it must lie in W. Thus the general line a is either parallel to a general line in W or else a has one single element in common with W. Definition. If a general line and a general threefold have one single element in common, they will be said to intersect in that element. REMARKS. Since a separation threefold contains neither an inertia nor an optical line it is evident that it can contain no general line which is parallel to either of these. Thus it follows from the last theorem that : every inertia and every optical line intersects every separation threefold. Again, an optical threefold does not contain any inertia line, and all the optical lines which it contains are parallel to one another. Thus : every inertia line and every optical line which is not parallel to a generator of an optical threefold intersects the optical threefold. Analogous results to these may be deduced from Theorem 153, with regard to the intersection of certain types of general lines with certain types of general planes. Thus, since a separation plane contains neither an inertia nor an optical line, it follows from Theorem 153 that: if W be a rotation three- fold every inertia and every optical line in W intersects every separation plane in W. Similarly : if W be a rotation threefold, every inertia line in W and every optical line in W which is not parallel to a generator of an optical plane in W intersects the optical plane. Again : if W be an optical threefold every optical line in W intersects every separation plane in W. THEOREM 165. If W be a general threefold and P be a general plane which does not lie in W, then if P has one element in common with W, it has a general line in common with W. Let P and W have the element A in common and let B be any element in P which does not lie in W. Let 6 be any general line in P which passes through B but is distinct from BA. Then, by Theorem 164, b must either intersect W in some element, say C, or else b must be parallel to some general line, say b', which lies in W. 256 A THEORY OF TIME AND SPACE In the first case P and W have the two distinct elements A and G in common and therefore have the general line AC in common. In the second case a general line b" passing through A and parallel to b' or identical with it must lie in W. But b" must be parallel to b and since it passes through the element A of P it must lie in P. Thus in this case P and W have the general line b" in common and so the theorem holds in general. THEOREM 166. If Wi an d ^2 be two distinct general three/olds having an element A in common, then they have a general plane in common. Let B be any element which lies in W l but not in W 2 . Then the general line AB lies in W. Let Q and R be any two distinct general planes which contain the general line AB and which lie in TTj. Then Q does not lie in W% but has the element A in common with W 2 and therefore, by Theorem 165, Q has a general line, say a, in com- mon with TF 2 . Similarly R has a general line, say b, in common with W 2 . Now both a and 6 must be distinct from the general line AB since the latter does not lie in W 2 and, since Q and R have only the general line AB in common, it follows that 6 is distinct from a. Thus a and b are two general lines intersecting in A and each of them lying both in TFi and W 2 and so they determine a general plane, say P. But P contains three elements in common both with Wi and W 2 and which do not all lie in one general line and so P lies both in W 1 and F 2 . Thus Wi and W 2 have a general plane in common. THEOREM 167. If P l and P 2 be two general planes having no element in common, then through any element of either of them there is at least one general line lying in that general plane which is parallel to a general line in the other general plane. Let 0! be any element in P x and let 2 be any element in P 2 and let the general line O l 2 be denoted by a. Then P l and a determine a general threefold, say W l} while P 2 and a determine a general threefold, say W 2 . A THEORY OF TIME AND SPACE 257 If W 2 should be identical with W l} then P 1 and P 2 lie in one general threefold and, since they have no element in common, it follows by Theorem 153 that any general line in P l is parallel to a general line in P 2 , and so P l and P 2 are parallel to one another. If W 2 be not identical with W l then, since Wi and W 2 have all the elements of a in common, it follows, by Theorem 166, that they have a general plane, say Q, in common which must contain a. But now Q must be distinct from both P l and P 2 , for otherwise P^ or P 2 would contain a and so P l and P 2 would have an element in common, contrary to hypothesis. But now Pj and Q both lie in W l and they have the element O l in common, and therefore, by Theorem 152, they have a general line, say 6j, in common, which passes through O l . Similarly P 2 and Q have a general line, say 6 2 , in common, which passes through 2 . But since 6 X and b 2 lie in P l and P 2 respectively they can have no element in common, and, since they both lie in the general plane Q, they must be parallel to one another. Thus the theorem is proved. REMAKES. It is easy to see that if two general planes P! and P 2 have one single element in common, then no general line in P l can be parallel to any general line in ? 2 . For let ttj and # 2 be two general lines in P x and P 2 respectively, then a l cannot be parallel to a 2 if both pass through 0. Further, they cannot be parallel if one passes through and the other does not, for then they could not lie in one general plane. Finally they cannot be parallel if neither of them passes through 0, for then a general line a/ passing through and parallel to a 1 would lie in Pj and so could not be parallel to a. 2 as it would require to be if a a were parallel to a lt THEOREM 168. If W be a general threefold and be any element outside it, and, if further, a and b be two distinct general lines intersecting in and each of them parallel to a general line in W, then : (1) The general plane containing a and b has no element in common with W. (2) The general plane containing a and b is parallel to a general plane in W. R. 17 258 A THEORY OF TIME AND SPACE Neither a nor b can have any element in common with W, since, by Theorem 150, if either of them had an element in common with W, it would require to lie entirely in W and so could not contain the element 0. But, if P be the general plane containing a and b, any element in P must lie either in a or in a general line parallel to a and intersecting b. But every general line of this character must be parallel to the general line in W to which a is parallel, and therefore can have no element in common with W. Thus P can have no element in common with W. In order to prove the second part of the theorem, let a and &' be general lines in W to which a and b are respectively parallel. Then a' and b' either intersect, in which case they lie in a general plane which lies in W and is parallel to P, or else a general line, say 6", parallel to b', may be taken through any element of a' and then b" must lie in W, by Theorem 150. Thus in this case a and b" will lie in a general plane which will lie in W and be parallel to P. Thus in all cases P will be parallel to a general plane in W. Definition. If IT be a general threefold and if through any element A outside W a general line a be taken parallel to any general line in W, then the general line a will be said to be parallel to the general three- fold W. Definition. If W be a general threefold and if through any element A outside W a general plane P be taken parallel to any general plane in W, then the general plane P will be said to be parallel to the general threefold W. THEOREM 169. If W be a general threefold and be any element outside it, and if through there pass three general lines a, b, and c, which do not all lie in one general plane and which are respectively parallel to three general lines in W, then a, b and c determine a general threefold W, such that every general line in W is parallel to a general line in W. Let P be the general plane containing b and c. Then since a, b and c do not lie in one general plane it follows that a can only have the one element in common with P. Now the general line a can have no element in common with W, for then, since it is parallel to a general line in W, it would, by Theorem 150, require to lie in W and so could not contain the element 0. A THEORY OF TIME AND SPACE 259 Again, by Theorem 168, the general plane P can contain no element in common with W, nor can any general plane which is parallel to P and which intersects a. But now any element in W must either lie in P or in a general plane parallel to P and intersecting a. Thus no element in W can lie in W, and so no general line in W can have an element in common with W. Thus, by Theorem 164, any such general line must be parallel to a general line in W. Similarly any general line in TFmust be parallel to a general line in W. Definition. If W be a general threefold and if through any element A outside W three general lines be taken not all lying in one general plane but respectively parallel to three general lines in W, then the three general lines through A determine a general threefold which will be said to be parallel to W. REMARKS. Since a general line can only be parallel to a general line of the same kind, and since if one general threefold be parallel to another, any general line in either of them is parallel to a general line in the other, it follows that a general threefold can only be parallel to a general threefold of the same kind. Again if W be a general threefold and A be any element outside it, while W is a general threefold through A parallel to TF, then since W contains the general line through A parallel to any general line in W, the general threefold W must be uniquely determined when we know W and A. Also, since two distinct general lines which are parallel to a third general line are parallel to one another, it follows that : two distinct general three/olds which are parallel to a third general threefold are parallel to one another. Again, from Theorem 164, it is evident that : if W be a general threefold and A be any element outside it, then any general line through A must either lie in the general threefold passing through A and parallel to W, or else must intersect W. It is also to be noted that if a general threefold W be normal to a general line a, then any general threefold W parallel to W must also be normal to a. OTHER CASES OF NORMALITY. We have already considered the normality of a general line to a general line, a general plane, or a general* threefold. 172 260 A THEORY OF TIME AND SPACE We have also considered the complete normality of a general plane to a general plane. These are the only cases in our geometry in which the normality of Ti-folds is complete. Thus it is not possible to have every general line in a general plane P normal to every general line in a general threefold W, for then we should have more than one general plane passing through any element of W and completely normal to P, which, as we have seen, is impossible. For a similar reason we cannot have every general line in a general threefold W-^ normal to every general line in a general threefold TF 2 . The most we can have in these directions is to have a general plane P through any element of which there is one single general line lying in P which is normal to a general threefold W ; or to have a general threefold W through any element of which there is one single general line lying in W l which is normal to a general threefold TF 2 . Again, we may have a general plane P l through any element of which there is one single general line lying in P which is normal to a general plane P 2 . In these cases we have what may be described as partial normality. In ordinary three dimensional geometry the normality of two planes is of this partial character. Since it is desirable, so far as is possible, to have our nomenclature in conformity with that employed in ordinary geometry, we shall find it convenient to describe the general planes and general threefolds in the above cases as normal to one another. Thus we may have general planes normal to one another or com- pletely normal to one another : the expression ' normal ' by itself being taken to mean partially normal. In the case of a general plane or a general threefold which is partially normal to a general threefold the word normal may be used by itself without any ambiguity. Thus we have the following definitions : Definition. A general plane P 1 will be said to be normal to a general plane P 2 if through any element of P l there is one single general line lying in P 1 which is normal to P 2 . Definition. A general plane P will be said to be normal to a general threefold W if through any element of P there is one single general line lying in P which is normal to W. Definition. A general threefold W l will be said to be normal to A THEORY OF TIME AND SPACE 261 a general threefold W 2 if through any element of W^ there is one single general line lying in W l which is normal to W 2 . It is evident in the above three definitions we might substitute the word every for the word any. It is easy to see that if a general plane P l be normal to a general plane P 2 , then P 2 will be normal to P l . It will be sufficient to consider the case where P l and P 2 have an element A in common. Let a be the one single general line lying in P A and passing through A which is normal to P 2 and let 6 be any other general line in P x which passes through A. Then by Theorem 146 there is at least one general line, say c, passing through A and lying in P 2 which is normal to 6. But c must also be normal to a and therefore c must be normal to Pj. Further, there cannot be more than one general line passing through A and lying in P 2 which is normal to 6, unless P! fe completely normal and not merely normal to P 2 . Again, a separation line a may be normal to all three types of general plane and also may lie in all three types of general plane. If then a be normal to any general plane P l and if P 2 be any general plane containing a but not completely normal to Pj then P 2 will be normal to P lt Thus any type of general plane may be normal to any type of general plane. In particular, since an optical plane contains a series of optical lines which are normal to it, it follows that an optical plane is normal to itself. It is evident from the definitions that, if a general plane P be normal to a general threefold W, then P will be either simply normal or completely normal to any general plane in W. THEOREM 170. If a general plane P be normal to a general threefold W, then through any element of W there is one single general plane lying in W and completely normal to P. It will be sufficient if we prove the existence of one general plane, say Q, which lies in W and is completely normal to P. For let 0' be any element in W but not in Q and let Q' be a general plane through 0' parallel to Q. 262 A THEORY OF TIME AND SPACE Then, by Theorem 150, if Q lies in W, Q' must also lie in W, and if Q be completely normal to P then Q' must also be completely normal to P. Further, we know that there cannot be more than one general plane passing through a given element and completely normal to P. We shall therefore proceed to show that a general plane such as Q exists. Since P is normal to W, there is one single general line passing through any element of P and lying in P which is normal to W. Let a be any such general line. Then, if W be a rotation or separation threefold, a will intersect W in some element, say 0, while if W be an optical threefold, a will be an optical line either entirely in W, or entirely outside W. We shall first consider the case where W is a rotation or separation threefold. In this case a cannot lie in W, but has the element in common with it. It follows, by Theorem 165, that P and W have a general line in common which we shall call b and which must be distinct from a. But now, by Theorem 159, there is a general plane, say Q, passing through and lying in W to which b is normal. Since however a is normal to W, it follows that a also is normal to Q. f Thus we have the two intersecting general lines a and b both lying in P and both normal to every general line in Q. It follows that Q is completely normal to P. Consider next the case where W is an optical threefold. Here a must be either a generator of W, or else be parallel to a generator and lie completely outside W. Suppose first that a is completely outside W. If we take any general line in P which intersects a, then this general line either intersects W, or does not intersect W. If it does not intersect W, then, by Theorem 168, P must be parallel to some general plane, say P', lying in W. Now, since P contains an optical line, P' must contain an optical line, and since P' lies in the optical threefold W, it follows that P' (and therefore P) is an optical plane. Let a be any generator of P' and let be any element in a'. Then, by Theorem 162, there is one and only one general threefold passing through and normal to a' and this must be identical with W. Thus every general line which passes through and is normal to a must lie in W. A THEOKY OF TIME AND SPACE 263 But through there passes one single optical plane, say Q, which is completely normal to P 1 , and, since therefore a must be normal to every general line in Q, it follows that Q must lie in W. Then, since P is parallel to P', it follows that Q is completely normal to P. Next suppose that a lies outside W, as in the last case, but that P is not parallel to any general plane in W. In this case any general line in P which intersects a must inter- sect W. Thus P has an element, and therefore must have a general line, say a', in common with W. Further, since a has no element in common with W, it follows that a must be parallel to a and therefore must be an optical line. Now let be any element in. a' and let b be any general line passing through and lying in P, but distinct from a'. Then, by Theorem 159, there is at least one general plane, say Q, lying in W and passing through to which b is normal. But a is normal to W and therefore normal to Q, and so we have the two intersecting general lines a and 6 both normal to Q and both lying in P. Thus Q must be completely normal to P.' Next consider the cases where a is a generator of W. Here P may either lie in W or not lie in W. Suppose first that P lies in W. Then P contains an optical line and therefore is an optical plane, and, as we have already proved in connection with the second case, there is an optical plane, say Q, passing through any element of P and lying in W which is completely normal to P. Finally let us take the case where a is a generator of W, but where P does not lie in W. Here we have only to notice that if we take any optical line a lying in P and parallel to a, then a must lie entirely outside W, and this case becomes identical with the third one considered. Thus in all cases a general plane, such as Q, exists, and so the theorem is proved. THEOREM 171. If a general threefold W l contain a general line which is normal to a general threefold W 2 , then W z contains a general line which is normal to W l . Let a be a general line lying in Wi and which is normal to W 2 . 264 A THEOKY OF TIME AND SPACE Suppose first that W 2 is a rotation or separation threefold ; then a will be a separation or inertia line which intersects W 2 in some element, say 0. Then W l has the element in common with W 2) and so, by Theorem 166, Wi and W 2 have a general plane, say P, in common. Now, by Theorem 160, there is one single general line, say b, passing through and lying in W 2 which is normal to P. But, since a is normal to W 2 , it follows that b is normal to a. Further, since a is a separation or inertia line it cannot lie in the general plane P, to which it is normal. Thus b will be normal to three general lines in Wi all passing through and which do not all lie in one general plane. Thus b must be normal to W 1 . Next suppose that W 2 is an optical threefold. Then a must be an optical line and must either lie entirely in W 2 or else entirely outside W 2 , but parallel to a generator of W 2 . If a should lie in W 2 then clearly we might have W l identical with W 2 . If however a lies in W 2 but if Wi be not identical with W 2 , then there will be elements of W l which do not lie in W 2) and through any such element there would be a general line, say a', parallel to a and lying entirely outside W 2 but inside W^ Since a! would also be normal to W 2) it follows that this case may always be treated as a case in which there is a general line lying in W^ which is normal to W 2 but entirely outside W 2 . We shall therefore consider the case where a is an optical line lying entirely outside W 2 . In this case we may have W l either parallel to W 2 or not parallel to W 2 . If Wi be parallel to W 2 then W l must be an optical threefold and a must be normal to it. Thus any generator b of W 2 will be parallel to a and therefore normal to TFj. Thus, since b lies in W 2) the theorem holds also in this case. Next suppose a lies entirely outside W 2 , but that Wi is not parallel to W 2 . Then there must be some general line in Wi which is not parallel to any general line in W z and which must therefore intersect W 2 . Thus Wi and W 2 have an element, and therefore, by Theorem 166, have a general plane, say P, in common. Since a has no element in common with W 2) it can have no element A THEORY OF TIME AND SPACE 265 in common with P, and, since a and P both lie in W lt it follows, by Theorem 153, that a is parallel to a general line in P. Thus, since a is an optical line, P must contain an optical line, and, since P lies in W 2 , it follows that P must be an optical plane. Now let of be any generator of P and let be any element in of. Then we saw in the course of proving the last theorem that the one single optical plane which passes through and is completely normal to P must lie in the optical threefold W 2 . Let Q be this optical plane and let c be any general line which passes through and lies in Wi but not in P. Then, by Theorem 146, there is at least one general line, say 6, passing through and lying in Q which is normal to c. But since b lies in Q it is normal to every general line in P. Thus b is normal to three general lines all passing through and lying in W 1 but not all lying in one general plane. It follows that b is normal to W l and lies in W 2 . Thus in all cases there is a general line lying in W 2 which is normal to W,. The above theorem might also be stated in the form : If a general threefold W l be normal to a general threefold W 2 , then W 2 is normal to Wi. SOME ANALOGIES. Before proceeding with the next part of our subject we shall point out a few analogies which exist between an acceleration plane, a rotation threefold and the whole set of elements. We have seen that : if P be an acceleration plane and A be any element in it there are 'two and only two optical lines passing through A and lying in P. We have an analogue to this in the case of a rotation threefold. We shall show that if W be a rotation threefold and a be any separation line in it there are two and only two optical planes containing a and lying in W. In order to prove this : let be any element in a. Then, by Theorem 159, there is at least one general plane lying in W and passing through to which a is normal. Further, there cannot be more than one such general plane, for otherwise a would require to be normal to the rotation threefold W and would therefore intersect W contrary to the hypothesis that a lies in W. Let P be this one general plane. 