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IN LIB.
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1
ARTES
1837
SCIENTIA
VERITAS
LIBRARY
OF THE
UNIVERSITY OF MICHIGAN
P
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QUALIS PENINSULAN AMOS NAM
CIRCUM
DEPARTMENT
OF
ENGINEERING
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v. 1
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.
A TREATISE
ON
34791

HYDRODYNAMICS
With numerous Examples.
Effred
AV
maid
Barna
BY
BASSET, M. A.
OF LINCOLN'S INN, BARRISTER AT LAW; FELLOW OF THE CAMBRIDGE PHILOSOPHICAL
SOCIETY; AND FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE.
VOLUME I.
CAMBRIDGE:
DEIGHTON, BELL AND CO.
LONDON: GEORGE BELL AND SONS.
1888
[All Rights reserved.]
1
Cambridge:
PRINTED BY C. J. CLAY, M.A. AND SONS,
AT THE UNIVERSITY PRESS.
Reclass,
D&H. Apr.
PREFACE.
IN the present Treatise I have endeavoured to lay before the
reader in a connected form, the results of the most important in-
vestigations in the mathematical theory of Hydrodynamics, which
have been made during modern times. The Science of Hydro-
dynamics may properly be considered to include an enquiry into
the motion of all fluids, gaseous as well as liquid; but for reasons
which are stated in the introductory paragraph of Chapter I.,
the present treatise is confined almost entirely to the motion
of liquids. The progress of scientific knowledge in all its
branches has been the peculiar feature of the present century,
and it is therefore not surprising that during the last fifty years
a great increase in hydrodynamical knowledge has taken place;
but many of the most important results of writers upon this
subject have never been inserted in any treatise, and still lie
buried in a variety of British and foreign mathematical periodicals
and transactions of learned Societies; and it has been my aim to
endeavour to collect together those investigations which are of
most interest to the mathematician, and to condense them into a
form suitable for a treatise.
The present work is divided into two volumes, the first
of which deals with the theory of the motion of frictionless
liquids, up to and including the theory of the motion of solid
bodies in a liquid. In the second volume, a considerable portion
of which is already written, it is proposed to discuss the theory of
rectilinear and circular vortices; the motion of a liquid ellipsoid
+
iv
PREFACE.
i
under the influence of its own attraction, including Professor
G. H. Darwin's important memoir on dumb-bell figures of equi-
librium; the theories of liquid waves and tides; and the theory
of the motion of a viscous liquid and of solid bodies therein.
References have been given throughout to the original autho-
rities which have been incorporated or consulted; and a collection
of examples has been added, most of which have been taken from
University or College Examination Papers, which have been set
during recent years.
The valuable report of Mr W. M. Hicks on Hydrodynamics, to
the British Association in 1881-2, has proved of great service in
the difficult task of collecting and arranging materials. I have
also to express my obligations to the English treatises of Dr
Besant and Professor Lamb, from the latter of which I have
received considerable assistance in Chapters IV. and VI.; and also
to the German treatise of the late Professor Kirchhoff.
I am greatly indebted to Professor Greenhill for his kindness
in having read the proof sheets, and also for having made many
valuable suggestions during the progress of the work.
In a treatise which contains a large amount of analytical
detail, it is probable that there are several undetected errors;
and I shall esteem it a favour if those of my readers who discover
any errors or obscurities of treatment, or have any suggestions to
make, will communicate with me.
UNITED UNIVERSITY CLUB,
PALL MALL, EAST.
ART.
1.
Introduction
CONTENTS.
CHAPTER I.
HYDROKINEMATICS.
2. Definition of a fluid.
3. Lagrangian method, and Eulerian or flux method
4. Velocity and acceleration-Lagrangian method
5-6.
""
7. Analytical lemma
-flux method
•
8-9. The equation of continuity
•
10. Definition of the velocity potential, and the forms of Laplace's operator
in polar and cylindrical coordinates
11. Lagrange's equation of continuity
12. The bounding surface
13. Lines of flow and stream lines
14-15. Properties of lines of flow in a liquid .
16.
17.
Earnshaw's and Stokes' current function
Molecular rotation
•
18-19. Formulae of transformation
Examples
•
CHAPTER II.
ON THE GENERAL EQUATIONS OF MOTION OF A
PERFECT FLUID.
20. Pressure at every point of a fluid is equal in all directions
21. Equations of motion of a perfect fluid
22. Pressure is a function of the density
23. Equations of motion referred to moving axes
B.
PAGE
1
3
5
447AA C
8
9
9
10
11
11
12
13
15
19
20
21
21
b
vi
CONTENTS.
ART.
PAGE
29.
24. Equations satisfied by the components of molecular rotation
25. Stokes' proof that a velocity potential always exists, if it exists at any
particular instant
26. Physical distinction between rotational and irrotational motion
27. Lagrange's hydrodynamical equations of motion
28. Weber's transformation
Proof of theorem that if the pressure is not a function of the density,
vortex motion can be generated or destroyed in a perfect fluid
30. Cauchy's integrals
31. Integration of equations of motion when a velocity potential exists.
32. Definition of a vortex line
33. Clebsch's transformation.
34. Proof that
[ " at f f f (pp¹+ V) de dy de is a maximum or minimum
te
35-36. Energy and least action
37. Steady motion-Bernoulli's theorem
22
22
23
25
25
26
•
26
27
27
28
30
31
33
38.
Conditions of steady motion which is symmetrical with respect to an axis,
or is in two dimensions
39-40. Steady motion of a liquid-Clebsch's method
34
36
41.
Maximum and minimum theorem
38
•
42. Impulsive motion
38
43.
44.
Motion of a liquid surrounding a sphere which is suddenly annihilated
Torricelli's theorem.
39
•
40
45. Application of the hypothesis of parallel sections, to the motion of liquid
flowing out of a vessel
Examples.