266 A THEORY OF TIME AND SPACE Then P cannot be a separation plane, for, since a is a separation line, this would require W to be a separation threefold, contrary to hypothesis. Again, P cannot be an optical plane for this would require W to be an optical threefold, contrary to hypothesis. Thus P must be an acceleration plane and so there are two and only two optical lines, say Cj and c 2 , which pass through and lie in P. Thus a must be normal to both c a and c 2 and it cannot be normal to any other optical line passing through and lying in W', for such an optical line could not lie in P, and if a were normal to such an optical line in addition to c x and c 2 , it would require to be normal to W t which we know to be impossible. But now Cj and a lie in one optical plane, say R 1 , while c 2 and a lie in another optical plane, say R 2 . Now R l and R z are the only optical planes in W which contain a; for the existence of a third would require the existence of a third optical line passing through 0, lying in W and normal to a, which, as we have seen, is impossible. This proves the required result. Again we have a corresponding result for the whole set of elements. We shall show that if 8 be any separation plane there are two and only two optical threefolds containing S. For let be any element in $ and let P be the one single accelera- tion plane which passes through and is completely normal to S. Further let Cj and c 2 be the two generators of P which pass through 0. Then c x and c 2 are each normal to 8, and accordingly c x and 8 determine one optical threefold, say W lt while c 2 and S determine another optical threefold, say W 2 . Now W l and W 2 are the only optical threefolds which contain S, for the existence of a third would require the existence of a third optical line passing through and normal to 8. But if there were three optical lines passing through and normal to S, there would be more than one acceleration plane passing through and completely normal to S, which we have seen is impossible. Thus there are two and only two optical threefolds containing S, and so we see that we have here a certain analogy between an acceleration plane, a rotation threefold, and the whole set of elements. It was pointed out in another part of this work that if W be a rotation threefold and A be any element in it, then there are an infinite number of optical lines which pass through A and lie in W. A THEORY OF TIME AND SPACE 267 It is easy to show that if a be any separation line, there are an infinite number of optical planes which contain a, although, as we have seen, there are only two in any one rotation threefold containing a. Thus let be any element in a and let W be the one single rotation threefold which passes through and is normal to a. Then there are an infinite number of optical lines passing through and lying in W, and each of these must be normal to a. Thus each of these optical lines and a determines an optical plane and all these latter must be distinct. It follows that there are an infinite number of optical planes con- taining any separation line. It is easy to show that if W be a rotation threefold and a be any optical line in it, then there is one and only one optical plane containing a and lying in W. For let be any element in a ; then, since there are an infinite number of optical lines passing through and lying in W, there are an infinite number of acceleration planes lying in W and containing a. Let P be any such acceleration plane. Then by Theorem 160 there is one and only one general line, say 6, passing through and lying in W which is normal to P. Then b must be normal to a, and so a and b determine an optical plane, say R, which lies in W. Now R is the only optical plane which contains a and Kes in W ; for let R' be any other optical plane containing a. Then any element X lying in R but not in a would be neither before nor after any element of R, and so X and R would lie in an optical threefold and could not lie in W. This proves that R is the only optical plane containing a and lying in F. If now we consider the whole set of elements we can easily show that if a be an optical line there is one and only one optical threefold containing a. In order to prove this we have only to remember that a is normal to any optical threefold containing it and, by Theorem 162, if be any element in a, there is one and only one optical threefold passing through and normal to a. Again, if P be an optical plane, there is one and only one optical threefold containing P ; for if A be any element which is neither before nor after any element of P, then P and A determine an optical three- fold, say W, 268 A THEORY OF TIME AND SPACE Also W is the only optical threefold containing P, for otherwise we should have more than one optical threefold containing any optical line in P. THEOREM 172. If A, B, C, D be the corners of an optical parallelogram, (AC being the inertia diagonal line) and if A, B', 0, D' be the corners of a second optical parallelogram, while A', B', C', D' are the corners of a third optical parallelogram whose diagonal line A'G' is conjugate to BD, then A, B, C', D will be the corners of a fourth optical parallelogram. In order to prove this important theorem, we shall first prove the following lemma. If 0, C and C' be three distinct elements in an acceleration plane P such that 00 and OC' are inertia lines while CC' is a separation line, and if further CC" be another separation line intersecting OC' in C", and if M be the mean of C and C' while N is the mean of C and C", then if MO be conjugate to CC' we cannot have NO conjugate to CC". It will be sufficient to consider the case where is before C, since the case where is after C is quite analogous. Since CC' is a separation line, while OC' is an inertia line, and since is before C it follows that must also be before C'. Let E, C, F, C' be the corners of an optical parallelogram in the ac- celeration plane P and let F be after E. Then FE is conjugate to CC' and intersects it in M and must therefore by hypothesis be identical with MO. Now E must be after 0, for in the first place E cannot be identical with since EC' is an optical line while OC' is an inertia line. Again, cannot be after E for then we should have after one element of the optical line EC' and before another element of it contrary to Theorem 12. Thus since OE is an inertia line we must have E after 0. Fig. 38. A THEORY OF TIME AND SPACE 269 Now the element C" is distinct from C', and since C and C' lie in an inertia line we must have one after the other. Suppose first that C' is after C". Let the optical line through C" parallel to C'E intersect GE in E' and let the optical line through C" parallel to E'C intersect CF in F'. Then E', C, F', C" are the corners of an optical parallelogram, and, since CC" is a separation line, E'F' must be an inertia line conjugate to it and intersecting it in the element N. But now C"E' is a before-parallel of C'E while C"F' is a before- parallel of C'F. Thus we must have E' before E and F' before F. Now let the inertia line E'F' intersect the optical line EC' in the element G. Then E' is before E and is therefore in the ft sub-set of E and so G must be in the a sub-set of E. Thus G must be after E. But since we have also F after F' it follows that EF and GF' intersect in an element, say H, which is between EC' and CF. Thus H is linearly between E and F and is therefore after E. But E is after and therefore H is after 0. Thus the conjugate to CC " through N in the acceleration plane P intersects MO in an element which is after and so NO cannot be conjugate to CC" . This proves the lemma provided that G' is after C". Next consider the case where C" is after C'. Suppose, if possible, that NO is conjugate to CC". .Then by the case already proved MO could not be conjugate to CC', contrary to hypothesis, and so the lemma is proved in general. We shall now make use of this lemma in order to prove the theorem. We shall suppose that C is after A and C' after A'. Now since the first and second optical parallelograms have the pair of opposite corners A and C in common it follows, by Theorem 59, that they have a common centre, say 0. Further, since the second and third optical parallelograms have the pair of opposite comers B' and D' in common, they have also the same centre 0. Thus AC and AC' intersect in the element 0, and since they are both inertia lines they must lie in one acceleration plane, say P. But C and C' are distinct elements lying in the a sub-sets of B' and 270 A THEORY OF TIME AND SPACE D f , and therefore G' is neither before nor after G, and, in an analogous way, A' is neither before nor after A. Thus CC' and A A' are both separation lines. Let M be the mean of G and G': Then B', G and G' are three corners of an optical parallelogram having M as centre, while D', G and G' are three corners of another optical parallelogram of which M is the centre. Fig. 39. Further MB' and MD' must both be inertia lines and are each con- jugate to GG'. Thus, by Theorem 102, CC' is conjugate to every inertia line which passes through M and lies in the acceleration plane containing MB' and MD'. But is linearly between B' and I)' while M is after both B' and D r but is not in the general line B'D' and so, by Theorem 129 (b\ MO is an inertia line. A THEORY OF TIME AND SPACE 2*7 1 Thus, since MO is in the acceleration plane containing MB' and MD', it follows that CG' is conjugate to MO. If now we consider the optical parallelogram having B and D as opposite corners and lying in the acceleration plane containing BD and A'G' it follows, since is the mean of B and D, that must be the centre of this optical parallelogram. Further, since by hypothesis A'G' is conjugate to BD, it follows that the remaining two corners of this optical parallelogram must lie in A'G'. Let A" and G" be these remaining corners and let G" be after A". Then just as CG' was shown to be a separation line, we may show that GG" is a separation line, and if N be the mean of G and G" we may show that GG" is conjugate to NO, which may be proved to be an inertia line as was MO. But if GG" were distinct from GG' , our lemma shows that this would not be possible, and so GG" must be identical with CG'. Thus, since G" lies in A'G 1 , it follows that G" is identical with G'. Similarly A" is identical with A and therefore A , B, G', D are the corners of an optical parallelogram as was to be proved. THEORY OF CONGRUENCE. We are now in a position to consider the problems of congruence and measurement in our system of geometry. The first point to be examined is the congruence of pairs of elements, and we shall find that there are several cases which have to be considered separately. Two distinct elements A and B will be spoken of briefly as a pair and will be denoted by the symbols (A, B) or (B, A). The order in which the letters are written will be taken advan^ge of in order to symbolize a certain correspondence between the elements of pairs, as we shall shortly explain. Since any two distinct elements determine a general line, there will always be one general line associated with any given pair, but different pairs will be associated with the same general line. If we set up a correspondence between the elements of a pair (A, B) and a pair (G, D) we might either take G to correspond to A and D to B, or else take D to correspond to A and G to B. The first of these might be symbolized briefly by : (A, B) corresponds to (C, D\ or (B, A) corresponds to (D, C). 272 A THEORY OF TIME AND SPACE The second might be symbolized by : (A y B) corresponds to (D, C), or (B, A) corresponds to (C, D). If we consider the case of pairs which have a common element, say (A t B) and (A, C}, and if (A t B) corresponds to (A, C), then the element A corresponds to itself and will be said to be latent Now the congruence of pairs is a correspondence which can be set up in a certain way between certain pairs lying in general lines of the same type. In dealing with this subject it will be found convenient to have a systematic notation for optical parallelograms, so that we may be able to distinguish how the different corners are related. If A, B, C, D be the corners of an optical parallelogram we shall use the notation A BCD when we wish to signify that the corners A and D lie in the inertia diagonal line and that A is before Z), while B and C lie in the separation diagonal line so that the one is neither before nor after the other. If be the centre of the optical parallelogram A B CD, then it is obvious that will be after A and before D. Definition. A pair (A, B) will be spoken of as an optical pair, an inertia pair, or a separation pair according as AB is an optical, an inertia, or a separation line. We shall first give a definition of the congruence of inertia pairs having a latent element. Definition. If A l BGD l and A 2 BCD 2 be optical parallelograms having the common pair of opposite corners B and C and the common cenUre 0, then the inertia pair (0, Dj) will be said to be congruent to the inertia pair (0, D 2 ). This will be written : (0, A)(H)(0,A). Similarly the inertia pair (,0, A^ will be said to be congruent to the inertia pair (0, A 2 ). If (0, A) be any inertia pair and a be any inertia line intersecting OD l in 0, then the above definition enables us to show that there is one and only one element, say X, in a which is distinct from and such that: A THEORY OF TIME AND SPACE 273 For, by Theorem 105, there is at least one separation line, say c, which passes through and is conjugate to both OD l and a. Thus OA and c determine an acceleration plane, say P lt while a and c determine an acceleration plane, say P 2 . Now if DI be after there is one single optical parallelogram in P l having as centre and D l as one of its corners. If A l be the corner opposite D l and if B and C be the remaining corners, this optical parallelogram will be A l BCD l , where B and C will lie in c. Again in the acceleration plane P 2 there will be one single optical parallelogram having B and C as a pair of opposite corners and as centre. If A 2 and D 2 be the remaining corners they will lie in a, and if D 2 be after A 2 , this optical parallelogram will be A Z BCD*. Thus we may identify D 2 with X and can say that there is at least one element X lying in a and distinct from and such that : We have now to show that the element X is unique in this respect in the general line a. Let c' be any other separation line distinct from c which passes through and is conjugate to both OD l and a. Then OA and c determine an acceleration plane, say P/, while a and c' determine an acceleration plane, say P/. There is one single optical parallelogram in P/, having A l and D 1 as a pair of opposite corners, and this optical parallelogram has also as its centre. If B' and C' be the remaining corners this optical parallelogram will be AjFC'Di. But now we have the optical parallelograms A l B'C'D l) A 1 BCD 1 , A 2 BCD 2 and the diagonal line A 2 D 2 of the last of these is conjugate to B'C' and so it follows, by Theorem 172, that the elements A 2) B', D 2 , C' form the corners of a fourth optical parallelogram A 2 B'C'D 2 . Now A^B'C'Dc, will lie in the acceleration plane P 2 ' and will have as its centre, and further A 2 B'C'D 2 is the only optical parallelogram which lies in P 2 ' and has B' and C' as a pair of opposite corners. Thus the element D 2 or X is independent of the particular separation line passing through and conjugate to both OD l and a, which we may select as the separation diagonal line of our optical parallelograms. It follows that there is one and only one element X in a which is such that : R. 18 274 A THEORY OF TIME AND SPACE The same result follows if D l be before instead of after it. Again if (0, A), (0, A) and (0, A) be inertia pairs such that : and (0,A) we may easily show that : (0,A)(=)(0, D s ). In order to see this we have only to remember that whether the inertia lines OD l) OD 2 , OA all lie in one acceleration plane or in one rotation threefold, there must be at least one general line passing through and normal to all three. Since only a separation line can be normal to an inertia line, this separation line will be conjugate to OA, OA and OA, and if we call it c, then OD l and c will determine an acceleration plane, say P lt OA and c will determine an acceleration plane, say P 2 , and OA and c will deter- mine an acceleration plane, say P 3 . Now in P 1 there will be one single optical parallelogram having as centre and D l as one of its corners, while in P 2 there will be one single optical parallelogram having as centre and D 2 as one of its corners, and finally in P s there will be one single optical parallelogram having as centre and A as one of its corners. Since (0, A) (=) (0, A) and (0, A) (=) (0, A) these three optical parallelograms will have a common pair of opposite corners, and so it follows from the definition that : Thus for inertia pairs having a latent element, the relation of con- gruence is a transitive relation. It is to be observed that if (0, A) be an inertia pair we may write : (0, A)(0, A), or an inertia pair is to be regarded as congruent to itself. We shall next consider the congruence of separation pairs having a latent element. This case differs somewhat from the one we have considered. While two intersecting inertia lines always lie in an acceleration plane, two intersecting separation lines may lie either in a separation plane, an optical plane, or an acceleration plane. An inertia line can only be conjugate to two intersecting separation lines if these lie in a separation plane, as follows from Theorem 99. A THEORY OF TIME AND SPACE 275 Thus if we were to give a definition of the congruence of separation pairs having a latent element which was strictly analogous to that given for inertia pairs, such a definition would be incomplete. It is however possible, by a slight modification, to give a definition which will hold for all cases. In order to avoid complication we shall first' explain what we mean by an inertia pair being " conjugate " to a separation pair or a separation pair being "conjugate" to an inertia pair. Definition. If A BCD be an optical parallelogram and be its centre, then the inertia pairs (0, D) and (0, A) will be spoken of as conjugates to the separation pairs (0, B) and (0, C) and also conversely. The pair (0, D) will be called an after-conjugate to the pairs (0, B), (0, 0), while (0, A) will be called a before-conjugate to the pairs (0, B) } (0, C). Further, either of the separation pairs (0, B), (0, C) will be called an after-conjugate to (0, A) and a before-conjugate to (0, D). Now we know that there are an infinite number of acceleration planes which contain any given separation line, and so there are always inertia pairs which are conjugate to any given separation pair. Knowing this we can give the following definition of the "congruence" of separation pairs having a latent element. Definition. If (0, B^ and (0, B 2 ) be separation pairs and if (0, A) and (0, A) be inertia pairs which are after-conjugates to (0, BJ and (0, B 2 ) respectively, then if (0, D,) (=) (0, A) we shall say that (0, B,) is congruent to (0, $2) and shall write this : (0, B,) {=} (0, B 2 ). If (0, A') be any inertia pair which is an after-conjugate to (0, B^), but is distinct from (0, A), then it is obvious by definition that : But since (0, A) (=) (0, A), and, since these are inertia pairs, it follows that : (0, D 1 ')(=)(0,D,). Thus the congruence of (0, BJ to (0, B 2 ) is independent of the particular after-conjugate to (0, Bj) which we may select, and similarly, it is independent of the particular after-conjugate to (0, B 2 ) which we may select. 182 276 'A THEORY OF TIME AND SPACE Again if (0, ft), (0, ft) and (0, ft) be separation pairs such that : (0, ft) {EE} (0, ft), and (0, ft) {=} (0, ft), we may easily show that : (o, ft) {=} (o, ft). In order to prove this, let (0, A), (0, A) and (0, D 8 ) be inertia pairs which are after-conjugates to (0, ft), (0, ft) and (0, ft) respectively. Then we must have (0,A)(=)(0,A), and (0, D 2 )(=)(0, A), and, since these are inertia pairs, it follows, as previously shown, that : (0, A)0=)(0,A), Thus, by the definition : (0, 3.) {=} (0, 3), and so, for separation pairs having a latent element, the relation of con- gruence is a transitive relation. Again, if (0, B) be any separation pair and a be any separation line passing through 0, there are two and only two elements, say X l and F 1} in a which are distinct from 0. and such that : (0, fl) {=}(0, *,), and (0, B) {=} (0, TO- This may be easily shown as follows. Let (0, D) be any inertia pair which is an after- conjugate to (0, B) and let b be any inertia line which passes through and is conjugate to a. Then, as we have already seen, there is one and only one element, say A, ty m g i n & an d distinct from and such that : (0, D) (=) (0, A). But now a and b determine an acceleration plane and in this acceleration plane there is one and only one optical parallelogram having as centre and A as one of its corners. If this optical parallelogram be A 1 B 1 C 1 D 1 , then the elements B l and d will lie in a and the inertia pair (0, A) will be an after-conjugate to each of the separation pairs (0, ft) and (0, C\). Thus since (0, D) (=) (0, A) it follows that : and (0,B){=}(0,CJ. A THEORY OF TIME AND SPACE 277 Again if there were any other element, say B 2 , lying in a and distinct from both B 1 and C l and such that we had (0, B) {=} (0, B 2 ), then there would be an element, say D 2 , lying in b and such that (0, D 2 ) was an after-conjugate to (0, B 2 \ Since B 2 is supposed distinct from both B l and G lt therefore D 2 would require to be distinct from Dj. But since we have supposed (0, B) {=} (0, B z \ therefore we should have (0. D) (=) (0, D 2 ) and so we should have the two distinct elements D l and D 2 lying in the inertia line b and such that : (0,D)(=)(0,A) and (0, D) (=) (0, D a ), which we have already shown to be impossible. Thus we may identify B l with X 1 and Q with Fj and say that there are two and only two elements X 1 and Y l lying in a and distinct from and such that : (0, 5) {=) (0, X,) and (0, 5) {=} (0, F,). If A BCD be an optical parallelogram and be its centre, we observe that according to our definitions we have (0, B) {=} (0, (7), but not (0, A)(=)(0, D). The reason why we make this distinction is that in the separation pairs we have neither before nor after B and also neither before nor after C, while in the inertia pairs we have after A and before D. Thus in the first case the relations are alike in respect of before and after, while in the second case the relations are different. The question now arises as to the " congruence " of optical pairs. In this case constructions such as those by which we defined the congruence of inertia and separation pairs having a latent element, entirely fail and there is nothing at all analogous to them. We are thus led to regard optical pairs as not determinately com- parable with one another in respect of congruence, except when they lie in the same, or in parallel optical lines. As regards the " congruence " of pairs lying in the same general line, we have as yet given no definition, except for the very special case of inertia or separation pairs having a latent element ; while no definition whatever has been given of the " congruence " of pairs lying in parallel general lines. A definition covering all these omitted cases can be given, which applies to all three types of pair. 278- A THEORY OF TIME AND SPACE We must first however define what we mean when we say that one pair is opposite to another. Definition. A pair (A, B) will be said to be opposite to a pair (G, D) if and only if the elements A, B, G, D form the corners of a general parallelogram in such a way that AB and CD are one pair of opposite sides, while AC and BD are the other pair of opposite sides. This will be denoted by the symbols (A, B) D (G, D). It will be observed that the use of the symbol Q implies that the pairs (A, B) and (G, D) lie in distinct general lines which are parallel to one another. If however we have (A, B) D (C, D\ and (E, F) D (C, D), then the pairs (A, B) and (E, F) may lie either in the same or in parallel general lines. If (A y B) and (E, F) do not lie in the same general line, it follows from Theorem 127 that we may write (A, B) D (E, F). We have now to prove the following theorem : THEOREM 173. // (A, B), (A', B') and (C, D) be pairs such that : (A, B) D (C, D), and (A, B') Q (C, D\ and if (G', D') be any other pair such that: (A, B) D (G', D'), and which does not lie in the general line AB', then we shall also have (A', B'} D (C", D'}. We shall first consider the case where (A, B) and (A', B') do not lie in one general line. In this case since (A, B) D (C, D), and (A' t B') Q (C, D), it follows by Theorem 127 that : (A, B') D (A, B). But (C', D') D (A, B) A THEORY OF TIME AND SPACE 279 by hypothesis, and so, since (C' } D') and (A', B') do not lie in one general line, it follows that: (A', B') D (0', D'). Next consider the case where (A, B) and (A', B') He in one general line. There are two sub-cases of this : (1) (C, D) and (C 1 ' , D') do not lie in one general line. (2) (C, D) and (C", D') do lie in one general line. Consider first sub-case (1). Here since (C, D) Q (A, B\ and (C", D') D (A, B\ and since (C, D) and (C', D') do not lie in one general line, it follows that: (C", V) D (0, D). But (A', B') D (C, D), and so, since (C',- D') and (A, B') do not lie in one general line, it follows that: (A 1 , ') D (C", D'). Next consider sub-case (2). Let E be any element in the general line A C r distinct from both A and C' and let a general line through E parallel to AB intersect D'B in the element F. Then we shall have (E, F) D (A, B), and also (E, F) Q (C', D'). But now since E is distinct from A and also from C' it follows that the general line EF must be distinct from the general line containing (A, B) and (A', B') and must also be distinct from the general line containing (C", D') and (C, D). Thus since (E, F) Q (A, B), and (C, D) D (A, B\ and since (E, F) and (C, D) do not lie in one general line, it follows that: (E, F) D (C, D\ Also since (A', B'} Q (01 D), and since (E, F) and (A', B') do not lie in one general line, it follows that : (A', B') D (E, F). 