41
43
CHAPTER III.
ON SOURCES, DOUBLETS AND IMAGES.
46. Velocity potential due to a source or sink
48
47.
do.
due to a doublet
•
49
48.
do.
due to a doublet sheet.
49
49.
do.
due to a source and doublet in two dimensions
49
50. Theory of images
50
51. Image of a source in a liquid bounded by a fixed plane
52. Image of a source in a sphere
51
•
51
53. Image of a doublet in a sphere, whose axis passes through the centre of
the sphere.
53
54. Image of a doublet in a sphere, whose axis is perpendicular to the line
joining it with the centre of the sphere
54
55. Image of a source and line sink in a sphere
56. Image of a source and doublet in a cylinder
54
56
57. Image of a source between two parallel planes.
Examples.
56
59
CONTENTS.
CHAPTER IV.
vii
ART.
VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
58. Statement of problem
59. Fundamental properties of vortex filaments
60. Integration of the equations which determine the components of
molecular rotation in terms of the velocities
61. Velocity due to a vortex
•
62. Velocity potential due to a vortex
•
•
PAGE
62
62
64
=
66
•
66
63. Vortex sheets
68
64. Surfaces of discontinuity.
68
65. Surfaces of discontinuity possess the properties of vortex sheets
66. Cyclic and acyclic irrotational motion
68
70
70
70
67. Flow and circulation
68. Stokes' theorem
69. Circulation due to a vortex filament is equal to twice the product of its
angular velocity and its cross section
70. Polycyclic velocity potentials
•
71. Stream lines cannot form closed curves unless the motion is cyclic .
72. Circulation is independent of the time
•
73. Irrotational motion which is acyclic cannot become cyclic
74.
Reconcileable and irreconcileable lines
75. Simply and multiply connected regions
76. A multiply connected region is reducible to a simply connected one
77-78. Reduction of polycyclic velocity potentials to monocyclic functions
79. Vorticity
71
72
72
72
~ ~ ~ ~ ~ ~~**P
73
73
73
74
74
75
80. Green's theorem
76
81. Deductions from Green's theorem-Kinetic energy of a liquid
82. Physical interpretation of Green's theorem
77
78
83. Liquid whose motion is acyclic and irrotational, is reduced to rest if the
motion of the bounding surface is destroyed
78
84. Extension of Green's theorem to spaces bounded internally by several
closed surfaces
85. Kinetic energy of a liquid occupying such a region'.
86. Adaption of Green's theorem to two-dimensional space
•
87.
88.
Stokes' theorem a particular case of Green's theorem
Thomson's extension of Green's theorem.
89. Kinetic energy of an infinite liquid occupying a multiply connected space
90. Kinetic energy of a liquid contained within a closed surface, is less
when the motion is irrotational and acylic, than if the liquid had any
other possible motion
**:IIZ
78
78
79
80
81
82
91. Kinetic energy when the motion is rotational
888
83
83
•
92. Kinetic energy due to two vortex filaments is proportional to the electro-
kinetic energy due to two electric currents
84
93. Another expression for the kinetic energy
85
94. Kinetic energy in terms of Stokes' current function
85
95. Connection between vortex motion and electromagnetism
Examples
86
89
น
viii
CONTENTS.
2
CHAPTER V.
ART.
ON THE MOTION OF A LIQUID IN TWO DIMENSIONS.
96. Statement of problem
97. Boundary conditions for a cylinder moving in a liquid
98-101. Conjugate functions, and their properties.
102. Examples of conjugate functions-Circular cylinder moving in an in-
finite liquid-Initial motion due to a circular cylinder in a liquid
bounded by a fixed concentric cylinder-Motion of a liquid con-
tained in a rotating equilateral prism-do. in a rotating elliptic
cylinder.
103. Motion of a liquid contained in a rotating rectangular prism
104-105. Motion of a liquid contained in a rotating sector
106. Further applications of conjugate functions
•
107. Motion of an elliptic cylinder in an infinite liquid.
108. Motion of translation of a cylinder whose cross section is the inverse of
an ellipse with respect to its centre.
PAGE
90
90
91
93
96
98
100
100
102
•
104
105
do.
this form.
when liquid is contained within a cylindrical cavity of
105
112. Application of results to the theory of conduction of heat and of
electrified cylinders
106
106
•
109. Expression of results in terms of elliptic functions
110. Current function due to the rotation of the cylinder
111.
•
113. Motion of a cylinder whose cross section is the inverse of an ellipse
with respect to its focus
•
114. Motion of translation of a cylinder whose cross section is a lemniscate. 106
115. Coefficients of cos ne in the expansion of (1+2c cos 0+c²)
116. Motion of rotation of a cylinder whose cross section is a lemniscate
117-119. Motion of a cylinder whose cross section is a lemniscate of Bernoulli
120-121. Dipolar coordinates
122. Motion of two circular cylinders in an infinite liquid
123. Kinetic energy of an infinite liquid in which two circular cylinders are
moving
124. Expressions for the coefficients of the velocities
Examples
107
109
•
109
110
112
113
•
114
115
•
CHAPTER VI.
DISCONTINUOUS MOTION.
125-127. Statement of problems to be solved
120
•
128-129. Representation of a vector by means of a complex quantity, and the
properties of the latter
121
•
134.
130. Every complex has a differential coefficient
131. Kirchhoff's method of solving problems of discontinuous motion
132-133. Transformation by means of complex variables
Particular cases of transformation .
135. Motion of a jet escaping from a slit
122
123
124
•
126
•
127
ง
CONTENTS.
ix
ART.
PAGE
136. Motion of a jet escaping through a small tube
129
137. Coefficient of contraction of a jet can never be less than
130
138. Stream of liquid flowing past a rectangular lamina. Pressure on the
lamina
131
139. Conditions of stable and unstable equilibrium of the lamina.
140. Intrinsic equation of the surface of discontinuity
134
135
•
Examples
135
CHAPTER VII.