280 A THEOEY OF TIME AND SPACE But (C 1 , D') D (E, F\ and since (A, B') and (C', D') lie respectively in the distinct general lines AB and CD, it follows that : (A', B') D (C', D'). Thus the theorem holds in all cases. We are now in a position to introduce the following definition : Definition. A pair (A, B) will be said to be co-directionally con- gruent to a pair (A, B') provided a pair ((7, D) exists such that : (A, B) D (C, D), and (A, B') Q (C, D). The theorem just proved shows that we are at liberty to replace the pair (C, D) by any other pair ((?', D') such that : (A,B)[3 (C', D'\ provided (C', D') does not lie in the general line A'B'. It is evident that (A, B) Q (C, D) implies that (A, B) is co- directionally congruent to (C, D), but (A, B) being co-directionally congruent to (C, D) does not imply that (A, B) Q (C, D), since (A, B) and (C, D) might lie in the same general line. It is also obvious that : (A, B) is co-directionally congruent to (A, B). We shall ultimately represent co-directional congruence by the same symbol = as we shall use for the other cases of congruence, but when we wish to make it clear that the congruence is co-directional we shall use the symbol | = |. Thus we see that: (A, B) Q (C y D) implies (A, B) = \ (C, D\ but (A, B) | = | (C, D) does not imply (A, B) D (C, D), except when AB. and CD are distinct general lines. We have next to show that if (A,B)\= (C,D\ and (C t D)\= (E,F), then must (A, B) =| (%, F). This is easily proved ; for if a be any general line parallel to AB but distinct from CD and Efi and therefore also parallel to them, we may select any pair (G, H) in a, such that : (A,B) D (G,H) (1). Then since (A, B) = (C, D), it follows, by Theorem 173, that : (C, D) D (, R). A THEORY OF TIME AND SPACE 281 Similarly since (<7, D) \ = \ (E t F), it follows that: (E, F) Q (#, H) (2). Thus from (1) and (2) it follows that : (A, B) = (E, F), and so we see that : the relation of co-directional congruence of pairs is a transitive relation. If (A, B) D (0, D) and if B be after A then it is easy to see that D must be after G. In the first place AB must be either an optical or inertia line and, since CD is parallel to AB, it follows that CD must be the same type of general line as AB. Suppose first that A B is an optical line. Then C could not be after D, for then by Theorem 57 or Theorem 91 AC and BD would intersect, contrary to the hypothesis that they are parallel. Thus, since C and D are distinct, and since CD is an optical line, it follows that D must be after C. Next suppose that A B is an inertia line. Then AB and CD must lie in an acceleration plane, say P. If AC and BD should happen to be optical lines then, since B is after A, it follows that BD would be an after-parallel of AC and so, since CD is an inertia line, it would follow that D must be after C. Next suppose that AC and BD are not optical lines. Let AE arid BE be generators of P of opposite sets passing through A and B respectively and intersecting in E. Let CF be an optical line through C parallel to AE and let it intersect the general line through E parallel to AC in F. Then EF must be parallel to BD and so, by Theorem 127, DF must be parallel to BE and therefore must be an optical line. But now since B is after A we must have E after A and B after E. Thus by the first case we must have F after C and D after F and therefore D after C as was to be proved. Thus in all cases if B be after A we must have D after C and similarly if B be before A we must have D before C. It follows directly from this that if (A, B) D (C, D\ and (A' t B') Q (C, D\ then if B be after A we must have D after C and therefore B' after A. 282 A THEORY OF TIME AND SPACE Thus if (A, B) =| (A', B?) and if B be after A we must have B after A' while if B be before A we must have B' before A'. Again if three corners of a general parallelogram A, G and D be given and if we know that two of the side lines are A'C and CD, then the general parallelogram is uniquely determined. If then any pair (A', X) be co-directionally congruent to a pair (A, B), where A, B and A' are given, it is easy to see that X is uniquely determinate, provided we know that A corresponds to A. For let a be a general line parallel to AB, but which does not pass through A', and let (C, D) be any pair in a such that : (A, B) D (C, D). Then there is one single general parallelogram having A', C and D as three of its corners and A'G and CD as two of its side lines. If B' be the remaining corner we shall have (A', B') D (C, D\ and so (A', B')\ = \(A, B). Thus X must be identified with J9', which is a definite element. THEOREM 174. If (0i, A^ and (0 2 , A 2 ) be inertia pairs such that : and if (O lt B^ be any separation pair which is conjugate to (O l5 A^, then there is a separation pair, say (0 2 , B 2 ), which is conjugate to (0 2 , A 2 ) and such that : (0 1( A)N!(0 2)J B 2 ). It is evident that (O n , Aj) and (0,, B^ must lie in an acceleration plane, say Pj. Since the inertia line 2 ^l 2 must be either parallel to the inertia line O l A l , or else identical with it, it follows that 2 A 2 must either lie in P l or in an acceleration plane parallel to Pj. We shall first consider the case where 2 A 2 lies in an acceleration plane P 2 parallel to P l . If now we take the one single optical parallelogram in P l having O a as centre and A l as one of its corners, then B will be another corner. If (Oi, BT) be an after-conjugate to (O lt A^ we may take this optical parallelogram to be A^B^D^ while if (O lt B^) be a before-conjugate to (Oi, AI)WQ may take the optical parallelogram to be D 1 B 1 C 1 A 1 . Now the acceleration plane P l and the general line 1 2 determine a general threefold containing P 2 and so, as we have already seen, if A THEORY OF TIME AND SPACE 283 through any element of P l distinct from 1 a general line be taken parallel to 1 2 then this general line will intersect P 2 . Now through the elements A lt B 1} GI and D 1 let general lines be taken parallel to 1 2 and let these intersect P 2 in the elements A 2 , B 2 , C 2 and D 2 respectively. Then any two of the general lines Oi0 2 , A-^A^ B^B 2 , C-iC 2 , A A are parallel to one another and therefore any two of them lie in a general plane. Since however the elements A lt O l and D l lie in one' general line, the three general lines A 1 A 2 , Oj0 2 and D 1 D 2 lie in one general plane, and since the elements B lt and C l lie in one general line, the three general lines B 1 B 2 , 1 2 and GiC a lie in one general plane. Thus the elements A 2 , 2 and D 2 lie in one general line parallel to the general line containing A l} O l and D lt while the elements B 2) 2 Fig. 40. and G a lie in another general line parallel to that containing B l} O l and GI. Further the general lines A 2 B 2 , A 2 G 2) B 2 D 2 , C 2 D 2 must be re- spectively parallel to A 1 B 1 , A l C l) B l D l , C 1 D 1 and, since these latter are all optical lines, it follows that A 2 B 2 , A 2 C 2 , B 2 D 2 , G 2 D 2 are all optical lines. Thus A 2 , B 2) C 2) D-2, form the corners of an optical parallelogram having 2 as centre. Further the diagonal line A 2 D 2 is an inertia line, while the diagonal line B 2 C 2 must be a separation line. Thus the separation pair (0 2 , B 2 ) is conjugate to the inertia pair (0 2 , A 2 ), and, since 2 is after or before A 2 according as O l is after or before A^ it follows that (0 2> B 2 ) is an after- or before-conjugate to (0 2 , A 2 ) according as (Oi, B^ is an after- or before-conjugate to (O^^). 284 A THEORY OF TIME AND SPACE Also we have (ft, BJ Q (O a , ^ 2 ) and so (ft, B^\= (0 2) B 2 ). This proves the theorem provided 2 ^L 2 does not lie in P lm Consider next the case where 2 -A 2 does lie in P lt Let P be any acceleration plane parallel to P l and let (0', A') be any inertia pair in P' such that : (0,, A,) D (0', A'). Then, by Theorem 173, since (0 1? A^ \ = \ (0 2 , A 2 ) we must have (0,, AJ (0', A'). Thus, by the case already proved, it follows that there is an optical parallelogram lying in P which has 0' as centre and A' as one of its corners and such that, if we denote it by A'B'C' D' or D'B'C'A' (ac- cording as the optical parallelogram in P l is A 1 B 1 C 1 D 1 or D l BiC l A 1 ), then: (0,, ,) G (0', B'). But in a similar manner we can show that there is an optical parallelogram lying in P 1 which has 2 as centre and A 2 as one of its corners and such that, if we denote it by A. 2 B. 2 C 2 D 2 or D 2 B 2 C 2 A 2 (according as the optical parallelogram in P' is A' B'C' D' or D'B'C'A'), then: (0 2 , 2 )D(0','). Thus it follows from definition that : Further (0 2 , B 2 ) is conjugate to (0 2 , A 2 ) and will be an after- or before-conjugate to (0 2 , A z ) according as (0 1? BJ is an after- or before- conjugate to (0 I? Aj). Thus the theorem holds in all cases. THEOREM 175. If (0 1} Ai) and (0 2 , A 2 ) be separation pairs such that : (O^OI-EE (0 2 ,A 2 ) and if (O l} B^ be any inertia pair which is conjugate to (O l , A^ then there is an inertia pair, say ,(0 2 , B 2 ), which is conjugate to (0 2 , A 2 ) and such that : (ft, B,) | = I (0 a , B 2 ). The proof of this theorem is quite analogous to that of Theorem 174. Also it will be seen that (0 2 , B 2 ) will be a before- or after-conjugate to (0 2 , AZ) according as (O l} B t ) is a before- or after-conjugate to A THEORY OF TIME AND SPACE 285 We have now to prove certain theorems involving both the co- directional congruence of pairs and the congruence of pairs having a latent element. We shall make use of the symbols (=), { = } and = | in the manner already explained in order to show clearly the types of congruence to which we refer. THEOREM 176. If (0i, AI), (0i, BI) and (0 2 , At) be inertia pairs such that : and (0i, ^ then there is an inertia pair (0 2 , B 2 ) such that : (0 2 ,A 2 )(=)(0 2 ,B 2 ) and OB Let c be any separation line passing through O l and normal to both the inertia lines 0i^n and 1 B l , and let C\ be an element in c such that the separation pair (O lt Q) is conjugate to (O l} AJ. Then since (O lt A,) (=) (O lf BJ it follows that (0 1? C,) must also be conjugate to (0 1; BI). But since (O l> AJ = (0 2 , A z ) it follows, by Theorem 174, that there is a separation pair, say (0 2 , C 2 ), which is conjugate to (0 2 , A 2 ) and such that: But now, by Theorem 175, since (0 1} B^ is conjugate to (0 X , d), it follows that there is an inertia pair, say (0 2 , B 2 ), which is conjugate to (0 2 , C 2 ) and such that : But now since (0 l5 A 1 )(=)(0 lt B^ and these are inertia pairs we must have A l and B l either both after O l or both before 0^. Further A 2 must be after or before 2 according as A l is after or before O l} while B 2 must be after or before 2 according as B 1 is o/j^e?' or before 0^. Thus J.2 and 5 2 are either both after 2 or both before 2 . Since therefore (0 2 , (7 2 ) is conjugate to both (0 2 , A 2 ) and (0 2 , B 2 ) it follows that : Thus the theorem is proved. 286 A THEORY OF TIME AND SPACE THEOREM 177. If (Oi f AJ, (Oi, Si) and (0 2 , A 2 ) be separation pairs such that: (O lt A,) {=} (O lf Bi) and (O l} A,) = | (0 2 , A 2 ), then there is a separation pair (0 2 , B 2 ) such that : (0 2 , A) {=} (0 a , 5 2 ) and (01,501= (0 a ,5 a ). Let (0 lf A) and (Oj, #0 be inertia pairs which are after-conjugates to (O lt AI) and (0 1} 50 respectively. Then since (0,, A,) {=} (O lt BJ, we must have (O lt A) (=) (0i, E^). But now since (0 15 A^) = | (0 2 , J. 2 ), and since (0 X , A) is an inertia pair which is an after-conjugate to (Ou J-0, it follows, by Theorem 175, that there is an inertia pair, say (Ojj, AX which is an after-conjugate to (0 2 , A z ) and such that : (O lf A) = i (0 2 , A). But now (Oj, A), (0i, E-^ and (0 2 , A) are inertia pairs such that : and (0i, A) l = | (0 a , A) and so, by Theorem 176, there is a pair (0 2 , ^2) such that : and (0i,^i)| = | (0 2 ,^ a ). Since however (0^ B^) is -a separation pair which is a before-conjugate to the inertia pair (0^ E^), it follows, by Theorem 174, that there is a separation pair, say (0 2 , 5 2 ), which is a before-conjugate to (0 2 , E 2 ) and such that : But since (0 2 , A) and (0 2 , j^ 8 ) are after -conjugates to (0 2 , ^-2) and (0 2 , 5 2 ) respectively, and since it follows by definition that : (0., A) N (0,, S 2 ). Thus the theorem is proved. A THEORY OF TIME AND SPACE 287 THEOREM 178. If (A, B) and (A, C) be inertia pairs such that: then there is an inertia pair (C, D) such that : and (B,A)\ = (C, D). Let a be any separation line which passes through A and is normal to both AB and AC. Let A! be an element in a such that the separation pair (A, AJ is conjugate to the inertia pair (A, C). Then since (A, B) (=) (A, C), it follows that (A, A^) is also conjugate to (A, B). Let the general line through C parallel to AA l intersect the general line through A, parallel to A C in the element C lt and let the general line through B parallel to AA l intersect the general line through A, parallel to AB in the element S lt Thus (C, CO D (A, 40 and (B ) B l }^(A,A,\ and therefore (B, B,) | = j (C, Ci). Now since (A, A,) is conjugate to (A, C) it follows that A,C is an optical line and it is easy to show that AC\ is also an optical line as follows. 288 A THEORY OF TIME AND SPACE Since A-^G and AC l are diagonal lines of the general parallelogram whose corners are A, A lf C 1} C, it follows that they intersect in an element, say E. which is the mean of A l and C. If F be the mean of A 1 and A, then EF is parallel to GA and there- fore EF is conjugate to the separation line AAi. Thus A, A i and E are three corners of an optical parallelogram whose centre F lies in A A 1 and therefore AE (that is ACi) is an optical line. Similarly, since (A, A-^ is conjugate to (A, B), it follows that AB l is an optical line. But since GC l and BB 1 are parallel to AA l we have GG- conjugate to CA, and BB l conjugate to BA. Thus the inertia pairs (C, A) and (B, A) are conjugate to the separation pairs (C, Q) and (B, B^ respectively. 'But now since (B, B,) = \ (C, C,), and since (B, A) is an inertia pair which is conjugate to (B, B^, it follows, by Theorem 175, that there is an inertia pair, say (C, D), which is conjugate to (0, C^) and such that : (B, A)\ = \ (C, D). But from this it follows that D must be before or after C according as A is before or after B. Since however (A, B) (=) (A, C), we have A before or after B according as A is before or after C. Thus we must have D before or after C according as A is before or after C. Since therefore the inertia pairs (C. A ) and (C, D) are both conjugate to the separation pair (C, C^, it follows that : (G,A)(=)(C,D). Thus the theorem is proved. THEOREM 179. If (A, B) and (A, C) be separation paws such that: (A*, B) {=} (A, C), then there is a separation pair (C, D) such that : (G, A) {=} (C, D) and (B,A)\= (G, D). A THEORY OF TIME AND SPACE 289 Let a be any inertia line which passes through A and is normal to AB, and let a be any inertia line which passes through A and is normal to AC. Let A-L be an element in a such that the inertia pair (A, A^) is an after-conjugate to the separation pair (A, B) and let A' be an element in a' such that the inertia pair (A, A') is an after-conjugate to the separation pair (A, C). Then since (A, B) {-} (A, C) it follows that : (A, A^ (=) (A, A'). Let the general line through C parallel to A A' intersect the general line through A' parallel to AC in the element C' and let the general Fig. 42. line through B parallel to AA l intersect the general line through parallel to AB in the element B^ Then (C, C') Q (A, A') and (B, BJ Q (A, A,), and so we may write (C, G')\ = \ (A, A') and (B, B,)\ = \ (A, A,). But since (A, A'), (A, A^ and (G, C') are inertia pairs such that : (A,A)(=)(A ) A l ) and (A, A')\ = \ ((7, C'\ R. 19 290 .' A THEOKY OF TIME AND SPACE therefore, by Theorem 176, there is an inertia pair (C, 0^) such that : and (A,A,)\ Thus since (B, B 1 )\ = \ (A, AJ, it follows that : (B, B^)\ = \ (C, CJ. But now we may show in the manner employed in the last theorem that since (A, A^ is an after-conjugate to (A, B), therefore (B, B^ is an after-conjugate to (B, A) } and since (A, A') is an after-conjugate to (A, C) therefore (C, C') is an after-conjugate to (C, A). Since however we have the inertia pairs (B, B^ and (C, C-,) such that: and since (B, A) is a, separation pair which is conjugate to (B, B^, it follows, by Theorem 174, that there is a separation pair, say (C, D), which is conjugate to (C, G^ and such that : (B, A)\ = \(C, D). But now (A, A-L) is an after-conjugate to (A, B) and so A 1 is after A. Thus since (A, A^\ = \ (G, 0,) it follows that <7 r is after G, and so since (G, Cj) is conjugate to (G, D) it must be an after-conjugate. But (G, G') is an after-conjugate to (C, A) and so since (G,G')(=)(G,C l \ it follows from the definition that : (0, A) {=} (C, D). Thus the theorem is proved. THEOREM 180. (1) If A, B and G be three distinct elements and the pairs (A, B) and {B, G) be such that : (A, B) = \ (B, 0), then B is the mean of A and G. (2) If A, B and G be three distinct elements such that B is the mean of A and G, then the pairs (A, B) and (B, G) are such that: (A,B) First suppose that : (A, B)\ = \(B, 0). A THEORY OF TIME AND SPACE 291 Then by the definition of co-directional congruence there must be a pair, say (D, E), such that : (A, B) D (D, E) and (B, C) D (D, E). Now since the pairs (A, B) and (B, C) have a common element B they cannot lie in parallel general lines and so must lie in the same general line. Then BE and CD must be the diagonal lines of the general parallelo- gram whose corners are B, C, D and E and so BE and GD must intersect in an element F which is the mean of D and C. But D does not lie in the general line AC, and so since BF is parallel to AD it follows, by Theorem 81, 92, or 116, that B is the mean of A and C. Next, to prove the second part of the theorem suppose that B is the mean of A and C. Let (D, E) be any pair such that : (B, C) H (D, E). Then the diagonal lines BE and CD of the general parallelogram, whose corners are B, C, D and E, must intersect in an element F which is the -mean of D and C. But since D does not lie in the general line AC it follows, by Theorem 81, 92, or 116, that BF (that is BE) is parallel to AD. Thus since also AB is parallel to DE it follows that : (A, B) D (D, E). Thus by the definition of co-directional congruence we have (A,B) = \(B,C). Thus both parts of the theorem are proved. We are now in a position to introduce general definitions of the congruence of inertia and separation pairs. This is done by combining co-directional congruence with congruence in which an element is latent, in the following manner. Definition. An inertia pair (A ly B^) will be said to be congruent to an inertia pair (A z , B 2 ) provided an inertia pair (A 2 , C 2 ) exists such that: (A lt B,) | = | (A C 2 ) and (A 2) B 2 ) (=) (A 2 , C 2 ). 192 292 A THEORY OF TIME AND SPACE Definition. A separation pair (A 1} Bj) will be said to be congruent to a separation pair (A 2 , B 2 ) provided a separation pair (A 2) C 2 ) exists such that : and (A 2> B 2 ) {=} (A 2 , C 2 ). We shall denote the generalized congruence of inertia or of separation pairs by the symbol =, thus : (A,, ,) = (A B 2 ). We shall also use the same symbol to denote the congruence of optical pairs, but in the latter case it is to be regarded as simply equivalent to the symbol | = | , since the only congruence of optical pairs is taken to be co-directional. Let us consider now two inertia pairs (A lt B^ and (A 2t B 2 ) such that: (A lt B,) = (A 2> B 2 ). Then there exists an inertia pair (A 2) C 2 ) such that : and (A 2) B 2 )(=)(A 2 ,C 2 ). But by Theorem 176 there exists an inertia pair (A ly Q) such that : (A., B 2 )\ = \ (A lt 00 and (A 1 ,B 1 )(=)(A 1 ,O l ). Thus we may write (A 2 , B 2 ) = (A lt B^. Again, by Theorem 178, there is an inertia pair (5 2 , Z* 2 ) such that: and (C s ,A 2 )\=\(B i ,D,). Since however (0 2 , A 2 ) \ = (B^, A^, we have (B l} A,) =\(B 2 ,D 2 ), which together with the relation (B 2) A 2 )(=)(B 2> D 2 ) gives us (B l} AJ = (B 2 , A 2 ). If now we take instead two separation pairs (A lf B^ and (A 2 , B 2 ) such that: A THEORY OF TIME AND SPACE 293 then, by using Theorem 177 in place of Theorem 176, we may prove that: (A 2) B 2 )~(A l) B l ). Also, by a similar method to that employed in the case of inertia pairs, but using Theorem 179 in place of Theorem 178, we may prove that: (B l ,A l ) = (B 2) A 2 ). Again, if (A, E) be an inertia pair, we have (A,B) =\(A,B) and (A, B) (=) (A, B). Thus we have (A, B) = (A, B). t A similar result obviously holds if (A, B) be a separation pair. Again if (A, B) and (A, C) be inertia pairs such that: (A,B)(=)(A, C), then since (A, C) \ = (A, C) we may write (A, B)=(A, C). A similar result holds if (A, B) and (A, 0) be separation pairs such that: (A,B){=}(A,C). Further it is also clear that : (A lt B,) | = | (A 2 , B 2 ) implies (A 19 B l ) = (A 2 ,B 2 ), both when (A l , B^ and (A 2 , B 2 ) are inertia pairs and when they are separation pairs. Again if (A lt B^, (A 2) B 2 ) and (A 3 , B 2 ) be inertia pairs such that : (4i,A)s<4 a ,Zy and (A 2 ,B 2 )=(A S ,B 3 ), then, by the definition of congruence, there is an inertia pair (A 2 , <7 2 ) such that : (A l ,B l )\ = \(A 2 ,C 2 ) (1) and (A 2) B 2 )(=)(A 2 ,C 2 ) (2). Also there is an inertia pair (A 3 , C 3 ) such that : (A 2 ,B,)\ = \(A S ,C 3 ) (3) and (A t , ,)(=) (A,, G,) . ...(4,1 294 A THEORY OF TIME AND SPACE Now from (2) and (3) it follows, by Theorem 176, that there is an inertia pair (A 3 , A) such that : (AC t )(s)(AD,) (5) and (A i ,G i )\ = \(A,,D t ) (6). But from (1) and (6) it follows that : (4,, 5,) = (AD.) (7), while from (4) and (5) it follows that : (A 9 ,B,)(=) (A 3 , A) (8). Thus from (7) and (8) it- follows that : (A l} B l ) = (A 3) B 3 ). A similar result may be proved for the case of separation pairs ; using Theorem 177 in place of Theorem 176. Thus for inertia or separation pairs the general relation of congruence is a transitive relation. Again if (A, B) be any separation pair and P be any acceleration plane containing the separation line AB, there is one single optical parallelogram in P having B as centre and A as one of its corners. If C be the corner opposite to A then, by definition, B is the mean of A and C. Thus by Theorem 180 we have (A,B)\ = \(B,C). But also by definition we have (B, A) {=} (B, C). And so (A,B) = (B,A). We have not however a corresponding result in the case either of inertia or optical pairs since the elements in such pairs are asymmetri- cally related. .THEOREM 181. If (0 1? A) and (0 2 , A) be inertia pairs while (O lt A) and (0 2 , B 2 ) are separation pairs which are before-conjugates to (O l , A) and (0 2 , A) respectively or else after-conjugates to (0 1? A) and (0 2 , A) respectively ; then : (l) If (0,,A) = (0 2 , A) we shall also have (0,, 5,) = (0,, 2 ). A THEORY OF TIME AND SPACE . 295 (2) // (ft, 5.) =<0,, ft) we shall also have (O l} A) = (0 2 , A)- Let us consider the first part of the theorem. Since (0!, A) = (0, A), it follows by definition that there is a pair (0 2 , IT) such that : and (0 a , A)(=)(0 2 , Then (0 2 , D') is an inertia pair and so, since (O lf A) is a separation pair which is conjugate to (0 1? A), it follows, by Theorem 174, that there is a separation pair, say (0 2 , 5'), which is conjugate to (0 2 , D') and such that : (ft, ft) | 3= | (ft, J^ Now if A be q/ter 2 we shall also have D' after 2 and so (0 2 , A) and (0 2 , DO will be after-conjugates to (0 2 , B 2 ) and (0 2 , B') respectively. Thus we shall have (0,, ft) {=} (0,, 5')- If on the other hand A be before 2 we shall also have D 7 before 2) and so (0 2 , A) an d (02, D') will be before-conjugates to (0 2 , B 2 ) and (0 2 , B') respectively. Now by completing the optical parallelograms implied in the relation (ft, A)(=)(0 2 ,D'), we see that in this case there will be inertia pairs, say (0 2 , A 2 ) and (0 2 , A'), which will be after-conjugates to (0 2 , B 2 ) and (0 2 , B') respect- ively, and such that : (0 2> A 2 )(=)(0 2) A'). ' Thus we have also in this case (0 2 ,B 2 ){=}(0 2 ,B'). Combining this with the relation (ft, 501 = 1(0,. B'), it follows by definition that : (O lt A) = (0 2 , 5 a ). Thus the first part of the theorem is proved. 296 A THEORY OF TIME AND SPACE Consider now the second part of the theorem. Since (0 l3 A) ~ (0 2 , B 2 ), it follows by definition that there is a pair (0 2 , B') such that : (O^B^l- (0 2) B f ) and (0 2 , B 2 ) {=} (0 2 , B'). Then (0 2 , B') is a separation pair and so, since (0 1? A) is an inertia pair which is conjugate to (O lt BJ, it follows, by Theorem 175, that there is an inertia pair, say ( 2 , D'\ which is conjugate to (0 2 , B') and such that : (On A) 1 = 1(0,, DO- Now if (0i, J3j) and (0 2 , B 2 ) be before-conjugates to (O l} A) and (0 2 , A) respectively we shall have A after 0j and therefore D' a/ter 2 , and also we shall have A o/fer 2 . Thus (0 2 , A) and (0 2 , D') will be after-conjugates to (0 2 , B 2 ) and (0 2 , B') respectively and s.o, since it follows that : (0 2 , A) (=) (0 2 , D'). If, on the other hand, (0 1? B^ and (0 2 , B 2 ) be after-conjugates to (0j, A) and (0 2 , A) respectively, we shall have A &e/ore 0i and there- fore D' before 2 and also we shall have A before 2 . Thus (0 2 , A) and (0 2 , D') will be before-conjugates to (0 2 , B 2 ) and (0 2 , B') respectively. Now by completing the optical parallelograms implied in these relations we see that there are inertia pairs, say (0 2 , A 2 ) and (0 2 , A'), which are after-conjugates to (0 2 , .B^'and (0 2 , B') respectively and such that A, 2 and A 2 lie in one inertia line and also D' y 2 and A' lie in one inertia line. Now since . (0 2 , B 2 ) {=} (0 2 , ff), we must have (0 2 , A 2 )(=)(0 2 , A'), and therefore also in this case Combining this with the relation (O..A) H it follows by definition that : (O.DO^CO^A). Thus the second part of the theorem is proved. A THEORY OF TIME AND SPACE 297 From Theorems 180 and 181 it follows that: if (O lf A) and (0 2 , A) be inertia pairs while (O l} A) and (0 2 , B 2 ) are separation pairs such that (O lt Bj) is a before-conjugate to (0 1} A) and (0 2) B 2 ) is an after-conjugate to (0 2 , A), then : (1) If (0,,A)s(A, O,), we shall also have (O lf A) s (0 2 , 5 2 ). (2) If (Oi,A) = (0fl), we shall also have (0 1? A) = (A. 2 ). THEOREM 182. ,\ (A 2 , B 2 ), (B 1} Q, (B 2) C 2 ) = | (B 2 , 2 ), i/ C l be distinct from A lt then we shall also have The elements A lt B l and G l must lie in at least one general plane, say P!, and since A 2 B 2 must either be parallel to A^ or identical with it, while AQj must either be parallel to B^ or identical with it, it follows that there is a general plane, say P 2 , either parallel to P! or identical with it which contains the elements A 2 , B 2 and C 2 . Let P' be a general plane parallel to P z and P 2 , and therefore distinct from both, and let (A', B') and (B f , C') be pairs in P' such that: (A,, A) D (A', B') and (A, CO D (#, 0'). Then, by Theorem 127, since A 1 A' and OjO 7 cannot lie in the same general line (owing to (^ being distinct from A^ and both of them lying in Pj), it follows that : (A,, C.) D (A', C'). Thus C' must be distinct from A. But now, since A 2 , B 2 and (7 2 lie in P 2 while A', B' and C' lie in the parallel general plane P', it follows that A 2 B 2 cannot be identical with A'B , and B 2 C 2 cannot be identical with B'C'. Thus since (A lt B,) \ = (A 2) B 2 ) and C 298 A THEORY OF TIME AND SPACE it follows that we must have (A,, Bi) D (A', ') and (B,, C 2 ) Q (B', O'). Thus since C' is distinct from A', it follows that : (A 2 , C 2 ) D (A', C'}. But we have seen that : (A lf C,) D (A', C"), and so (A^CJ = \(A 2) G 2 ). Thus the theorem is proved. REMARKS. One special case of this theorem deserves attention. If j5j be linearly between A l and G l} it follows, by Theorems 79, 95 and 119, that B r must be linearly between A' and C" and similarly B 2 must be linearly between A 2 and C 2 . We shall require this result in proving the next theorem. Again, since the only congruence of optical pairs is co-directional we may state the following result : If (A lt BJ, (A* B 2 ), (B l} d), (B Z) (7 2 ) be optical pairs such that : (44)s(4JU and (B lt C,) = (B t , C a ), then if B l be linearly between A l and G l we shall have B 2 linearly between A z and C 2 and also have THEOREM 183. If (A-i, B^, (A 2 , B 2 ), (B l} (7j), (B 2 , C 2 ) be inertia pairs such that: and (B l ,C 1 ) = (B 2) G 2 ) > then if B l be linearly between A l and G l while B 2 is linearly between A and G 2 , we shall also have (A,, C,) ~ (A,, C 2 ). A THEORY OF TIME AND SPACE 299 If (A St B') be an inertia pair such that : (A,, B l )\ = \(A 2 , ff), then we may take a separation line b which passes through A% and is normal to both A 2 B 2 and A 2 B'. Let D 2 be an element in b such that (A 2 , A) is conjugate to (A 2 , B 2 ) and let (A lt A) be a separation pair such that: Now A l D 1 must be either parallel to A 2 D 2 or identical with it, while A^ must be either parallel to A 2 B f or identical with it, and so J. x A must be normal to A&. Since A^ and A 2 B 2 are inertia lines, it follows that A^ and A l D l lie in an acceleration plane and also A 2 B 2 and A 2 D 2 lie in an acceleration plane. Then since (A 2 , A) is conjugate to (A 2 , A)> it follows that D Z B 2 is an optical line. Let B l oe an element in A^B^ such that (A l} BJ) is an after- or before-conjugate to (A lt A) according as (A 2 , B 2 ) is an after- or before- conjugate to (A^ A)- Then, by Theorem 181, since (A D 2 ) = (A l , A) we must have (A 2 , B 2 ) = (A l} #/) But (A l ,B 1 )~(A 2) B 2 ) and so (A lf B l ) = (A l) /) Thus, since A^ is an inertia line we must have BI identical with B l and so, since (J.,, BJ is conjugate to (A ly A), it follows that A A is an optical line. Now let the optical line through (7 X parallel to B l A intersect A l D l in F l and let the separation line through Bj_ parallel to A^ intersect C& in E lt Then, since B l is linearly between A^ and (7 X , it follows, by Theorem 78, that A is linearly between A l and F lt Let (A, ^a) be a pair such that : Then, by the remarks at the end of Theorem 182, A will be linearly between A a and F 2 and we shall also have 800 A THEORY OF TIME AND SPACE Now let the optical line through F 2 parallel to D 2 B 2 intersect A 2 B 2 in ft', and let the separation line through B 2 parallel to A 2 F 2 intersect F 2 C 2 ' in E 2 . Then we have (B 19 EJ\ = \ (A, ^i) and (A, F l ) I = I (A, F*\ and therefore (B lt EJ \ = \ (A, F 2 ). But we have also (A, F a ) \ = \ (B 2 , E 2 ), and so (1,^1) = (# 2 , #2). But now since B l E l is parallel to A l D l it must be normal to B l C l , and since E^C^ is an optical line, it follows that (B l , E^ is conjugate to (A, ft). Similarly, since A^a is parallel to J. 2 D 2 it must be normal to B 2 C 2 and, since E Z C 2 is an optical line, it follows that (#2, E 2 ) is conjugate to (B 2 , C/). But now, since A i g linearly between ^. 2 and ^ 2 > it follows that B 2 is linearly between A 2 and 2 '. If then B 2 be a/fer J. 2 we must have C 2 after B 2 , while if B 2 be before A 2 we must have ft 7 before B 2 . Bulb, since .Z? 2 is linearly between A 2 and (7 2 , it follows that if B 2 be q/i56r ^1 2 we must have C 2 after B 2) while if B 2 be before A 2 we must have C 2 before B 2 . Thus ft' is a/ter or before B 2 according as ft is ci/zter or before B 2 . But since (A, ft) an d (5 2 , ft) are inertia pairs such that : (JJ,, CO = (ft, (7 2 ) ; it follows that C 2 is after or before B 2 according as ft is o/iter or before J?j and therefore ft 7 is after or before B z according as ft is after or before B. Thus since (B l} E,) = (B 2 , E 2 ), it follows, by Theorem 181, that : (B 1} ft) = (ft, ft'), and since (5^, ft) = (B 2 , ft) it follows that : (B 2 , ft) = (5 2 , ft'). Thus since these pairs lie in the same inertia line we must have ft' identical with ft. But now ft will be a/fer or before A 2 according as ft is after or fte/brc'^! and so (A 2 , ft) will be an after- or before-conjugate to (A 2 , F 2 ) according as (A lt ft) is an after- or before-conjugate to (A ly FJ. A THEORY OF TIME AND SPACE 301 Thus since (A lt F,) = (A 2) F 2 ) it follows, by Theorem 181, that : (A 1} 00 = (A 2 , C 2 ), and so the theorem is proved. THEOREM 184. If(A l} B}), (A 2 , B 2 ), (B l} Ci), (B 2 , C 2 ) be separation pairs such that: and (B lt ft) = (B 2) 2 ), then if B l be linearly between A and Ci while B 2 is linearly between A z and 2 we shall also have Let (A lt A) and (A 2 , A) be inertia pairs which are after-conjugates to (A lt Bi) and (A 2 , B 2 ) respectively. Then since A l A and A 2 D 2 are inertia lines it follows that A l D l and J-i-Sj lie in an acceleration plane and A 2 D 2 and A 2 B 2 lie in an acceleration plane. Since (A 1} A) is conjugate to (A lt B^ it follows that B 1 D 1 is an optical line, and similarly B 2 D 2 is an optical line. Now let the optical line through G l parallel to B 1 D l intersect A l A in F l} and let the optical line through C 2 parallel to B 2 D 2 intersect A 2 D 2 inF 2 . Then, since B 1 is linearly between A l and Ci, it follows, by Theorem 78, that A is linearly between A l and F^. Similarly D 2 is linearly between A 2 and F 2 . Let the inertia line through B^ parallel to A l F l intersect Ci^i in E lt and let the inertia line through B 2 parallel to A 2 F 2 intersect G 2 F 2 in E 2 . Then since (A i} A) is an after-conjugate to (A lt Bj) we must have D l after A 1 and since D 1 is linearly between A^ and F l we must have ^i after D l . Thus we must have E after B l and in a similar manner we can show that E 2 must be after B 2 . But now, since (A ly A) is conjugate to (A lt B^, it follows that A 1 D l is normal to A l B l , and since B l E l is parallel to A 1 A while A l} B 1 and C-L lie in one general line, it follows that B 1 E l is normal to B^ Q. Thus since C-^E^ (that is C l F l ) is an optical line it follows that (B ly E^) is conjugate to (B lt Ci) and (A lf F^ is conjugate to (A lt Ci). Further A is after A, and F l is after A and so F l is after A^ 302 - A THEORY OF TIME AND SPACE Thus (Si, EI) and (A lt F^) are after-conjugates to (B l} GI) and (Ai t GI) respectively. Similarly (B 2 , E 2 ) and (A 2 , F 2 ) are after-conjugates to (JB a > G a ) and (A 2 , C 2 ) respectively. But now since (Ai, Si) = (A 2 , B 2 ), while (Ai, A) and (A z , A) are after-conjugates to (A^ Si) and (A 2 , B 2 ) respectively, it follows, by Theorem 181, that : (A lt A) = (A a , A). Similarly, since (B l} GI) = (B 2 , (7 2 ), while (Si, EI) and (B 2 , E 2 ) are after- conjugates to (B it GI) and (B 2) (7 2 ) respectively, it follows that : But we clearly have (B lt E,) = \ (A, ^i) and (B 2) E 2 )\ = \(D 2) F Z ). Thus we have (A, ^i) = (A, F*). But since A is linearly between AI and F it while A i g linearly between A 2 and F 2) it follows, by Theorem 183, that (A lf FJ = (^1 2 , ^ 2 ). We have however seen that (A iy FI) and (A a , F 2 ) are after-conjugates to (Ai, GI) and (J. a , Ca) respectively, and so it follows, by Theorem 181, that: (A l) G l ) = (A, ) G 2 }. Thus the theorem is proved. THEOKEM 185. If A and B be two distinct elements and E be any element in AB distinct from A and B, while F is an element in AB such that: then we shall have (A', F) = \ (E, B). Let a- be a general line parallel to AB and let a general line through A intersect a in A' while parallel general lines through B and E inter- sect a in B' and E' respectively. Finally let a general line through F parallel to A A' or BE' intersect a in F'. A THEORY OF TIME AND SPACE 303 Now we clearly have (E, E') D (B, B'\ and so (E, E')\ = \ (B, B'). But, since E' and A lie in parallel general lines, they must be distinct, and so, by Theorem 182, (A,E') = \(F,B'). Now F cannot coincide with A for then E would require to coincide with B, contrary to hypothesis, and so FB' must be parallel to AE'. Thus we must have (A, F) D (E', B'). But we also obviously have (E, B) Q (E' 9 B'\ and so (A, F)\ = \ (E, B). Thus the theorem is proved. THEOREM 186. If B and G be two distinct elements in a separation line and be their mean, and if A be any element in a separation line a which passes through and is normal to BC, then : (A,B) = (A,C). Since a is normal to BG and since they are both separation lines, it follows that a and BC lie in a. separation plane, say & If the element A should happen to coincide with 0, then since BG is a separation line, the theorem obviously holds. Suppose next that A does not coincide with and let d be an inertia line passing through A and normal to S. Let P be the acceleration plane containing a and d. Now since d is normal to 8 t it follows that BC is normal to d, and since BG is also normal to a, and since a and d intersect and lie in P, it follows that BG is normal to P. Let D be the one single element common to d and the a. sub-set of B. Then BG and BD determine a general plane, say Q, which must be either an optical plane or an acceleration plane, since BD is an optical line. But now P and Q have the general line OD in common, and since BG is normal to P it follows that BG is normal to OD. If Q were an 'optical plane OD would require to be an optical line, while if Q were an acceleration plane OD would be an inertia line. 304 .- A THEORY OF TIME AND SPACE But since ED is an optical line in Q and since BD and OD intersect, it follows that Q cannot be an optical plane. Thus Q must be an acceleration plane and OD must be an inertia line normal to the separation line BG. Thus, since is the mean of B and C, it follows that B, C and D are three corners of an optical parallelogram of which is the centre. Thus CD is an optical line. But since AD is normal to S it must be normal to both AB and AC. Also, since D is in the a sub-set of B and is distinct from B it follows that D is after B. Thus, since AB is a separation line while AD is an inertia line, it follows that D is after A, and accordingly (A, D) is an after-conjugate to both (A, B) and (A, C). Thus we have (A, B) = (A, C), and so the theorem is proved. THEOREM 187. If A, B and C be three distinct elements which lie in a separation plane S, but do not all lie in one general line, and if be the mean of B and C while (A, B) = (A, C), then AO is normal to BC. Let d be an inertia line passing through A and normal to S and let P be the acceleration plane containing d and AO. Then d is normal to both AB and AC and, since there is one definite element, say D, in d such that (A, D) is an after- conjugate to both (A, B) and (A, C). Thus BD and CD are optical lines and since they intersect they must lie in an acceleration plane, say Q. But now since is the mean of B and (7, it follows that B, C and D are three corners of an optical parallelogram whose centre is 0, and therefore BC is normal to QD. But OD is common to both Q and P, while BC (since it lies in 8) is normal to AD, which also lies in P. Thus BC is normal to two intersecting general lines in P and there- fore BC is normal to P. But AO lies in P and therefore AO is normal to BC. Thus the theorem is proved. A THEOKY OF TIME AND SPACE 305 THEOREM 188. If A, B and C be three distinct elements which lie in a separation plane S, but do not all lie in one general line, and if and if be an element in BC such that AO is normal to BC, then is the mean of B and C. Let 0' be the mean of B and C. Then, by Theorem 187, AO' is normal to BC, and, by hypothesis, AO is normal to BC. But both AO' and AO pass through the element A and lie in the separation plane S and we have already $een that there is only one general line in a given separation plane which passes through a given element and is normal to another general line in the separation plane. Thus AO must be identical with AO' and therefore must be identical with 0'. It follows that is the mean of B and C and so the theorem is proved. Definitions. If A and B be two distinct elements, then the set of all elements lying linearly between A and B will be called the segment AB. The elements A and B will be called the ends of the segment, but are not included in it. The set of elements obtained by including the ends will be called a linear interval. If A and B be two distinct elements, then the set of elements such as X where B is linearly between A and X may be called the pro- longation of the segment AB beyond B. Such a set of elements will also be spoken of as a general half-line. The element B will be called the end of the general half-line. We shall describe segments and general half- lines as optical, inertia, or separation, according as they lie in optical, inertia, or separation lines. It is easy to see that any element .B in a general line divides the remaining elements of the general line into two sets such that B is linearly between any two elements of opposite sets, but is not linearly between any two elements of the same set. For let A and G 'be two elements in the general line such that B is linearly between A and C, and let A be another element in it distinct from A and such that B is linearly between A and C. R. 20 306 A THEORY OF TIME AND SPACE Then by the analogue of Peano's Axiom 10 since A and A' are distinct we must either have A' linearly between B and A or A linearly between B and A'. Thus we cannot have B linearly between A and A'. Again if C' be an element distinct from C and such that B is linearly between A and C' \t follows similarly that we must have either C' linearly between B and C or C linearly between B and C' and cannot have B linearly between C and C f . If now A' be linearly between A and B then since B is linearly between A and C' it follows, by the analogue of Peano's Axiom 8, that A' is linearly between A and C'. But now since B is linearly between A and C' and also A' is linearly between A and C f it follows, by the analogue of Peano's Axiom 9, that either B is linearly between A and A' or B is identical with A' or B is linearly between A' and C'. The first and second of these alternatives are impossible since by hypothesis A' is linearly between A and B. Thus we must have B linearly between A' and C'. Next take the case where A is linearly between A' and B. Since we also have B linearly between A and C', it follows by the analogue of Peano's Axiom 11 that B is linearly between A' and C". This shows that B divides the remaining elements of the general line AC into two sets in the manner above stated. It is clear that these sets are general half-lines. If elements X and Y lie in the same general half-line whose end is B, they will be said to lie on the same side of B. If, on the other hand, B be linearly between X and Y, then these elements will be said to lie on opposite sides of B. A general half-line whose end is B and which contains an element X may be denoted by general half-line BX. It is also easy to see that any general line b in a general plane P divides the remaining elements of P into two sets such that if A and C be any two elements of opposite sets then b will intersect AC in an element linearly between A and C', while if A and A' be two elements of the same set, then b will not intersect A A' in any element linearly between '^4. and A. For let A and C be two elements in P and let them be such that b intersects A C in an element B linearly between A and C. Also let A' be another element in P distinct from A and such that b intersects A'C in an element B' linearly between A' and (7. A THEORY OF TIME AND SPACE 307 If A' should happen to lie in AC then E' will coincide with B, and since b cannot intersect AC in any other element besides B, it follows that b does not intersect A A' in an element linearly between A and A. If A' does not lie in AC then it follows from Theorem 128 (2) that b cannot intersect AA in an element linearly between A and A. Again if C' be an element lying in P and distinct from C and such that b intersects AC' in an element B" linearly between A and C', it follows in a similar manner that b cannot intersect GO' in an element linearly between C and C'. Now C' either does or does not lie in A' A ; if C' does lie in A' A then since B'' is linearly between A and C', but is not linearly between A and A', it follows that B" must be linearly between A' and C'. If, on the other hand, C' does not lie in A A, then since B" is linearly between A and C', and since b does not coincide with either AC' or B"A', it follows by Theorem 128 (1) that b either intersects A' A in an element linearly between A' and J., or else intersects A'C' in an element linearly between A and (7 X . But the first of these alternatives is excluded and so the second must hold. This shows that the general plane P is divided by the general line b in the manner above stated. If elements X and Y lie in the general plane P, but not in the general line 6, they will be said to lie on the same side of b if they both lie in the same set and will be said to lie on opposite sides of b if X lies in one of the sets and Y in the other set. Definition. If a general line b lies in a general plane P then either of the sets of elements on one side of b will be called a general half-plane. The general line b will be called the boundary of the general half- plane. The following important result which may be conveniently expressed in the nomenclature of general half-lines can be easily proved. If (A lt B^, (A 2 , B 2 ), (A ly A), (A 2 , C 2 ) be inertia, optical or separation pairs such that : (A l ,B J ) = (A t ,B s ), and (A,, C,) = (A,, 2 ), then if B 1 be linearly between A x and C l and if C 2 lies in the general half-line A 2 B 2 , we shall also have B 2 linearly between A 2 and C 2 . In the case of optical pairs the above congruences imply that A l B l and A 2 B 2 are the same or parallel optical lines, but nothing of this sort is implied in the case of inertia or separation pairs. 