ON THE KINEMATICS OF SOLID BODIES MOVING IN A
LIQUID.
141. Conditions to which the velocity potential must be subject
142. Boundary conditions for the case of a single solid.
143–144. Velocity potential due to the motion of a sphere
145.
do. due to the motion of the solid formed by the revolution
of two spheres cutting orthogonally
137
137
138
139
146.
do.
147.
do.
due to the initial motion of two concentric spheres
due to the motion of an ellipsoid .
140
140
148.
do.
due to the motion of an ellipsoid of revolution
143
149.
do.
due to the motion of a circular disc
144
150.
do.
151.
do.
due to liquid contained in a rotating ellipsoidal cavity.
due to liquid contained between two confocal ellipsoids
152. Magnetic potential of a spherical bowl .
145
145
146
154.
do.
153. Velocity potential due to the motion of liquid about a spherical bowl
due to a source situated on the axis of the bowl
155.
do. due to the motion of the bowl in an infinite liquid
156. Electro-static potential of a bowl placed in a field of force symmetrical
with respect to the axis of the bowl
147
149
•
149
150
157. Electro-static potential when the bowl is placed in a uniform field of
force parallel to the axis
152
•
158. Interpretation of the result
153
•
159. Electro-static potential when the bowl is placed in a uniform field of
force perpendicular to a plane containing the axis
160. Current function due to the motion of a solid of revolution parallel to
its axis
154
155
Examples
156
CHAPTER VIII.
ON THE GENERAL EQUATIONS OF MOTION OF A SYSTEM
OF SOLID BODIES MOVING IN A LIQUID.
161. The motion can be determined by Lagrange's equations
159
162. Acyclic motion
159
163. Kinetic energy is a homogeneous quadratic function of the velocities of
the solids alone
160
X
CONTENTS.
“་
ART.
164. Proof that Lagrange's equations can be employed
165. Impulse of the motion
166. Hamiltonian equations
167. Kirchhoff's equations
168-9. Geometrical equations
170. Cyclic motion
•
•
PAGE
•
161
162
163
164
165
•
167
•·
169
170
•
171
175
171. Kinetic energy is the sum of a homogeneous quadratic function of the
velocities of the solids and a similar function of the circulations
172. Expressions for the generalised components of momentum due to the
cyclic motion.
173. The modified Lagrangian function.
174. Interpretation of the result. Product of the circulation and density is
a generalised component of momentum .
175. The flux through an aperture relative to the solid is, the generalised
velocity corresponding to the product of the density and the cir-
culation through that aperture
176. Modified function for a single solid having one aperture
177-178. Modified function for a system of cylinders
179. Explanation of results
•
176
•
177
178
181
CHAPTER IX.
ON THE MOTION OF A SINGLE SOLID IN AN INFINITE
LIQUID.
180. General expression for kinetic energy
•
182
181. Kinetic energy due to the motion of an ellipsoid and of a solid of
revolution
183
•
182. Motion of a sphere under the action of gravity
183
•
183. Motion may become unstable owing to the formation of a hollow
184. Initial motion of a sphere, when the liquid is enclosed within a con-
centric spherical envelope
185
185
•
185. Motion of a circular cylinder when there is circulation .
186. Determination of the motion by means of Lagrange's equations
187. Motion of an elliptic cylinder
188. Motion of a cylinder whose cross section is a curve such as a cardioid
189. Motion of an ellipsoid
186
•
188
189
193
•
193
•
190. Stability and instability of steady motion parallel to an axis
191. Calculation of the coefficients of inertia of an ellipsoid
192. Motion of an ellipsoid when two of its axes remain in a plane
193. Motion of a ring-shaped solid of revolution through whose aperture
194
•
195
•
195
•
there is circulation .
196
functions.
194. Motion of a ring produced by an impulsive couple about a diameter
195. Angular motion expressible in terms of the time by means of elliptic
196. Explanation of results by means of general principles
197
•
199
202
197. Steady motion and stability of a ring moving parallel to its axis
198. Steady motion and stability when the centre of inertia describes a circle 203
202
•
199-200. Helicoidal steady motion
205
201. Stability of helicoidal steady motion
207
CONTENTS.
xi
ART.
202. Expressions for the kinetic energy in the case of an isotropic helicoid
and other solids
203. Three directions of permanent translation for every solid
PAGE
208
209
204.
Wrenches and screws
210
205. Infinite number of steady motions when the impulse consists of a twist
about a screw
210
206. Motion of a solid is determinate when the impulse consists of a couple 212
Examples
213
CHAPTER X.
ON THE MOTION OF TWO CYLINDERS.
207. Expression for kinetic energy.
208. Motion of a cylinder in a liquid bounded by a fixed plane
209. Ratio of initial to terminal velocity when the cylinder is in contact
with the plane and is projected from it
•
210. Conditions that the cylinder may or may not strike the plane
211-212. Motion of one cylinder when the other is fixed
213. Motion of a cylinder in a liquid bounded by a fixed plane, when there is
circulation
214. Steady motion of a cylinder when the liquid is bounded by a horizontal
plane
215-216. Deduction of results from general reasoning
Examples
CHAPTER XI.
219
219
221
221
222
223
224
•
226
·
227
ON THE MOTION OF TWO SPHERES.
217. Kinetic energy consists of nine terms only
229
218. Motion along the line of centres
230
219. Calculation of the coefficients of the velocities in the expression for the
kinetic energy, by the method of images .
231
220-221. Complete values of the coefficients
232
·
222. Approximate values of the coefficients, in terms of series of powers of
the reciprocal of the distance between the spheres
236
223. Motion perpendicular to the line of centres
237
the coefficients.