202 308 A THEORY OF TIME AND SPACE In all cases there is an element, say C 2 , in A 2 B 2 and on the opposite side of B 2 to that on which A 2 lies and such that : ( C,) = OB 2 , 0,'). Then in all cases it follows that : (A l) C l } = (A,,C^, and so (A 2 , C 2 ) = (A 2 , C 2 ). But C 2 and C 2 both lie in A 2 B 2 and on the same side of A 2 , and must therefore be identical. Thus B 2 is linearly between A 2 and C 2 . Definition. If A, B, C be three distinct elements which do not all lie in one general line, then the three segments AB, BC, CA, together with the three elements A, B, C, will be called a general triangle, or briefly a triangle in an acceleration, optical, or separation plane, as the case may be. The elements A, B, C will be called the corners while the segments AB, BC, CA will be called the sides of the general triangle. THEOREM 189. If A lf B l , d be the corners of a triangle in a separation plane P l and A s , B 2 , C 2 be the corners of a triangle in a separation plane P 2 and if further (C^A,) = (C 2 ,A Z \ (4,5,) = (ft, ft), while B l C l is normal to AiC l} and B 2 C 2 is normal to A 2 C 2t then we shall also have (A l ,B l ) = (A 2 ,B 2 ). In order to prove this theorem we shall consider a number of special cases on which the general proof is made to depend. CASE I. B 2 identical with B l and C 2 identical with C l} while P 2 is identical with P lt In this case since the separation lines A& and A 2 C-^ are both normal to B l G l and both lie in the separation plane P 1 and pass through the element C l} they must be identical. If further A 2 should coincide with A 1 the result is obvious, and so we shall suppose that A 2 does not coincide with A lt Now since (C lf A^, &c. lie in the separation plane P l they must all be separation pairs and since in this case (ft, A 2 ) = (A 2 , C,), it follows that : (A 2 , C,) = (ft, AJ. A THEORY OF TIME AND SPACE 309 Thus G! must be the mean of A l and A 2 and therefore, by Theorem 186, we have or (A lt #0 = (A 2) B 2 ). CASE II. B 2 identical with B l and G 2 identical with GI, while PJ and P 2 lie in the same separation threefold W. If P 2 should be identical with P l this case reduces to Case I, and so we shall suppose them distinct. Now since A 1} B^ and A 2 are three distinct elements in W which do not lie in one general line, it follows that A lt B l and A 2 lie in a separation plane, say R, which must be distinct from both P l and P 2 , since these latter two separation planes are supposed distinct. Similarly A l ,G l and A 2 lie in a separation plane, say 8, which is also distinct from P 1 and P 2 . Now let be the mean of AI and A%. Then, by Theorem 187, since (0,, A,) = (C,, A,), it follows that G 1 is normal to A 1 A 2 . But since B^G l is normal to A^ and to A^G^ which are distinct intersecting separation lines, it follows that B l G l is normal to S and therefore must be normal to AiA 2 . Thus A 1 A 2 is normal to the two intersecting separation lines BiGi and (7x0 and must therefore be normal to every general line in the general plane containing them. It follows that A 1 A 2 is normal to B 1 0. But now, by Theorem 186, since is the mean of A l and A* and since B 1} A l} A 2 lie in a separation plane, it follows that : (B l) A l )^(B 1) A 2 ), or (A lt B,) = (A 2) B 2 ). CASE III. G 2 identical with G l and P 2 identical with P l . Let b be a separation line passing through C l and normal to P! and let B ' be an element in 6 such that : Then we shall also have Now the separation plane P and the separation line b determine a separation threefold, say W, which contains A lt B lJ C l} A z , B 2) B'. 310 ' A THEORY OF TIME AND SPACE Again, since B'C l is normal to P, it must be normal to G^A l} C^ C.B, and G,B 2 . Then since (C lt B,) = (C lt B'), it follows by Case II that : (A li B 1 ) = (A li B f ) ..... ..... ......... . ....... (1). Again since (C lt A,) = (C lt A a ), it follows by Case II that : (A lt B / ) = (A, f B f ) ........................... (2). Further, by Case II, since (Ci,5') = (0i,*X it follows that: (A 2) B') = (A 2 , B 2 ) ............. . .......... (3). Thus from (1), (2) and (3) it follows that : CASE IV. P 2 either identical with P l or parallel to P 3 . There is, as we have already seen, one single element, say A', such that: Similarly there is one single element, say B', such that : (C ly B,)\ = \ (ft, B'). Now, since P 2 is either identical with P l or parallel to P lt it follows that P 2 must contain ftJ/ and C a B'. Also since G a A f must be either parallel to C 1 A 1 or identical with it, and since C Z B' must be either parallel to C 1 B l or identical with it, then since B^ is normal to A^G^ it follows that B'C 2 is normal to A'C 2 . But now, by Theorem 182, we must have Also since (G lt A^)= (C 2 , A 2 ), it follows that : (G 2 , A 2 ) = (C 2 , A'), and since (C l} BJ = (C 2 , B 2 ), it follows that : (6V, B 9 ) = (C 2) B'). Thus, by Case III, it follows that : (A',B') = (A,,B 1 ). Since however we have (^l,, ,) = (A', B'), it follows that : (A lt B,) = (A 3 , B s ). A THEORY OF TIME AND SPACE 311 CASE V. P l and P 2 lie in the same separation threefold W. If P l and P 2 have no element in common, then since they both lie in W they must be parallel to one another and the result follows from Case IV. We shall therefore suppose that P l and P 2 have an element in common, but are distinct. Then, by Theorem 152, they have a second element in common, and therefore have a general line in common which we shall call 6. Let G be any element in b and let a 1 and a 2 be separation lines passing through C and normal to b and lying in P l and P 2 respectively. Let B be an element in b such that : (O u B l ) = (G, B). Then we shall also have Let AI and A 2 be elements in a 1 and a 2 respectively such that : (Q^AteWA^ and (C 2 , A*) = (C, A 2 '). Then since (Ci, A,) = (G 2 , A 2 ), we have (C, A,') = (G, A 2 '). Thus by Case II it follows that : (A 1 ',B) = (A 2 ',B) ........................ (1). But by Case IV it follows that : (A l ,B l ) = (A l ',B) ........................... C2), and similarly it follows that : (At,B,) = (At',B) ..... . ..................... (3). Thus from (1), (2) and (3) it follows that : (A 19 BJ = (A 2) B a ). Thus whether P l and P 2 are identical, or parallel, or whether they have a general line in common, the theorem holds provided P l and P 2 lie in the same separation threefold W. CASE VI. P! and P 2 do not lie in one separation threefold. In this case we may take one separation threefold, say TFj, which contains P l and another separation threefold, say W 2 , which contains P a . Now W 2 may be either parallel to TF 15 or else not parallel to it. 312 A THEORY OF TIME AND SPACE If X be any element in W 2 but not in W l then, if W 2 be not parallel to W l , there must be at least one separation line passing through X and lying in W 2 which is not parallel to any separation line in Wi and which therefore, by Theorem 164, must intersect W l . Thus if W 2 be not parallel to W l it must have an element in common with it and so, by Theorem 166, W l and W 2 must have a general plane in common. Suppose then first that W 2 is parallel to W l and let C be any element in W 2 , Then there is a general line, say a, passing through C and lying in W 2 which is parallel to C l A l . Similarly there is a general line, say b, passing through C and lying in W z which is parallel to CiBi. Thus since B& is normal to A l G l , it follows that 6 is normal to a. Now let A and B be elements in a and b respectively such that : and But since (B lt A) = (B 2) A), it follows that : (B 2 , D 2 ) = (B 2 , D 2 ). Thus since D 2 and A lie on the same side of B 2 they must be identical, and so (N lt A) = (A r 2 ,A). In the case where D l is linearly between B l and N! we may use a similar method to prove the same result. Thus in all cases where Nj, does not coincide with A, B l or C l we have (J\r,A) = (AT 2 ,A) and (A lt N l ) = (A^N s ). Thus, by Theorem 189, it follows that: (A ly A) = (A*, A), and so the theorem holds in all cases. Definitions. If and X be two distinct elements in a separation plane S, then the set of all elements in 8 such as X where (0,Z)5(0,Z ) will be called a separation circle. The element will be called the centre of the separation circle. Any one of the linear intervals such as OX will be called a radius of the separation circle. If X l and X 2 be two elements of the separation circle such that XiX 2 passes through 0, then the linear interval X 1 X 2 will be called a diameter of the separation circle. Any element which lies in a radius but which is not an element of the separation circle itself will be said to lie inside or in the interior of the separation circle. Any element which lies in S but not in a radius will be said to lie outside or exterior to the separation circle. 320 A THEOEY OF TIME AND SPACE THEOREM 194. If a separation circle and a separation line both lie in a separation plane S, they cannot have more than two elements in common. Let a be the separation line and the centre of the separation circle which we shall suppose to have at least one element A in common with a. Then a either does or does not pass through 0. If a passes through then we know that there is one single element, say B, lying in a and distinct from A and such that : (0,A) = (0,B). Thus B is a second element common to the separation circle and the separation line and in this case no other such element exists. Next suppose that a does not pass through 0. Two cases are possible : (1) OA is normal to a, (2) OA is not normal to a. In Case (1) suppose, if possible, that B is a second element common to the separation circle and separation line and let M be the mean of A and B. Then we should have (0,A) = (0, B\ and since A, B and would not lie in one general line, it would follow, by Theorem 187, that ON must be normal to AB. But by hypothesis OA is normal to AB, and, since S is a separation plane, we cannot have a second separation line passing through and lying in S and which is normal to AB. Thus no such element as B exists and so in this case the separation circle and the separation line have only one element in common. Consider next Case (2) where A is not normal to a, and let N be an element in a such that ON is normal to a. Let B be an element in a and on the opposite side of N to that on which A lies and such that : (N, A) = (N, B). Then N will be the mean of A and B, and so, by Theorem 186, Thus B is a second element common to the separation circle and the separation line. A THEOKY OF TIME AND SPACE 321 Suppose now, if possible, that B' is another element in a distinct from A and B and which is an element of the separation circle. Then we should have and (0, B) = (0, B'). Thus since ON is normal to AB' it would follow, by Theorem 188, that N must be the mean of A and B f and of B and B', in addition to being the mean of A and B, which is clearly impossible, since in any acceleration plane containing a there is only one optical parallelogram having N as centre and A as one of its corners. Thus no such element as B' can exist and so the separation circle and separation line cannot in any case have more than two elements in common. THEOREM 195. If a separation circle in a separation plane S pass through an element A which is inside, and another element B which is outside a second separa- Fig. 43. tion circle in S, then the two separation circles have two elements in common which lie on opposite sides of the separation line AB. Let Oj be the centre of the separation circle inside which A lies and outside which B lies, and let 2 be the centre of the other separation circle. R. 21 322 A THEORY OF TIME AND SPACE Let W be a rotation threefold containing 8. Then, by Theorem 160, through any element of S there is one and only one general line which lies in W and is normal to S. Further such a general line must be an inertia line, since otherwise W would be an optical or a separation threefold. Let Cj and c 2 be inertia lines passing through 1 and 2 respectively and which lie in W and are normal to S. Then c x and c 2 must be either parallel or identical, but we shall show in the course of our proof that they cannot be identical. Let X 1 and X 2 be any elements of the separation circles whose centres are 0j and 2 respectively and let elements Oj and G 2 be taken in G! and c 2 respectively such that (0 1} (?i) is an after-conjugate to (0j, X^), and (0 2 , C 2 ) an after-conjugate to (0 2 , X z ). Then if XJ be any element of the separation circle whose centre is O l we shall have (O l ,X^m.(p l ,X{). and so, since 0^ is normal to 1 X l ' ) it follows that (0 1? (7j) is also an after-conjugate to (0 l5 Xi) and therefore XiG 1 is an optical line. Similarly if X 2 be any element of the separation circle whose centre is 2 , then X 2 G 2 will be an optical line. Thus AC 2 and BC 2 are optical lines. Now if we take a separation line in S which passes through O l and A there will be two elements in it, say Fand Y', such that: (O lt 7)^(0,,^) and (ft, F') 5(0^). Then since J. lies inside the separation circle whose centre is Oj it must lie in one of the radii O^Y or 0jF' and be distinct from F and F'. Thus A must lie linearly between F and Y'. But since (0!, Q) is an after-conjugate to (0 13 F) and (O l} Y') it follows that G! is after both F and F' and so, by Theorem 129 (6), C^A is an inertia line and A is before Oj. Again if we take a separation line in S which passes through O l and B there will be two elements in it, say Z and Z', such that : and (0,,^')s(0,,ir i ), and since J9 lies outside the separation circle whose centre is O lt it does not lie in either of the radii Q^Z or O^Z'. Thus B does not lie linearly between Z and Z' and does not coincide with either Z or Z '. A THEORY OF TIME AND SPACE 323 It follows that we must either have Z linearly between Z' and B, or else Z' linearly between Z and E. Now it is evident (as in the case of Y and F') we must have Ci after both Z and Z '. But B cannot be after Ci for then B would be after Z which is impossible, since ZB lies in $ and is therefore a separation line. Further Ci cannot be after B for then, by Theorem 129 (6), either CiZ or G 1 Z / would require to be an inertia line, whereas we know that they are both optical lines. Thus B is neither before nor after G 1 and so G 1 B is a separation line. Now C 2 cannot be identical with or before G l} for, since B must be before C 2 , it would follow that B was before G l} which we have already seen is not so. Neither can G 2 be after Ci, for we have seen that C l is after A and that CiA is an inertia line, and so we should have C l after one element of the optical line AC 2 and before another element of it and yet not itself an element of it, which is impossible by Theorem 12. Thus (7 2 is neither before nor after Ci, and is distinct from it, which proves that c and c 2 are distinct inertia lines, and so Ci C 2 is a separation line which must lie in W, since Ci and C 2 are distinct elements of W. Now let M be the mean of Ci and C 2 . Then as was shown in the remarks at the end of Theorem 171, since GiC 2 is a separation line lying in W, there is one and only one general plane, say P, passing through M and lying in W to which CiC^ is normal. Also it was proved that this must be an acceleration plane and therefore contains optical and inertia lines. But an optical or inertia line cannot either lie in a separation plane or be parallel to it and so, by Theorem 153, P and $ must have at least one element in common. Thus, by Theorem 152, P and S must have a general line in common which we shall call a. Now we have seen that AC 2 is an optical line and G 2 is after A, while ACi is an inertia line and Ci is after A. Thus Ci is not an element of the optical line AG 2 but is after an element of it and so there is one single element, say F, which is an element both of the optical line AG, 2 and the ft sub-set of Ci. Thus, since F cannot coincide with Ci, it must be before G 1) and FC l must be an optical line. But, since M is the mean of G 1 and G 2 , and since CiC a is a separation line, it follows that FM is an inertia line which is normal to CjC^. 212 324 A THEORY OF TIME AND SPACE Thus, since FM clearly lies in W, it must lie in P. Now the general line a lies in S and is therefore a separation line and, since it also lies in P, it must intersect FM in some element, say G. But now since AC l is an inertia line while FC is an optical line, it follows that F cannot coincide with A. Also since AF is an optical line and Ci lies in the a sub-set of F, it follows that A must lie in the ft sub-set of F and so, since A and F are distinct, we must have F after A. But now since A and G lie in the separation plane S the one is neither before nor after the other. Now G could not either coincide with F or be after F, for in either case it would be after A, which is impossible. Thus, since G and F both lie in the inertia line FM, it follows that G is before F. But, since F is before Ci, it follows that G is before G : . The element F, however, is the only element common to the inertia line FM and the 13 sub-set of Ci and so G does not lie in the /3 sub-set of C,. It follows that GC-i is not an optical line and must therefore be an inertia line. Thus the separation line a and the inertia line GG l determine an acceleration plane, say Q. But now through the element Ci of the acceleration plane Q there pass two and only two optical lines lying in Q and both of these must intersect a. Since Ci does not lie in a, these elements of intersection which we shall call D and E must be distinct, and since O l C l is normal to 8 it must be normal to 1 D and O^E. Thus (0 15 Ci) is an after-conjugate to both (O lt D) and (0 1} E} and so It follows that D and E are elements of the separation circle whose centre is O x . But now since CiD is an optical line it follows that C 1 D and Ci(7 2 must lie either in an optical or an acceleration plane. Since, however, MD lies in P it must be normal to Ci(7 2 and must therefore be either an optical or inertia line. If, however, MD were an optical line normal to Ci0 2 it could not A THEORY OF TIME AND SPACE 325 intersect any other optical line in the general plane containing MD and OjC'g and therefore could not intersect C 1 D. It follows that MD cannot be an optical line and therefore must be an inertia line, and so C 1 D and C f 1 (7 2 lie in an acceleration plane. Thus, since M is the mean of G 1 and G 2 , it follows that C 2 D is an optical line. Similarly C 2 E is an optical line. Thus since 2 G 2 is normal to S and therefore normal to 2 D and 2 E, it follows that (0 2 , C 2 ) is an after-conjugate to both (0 2 , D) and (0 2 , E). Thus we have (0 2 , D) = (0 2 , Z 2 ) and (0 2 , E) = (0 2 , X 2 \ and so D and ^ are also elements of the separation circle whose centre is 2 . It follows that the two separation circles have the two elements D and E in common. We can easily see that these are the only elements which they have in common, for if D' be any element common to the two separation circles, then D' must lie in two optical lines passing through C\ and C 2 and it must also lie in S. Thus since M is the mean of C and C 2 we must have D'M normal to G^G 2 and, since D'M lies in TT, it must lie in P. Thus D' must lie in P and, since it also must lie in S, it follows that it must lie in a. But (7jZ) and G 1 E are the only optical lines which pass through C l and intersect a, and so D' must be identical with either D or E. Thus D and E are the only elements common to the two separation circles. We have next to prove that D and E lie on opposite sides of AB. Since A lies inside the separation circle whose centre is O l while B lies outside it, it follows that neither A nor B can coincide with either Dor E. Also, by Theorem 194, neither A nor B can lie in DE. But A and B must both lie in S and, since DE contains all elements common to P and S, it follows that neither A nor B can lie in P. Let R be the acceleration plane containing AC 2 and BG 2 . Then since A, B and G 2 all lie in R, but not all in one general line, and since they all lie in W, it follows, by Theorem 149, that R lies in W. But P and R have the element F in common and therefore, by Theorem 152, they have a general line in common which we shall call/. 326 A THEORY OF TIME AND SPACE Now F is after A and before 2 and lies in AC> 2 and so F is linearly between A and (7 2 . The general line / passes through F, and since neither A nor B lie in P, while / does lie in P, it follows that/ is distinct both from AC 2 and #F. Thus, since / lies in R, it follows that either /intersects BC 2 in an element linearly between B and (7 2 , or else must intersect BA in an element linearly between B and A. Suppose first, if possible, that / intersects BG 2 in an element K linearly between B and (7 2 . This would mean that BC Z intersected P in an element K which was after B and before (7 2 . Then KM would be normal to C^C^, and since it would intersect the optical line BC 2 , it would follow that KM must be an inertia line. Thus KC l would be an optical line, since M is the mean of C l and C Zt and since K is supposed to be before G 2 , it would follow that K must be before Q. But, since K is supposed to be after B, it would follow that Ci was after J5, which, as we have already seen, is not so. Thus / cannot intersect BG 2 in any element linearly between B and C s and therefore / must intersect BA in an element H linearly between Band A. Since H lies in B A and also in P it must lie in DE. Thus DE intersects BA in an element linearly between B and A. But now, since C 2 is after both A arid B, it follows, by Theorem 129 (6),- that C 2 H is an inertia line and H is before G 2 . Since C 2 D and Oa-E' are optical lines, H cannot coincide with either D or E and so we must have either : D linearly between H and E, or E linearly between H and D, or H linearly between D and E. But since C 2 is a/fer D, # and H it would follow, by Theorem 129 (6), in the first case that C^D must be an inertia line, which we know is not so ; while in the second case it would follow that G 2 E must be an inertia line which we also know is not so. Thus we are left with the third alternative, namely that H is linearly between D and E. It follows that D and E are on opposite sides of AB and so the theorem is proved. A THEORY OF TIME AND SPACE 327 Definition. If and X be two distinct elements in a separation three- fold W, then the set of all elements in W such as X where (0, X) = (0, X.) will be called a separation sphere. The element will be called the centre of the separation sphere. The terms radius, diameter, inside, outside, &c. may be defined in a similar manner to the case of a separation circle. THEOREM 196. If A , AI and C be three distinct elements such that A l is linearly between A and C and if A 2 , A 3 , A 4 , ... be elements such that: A l is linearly between A Q and A 2 , A 2 is linearly between A and A S) and such that : (A , AJ = (A ly A 2 ) = (A 2 , A s ) . . ., then there are not more than a finite number of the elements A l , A 2 , A 3 , . . . linearly between A and C. It is evident that all the series of elements A lt A 2 , A 3 , ... lie in the general line A G which we shall call a. We shall first prove the theorem for the case where a is an inertia line and C is after A . We shall suppose that a lies in an acceleration plane P. Now since A l is linearly between A and C we must have A l after A Q , and so if we take two generators of P of opposite sets passing through A and A t respectively, they will intersect in some element, say B , which will be after A Q and before A l and must lie outside a. Let b be an inertia line passing through B and parallel to a and let optical lines parallel to AoB and passing through A l} A 2) A 3t ... intersect b in the elements B^ B 2 , B 3 , ... respectively. Now since A- is after A and since further : A l is linearly between A and A 2 - t A 2 is linearly between A and A 3 , it follows that : A! is after A ; A 2 is after A l \ A 3 is after A 2 ] .... 328 - A THEORY OF TIME AND SPACE Thus since (A 0) AJ = (A lt A 2 ) = (A 2 , A 3 ) . . . it follows that : A^ is the mean of A and A 2 , A 2 is the mean of AI and A s . But now by construction we have (A., A,) D (5 , B 2 \ and so, since A l is the mean of A and A 2t and since A l B l is parallel to A Bo and A 2 B 2 , it follows that 5j is the mean of B Q and B 2 and so, by Theorem 98, A 2 B t is parallel to A 1 B . Similarly A 3 B 2 is parallel to A 2 B l and so on. Thus, since A^B^ is an optical line, it follows that A 2 B lf A 3 B 2 , ... are all optical lines and so A lt A 2 , A 3) ... mark steps taken along a with respect to b. But since a and b do not intersect, it follows by Post. XVII that C may be surpassed in a finite number of steps taken from A . Thus there cannot be more than a finite number of the elements A lt A 2 , A 3) ... linearly between A and (7. Similarly if C be before A the same result follows by using Theorem 64 in place of Post. XVII, and so the theorem is proved for the case where a is an inertia line. Consider next the case where a is a separation line and let b be any inertia line which passes through A . Then a and 6 determine an acceleration plane which we shall call P. Now one of the generators of P which pass through A l intersects b in an element which lies in the a sub-set of A lt while the other generator intersects b in an element which lies in the ft sub-set of A lt Let the former of these generators be called / a , and let it intersect b in the element AI. Then since A l does not lie in b, it follows that A t ' is after A l and so, since A A l is a separation line, we must also have AI after A . Let/o^j/s,/^ ... and/ c be generators of P parallel to/I and passing through A , A 2 , A S) A 4 , ... and C respectively. Further let / 2 ,/ 3 ,/ 4 , ... and/ c intersect b in A 2 , A 3 ', A 4 ', ... and C' respectively. Then, since AI is after A , it follows that/I is an after-parallel of/ , and since A l is linearly between A and C, it follows that f c is an after- parallel of fa and so C 1 ' is after A^. A THEORY OF TIME AND SPACE 329 Further : A l is linearly between A and A 2 , A 2 is linearly between A and A 3t Thus we have / x is an after-parallel of/ f 2 is an after-parallel of/i, / 3 is an after-parallel of f 2 , Thus we have A! is linearly between A Q and A 2 , A 2 is linearly between A 1 and A 3) A s is linearly between A 2 and A 4) But since (J. 0j ^4.i) = (A lt A 2 ) = (A 2 , A 3 ) . it follows that: A l is the mean of A and A 2t A 2 is the mean of A l and A 3 , Thus, by Theorem 82, AI is the mean of A and J./ is the mean of AI and and so ( J. , J./) = (AS, A 2 ') = (A 2 ' t A 3 ') .... Thus by the first case of the theorem, there cannot be more than a finite number of the elements AS, A 2) A 3 ' linearly between A Q and G r . But each of these elements which is linearly between A and G' corresponds to one of the series A lt A 2 , A 3 , ... which is linearly between AQ and (7, while any one which is not linearly between A and C' corre- sponds to one of the series A lt A 2 , A 3 , ... which is not linearly between A Q and G. Thus there are not more than a finite number of the elements AI, A 2 , A 3 , ... linearly between A and G, and so the theorem holds when a is a separation line. As regards the case where a is an optical line and C is after A we may proceed just as we have done for the case where a is a separation line. 330 A THEORY OF TIME AND SPACE In this case a is one of the generators of the acceleration plane P, while /o,/i,/ 2 , .../ c will be generators of the opposite set. The result then follows in a similar manner. In the case where a is an optical line and C is before A we also make use of a similar method except that the element C' in the inertia line b will be before A instead of after it. Thus the theorem holds in all cases. REMARKS. It will be observed that the above theorem is equivalent to the Axiom of Archimedes and has been deduced by the help of Post. XVII. In our remarks on the introduction of this postulate, its analogy to the Axiom of Archimedes was pointed out together with the fact that the postulate contains no reference to congruence. Having denned congruence of pairs we are able to deduce the Axiom of Archimedes in the usual form as given above. We shall now give the final postulate of our system which is equi- valent to the Axiom of Dedekind. POSTULATE XXI. If all the elements of an optical line be divided into two sets such that every element of the first set is before every element of the second set, then there is one single element of the optical line -which is not before any element of the first set and is not after any element of the second set. Since an element is neither before nor after itself, it is evident that this one single element may belong either to the first or second set. Again, if a be an optical line in an acceleration plane P, then through each element of a there passes one single generator of P of the opposite system to that to which a belongs. Also every such generator intersects a. Thus there is a one-to-one correspondence between the elements of a and the generators of P of the other system and so it follows that : if either system of generators of an acceleration plane be divided into two sets such that every generator of the first set is a before-parallel of every generator of the second set, then there is one single generator of the system which is not a before-parallel of any generator of the first set and is not an after-parallel of any generator of the second set Again if b be any inertia or separation line and if P be an acceleration plane containing it, then if we select either system of generators of P, there is a one-to-one correspondence between the elements of b and the generators of the selected system which pass through these elements. A THEORY OF TIME AND SPACE 331 If b be an inertia line and X and Y be any two elements of b, then X will be before or after Y according as the generator through X is a before- or after-parallel of that through Y. Thus the property formulated in Post. XXI holds for an inertia line as well as for an optical line. It is also clear that a corresponding result holds in the case of a separation line, but since here no element is either before or after another, the property must be formulated somewhat differently. In order to state the result when b is a separation line we may make a perfectly arbitrary convention with regard to the use of the words right and left. Thus if X and Y be any two elements of 6, we may say that X is to the left or right of Y according as the generator of the selected system which passes through X is a before- or after-parallel of that through Y. We may therefore state the property as. follows : If all the elements of a separation line be divided into two sets such that every element of the first set is to the left of every element of the second set, then there is one single element of the separation line which is not to the left of any element of the first set and is not to the right of any element of the second set. Definitions. If (A, B) and (G, D) be inertia or optical pairs in which B is after A and D after C, or if (A, B) and (C, D) be separation pairs, then: (1) If (A, B) = (C, D) we shall say that the segment AB is equal to the segment CD. (2) If (Ay B) = (C, E), where E is any element linearly between C and D, we shall say that the segment AB is less than the segment CD. (3) If (A, B) = (C, F), where F is any element such that D is linearly between C and F, we shall say that the segment AB is greater than the segment CD. In the case of separation or inertia segments we must always have either : AB is equal to CD, or AB is less than CD, or AB is greater than CD. In the case of optical segments, however, this is only true provided they lie in the same, or in parallel optical lines. Again if (A, B) and (C, D) be inertia or optical pairs in which B is 332 A THEOEY OF TIME AND SPACE after A and D after C, or if they be separation pairs, and if E, F, G be elements such that F is linearly between E and G while (A, B) = (E, F) and (C, D) = (F, G), we shall say that the length of the segment EG is equal to the sum of the lengths of the segments AB and CD. It is evident that the lengths of two optical segments can only have a sum in this sense provided they lie in the same or in parallel optical lines, whereas the lengths of two inertia segments or two separation segments always have a sum. Having thus introduced the idea of the length of a segment being equal to the sum of the lengths of two others we can obviously have any multiple and also (as follows from the remarks at the end of Theorem 82) any sub-multiple of a given segment : using the terms " multiple " and " sub-multiple " in the ordinary sense. We may also clearly have a segment equal to any proper or improper fractional part of the given segment. Again, if A and A 1 be two distinct elements in a general line I, it may easily be shown that there are elements A z , A 3 , A 4 ... A n ... in I on the same side of A as is A l and such that the segment A A n is equal to n times the segment A Q A l . Similarly there are elements A^, A^, J._ 3 ... A_ n ... lying in I but on the opposite side of A and such that the segment A_ n A Q is equal to n times the segment A A l . Again, it may easily be shown that corresponding to any positive rational number r=- there is an element A r in I on the same side of ? A Q as is A l and such that q times the segment A A r is equal to p times the segment A A lt Similarly, corresponding to any negative rational number r = - , it may be shown that there is an element A_ r in I, but on the opposite side of A and such that q times the segment A_ r A is equal to p times the segment A A l . By making use of our equivalents of the Archimedes and Dedekind axioms for the elements of a general line along with the corresponding properties of real numbers, it is possible in this way to set up a one-to- one correspondence between the elements of a general line and the aggregate of real numbers. The logical steps involved in setting up such a correspondence have A THEORY OF TIME AND SPACE . 333 been carefully investigated by others and it is unnecessary to go into further details here. These may be found, for instance, in Pierpont's Theory of Functions of Real Variables, vol. I, chapters I and II, and in other works. The absolute value of the difference of the real numbers correspond- ing to the two ends of any segment of I gives us a real number which may be called the numerical value of the length of the segment in terms of the unit segment A^A^. If I be an inertia or separation line, the length of any segment of a general line of the same kind as I is always expressible in terms of our selected unit segment ; but this is not true in general if I be an optical line. It is to be observed that the length of an inertia or optical segment is independent of which end of the segment is after the other. In this respect the question of the equality of inertia or optical segments differs from the closely related question of the congruence of the inertia or optical pairs forming the ends of the segments. The criterion of proportion given by Euclid and which, according to Sir T. L. Heath, is probably due to Eudoxos, is clearly applicable in our geometry. Certain results regarding the proportion of segments may easily be shown to hold for all types of general line. Thus if 0, A, B be the corners of any general triangle, and if G be any element distinct from A in the general half-line OA, then if a general line through G parallel to AB intersect OB in the element D, we must have segment A : segment OG = segment OB : segment OD. This may be proved by the method of De Morgan described by Heath in his edition of Euclid, vol. n, p. 124. If OA should be a separation line and OB an inertia line normal to OA (or conversely), while AB is an optical line, then CD will also be an optical line and so it follows that : separation segments are proportional to their conjugate inertia segments. Again if through D a general line be taken parallel to OA and inter- secting AB in the element E, while through E a general line is taken parallel to OB and intersecting OA in the element F, we clearly have (C,D)\= (A,E) and (A, C)\ = \ (F, 0). Thus by Theorem 185 (A, F)\ = \(C, 0). 334 A THEORY OF TIME AND SPACE But since EF is parallel to BO it follows that : segment AO : segment AF= segment AB : segment AE. Thus since segment AF = segment 00. and segment AE segment CD, we have segment AO : segment 00 = segment AB : segment CD. THEOREM 197. The geometry of a separation threefold is formally identical with the ordinary (Euclidean) geometry of three dimensions. This theorem may be proved by showing that in a separation three- fold a set of propositions hold which have already been demonstrated to be sufficient as a basis for the ordinary three dimensional (Euclidean) geometry. We have already seen that various axioms of Peano, not involving the idea of congruence, hold generally in our system, and not merely in a separation threefold. A separation threefold, however, being the only .form of general threefold in which all general lines are of one type and all general planes also of one type, is evidently the only one in which we could hope to find formal identity with Euclidean geometry. It is proposed here to give a set of propositions which will be found to be equivalent to a set of assumptions given by Veblen, and shown by him to be sufficient for the deduction of ordinary geometry. Yeblen speaks of three points being in the '"order" [ABC], meaning thereby that the three points are related in a manner analogous to that of three elements A, B, C in our geometry where B is "linearly between" A and 0. The existence, however, of optical and inertia lines in which two elements are asymmetrically related makes it undesirable to speak of elements in a separation line being in the " order " {ABC} ; since this seems to suggest that B is after A and C after B. The expression " linearly between " carries no such suggestion and, as it serves the purpose equally well, it will be employed here. Let us consider the set of elements lying in a separation three- fold W. We know that every general line in W is a separation line and every general plane in W is a separation plane. A THEORY OF TIME AND SPACE 335 If now we consider the definition of an element B being "linearly between " the elements A and C, we see that : (1) If the element B be linearly between the elements A and C, then J., B and C must be distinct and also B is linearly between the elements C and A. From this same definition in conjunction with the remark at the end of Theorem 42, it follows that : (2) If the element B be linearly between the elements A and (7, then G is not linearly between the elements B and A. From the definition of a separation line, it follows at once by Post. XIII that: (3) If the two distinct elements G and D lie in the separation line AB, then A lies in the separation line CD. From Theorem 43 and the definition of " linearly between" it follows that: (4) If A and B be two distinct elements in W, there exists in W at least one element C such that B is linearly between A and G. If three distinct elements A, B, G do not all lie in one general line, and if D be an element such that G is linearly between B and D, it is evident that A, B and D cannot lie in one general line. From this, in conjunction with Theorem 115, it follows directly that : (5) If three distinct elements A, B and G lie in TF, but not in the same separation line, and if D and E are two elements such that G is linearly between B and D and E is linearly between G and A, then there exists in W an element F, such that F is linearly between A and B, and such that D, E and F lie in the same separation line. Since all the elements of W do not lie in one separation line, it follows that : (6) There exist three distinct elements in W, say A, B, G, such that B is not linearly between A and (7, and C is not linearly between B and A, and A is not linearly between G and B. If A and B be two distinct elements in W, and if a be any separation line in W while C is any element in a, then we have seen that there is one single element, say D', such that : (A,B)\ = \(G,D'); also we have seen that there are two and only two elements in a, say D l and D 2 , such that : (a,A){ = }(0;#) and (C, D 2 ){ = ] (G, D'), 336 A THEORY OF TIME AND SPACE and accordingly (A, B) = (C, DJ and (A, B) = (C, D 2 ). But these elements D l and D 2 lie on opposite sides of C, and therefore there is only one of these elements in either of the separation half-lines into which C divides a. Thus : (7) If A and B be any two distinct elements in W and if C be the end of any separation half-line in TF, there exists one and only one element D in this separation half-line, such that : (A, B) = (0, D). Since we have shown that the relation of congruence for separation pairs is a transitive relation, it follows that : (8) If (A, B), (C, D\ (E, F) be separation pairs in W, such that: (A, B) = (C, D) and ((7, D) = (E, F), then we have (A, B) = (E, F). In Theorem 184 we proved that : (9) If (A lt B,), (A 2 , J5 a ), (B lt 0,), (B,, C 2 ) be separation pairs, such that : (A,, A) = (A,, B,) and (B,, CO = (B,, 2 ), then if B 1 be linearly between A and C l} and if B^ be linearly between A 2 and C 2 , we shall also have (X* . and if further B l C l be a separation line which is normal to the inertia line A l C l , while B 2 C 2 is a separation line which is normal to the inertia line A 2 C 2 , then : (1) // (d, A,) = (C 2> A 2 ) and (d, BJ = (C*, B,) t we shall either have (A T , J?j) = (A 2 , J9 2 ), or else both A l B l and A 2 B 2 will be optical lines. (2) // (A 1} G\) = (C 2 , A 2 ) and (d, BJ = (C 2 , B a ), we shall either have (A 1} B^ = (B 2 , A 2 ), or else both A l B l and B 2 A 2 will be optical lines. Consider first part (1) of the theorem. Since (Ci, AT) = (C a , A 2 ), and since these are inertia pairs, we must have either A l before G l and A 2 before C 2 , or else have A^ after C l and A 2 after C 2 . We shall only consider the case where A 1 is before C l and A 2 before (7 2 , since the other case is quite analogous. If A l B l were an optical line we should have (C lt A^ a before- conjugate to ((?!, B^, and if A 2 were an element in C. 2 A 2 , such that 344 A THEORY OF TIME AND SPACE (C 2 , A,?) were a before-conjugate to (0 2 , B^), then it would follow by Theorem 181 that we must have (C l} A,) = (G 2) A,'), and so we should have ((7 2 , A*) = (0 2 , A')- Thus A 2 ' would be identical with A Zt and so A 2 B. 2 would also be an optical line. We are not, however, at liberty to assert in this case that (A,, BJ = (A t , A), but only that they are both optical pairs. We shall suppose next that A l B l is not an optical line, and that accordingly A 2 B 2 is not an optical line. Let A and A be elements in C 1 A l and G Z A Z respectively, such that (Ci, A) is a before-conjugate to (G lt B^) y and (C 2 , A) is a before- conjugate to (C 2 , B 2 ). Then, by Theorem 181, we must have (C lt A) = (C*, A). Now two cases occur ; we may have (1) A^ linearly between D x and C lt or (2) A linearly between A 1 and C l . In the first case, since we also have (C,, A t ) = (C,, A,), it follows that we must also have A 2 linearly between A an d G 2) as was shown in the remarks at the end of Theorem 188. Similarly, in the second case we must also have A linearly between A 2 and (7 2 . Again, in the first case we have D l before C lt and must therefore have A 1 after D l and before C l . But A l could not be before B 1} for then A l would require to lie in the optical line A -#1 , which we know is not the case. Further A l could not be -after B ly for then, since G l is after A lt we should have G l after B l contrary to the hypothesis that B G^ is a separation line. It follows that in case (1) A l B l is a separation line, and similarly A 2 B 2 is a separation line. In case (2), on the other hand, we must have A l before D l and so, since A is before B l} we must have A^ before B l . A THEORY OF TIME AND SPACE 345 Thus, since D l A l is an inertia line, and since D l is the only element common to it and the /3 sub-set of B^ it follows in this case that A l B l is an inertia line, and similarly A 2 B 2 is an inertia line. We shall consider cases (1) and (2) separately. Case (1). We have here got A l B l and A 2 B 2) both separation lines. Now let W l and W 2 be rotation threefolds containing P : and P 2 respectively, and let S l and S 2 be the separation planes in W 1 and W 2 which pass through Ci and C 2) and are normal to the inertia lines A l C l and A 2 (7 2 respectively. Then since B l C l is normal to A 1 C l , it follows that B l C l must lie in Si and similarly B 2 C z must lie in $ 2 . Now since A^ BI is a separation line, there is an acceleration plane which passes through AI, lies in W l and is normal to AiBi. This acceleration plane contains two optical lines which pass through A l and must be normal to A l BI and which must intersect S lt since Si is a separation plane in the same rotation threefold alone: with these r JT o optical lines. Let one of these optical lines intersect Si in the element EI. Similarly we can show that there are two optical lines passing through A 2 and lying in TF 2 , and which are normal to A 2 B 2 . These optical lines may be shown in a similar manner to intersect S 2 , and we shall suppose that one of them intersects S 2 in the element E 2 . Now since the optical line AI EI is normal to the separation line AiBi, it follows that A^Ei and AiB l lie in an optical plane. Similarly A. 2 E 2 and A 2 B Z lie in an optical plane. But, since an optical line in an optical plane is normal to every general line in the optical plane, it follows that A^Ei is normal to EiB lf and similarly A 2 E 2 is normal to E 2 B 2 . Again, since E l Bi lies in Si and since Si is normal to Aid, it follows that A l Ci is normal to E^B^. Similarly A 2 C 2 is normal to E 2 B 2 . Thus E l B l is normal to the two intersecting general lines AI E l and AiGi, and is therefore normal to the general plane containing them. Similarly E 2 B 2 is normal to the general plane containing A 2 E 2 and A 2 C 2 . It follows that EiBi is normal to EiC lt and that E 2 B 2 is normal to E 2 G*. Again since Si is normal to AiQ it follows that EiCi is normal to AiGu and similarly it follows that E 2 C 2 is normal to A 2 C 2 . 346 A THEORY OF TIME AND SPACE Thus since AC l and A z C a are inertia lines while A 1 E 1 and A Z E 2 are optical lines, it follows that (C lt EJ and (C 2 , E 2 ) are after-conjugates to (Ci, AI) and (C 2 , ^i a ) respectively. But since (d, ^) = (0 a , ^ 2 ), it follows by Theorem 181 that : (C lf EJ = (C 2} E 2 ). Thus C l} B!, E are the corners of a triangle in the separation plane 81 and C 2 , B 2 , E 2 are the corners of a triangle in the separation plane S 2 , while further (E,, 0,) = (E 2 , 2 ), (0,, A) = (C t , BJ, and also B l E l is normal to C 1 E 1 and B 2 E Z is normal to C 2 E 2 , and so, by Theorem 190, (E l} A) = (#., A). But since ^JSj and A l B l are separation lines lying in an optical plane, of which ^.E'! is a generator, it follows from the remarks at the end of Theorem 199 that: (E lt A) = (-4i, ft). Similarly (E a , B 2 ) = (A 2 , B^. Thus we get finally (A lt B,) = (A 2 , B 2 \ and so the theorem is proved in case (1). Case (2). We have here got A^B^ and A 2 B 2 , both inertia lines. As before, let W l and W 2 be rotation threefolds containing P l and P 2 respectively, and let 8 l and S. 2 be the separation planes in Wi and W 2 which pass through C l and G 2 and are normal to the inertia lines A 1 C\ and A 2 C 9 respectively. Then, as in the first case, B l C l lies in Sj, and B 2 C 2 lies in >S^ 2 . Let 6j be the separation line in Si which passes through BI and is normal to B l C l , and similarly let 6 2 be the separation line in $ 2 which passes through B 2 and is normal to B 2 C 2 . Then, since AiB-^ is an inertia line, it follows that AiBi and 6j lie in an acceleration plane, say Qi, and similarly A 2 B 2 and b 2 lie in an acceleration plane, say Q 2 . Let one of the generators of Qi which pass through A! intersect 6j in the element F l} and let one of the generators of Q 2 which pass through A 2 intersect b 2 in the element F 2 . A THEORY OF TIME AND SPACE 347 Now since AiCi is an inertia line, it follows that A^Ci and AiFi determine an acceleration plane, and similarly A 2 C 2 and A 2 F 2 deter- mine an acceleration plane. Since G^ lies in Si it must be normal to A- i C l , and since C 2 F 2 lies in S 2 it must be normal to A 2 C 2 . Thus, since A l F l and A Z F 2 are optical lines, it follows that (C l} FI), (C 2 , F 2 ) are after-conjugates to (C 1} AI) and (C 2 , A 2 ) respectively, and so since (0 1; A,) = (<7 2) A 2 ), it follows by Theorem 181 that : But now (7 1} F lf B 1 are the corners of a triangle in the separation plane Si and C 2 , F 2 , B 2 are the corners of a triangle in the separation plane $ 2 , while further (0 lf B,) = (C t , B,), and also F l B l is normal to C 1 B 1 and ^ 2 5 2 is normal to C 2 B 2 and so, by Theorem 190, (B l) F l ) = (B 2 ,F 2 ). Since FiB 1 lies in ^ it is normal to A l C l , and by hypothesis it is also normal to B l C l and so, since AiC L and BiCi are intersecting general lines in P l} it follows that F 1 B l is normal to P x . Thus F^Bi must be normal to A l B l and similarly ^ 2 ^a must be normal to A 2 B 2 . But AiFi and ^. 