231. Expressions for the components of the pressure upon the sphere
Oscillations of the second order. Thomson's theorem
224-225. Calculation of the coefficients by means of images
226. Calculation of the images µ1, 1, µ'ı
1
227. Transference theorem in spherical harmonics.
228. Kinetic energy depends solely upon harmonics of the first degree
229. Calculation of the velocity potential, and of the approximate values of
230. Motion of a sphere in a liquid bounded by a fixed plane
232. Small oscillations of two spheres
233.
234. Pulsations of two spheres
237
·
239
240
242
243
•
244
245
246
247
248
•
xii
CONTENTS.
ART.
235. Velocity potential due to pulsations, is equivalent to that due to an
infinite system of images.
236. Formula for determining the pressure upon the spheres
237-238. Determination of that portion of the pressure which does not
depend upon the square of the velocity
PAGE
249
249
250
•
239-240. Approximate values of the velocity potential, and of the portion of
the pressure which depends on the square of the velocity
241. The spheres attract one another when their phases differ by less than a
quarter of a period
Examples
•
•
252
255
256
APPENDIX.
I. Proof of the equation p=kpY
•
II. Value of a q-series in terms of elliptic functions
259
260
III. Determination of the azimuthal motion of a solid of revolution by
means of Weierstrass's functions
261
ERRATA.
Page 11 line 7 read, ff(lu+mv+nw)dS.
15
11
where for when.
""
31 § 35
99
fluid for liquid.
""
32 line 11 add, and taking account of (21).
39 20 read, v for V.
33
333
46
8
51
17
II for π.
S' for H.
54 In the figure read, OR'H'S' for OR'H'S.
55 line 21 read, strength for density.
68 § 63 read, fluid for liquid.
70 § 68 Insert the letter D in the figure.
80 line 14 read, πр for 2πр.
107
"9
108
110
2 from bottom read, Ln+1 for Ln+1
8, 14, 18, 19, 20 read, 2E and 2F for E and F.
8 read, π for .
114 multiply the values of P, Q and L by p.
116 line 16 add, with unit velocity.
99
163
187
195
""
""
202
""
11 and p. 170 line 15 read, impulsive pressure.
19 read, terms in p/p.
5,
is for be.
1 and 2 from bottom read, A for a.
221 13 read, ratio of the terminal to the initial velocity.
223 In the figure read, y for n.
""
241 footnote read, Kugelfunctionen.
247 line 13 read, 2B/C, B2/C2 for 2B/c, B²/c².
CHAPTER I.
HYDROKINEMATICS.
1. THE science of Hydrodynamics may be divided into two
separate branches, viz. the motion of liquids and the motion of
gases. The chief interest arising from the latter branch of the
subject is due to the fact that air is the vehicle by means of
which sound is transmitted, and consequently the discussion of
special problems relating to the motion of gases belongs to the
theory of sound rather than to hydrodynamics; it must also be
recollected that in order to deal satisfactorily with many problems
connected with the motion of gases, it is necessary to take into
account changes of temperature and other matters which properly
belong to the science of thermodynamics. In the earlier chapters
of the present treatise the general theory of the motion of fluids
is discussed, including those peculiarities of motion which are
alike common to liquids and gases; but the subsequent chapters
are limited almost entirely to the consideration of special problems
relating to the motion of liquids.
In ancient times very little advance in hydrodynamics appears
to have been made. In modern times the earliest pioneers were
Torricelli and Bernoulli, whose investigations were due to the
hydraulic requirements of Italian ornamental landscape gardening;
but the first great step was taken by D'Alembert and Euler, who
in the last century successfully applied dynamical principles to
the subject, and thereby discovered the general equations of
motion of a perfect fluid, and placed the subject on a satisfactory
basis. The discovery of the general equations of motion was
followed up by the investigations of the great French mathe-
maticians Laplace, Lagrange and Poisson, the first of whom has
left us a splendid memorial of his genius in his celebrated Theory
of the Tides.
B.
1
2
HYDROKINEMATICS.
The next advance was made by Poisson' and Green; the
former of whom in 1831 discovered the velocity potential due
to the motion of a sphere in an unlimited liquid, and the latter
of whom in 1833, without a knowledge of Poisson's work, discovered
the velocity potential due to the motion of translation of an
ellipsoid in an unlimited liquid. Green's investigation was com-
pleted for the case of rotation by Clebsch³ in 1856.
4
The velocity potential due to the motion of a variety of cylin-
drical surfaces has also been discovered during the last fifteen
years; but a similar advance has not been made as regards the
motion of two or more solids. The kinetic energy of a liquid due
to the motion of two cylinders whose cross sections are circular,
has been obtained by Hicks and Greenhills. The former has also
written several valuable papers on the motion of two spheres,
which have placed this problem in a perfectly satisfactory con-
dition. A complete discussion of the motion of two oblate or
prolate spheroids whose excentricities are nearly equal to zero or
unity, would be an attractive subject for investigation, and would
throw light on the motion of two ships sailing alongside one
another.
In 1845 Professor Stokes' published his well-known theory of
the motion of a viscous liquid, in which he endeavoured to account
for the frictional action which exists in all known liquids, and
which causes the motion to gradually subside by converting the
kinetic energy into heat. This paper was followed up in 1850 by
another³, in which he solved various problems relating to the
motion of spheres and cylinders in a viscous liquid. Previously to
this paper no problem relating to the motion of a solid body in a
liquid had ever been solved, in which the viscosity had been taken
into account.
Since the time of Lagrange the essential difference between
the motion of a fluid when a velocity potential exists and when it
does not exist had been recognised; and an opinion very generally
1 Mém. de l'Acad. des Sciences. Paris, vol. xI. p. 521.
2 Trans. Roy. Soc. Edinburgh, vol. xIII. p. 54.
3 Crelle, vol. LII. p. 119.
4 Quart. Journ., vol. xvi. pp. 113 and 193.
5 Ibid. vol. xvш. pp. 356–362.
• Proc. Camb. Phil. Soc., vol. III. p. 276, vol. iv. p. 29, and Phil. Trans., 1880.
7 Trans. Camb. Phil. Soc., vol. vi. p. 287.
8 Ibid. vol. ix. part 11. p. 8.
INTRODUCTION.
3
prevailed that if at any particular instant some particular portion
of the fluid were moving in such a manner that a velocity poten-
tial existed, the subsequent motion of this same portion of fluid
would always be such that the component velocities of its ele-
ments would be derivable from a velocity potential. The first
rigorous proof of this important proposition was given by Cauchy,
and a different one was subsequently given by Stokes', but until
the year 1858 no complete investigation respecting the peculiari-
ties of rotational motion had ever been made. This was effected
by Helmholtz² in his celebrated memoir on Vortex Motion, which
may perhaps be considered the most important step in hydro-
dynamics which has been made during the present century. The
same subject was subsequently taken up by Sir W. Thomson and
the theory of polycyclic velocity potentials fully investigated.
During the last six years important additional investigations on
the theory of vortex rings have been made by Hicks' and J. J.
Thomson5.
3
The last twenty years have witnessed a great advance in
hydrodynamics, and numerous important papers have been written
by many eminent mathematicians both British and foreign,
which will be considered in detail in the present work.
We shall now proceed to consider the definitions and principles
of the subject.
2. A fluid may be defined to be an aggregation of molecules,
which yield to the slightest effort made to separate them from
each other, if it be continued long enough. All fluids with which
we are acquainted may be divided into liquids and gases; the
former are so slightly compressible that they are usually regarded
as incompressible fluids, whilst the latter are very highly com-
pressible.
A perfect fluid is one which is incapable of sustaining any
tangential stress or action in the nature of a shear; and it will be
shown in the next chapter that the consequence of this property
is, that the pressure at every point of a perfect fluid is equal
in all directions, whether the fluid be at rest or in motion. A
1 Trans. Camb. Phil. Soc., vol. vi. p. 305.
2 Crelle, vol. LV. p. 25; translated by Tait, Phil. Mag. (4) XXXIII. p. 485.
3 Trans. Roy. Soc. Edin., vol. xxv. p. 217.
4 Phil. Trans., 1881, 1884 and 1885.
↳ Adams' Prize Essay, 1882.
1-2
4
HYDROKINEMATICS.
perfect fluid is however an entirely ideal substance, since all fluids
with which we are acquainted are capable of offering resistance to
tangential stresses. This property, which is known as viscosity,
gives rise to an action in the nature of friction, by which the
kinetic energy is gradually converted into heat.
In the case of gases, water and many other liquids, the effects
of viscosity are small; such fluids may therefore be approximately
regarded as perfect fluids. It will therefore be desirable to com-
mence with the study of the motion of perfect fluids, reserving
the consideration of viscous fluids for the second volume.
There are certain kinematical propositions which are true for
all fluids, and which it will be convenient to investigate before
entering upon the dynamical portion of the subject. These
propositions form the subject of the present chapter.
3. The motion of a fluid may be investigated by two different
methods, the first of which is called the Lagrangian method, and
the second the Eulerian or flux method, although both are due to
Euler.
In the Lagrangian method, we fix our attention upon an
element of fluid, and follow its motion throughout its history.
The variables in this case are the initial coordinates a, b, c of the
particular element upon which we fix our attention, and the time.
This method has been successfully employed in the solution of
very few problems.
In the Eulerian or flux method, we fix our attention upon a
particular point of the space occupied by the fluid, and observe
what is going on there. The variables in this case are the
coordinates x, y, z of the particular point of space upon which we
fix our attention, and the time.
Velocity and Acceleration.
4. In forming expressions for the velocity and acceleration of
a fluid, it is necessary to carefully distinguish between the
Langrangian and the flux method.
I The Langrangian Method.
Let u, v, w be the component velocities parallel to fixed axes,
of an element of fluid whose coordinates are x, y, z and x+dx,
y+dy, z+dz at times t and t + St respectively, then
u = dx/dt=x, v = ÿ, w = ż
(1),
VELOCITY AND ACCELERATION.
5
where in forming i, j, ż we must suppose x, y, z to be expressed in
terms of the initial coordinates a, b, c and the time.
If the axes, instead of being fixed, were moving with angular
velocities 0₁, 02, 0, about themselves, the component velocities
would be given by the equations,
น
1)
= x − y¦¸ + z02, v = ÿ− z0¸+x0¸, w = ż — xе¸ + yo̟……….(2).
3
1
2
It should be noticed that x, y, ż are the velocities of the fluid
relative to the moving axes.
The expressions for the component accelerations are
ƒx=ú = ï‚ ƒ„=Ÿ‚ ƒ₂ = ï
when the axes are fixed, and
3
1
2
………….(3),
fx = = ú− və₂+w02, ƒv=v — w✪₁+ u03› ƒz = ŵ − u0₂+v03…….(4)
when the axes are in motion. Here u, v, w must be supposed to
be expressed in terms of a, b, c and t.
II. The Flux Method.
5. Let &Q be the quantity of fluid which in time St flows across
any small area A, which passes through a fixed point P in the
fluid; let p be the density of the fluid, q its resultant velocity, and
e the angle which the direction of q makes with the normal to A,
drawn towards the direction in which the fluid flows. Then
therefore
dQ = pq Adt cos e,
q=
P
1
dQ
A cos e dt
Now A cos e is the projection of A upon a plane passing
through P perpendicular to the direction of motion of the fluid;
hence &Q is the independent of the direction of the area, and is
the same for all areas whose projections upon the above-mentioned
plane are equal. Hence the velocity is equal to the rate per unit
of area divided by the density, at which liquid flows across a plane
perpendicular to its direction of motion.
The velocity is therefore a function of the position of P and
the time.