2 ^2 are optical lines while A l B l and J. 2J B 2 are inertia lines and so (B 1) F-^) and (J5 2 , jP 2 ) are after- conjugates to (Bi, AI) and (B 2) AZ) respectively. Thus since (B lf FI) = (B 2) F 2 ), it follows, by Theorem 181, that : that is (A lf B l ) = (A at -B i ) t as was to be proved. Consider now part (2) of the theorem. Since (Ai, Ci) = (G 2 , A 2 ) and since these are inertia pairs we must either have A^ before C f 1 and A 2 after C 2 or else have A l after Ci and A 2 before C 2 . 348 A THEORY OF TIME AND SPACE There is then no difficulty in showing that : (A lt B,) = (B 2 , A 2 ), provided that A^B^ be not an optical line. The proof is quite analogous to that of the first part of the theorem except that we make use of the result given at the end of Theorem 181 in place of Theorem 181 itself. It is also evident that if A l B l be an optical line, then B 2 A 2 must also be an optical line. Thus both parts of the theorem hold. It will be observed that the two parts of Theorem 200 are the analogue for acceleration planes of Theorem 189. THEOREM 201. (1) If B and C be two distinct elements in a separation line and be their mean and if A be any element in an inertia line a which passes through and is normal to BC, then either (A, B) = (A, C\ or else both AB and AC are optical lines. (2) // B and C be two distinct elements in an inertia line and be their mean and if A be any element in a separation line a which passes through and is normal to BC, then either (B, A) = (A, C), or else both BA and AC are optical lines. The two parts of this theorem follow directly from the last and no further proofs are required. It will be observed that they form the analogue of Theorem 186. THEOREM 202. If A, B, C be three distinct elements in an acceleration plane P which do not all lie in one general line and if (A,B) = (A,C), or if (B, A) = (A, C), then BC cannot be an optical line. Since the only congruence of optical pairs is co-directional, it is evident that neither AB nor AC can be optical lines and must therefore be either inertia or separation lines. A THEORY OF TIME AND SPACE 349 Consider first the case where they are inertia lines and (A, B) = (A, G). It is evident that we must either have A before both B and G or after both B and C. Suppose A is before both B and G and let a be a separation line passing through A and normal to the acceleration plane containing AB and AC. Let D be an element in a such that (A, D) is a before-conjugate to (A, B). Then (A, D) will also be a before-conjugate to (A, C) since (A, B) = (A,C). Thus DB and DC will both be optical lines, and so BC cannot be an optical line*. If A be after both B and G the result follows in a similar manner. Next consider the case where AB and AC are inertia lines but where (B, A) = (A, C). We must then either have. A after B and C after A or else A before B and C before A. In either case it is evident that BC could not be an optical line, for otherwise A would be after one element of it and before another and yet not lie in the optical line ; which we know to be impossible. Consider next the case where AB and AC are separation lines and where accordingly (A,B) = (A,C) implies (B, A) = (A, C), and conversely. Now we know that there is one single optical parallelogram in P having A as centre and B as one of its corners. Suppose, if possible, that BC is an optical line which we shall denote shortly by 6. Then b would be one of the side lines of this optical parallelogram, and we shall denote the opposite side line by &'. Let B' be the corner opposite to B and let D and D' be the remain- ing two corners : D lying in b and D' lying in b'. Let CA intersect b' in the element C' and let optical lines passing through C and C' respectively and parallel to BD' intersect b' and b in E r and E respectively. Then E', C, E, C' would form the corners of an optical parallelogram having also b and b' as a pair of opposite side lines. * It is also to be noted that B is neither before n6r after C in this case. 350 A THEORY OF TIME AND SPACE Thus, since the diagonal line CC' passes through A, it follows, by Theorem 62, that these two optical parallelograms would have a common centre A. But now either (A, D) or (A, D') would be an after-conjugate to (A, B) while (A, t E) or (A, E f ) would be an after-conjugate to (A, C) and DE and D'E' would both be optical lines. Thus by the first case of the theorem it is impossible that we should have (A,D) = (A,E) or (A, D') = (A, E'\ If however we had (A, B) = (A, C), these other congruences would require to hold and so it is impossible to have BC an optical line if * (A,B) = (A,C). Thus the theorem holds in all cases. It is important to note that while this result holds for an acceleration plane, it does not, as we have already shown, hold for an optical plane. Thus since an optical line can only lie in an acceleration or optical plane, it follows that : If B and C be two distinct elements in an optical line while A is an element which does not lie in BC, then if (A, B) = (A, C) the elements A, B, C must lie in an optical plane. THEOREM 203. If A, B and C be three distinct elements which lie in an acceleration plane P, but do not all lie in one general line, then : (1) If BC be a separation line and be the mean of B and C and if (A,B) = (A,C), or if AB and AC be both optical lines, we must have AO normal to BC. (2) If BC be an inertia line and be the mean of B and C and if (B, A) = (A, C), or if BA and AC be both optical lines, we must have AO normal to BC. Let us consider the firSt part of the theorem. A THEORY OF TIME AND SPACE 351 If AB and AC are both optical lines, then A, B and C are three corners of an optical parallelogram of which is the centre and so AO is normal to BC by definition. We shall therefore suppose that : (A, B) = (A, <7), and accordingly that AB and AC are either both inertia lines or both separation lines. Let d be a separation line passing through A and normal to P. Then P and d determine a rotation threefold W. Let e be a separation line in W which passes through and is normal to BC, but which is not parallel to d. Then BC and e, being separation lines which intersect and are normal to one another, must lie in a separation plane, say S, which must lie in W. . Now since 8 does not contain the separation line which passes through and is parallel to d, and since S and d both lie in W t but d does not lie in 8, it follows that d must intersect S in some element, say D. Now AD being normal to P must be normal to both AB and AC and, since these are either both inertia lines or both separation lines, it follows that AD and AB on the one hand and AD and AC on the other must lie in acceleration planes or else in separation planes. Also, since S is a separation plane, it follows that DB and DC are both separation lines. Thus" since (A, B) = (A, C) and (A, D) = (A, D), it follows by Theorem 189 or by Theorem 200 (according as AB and AC are separation lines or inertia lines) that : (B, D) = (C, D). But now since D, B, and C are three distinct elements in the sepa- ration plane 8 which do not all lie in one general line, and since is the mean of B and C, it follows, by Theorem 187, that DO is normal to BC. Now, since AD is normal to P, it follows that AD is normal to BC. Thus BC is normal to the two intersecting general lines DO and AD and must therefore be normal to the general plane containing them and which we may call Q. But AO lies in Q and so it follows that AO must be normal to BC ; as was to be proved. 352 .A THEORY OF TIME AND SPACE Consider now the second part of the theorem. If BA and AC are both optical lines, then A, B and C are three corners of an optical parallelogram of which is the centre and so by definition it follows that AO is normal to EG. We shall therefore suppose that : (B, A) = (A, G), and accordingly that BA and AC are either both inertia lines or both separation lines. Let the general line through B parallel to AC intersect the general line through C parallel to AB in the element D. Then B, A, C, D are the corners of a general parallelogram in the acceleration plane P and since is the mean of B and C it follows that AD must intersect BC in the element 0. Thus A D must be identical with A and must be the mean of A and D. But now we clearly have (B,D) = (A,C), and so we must have (B, D) = (B, A). Thus by the first part of the theorem it follows that BO is normal to DA : that is AO is normal to BC. Thus both parts of the theorem are proved. It is evident that the two parts of this theorem form the analogue of Theorem 187. When however BC is an optical line, no analogue exists as is shown by Theorem 202. THEOREM 204. If A, B and C be three distinct elements which lie in an acceleration plane P, but do not all lie in one general line and if be an element in BC such that A is normal to BC, then : (1) If BC be a separation line and if (A,B) = (A,C\ or if AB and AC be both optical lines, the element must be the mean of B and C. (2) If BC be an inertia line and if (B, A) EE (A, C), or if BA and AC be both optical lines, the element must be the mean of B and C. A THEOKY OF TIME AND SPACE 353 The proofs of these two parts are quite analogous to the proof of Theorem 188, using the two parts of Theorem 203 in place of Theorem 187. THEOREM 205. // A!, B lt ft be the corners of a triangle in an acceleration plane Pj and A 2 , B 2 , C 2 be the corners of a triangle in an acceleration, plane P 2 , and if further B l C l be normal to A l C l and B 2 C 2 be normal to A 2 G 2 , then : (1) If AI.GI and A 2 G 2 be inertia lines and and if (A lt B 1 ) = (A 9f B 9 \ or if A 1 B 1 and A 2 B 2 be both optical lines we shall also have (ft, BJ = (ft, B 2 ). (2) If A 1 G l and A 2 C 2 be inertia lines and (A lt ft) = (ft, A 2 ), and if (A l) B 1 ) = (B 2) A 2 ), or if A^ and A 2 B 2 be both optical lines we shall also have (Ci, ,) = (C,, BJ. (3) If A l C l and A^C 2 be separation lines and (C l9 Ai) = (Gs,A,) t and if (A l) B l ) = (A 2 ,B 2 ) ) or if A 1 B l and A 2 B 2 be both optical lines we shall also have either (Oi,A)s(0,,j;x or (B lt ft) = (O a> B 2 ). (4) If A l C l and A 2 C 2 be separation lines and (A l ,G l ) = (G 2 ,A 2 \ and if (A 1) B l )=(B 2) A 2 ) ) we shall also have either (Bi, ft) = (C 2 , B 2 ), or (0 1( = (0,, ,). Cases (3) and (4) become equivalent if A l B l and A 2 B 2 are both separation lines and we have then an ambiguity in the result. R. 23 354 A THEORY OF TIME AND SPACE If on the other hand A 1 B 1 and A 2 B 2 are both inertia lines the first mentioned alternatives hold in cases (3) and (4). If we only know that ^ift and A 2 B 2 are optical lines we again have ambiguity. The proof of part (1) of this theorem is quite analogous to that of Theorem 190, using Theorem 200 (1) in place of Theorem 189, and Theorem 203 (1) in place of Theorem 187. Similarly the proof of part (2) is quite analogous to that of Theorem 190, using Theorem 200 (2) in place of Theorem 189, and Theorem 203 (1) in place of Theorem 187. As regards part (3), this is clearly equivalent to part (4) if A l B l and A 2 B 2 are separation lines, for then (A 2 , B 2 ) = (ft, A 2 \ and the proof is again analogous to that of Theorem 190, with a slight modification. This modification is required because in this case ft*?! and B 2 G 2 must be inertia lines and it is not possible to find an element B 2 in B 2 C 2 and on the same side of C 2 as is B 2 such that : unless either B l is before Cj and B 2 before C 2 , or else ft is after C l and B 2 after C 2 . If either of these conditions hold the proof is analogous to Theorem 190, using Theorem 200 (1) in place of Theorem 189, and Theorem 203 (2) in place of Theorem 187. If, on the other hand, we have either B l before C l and B 2 after C 2 , or else B l after Oj and B 2 before C 2) we must take an element B 2 in B 2 C 2 , and on the same side of C 2 as is B 2 , and such that : Then Theorem 200 (2) takes the place of Theorem 189, and Theorem 203 (2) takes the place of Theorem 187. In the former of these cases we get and in the latter we get (ft, 00 = (C\, ft). If (A lt ft) and (A 2 , ft) are inertia pairs, then in part (3) their congruence shows that either ft is after A l and ft after A^, or else A THEORY OF TIME AND SPACE 355 B l is before A l and B. 2 before A 2 , and in these cases we can always find an element B 2 in B 2 C 2) on the same side of C 2 as is B 2 , and such that : Thus, by Theorem 200 (1), we have and so we get (A zy B 2 ) = (A 2 , B 2 ). But, since B 2 and B 2 lie in an inertia line, it follows from the footnote on page 349, that they must be identical, and so we must have (fyity &(&,%).. In part (4), if (A lt B^) and (B 2 , A 2 ) are inertia pairs, their congruence shows that either A l is after B 1 and B 2 after A 2 or else A l is before B l and B 2 before A 2) and then, using Theorem 200 (2), we may show in a similar manner that : (B lt C 1 ) = (G, t 'B a ). If A 1 B l and A 2 B 2 are optical lines the various parts of the theorem follow directly from Theorem 181 and the remark appended to it. THEOREM 206. If AU BI, G 1 be the corners of a triangle in an acceleration plane PI, and A 2) B 2 , C 2 be the corners of a triangle in an acceleration plane P 2 , and if A l C l be a separation line which is normal to the inertia line BiC^, then : (1) If (A l) C l ) = (A 2) C 2 \ and if (A^B^-^A^B,), or if A 1 B 1 and A 2 B 2 be both optical lines, we must also have A 2 C 2 normal to B 2 C 2 . (2) // (A lt C l )^(A 2 ,G 2 ) ) (Q 9 4)s(jM^ and if (A,, BJ = (B 2) A 2 ), or if A l B l and A 2 B 2 be both optical lines, we must also have A 2 C 2 normal to B 2 C 2 . From the congruences it follows that since A^C^ is a separation line, A 2 C 2 must be a separation line, and, since B^ is an inertia line, B 2 C Z must be an inertia line. 232 356 A THEOKY OF TIME AND SPACE Thus any general line normal to B 2 C 2 must be a separation line. The proof of part (1) is then quite analogous to that of Theorem 191, using Theorem 200 (1) in place of Theorem 189, and Theorem 203 in place of Theorem 187, while remembering Theorem 202 and footnote. Similarly the proof of part (2) is analogous to that of Theorem 191, using Theorem 200 (2) in place of Theorem 189, and Theorem 203 in place of Theorem 187, again remembering Theorem 202 and footnote. Thus both parts of the theorem hold. It is easy to show that for the case of acceleration planes there are also theorems analogous to Theorems 192 and 193, but it is to be observed that optical lines are exceptional since the only congruence of optical pairs is co-directional congruence. It will be found however that this does not present much difficulty so far as the geometry is concerned. ANALOGUE OF THE THEOREM OF PYTHAGORAS IN AN ACCELERATION PLANE. We have seen that the geometry of a separation plane is formally identical with that of an ordinary Euclidean plane, and accordingly in a separation plane the theorem of Pythagoras connecting the sides of a right-angled triangle must hold. Consider now the constructions for the two cases of Theorem 200 merely as regards the triangle whose corners are A 1} BI, C l . In case (1) B 1 C 1 is a separation line, A l Ci is an inertia line normal to B l C l , and A^ is a separation line. But now we obtained a triangle whose corners were B lt GI and E l which lay in the separation plane S l and such that E^B-^ was normal to EiC-i and in which accordingly we must have the segment relation : This triangle was related to the one whose corners are A l , B l} O a in such a way that : (E 1 ,B l )=(B 1 ,A l ), while ((?!, Ei) was a before- or after-conjugate to (C ly A^. Thus taking segments instead of pairs we get (B,C^ = (B.A,) 2 + (conjugate C^) 2 . Thus the analogue of the theorem of Pythagoras is in this case : (B, Arf = (5 1 (7 1 ) 2 - (conjugate C,A^ ............... (i). Again if we consider case (2) we have B^C^ is a separation line, A 1 C 1 is an inertia line normal to B^, and A 1 B 1 is also an inertia line. A THEORY OF TIME AND SPACE . 357 In this case we obtained a triangle whose corners were C l} B l and F 1 which lay in the separation plane 81 and such that B^F^ was normal to M- Thus we must have the segment relation : This triangle was related to the one whose corners are A lt B lt C l in such a way that (C lt F^) was a before- or after-conjugate to (C l} A^ while (B 1} F) was a before- or after-conjugate to (B lt AJ, and so, taking segments instead of pairs, we get (conjugate C, A l )' 2 = (B l C l ) 2 + (conjugate B^A^. Thus the analogue of the theorem of Pythagoras is in this case : - (conjugate J^i) 2 = (#iCi) 8 - (conjugate C.A^ ...... (ii). In the case where A^ is an optical line we obviously have = (B, O x ) 2 - (conjugate QAtf .................. (iii). Thus (i), (ii) and (iii) constitute the complete analogue of the Pythagorean theorem in an acceleration plane. If we consider a triangle whose corners are A lt B l ,C\ and which lies in an optical plane, then if B l G l be a separation line and AiCi be normal to B^CI we know that A& must be an optical line, while A l B l must be another separation line. Now we have shown that : and so taking segments instead of pairs we see that : This is the analogue of the Pythagorean theorem in an optical plane. Considering now equations (i), (ii), (iii) and (iv) we observe that the modifications which take place in the theorem of Pythagoras are such that when any side of the triangle becomes an inertia segment the corresponding square is replaced by the negative square of the conjugate of this inertia segment, while if any side becomes an optical segment, the corresponding square is replaced by zero. Again if A lt B lt G l be the corners of a triangle whose sides are of the stated kinds and for which one of the relations (i), (ii), (iii) or (iv) holds, it is easy to show that B l C T must be normal to J.! (7j 358 A THEOEY OF TIME AND SPACE Take for instance the case where B^ is a separation line, ^. 1 (7 1 an inertia line and A^ a separation line, and where (A J.i) 2 = (B,C^ - (conjugate G^A^f. Let A z , B 2 , C 2 be the corners of another triangle in an acceleration plane, such that B 2 C 2 is a separation line normal to the inertia line ^-2^2? and where (B 2 , 2 ) EE (B 19 G,\ and (C 2 ,A,) = (G ly A,). Then the segment B 2 C 2 must be greater than the conjugate to the segment C 2 A 2 , from which it is easy to see that B 2 A 2 must be a separation line. Thus we must have (B 2 A 2 )* = (B 2 C 2 )* - (conjugate CU 2 ), and so we must also have (ft, A,) = (ft, A)- Then by Theorem 206, since B^C 2 is normal to A 2 C 2 , it follows that ^Oj is normal to A^. The cases where A^B^ is an inertia line and relation (ii) holds and where A^B^ is an optical line and relation (iii) holds may be treated in a similar manner, and we may prove that B^C^ is normal to A.C,. Again, if A lt B lt G^ be the corners of a triangle in which B l C l and B l A l are separation lines while A^ is an optical line, and if then (B lt A 1 ) = (B lf C l ) f and so, as was pointed out in the remarks at the end of Theorem 202, the elements A lt B 1} Q must lie in an optical plane. Thus since A^ is an optical line, it follows that B 1 C 1 is normal to A^. Let A, B, G be the corners of a triangle in an acceleration plane P, and let AB-, BG and GA be all -separation lines or all inertia lines. It is easy to see that triangles of both these kinds exist, although as Theorem 31 shows it is not possible for AB, BG and GA to be all optical lines. Let !, 6 15 G! be generators of P of one set, which pass through A, B, G respectively and intersect BG, GA, AB iuA l} B l , G respectively. Then we may show by a method similar to that employed in the A THEORY OF TIME AND SPACE 359 remarks at the end of Theorem 199, that one and only one of the elements A lt B l} Cj is linearly between a pair of the corners A, B, C. Similarly we may show that if a 2 , b 2 , c 2 be generators of P of the opposite set passing through the elements A, B, C respectively and intersecting BG, GA, AB in A 2 , B 2 , G 2 respectively, then one and only one of the elements A 2 , B 2 , C 2 is linearly between a pair of the corners A, B, C. Now consider the case, for instance, where A l is linearly between B and C, and suppose first that AB, BG, GA are all separation lines. Then B cannot be linearly between A 1 and A 2 for then, by Theorem 129 (a) and (b), AB would require to be an inertia line, contrary to hypothesis. Similarly G cannot be linearly between A-^ and A 2 . Thus, since obviously A 2 cannot be identical with either B or G, it follows that A 2 must be also linearly between B and C. Now let be the mean of A and A 2 . Then is linearly between A l and A 2 , and therefore clearly it must lie linearly between B and G. But now A lt A, A 2 are three corners of an optical parallelogram of which is the centre, and so AO must be normal to A^A Z : that is to BG. Again, if instead of AB, BG, GA being all separation lines they are all inertia lines, a similar result holds. Let us take the case where A 1 is linearly between B and C. Then clearly B cannot be linearly between A l and A 2) for then AB would require to be a separation line, and, for a similar reason, G cannot be linearly between A and A 2 . Thus, since A 2 cannot coincide with either B or G, it follows that A 2 must also be linearly between B and G. As in the former case, if be the mean of A l and A 2 , then must be linearly between B and C, and AO must be normal to BG. Now in the case where AB, BG, GA are all separation lines, it follows from relation (i) that segment BA is less than segment BO, and segment AC is less than segment OC. Thus it follows that the sum of the lengths of the segments BA and AG is less than that of the segment BG. Similarly, if AB, BC, CA be all inertia lines, we may show by the help of relation (ii) that the sum of the lengths of the segments BA and A C is less than that of the segment BG. 360 A THEOKY OF TIME AND SPACE Now we know that in ordinary Euclidean geometry the sum of the lengths of any two sides of a triangle is greater than that of the third, and a similar result must hold in a separation plane. Thus, remembering what was proved at the end of Theorem 199, we have the following interesting results : If A, B, C be the corners of a general triangle all whose sides are segments of one kind, then : (1) If the triangle lies in a separation plane, the sum of the lengths of any two sides is greater than that of the third side. (2) If the triangle lies in an optical plane, the sum of the lengths of a certain two sides is equal to that of the third side. (3) If the triangle lies in an acceleration plane, the sum of the lengths of a certain two sides is less than that of the third side. If A lt B 1} Cj be the corners of a triangle such that A l B l , B^, C l A l are all separation lines, and if it is evident that' the sum of the lengths of any two sides must be greater than that of the third side. Thus it follows from what we have just shown that such a triangle cannot lie in any type of general plane, except a separation plane. In this case, however, we know from ordinary geometry that we must have B l C l normal to A-^C^ On the other hand, since an inertia line cannot lie in any but an acceleration plane, it is not possible to have A l B l , B l C l , C l A l all inertia lines and also to have the relation Having thus investigated the analogues of the theorem of Pytha- goras and its converse, we are now in a position to introduce coordinates. INTRODUCTION OF COORDINATES. If we take any element of the set as origin, we have already seen that we may obtain systems of four general lines through 0, say OX, OF, OZ, OT, which are mutually normal to one another. Three of these, say OX, Y, OZ, will be separation lines, while the fourth, OT, will be an inertia line. The three separation lines OX, Y, OZ will determine a separation threefold, say W, and OT will be normal to it. A THEORY OF TIME AND SPACE < 361 If we select any arbitrary separation segment as a unit of length and associate the number zero with the element 0, we may associate every other element of OX, OY, OZ with a real number, positive or negative, corresponding to the length of the segment of which that element is one end and the origin is the other. In this way we set up a coordinate system in W which will be quite similar to that with which we are familiar. Since all the theorems of ordinary Euclidean geometry hold for a separation threefold, the length of a segment in W will be given by the ordinary Cartesian formula. Again, not confining our attention merely to the elements of W, let A be any element of the whole set. Then A must either lie in OT, or else there is an inertia line through A parallel to OT, and, as has already been proved, this inertia line will intersect W in some element, say N. Further, AN must be normal to W. Now if A does not lie in W there will be a separation threefold, say W' t passing through A and parallel to W, and the inertia line OT must intersect W in some element, say M. Further, since W is parallel to W, both OT and AN must be normal to W. Thus, if OM and NA are distinct, MA and ON must both be separation lines normal to OM, and so, since OM and NA lie in an acceleration plane, we must have MA parallel to ON. Now we may select a unit inertia segment, just as we selected a unit separation segment, and with each element of T distinct from we may associate a real number positive or negative corresponding to the length of the segment of which that element is one end and the origin is the other. We shall suppose tliis correspondence to be set up in such a way that a positive real number corresponds to any element which is after and a negative real number to any element which is before 0. As regards the relationship between the unit separation segment and the unit inertia segment, the simplest convention to make is to take the unit inertia segment such that its conjugate is equal to the unit separation segment. More generally, we may take the unit inertia segment such that : (conjugate of unit inertia segment) = v (unit separation segment), where v is a constant afterwards to be identified with what we call the " velocity of light." 