6. We may therefore put u = F(x, y, z, t); whence if the axes
are fixed, and if u+ du be the velocity parallel to x at time t + St
of the element of fluid which at time t was situated at the point
(x, y, z),
Su = F(x + ust, y + vồt, z+wdt, t + dt) − F(x, y, z, t).
6
HYDROKINEMATICS.
..
Therefore the acceleration,
Su du
f = lim
du du
+u +v + w
St dt dx dy
Hence if ǝ/ǝt denotes the operator
d/dt+ud/dx+ vd/dy+wd/dz,
du
dz'
the component accelerations will be given by the equations
Əv
ди
aw
ƒx
Ət
}
Ət
fr f₁ =
It
>
(5).
When the axes are in motion let u + du be the component
velocity at time t + St, parallel to the new position of the axis of x,
of the element which at time t was situated at the point x, y, z;
then if U, V, W be the component velocities relative to the axes,
Su= F(x+ Ust, y + Vst, z+ W&t, t + St) — F(x, y, z, t).
Therefore
Su
du
du
du
St
dt
+ U
du
+ V
dx
+ W
dy
dz
>
where the values of U, V, W are given by (2). Hence if ə/ət
denote the operator d/dt + Ud/dx + Vd/dy + Wd/dz, the com-
ponent accelerations parallel to the moving axes are given by the
equations
ди
Ət
fx= vəz + w02, ƒv
Əv
Ət
1
wo₁+u03, ƒx=
ди
Ət
2
u✪₂+v0¸...(6).
1
ვ
Similarly it can be shown that if ☎, 0, z be cylindrical coordi-
nates, and u, v, w be the component velocities measured in the
directions in which the former quantities increase,
ди
v2
foo
Ət
>
fo= It
Əv UV
+
ди
,
Sz
Ət
…………..(7),
where
მ
a
d
d
d
au v² + w²
r
Əv uv
fo= +
Ət γ
+u +
v d
+w
Ət dt do @ do dz'
If (r, 0, 0) be polar coordinates and u, v, w be the velocities
measured in the directions in which these quantities increase,
fr
123
гиг
cot 0,
g
ди
Uw
f$
+
Ət
ՈՐ
UV
γ
+ cot .........(8),
where
a d d
+u +
Ət dt dr
v d
+
r do
พ
d
r sin 0 do
THE EQUATION OF CONTINUITY.
The Equation of Continuity.
7
7. Before proceeding further, it will be convenient to intro-
duce the following lemma, which is a particular case of Green's
Theorem, which will be considered more fully in Chapter IV.
Let §, n, & be any functions of x, y, z, which are finite and
continuous at all points within a closed surface S, then
||
(de + dn + d) dx dydz = [[(1 + mn + ng) dS…..(9),
SSS dę
dx dy dz
where the triple integral extends throughout the volume enclosed
by S, and the double integral is taken over the surface of S, and
l, m, n and the direction cosines of the normal at any point of
S drawn outwards.
Integrating the left-hand side of (9) by parts we obtain
SSS
[[] de dødydz = [[[ € dyds],
&
where the brackets refer to the limits of integration. Now since
the surface S is closed, it follows that any line parallel to ∞ which
enters the surface a given number of times must issue from it the
same number of times, hence if is positive at the point of
entrance, it must be negative at the corresponding point of exit;
hence
[SS § dydz] = ss 1 § dS,
where the integration with respect to S extends over the whole
surface. Treating the other two terms in a similar manner we
obtain the theorem in question.
8. If the motion of a fluid be continuous, it is evident that
the increase in the amount of fluid within a fixed space, which
takes place during any given interval, must be equal to the amount
which flows in across the boundaries of that space.
Let p be the density of the fluid at time t, then the increment
during an interval St in the mass of the fluid bounded by any
fixed surface S,
=
∙SSS de
St dx dy dz.
The amount of fluid which flows into S across the boundary,
♫♫p (lu + mv + nw) St dS,
´d
--[[[{d (pu)+d(pv)+d(px)} stdædy dz,
dy
dx
dz
8
HYDROKINEMATICS.
...
by (9). Equating these two values of the increment, we obtain
dp d (pu), d (pv), d (pw)
+
+
dt dx dy
+
dz
= 0 .....
……..(10).
This equation is usually called the equation of continuity.
In the case of a liquid p is constant, whence
du dv dw
+ +
dx dy dz
= 0.....
............
.(11).
9. The same result is often obtained in a different manner,
which we shall illustrate by finding the equation of continuity of a
liquid referred to polar coordinates.
Let u, v, w be the velocities in the r, 0, & directions, and let
r² sin edr 808 be a small element of volume. The quantity of
liquid which in unit of time flows in across the face r² sin 0808¢
= pur² sin 080 84.
=
The quantity which flows out across the opposite face
d
= pur² sin 08084 + p sin 0 dr (r²u) Sr so sp.
Hence the total loss
= p sin 0 d (ru) dr8084.
dr
Equating the total loss due to the flow across all the faces of
the element to zero, we obtain
d (r²u) d (v sin 0)
dw
sin
+r
+r = 0.
dr
de
аф
If cylindrical coordinates are employed, the equation is
d (wu)
dw
dv
+
de
b+
dw
0......
dz
(12).
...
(13).
10. In a large and important number of problems the quan-
tity udx+vdy+wdz is a perfect differential do, whence
=
dp/dx, v=dp/dy, w=dp/dz;
hence if ds be a linear element drawn in any direction, and q be
the velocity in the same direction q=dp/ds. The function
is called the velocity potential.
or
Substituting the above values of u, v, w in (11), we obtain
đẹp đẹp đẹp
dx² dy²
+ +
dz
V=0.
0.....
.(14),
7
.
THE BOUNDING SURFACE.
9
This equation is usually known as Laplace's equation, and the
operator V² as Laplace's operator.
2
The values of ▼² in polar and cylindrical coordinates are re-
spectively,
and
d2
2 d
1 d²
cote d
1 ďⓇ
+
dra
+
+
r dr
+
² do
для 20
7ª sin³ 0 dó³·
(15),
d
1 d 1 d²
d2
=
+
+
do2
a do + w² do² + dz²
.(16).
@
These results may be readily obtained by substituting the
values of u, v, w in terms of & in (12) and (13).
11. The preceding forms of the equation of continuity are not
convenient when the Lagrangian method is employed. To find an
appropriate form, consider a small rectangular parallelopiped
whose diagonal is PQ. Let a, b, c, a + da, b+ 8b, c+dc be the
coordinates of P and Q respectively. At the end of a time t, the
fluid of which the parallelopiped is composed will form a dif-
ferently situated oblique-angled parallelopiped. The volume of
the latter
= Jsa Sb Sc,
where J is the Jacobian of x, y, z and is equal to
dx dy dz
da' da' da
dx dy dz
,
db' db db
dx dy dz
dc' de' dc
dc
Ро
Hence if p, be the initial density, and p the density at time t,
the required equation is
Jp = Po
In the case of a liquid p = p, and therefore
J = 1.......
The Bounding Surface.
.(17).
.(18).
12. Besides the equations which must be satisfied within the
interior of a fluid, it is necessary that certain other conditions
should be satisfied at the boundary, which depend upon the
special problem under consideration.
If the fluid is bounded by a surface whose equation referred to
axes fixed in space is F (x, y, z, t) = 0, the normal velocity of the
10
HYDROKINEMATICS.
fluid at the surface must be equal to the normal velocity of the
surface, hence the sheet of fluid of which the boundary is com-
posed must always consist of the same elements of fluid.
Hence
and therefore
F (x+uồt, y + vdt, z + wôt, t + dt) = 0,
dF
dF dF dF
+u + v +w
dt dx
= 0 .....
dy dz
.(19).
(20).
If the boundary is fixed, the condition becomes
lu + mv + nw = 0.....
If the axes be in motion, the condition is
dF
+ U
dt
da
dF dF dF
+Ꮴ + W
dy dz
0.
(21),
where U, V, W are the velocities of an element of fluid relative to
the axes.
It should be noticed that (19) or (21) must be satisfied by
every surface which is composed of the same elements of fluid.
Lines of Flow and Stream Lines.
13. DEF. A line of flow is a line whose direction coincides
with the direction of the resultant velocity of the fluid.
The differential equations of a line of flow are
dx _dy__dz
U
ข
W
Χι
X2
1
Hence if X, (x, y, z, t) = α₁, X₂ (x, y, z, t) = a, be any two in-
dependent integrals, the equations X₁ = const., X₂ = const., are the
equations of two families of surfaces whose intersections determine
the lines of flow.
DEF. A stream line, or a line of motion, is a line whose
direction coincides with the direction of the actual paths of the
elements of fluid.
The equations of a stream line are determined by the simul-
taneous differential equations,
¿ = u, ÿ = v, ż= w,
where x, y, z must be regarded as unknown functions of t. The
integration of these equations will determine x, y, z in terms
of the initial coordinates and the time.
•
Г
LINES OF FLOW.
11
14. If through every point of a small closed curve lines
of flow be drawn, they will enclose a mass of fluid which may be
called a tube of flow.
Let us apply the lemma of § 7 to a portion of liquid bounded
a tube of flow and two planes perpendicular to it. Putting
u = §, v = n, w = (, and taking account of (11), we obtain
0=
= SSS
du dv dw
dx + dy + dz) dxdyd z = [] (1
2
lu + mv + nw) ds.
nw) dɛ.
At every point of the curved surface of the tube of flow,
lu + mv + nw = 0; at the two ends this quantity is respectively
equal to q, and -q,, where q, and q, are the velocities of the
liquid at the ends. Hence the surface integral = q₁dS¸ — ¶„ds = 0;
whence the product of the velocity of a liquid and the cross
section of a tube of flow is constant throughout the length of
the latter.
1
In the next place, a line of flow cannot begin or end in any
portion of a liquid throughout which the velocity is finite, but must
either form a closed curve or have its extremities in the boundaries
of the portion of liquid.
in
A
For if a line of flow ended the liquid, it would be possible to
draw a closed surface cutting a tube of flow once only. Hence
lu+mv+nw would be zero at every point of the closed surface
excepting where it cuts the tube of flow, and therefore the surface
integral would not be zero.
15. When a velocity potential exists, the equation
udx+vdy+wdz=0
is the equation of a family of surfaces, at every point of which the
velocity potential has a definite constant value, and which be
may
called surfaces of equi-velocity potential.
If P be any point on the surface, = const., and dʼn be an
element of the normal at P which meets the neighbouring surface
+84 at Q, the velocity at P along PQ will be equal to do/dn;
hence do must be positive, and therefore a fluid always flows
from places of lower to places of higher velocity potential.
The lines of flow evidently cut the surfaces of equi-velocity
potential at right angles.
16. The solution of hydrodynamical problems is much sim-
plified by the use of the velocity potential (whenever one exists),
..
12
HYDROKINEMATICS.
since it enables us to express the velocities in terms of a single
function . But when a velocity potential does not exist, this
4.
cannot in general be done, unless the motion either takes place
in two dimensions, or is symmetrical with respect to an axis.
In the case of a liquid, if the motion takes place in planes
parallel to the plane of xy, the equation of the lines of flow is
udy-vdx=0............
The equation of continuity is
..(22).
du dv
dx + dy
: 0,
which shows that the
left-hand side of (22) is a perfect differ-
ential dy, whence
dy.
u =
dy
dy
ข
dx
..(23).
The function is called Earnshaw's current function.