362 A THEORY OF TIME AND SPACE Now the element N lies in W and is determined by three co- ordinates, say x l ,y l ,z l , taken parallel to OX, OY, OZ respectively in the usual manner. Further segment NA segment OM, and so if ti be the length of OM in terms of the unit inertia segment, then the element A will be determined by the four coordinates x^ t y lt *i, *i- Let the length of the segment ON be denoted by a. Then as in ordinary coordinate geometry a 2 = x* + ^ + zj*. Thus if OA should be an optical line, we must have or ac l * + y 1 * + z l *-tft 1 * = ........................ (1). Again, if OA should be a separation segment and if r x be its length, it follows from the analogue of the theorem of Pythagoras for this case that : a 2 - wV = r*, or x? + y?+z?-tft? = r? ..................... (2). Finally, if OA should be an inertia segment and r a its length, it follows from the corresponding analogue of the Pythagoras theorem that : a 2 - (5). While if A A l be an inertia segment and r a be its length we must have ( Xi _ ^ + ( yi _ yo)2 + ( Zi _ ZQ f _ tf (t, - 1 )* = -v*rS (6). Thus the expression (*i - ^o) 2 + (yi - 2/o) 2 + & - ^o) 2 - ^ 2 ft - 4) 2 is positive, zero, or negative according as A^A-^ is a separation line, an optical line, or an inertia line. Accordingly if A Q and A 1 be any elements of the set, the conditions that A! should be after A are : (1) ( Xl - X rf+( yi -ytf + ( Zl - Z tf-v*(t,-ttf } is zero or negative j- . and (2) ti 1 is positive J The conditions that A l should be before A are : (1) ( Xl -x^-^(y l -y^ + (z l -z Q f-^(t l -t^ \ is zero or negative I . and (2) ^ t Q is negative The conditions that A l should be neither before nor after A are (if we include the case where A and A^ are identical): x l - # = 2/i - 2/o = Zi - Z Q = f ! - ^ = or else (as, - # ) 2 + (yi - 2/ ) 2 + (*i - ^o) 2 - ^ 2 (, to) be the coordinates of an element A the equation of the combined a and /3 sub-sets of A is (x-^y + ^-ytf + ^-ztf-v^t-ttf^Q (7). The a. sub-set of A Q will then consist of all elements (x, y, z, t) for which this equation is satisfied and for which t t is zero or positive ; while the sub-set of A Q will consist of all elements for which the equation is satisfied and for which t t is zero or negative. Definition. The set of all elements whose coordinates satisfy equation (7) will be called the standard cone with respect to the element whose coordinates are (#, y Q , z Q , t ). Taking v equal to unity, for the sake of simplicity, it is evident that the equation tf 2 + f + Z* - t z = C 2 represents the set of elements such as A, where OA is a separation segment whose length is c. Similarly, the equation 3? + f + 2* - t* = - C 2 represents the set of elements such as A, where OA is an inertia segment whose length is c. If we put y = and z = in the first of these we obtain x*-t 2 = c 2 , which gives us the relation between x and t for the portion of the corre- sponding set which lies in the acceleration plane containing the axes of x and t. This then represents the analogue of a circle in the acceleration plane. Similarly for the case of inertia segments putting y = and z = we get # 2 - 2 = - c 2 . The two equations x 2 t 2 = c- and x 2 1* = c 2 are of the same forms as the equations of a hyperbola and its conjugate in ordinary plane geometry. The equation #* t 2 = along with y = and z = represents the two optical lines through the origin in the same acceleration plane, and these correspond to the common asymptotes of the hyperbolas. A THEORY OF TIME AND SPACE 365 EQUATIONS OF AN OPTICAL LINE. For the sake of simplicity we shall again take v equal to unity. Let (a?j, y lt 2j, tj) be the coordinates of an element A l and let (#2, 2/2 , ^2 > ts) be the coordinates of an element A 2 distinct from A l and lying either in the a or ft sub-set of A l . The equation of the standard cone with respect to the element J.j is (x-x^ + (y-y^ + (z-z^-(t-t^ = Q.... ........ (1). Since A 2 lies either in the a or /3 sub-set of A l we must have *-ztf-(t*-ttf = Q ......... (2). If A 2 lies in the a sub-set of A l we must have t 2 ^ positive, while if A z lies in the /9 sub-set of A l we must have t 2 t\ negative. The equation of the standard cone with respect to the element A 2 is . (x-x 2 f + (y-y^+(z-z^-(t-t 2 ^=^ ......... (3). Remembering now the definition of an optical line given on page 13 and observing that if A l and A 2 be distinct elements and if A 2 lies in j there can be no element common to 2 and /3 a , we see that any element of the optical line defined by A 1 and A 2 must be such that its coordinates satisfy equations (1) and (3). Further the set of elements whose coordinates satisfy (1) and (3) must be identical with the optical line containing A l and A 2 provided equation (2) holds. Adding equations (1) and (2) and subtracting equation (3) we get on dividing by 2 (a* - 00 (x - #0 + (y 2 - y,) (y - y,) + (z 2 - zj (z - zj - (t 2 - t,) (t - t,) = 0. Thus {(x 2 - x,} (x - x,) + (y 2 - y,) (y - yj}* = {(z, - z,) (z - z,} - (t, - *0 (t - O} 2 . . .(4). But from (1) and (2) we get {(x - xtf + (y- y,) 2 } {(* - ^i) 2 + (y, ~ ?A) 2 } = {(z - z,J -(t- t^} {(z 2 - ztf - (t, - t^} . . .(5). Thus from equations (4) and (5) we obtain {(y - yO 2 fa - ^) 2 - 2 (y - yO fa - *0 ( x - *ti (y* - yO + ( x - ^) 2 (y - ^) 2 = - {(* - ttf (z 2 - z^ -2(t- t,) (z 2 - z,) (z - z,} (t, - t,) + (z - z$ (t 2 366 A THEORY OF TIME AND SPACE Thus for all real values of the coordinates we must have (y - 2/0 (#2 - #0 - O - #0 (2/2 - yO = > and ( - t,) (z, - z,) -(z- z,) (t, - O = 0. By a similar method we find that : (z - Zi) (y a - y,) - (y - y,) (z 2 - z,) = 0, and (t - t,) (ar a - ^) - (a? - as,) (t 2 - t,) = 0. Thus all elements which lie in the optical line passing through A l and A 2 must be such that their coordinates (x, y, z t t) satisfy the equations x -x l = y -y l = s -*! = t -^ ^2-#l 2/2-2/1 *-*! *2-*l Conversely a set of equations of the form (6) will represent an optical line, provided that : (# 2 - xtf + (y 2 - y,) 2 + 2 - ^) 2 ~ (t* - O 2 = 0. The expression in coordinate form of the various results which we have obtained by geometrical methods is now merely a matter of straight- forward analysis. In carrying out this analysis (as we have just seen in connection with the equations of an optical line) the fact that the coordinates of any element must be real frequently plays an important part. Since our main object has been to show how a system of geometry may be built up from ideas of after and before, it is unnecessary to go into these matters in detail in the present volume. INTERPRETATION OF RESULTS. It is evident that any element whose coordinates are (a, b, c, 0) must lie in the separation threefold W and accordingly the three equations x = a, y = b, z c must represent an inertia line normal to W and therefore parallel to or identical with the axis of t. Again, any equation of the first degree in a?, y, z, together with the equation t = 0, will represent a separation plane in W, while any two independent but consistent equations of the first degree in x, y, z, together with the equation t 0, will represent a separation line in W. Thus any equation of the first degree in a?, y, z (leaving out the equation t = 0) will represent a rotation threefold containing inertia A THEORY OF TIME AND SPACE 367 lines parallel to the axis of t ; while any two independent but consistent equations of the first degree in x, y, z will represent an acceleration plane containing inertia lines parallel to the axis of t. Thus corresponding to any theorem concerning the elements of W there will be a theorem concerning inertia lines normal to W and passing through these elements. Conversely, if we consider the system consisting of any selected inertia line together with all others parallel to it, then any two such inertia lines will determine an acceleration plane, while any three which do not lie in one acceleration plane will determine a rotation threefold. Since these inertia lines must all intersect any separation threefold to which they are normal, it follows that they have a geometry similar to that of the separation threefold and therefore of the ordinary Euclidean type. If then we call any element of the entire set an "instant"; any inertia line of the selected system a "point "; any acceleration plane of the selected system a " straight line " ; and any rotation threefold of the selected system a "plane"; we can speak of succeeding instants at any given point, and have thus obtained a representation of the space and time of our experience in so far as their geometrical relations are concerned. The distance between two parallel inertia lines of the system will naturally be taken as the length of the segment intercepted by them in a separation line which intersects them both normally. This, then, will be the meaning to be attached to the distance between two points. Time intervals in the usual sense will be measured by the lengths of segments of the corresponding inertia lines : that is to say, by differences of the t coordinates. Since we have defined the equality of separation and inertia segments in terms of the relations of after and before and have assigned an inter- pretation to these, it follows that the equality of length and time intervals in the ordinary sense is rendered precise. It is to be observed that the particular system of parallel inertia lines which we may select is quite arbitrary although the set of elements or instants contained in the entire system is in all cases identical. The distinction between different systems is that while two parallel inertia lines represent the time paths of unaccelerated particles which are at rest relative to one another ; two non-parallel inertia lines repre- sent the time paths of unaccelerated particles which are in motion with uniform velocity with respect to one another. Thus we are able to give a definition of absence of acceleration, but, 368 A THEORY OF TIME AND SPACE since all inertia lines are on a par with one another, we can attach no meaning to a particle or system being at " absolute rest." The definition of absence of acceleration based upon the relations of after and before and as regards a finite interval of time, may be thus expressed : Definition. If A and B be two distinct elements of any inertia line (B being after A ), then a particle will be said to be unaccelerated from the instant A to the instant B provided it lies in the inertia line AB throughout that interval. The physical signification of an optical line is : that a flash of light or other instantaneous electromagnetic disturbance in going directly from one particle to another would follow this time path. As regards a separation line ; since no element of it is either before or after another, then if our view be correct, no single particle could occupy more than one element, and so particles which occupy distinct elements of any separation line must be separate particles. The above considerations indicate the reasons for adopting the names we have assigned to the three types of general line. We shall suppose that the time path of any particle is a continuous curve having at every element a tangent which is an inertia line. The velocity of any particle with respect to a system of coordinates such as we have described will, then, in all cases be limited by what we call the " velocity of light." Any acceleration of the particle will thus determine an acceleration plane ; that being the type of general plane which osculates the time path. Similarly if the acceleration plane varies from element to element of. the time path so that the latter is tortuous, then this tortuosity deter- mines a rotation threefold. The names optical plane and optical threefold have been adopted because of the analogies of these to optical lines, and similarly the names separation plane and separation threefold have been adopted on account of the analogies to separation lines. Results involving only three coordinates x, y and t may be visualized by means of the three-dimensional conical order described in the intro- duction, but a certain amount of distortion appears in a model of this kind, since equal lengths in the model do not in general represent equal lengths as we have defined them. The optical significations of Posts. I to XVIII are however made clear by such models, and it is easily seen that the assertions made in A THEORY OF TIME AND SPACE 369 these postulates, when interpreted in the manner described, are in accordance with the ordinarily accepted ideas. Post. XXI also finds an interpretation in such a model, but its sig- nificance is concerned rather with the logic of continuity than with any observable physical phenomenon. Since it is possible to define equality of lengths in terms of after and before it seems superfluous to introduce any other conception of length, since the effect of this would merely be to destroy the symmetry which otherwise exists. It is further to be noted that the formal development of the theory of conical order does not in itself require that the a and ft sub-sets should be determined by optical phenomena, but merely that there should exist some influence having the properties which we have ascribed to light. Accordingly if it should be found hereafter that some other influence than light possessed these properties we should merely require to substitute this influence for light and interpret our results in terms of it. CONCLUSION. Our task now approaches completion. We have shown how from some twenty-one postulates involving the ideas of after and before it is possible to set up a system of geometry in which any element may be represented by four coordinates x, y> z, t. Three of these : #, y, z, correspond to what we ordinarily call space coordinates, while the fourth corresponds to time as generally under- stood. Since however an element in this geometry corresponds to an instant, and bears the relations of after and before to certain other instants, it appears that the theory of space is really a part of the theory of time. Of the postulates used : nineteen, namely I to XVIII and Post. XXI, may easily be seen to have an interpretation in three-dimensional geometry by making use of cones as described in the introduction. It follows that if ordinary geometry be consistent with itself, these nineteen postulates must be consistent with one another. Of the remaining two postulates, Post. XIX has the effect of intro- ducing one more dimension, while Post. XX limits the number of dimensions to four. Since by means of these we have been enabled to set up a coordinate system in the four variables x, y, z, t, the question of the consistency of the whole twenty-one postulates is reduced to analysis. R. 24 370 -A THEORY OF TIME AND SPACE It is not proposed to go further into this matter in the present volume, having said sufficient to leave little doubt that they are all consistent with one another. The question as to whether the postulates are all independent is mainly a matter of logical nicety and is of comparatively little import- ance provided that the number of redundant postulates be not large. In the course of development of the present work the writer suc- ceeded in eliminating a considerable number of postulates which he had provisionally laid down : the redundancies being generally indicated by the possibility of proving some particular result from several sets of postulates. One known redundancy has been permitted to remain : namely Post. II (a) and (6), which might have been deduced directly from Post. V and Post. VI (a) and (6). By retaining Post. II however, our first four postulates will be seen to hold for the set of instants of which any one individual is directly conscious, and the subject is thus better exhibited as an extension of the commonly accepted ideas of time. A still further diminution of the number of postulates might have been made by combining Posts. VI and XI in the way mentioned on page 25, but to have done so would have complicated still further the initial part of the subject, since Post. VI implies merely a two-dimensional conical order, while its combination with Post. XI makes the set of elements at least three-dimensional from the very beginning. Apart from the above-mentioned, no further definite indications of redundancy have been observed, and, although some redundant postulates may still remain, it seems unlikely that there can be many. This opinion is confirmed by a comparison with the number of fundamental assumptions given by various writers on the foundations of ordinary geometry. We have now concluded the exposition of the argument by which we have been led to the view expressed in the introduction : that spatial relations are to be regarded as the manifestation of the fact that the elements of time form a system in conical order: a conception which may be analysed in terms of the relations of after and before. This view would appear to have important bearings on general philosophy, but into these we do not purpose here to enter. One point may however be mentioned : The fundamental properties of time must, on any theory, be regarded as possessing a character which is not transitory, but in some sense A THEORY OF TIME AND SPACE 371 persistent ; since otherwise, statements about the past or future would be meaningless. We here touch on the difficult problem as to the nature of " univer- sals": a problem which has been much discussed by philosophers, but appears to be still far from a satisfactory solution. Though space may be analyzable in terms of time relations, yet these remain in their ultimate nature as mysterious as ever ; and though events occur in time, yet any logical theory of time itself must always imply the Unchangeable. Thus may we conclude in the words of Carlyle : " Know of a truth that only the Time-shadows have perished, or are perishable ; that the real Being of whatever was, and whatever is, and whatever will be, is even now and forever." INDEX References in black type are to definitic Acceleration plane, 36 Acceleration planes, sets of elements which determine, 55-56 After, 1 a sub-set, 11 Archimedes, 87, 330 Before in terms of after, 10 Between a pair of parallel optical lines in an acceleration plane, 51 Between, linearly, 102 j8 sub-set, 11 Carlyle, 371 Circle, separation, 319 centre of, 319 radius of, 319 diameter of, 319 inside, outside, 319 Co-directional congruence of pairs, 280 Complete normality of general planes, 216 Cone, standard, 364 Congruence, co-directional, of pairs, 280 general, of inertia pairs, 291 general, of separation pairs, 292 of inertia pairs having a latent element, 272 of separation pairs having a latent element, 275 Conjugate, inertia and separation lines, 156 pairs, 275 Coordinates, introduction of, 360 Cylinder, optical circular, 343 Dedekind, 330, 337 De Morgan, 333 Einstein, 1, 2, 8 Euclid, 333, 337 Eudoxos, 333 First element of an inertia line which is after an element, 155 Fizeau, 7, 8 General, line, 47 plane, 185 threefold, 224 General threefold, sets of elements which determine different types of, 236- 239 Generator, of acceleration plane, 40 of optical plane, 137 of optical or rotation threefold, 240 Half-line, general, 305 end of, 305 Half-plane, general, 307 boundary of, 307 Heath, 333 Inertia line, 47 Interpretation of results, 366 Intersection, of optical lines, 25 of general lines, 62 of general line and general plane, 224 of general line and general threefold, 255 Interval, linear, 305 Larmor, 1 Last element of an inertia line which is before an element, 155 Latent element, 272 Length of segment, numerical value of in terms of unit segment, 333 Line, general, 47 inertia, 47 optical, 13 separation, 47 Linear interval, 305 Linearly between, 102 Lorentz, 1 Mean, of two elements in an inertia or separation line, 83 of two elements in an optical line, 83 Minkowski, 1 Normality, of general planes, complete, 216 of general lines having a common element, 204-205 of general lines having no common element, 215 of general line and general plane, 216 of general line and general threefold, 233 INDEX 373 Normality, of general plane and general plane, 260 of general plane and general threefold, 260 of general threefold and general three- fold, 260 Optical, line, 13 plane, 137 threefold, 239 Optical line, equations of, 365-366 Optical parallelogram, 61 corners, diagonal lines &c., 61 side lines of, 75 opposite side lines, 75 centre of, 75 Optical planes, sets of elements which determine, 149-151 Pairs, inertia, optical and separation, 272 conjugate, 275 Parallelism, of acceleration planes, 57 of general lines, 66 of general line and general plane, 189 of general line and general threefold, 258 of general plane and general threefold, 258 of general threefold and general three- fold, 259 of general planes, 189 Parallelism of optical lines, 35 after-parallel, 35 before-parallel, 35 neutral-parallel, 35 Parallelogram, general, in an acceleration plane, 121 side lines, diagonal lines &c., 121 Parallelogram, general, in an optical plane, 146 side lines, diagonal lines &c., 147 Parallelogram, general, in a separation plane, 181 side lines, diagonal lines &c., 181 Parallelogram, optical, 61 corners, diagonal lines &c., 61 side lines, 75 opposite side lines, 75 centre of, 75 Parallels, axiom of, 120, 142, 181, 185 Peano, 104-107, 119, 142, 181, 185, 334 Pierpont, 333 Plane, acceleration, 36 general, 185 optical, 137 separation, 178 Postulate I, 10 II, 10 Postulate III, 10 IV, 10 V, 10 VI, 10 VII, 11 VIII, 11 IX, 14 X, 23 XI, 25 XII, 28 XIII, 37 XIV, 47 XV, 50 XVI, 62 XVII, 86 XVIII, 155 XIX, 199 XX, 240 XXI, 330 Proportion, 333 Pythagoras, 339, 356, 357, 360, 362 Representative elements in parallel accelera- tion planes, 162 Kotation, threefold, 239 Segment, of general line, 305 end of, 305 prolongation of, 305 Segments, equality of, 331 greater than, 331 less than, 331 Separation, line, 47 plane, 178 threefold, 239 Separation planes, sets of elements which determine, 183 Simultaneousness, 2, 6 Sphere, separation, 327 centre, radius, diameter &c., 327 Standard cone, 364 Steps, taking, along an inertia line, 86 surpassing in finite number of, 86 Sub-set a, 11 ft 11 Threefold, general, 224 optical, 239 rotation, 239 separation, . 239 Triangle, general, 308 sides of, 308 corners of, 308 Unaccelerated particle, 368 Veblen, 334, 337, 338, 339 CAMBBIDGE : PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS SELECTION FROM THE GENERAL CATALOGUE OF BOOKS PUBLISHED BY THE CAMBRIDGE UNIVERSITY PRESS An Introduction to the Theory of Multiply-Periodic Functions.. 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