When the motion takes place in planes passing through the
axis of z, the equation of the lines of flow may be written
The equation of continuity is
W
(wda – udz) = 0 ……....
d(au)
do
dw
b+
= 0,
dz
.(24).
which shows that the left-hand side of (24) is a perfect differential
dy, whence
1 dy
W
1 dy
И
@ do
@ dz
.(25),
where is Stokes' current function.
17. The existence of a velocity potential function involves.
the conditions that each of the three quantities,
dw/dy - dv/dz, du/dz - dw/dx, dv/dx – du/dy,
should be everywhere zero; when such is not the case we
shall denote the above quantities by 2§, 2n, 25. The quantities
, n,, for reasons which will be explained in the following
chapter, are called the components of molecular rotation. They
evidently satisfy the equation
dę, dn, dę
dx¹ dy¹ dz
0.
.(26).
FORMULAE OF TRANSFORMATION.
13
ot
Formulae of Transformation'.
18. The equations connecting the components of molecular
relation with the velocities are,
2}=
dw dv
dy dz
2n=
du dw
dz dx
dv du
25=
dx dy
.(27).
T
In order to obtain the equivalent equations when polar
coordinates are employed, let r, 0, &
be the coordinates of P, and let
u, v, w and u+Su, v + Sv, w + Sw be
the velocities at the points r, 0, $
and r+dr, 0+80, $+84 respec-
tively, measured in the directions
in which these quantities increase ;
also let u+ du', v + Sv', w + Sw' be
the velocities at the last mentioned
point parallel to the directions of
U, V, W.
Let us choose the axes of x, y, z
to
X
so as to coincide with the directions of r, 0, and 4 respectively, then
dx = dr, dy = rde, dz = r sin edo,
and therefore we at once obtain
du du dv dv
dv dw dw
dx
dr' dx dr' dx dr
.(28).
Let Q be a point whose coordinates are r, 0+80, &; then
du
dy
+2)-
du
u +
do
80)
Se) cos de − (v +
dv
-
80 | sin 80 - u
de
rso
dv
dy
1 du
r do
dv
0
du
δε
(v + dy de) cos 80 + (u + dy 50) sin 80 – 1
rso
v
1 dv
r do
dw' 1 dw
dy
r do
+
¹ Besant, Mess. of Math., vol. xi. p. 63.
.(29),
………..(30),
..(31).
:

14
HYDROKINEMATICS.
Let R be a point whose coordinates are r, 0, +84; and let
POR=8x, PTR dx'; then
du
dz
Hence
u +
аф
δφ
=
dx = sin @dp, dx' = cos 084.
cos dx - (w
w+
du
84) c
dp cos dx
r sin 084
1 du
r sin 0 do
dv
2
r
dw
84):
аф
Sp sin dx - u

-
(+ du 84) 00s &x - (w + du 84) sin dy-u
аф
cos dx
co
v
...(32),
dw

dv
аф
dz
r sin edo
1 dv
พ
cot ◊ ...
r sin 0 do
.(33),
p
dw
84) a
du
So cos do+u+
cosô¢+(u+ 84) sindx+(v+84) sindy'-w
аф
dv
аф
84)sin
||
r sin ese
ह
น V
+
γ
γ
+ - cot 0 ....
..(34).
dw
dz
(w+

1
do
dw
r sin 0 do
Hence
=
dw'
dy
dv 1 dw
dz r do
+
พ
cot
1 du dw
1 dv
r sin e do
2Æ
du'
dw'
พ
2n
dz dx
r sin 0 do
dr
γ
dv
du' dv
V
1 du
23=
+
dx dy dr
r
r de
.(35).
19. If cylindrical coordinates a, e, z are employed; let u, v, w
and u + Su, v + Sv, w + Sw be the velocities at the points ☎, e, z
and w+dw, 0+80, z+dz respectively; and let u + du', v + dv
be the velocities at the last mentioned point parallel to u and v.
Then
dx = dw, dy = wd0,
du du
dv dv dw'
dv dw dw
and
also
du
dy
dx do' dx do'
dw dx do
(༧
du
u +ão
1 du
παθ
80) cos 80 – (v + do 80) sin 80 – u
| | S
або
.(36),
..(37),
dv
dy
+
dv
do
1 dv
+
@ de
dw dw
Ղ
EXAMPLES.
80)
se) cos
du
80+(26+ δε
80) sin 50 – u
de
обо
•
(38),
dy ado
.(39),
du' du dv dv
and
dz
dz' dz
dz
.(40).
Therefore
1 dw dv
2Æ
w do
&&
dz
du
dw
2n
•
(41).
dz
da
dv
บ
1 du
2t
=
do
+
♡
@ do
15
EXAMPLES.
re
1. Find the equation of continuity in a form suitable for air
in a tube, and prove that if the density be f(at - x) when t is the
time and x the distance from one end of a uniform tube, the
velocity is
af (at − x)+(V− a) ƒ (at)
f(at - x)
where V is the velocity at that end of the tube.
2. If the motion of a liquid be in two dimensions, prove that
if at any instant the velocity be everywhere the same in magni-
tude, it is so in direction.
3. If every particle of a fluid move in the surface of a sphere,
prove that the equation of continuity is
dp
d
dt
cos e + ão (pw cos () +
d
(pw' cos 0) = 0,
аф
where p is the density, and the latitude and longitude of any
element, and w, w' the angular velocities of the element in latitude
and longitude respectively.
16
HYDROKINEMATICS.
1
4. In the last example prove that if the motion is irrotational
the velocity potential is equal to
ƒ (log tan 10+14) + F (log tan10 – 14),
where -1 and ƒ and F are arbitrary functions.
6=
5. An infinite mass of liquid is bounded by the plane zx, on
which are small corrugations given by y = (x). The velocity of
the liquid at an infinite distance from the plane is parallel to x
and equal to V. Prove that the velocity potential is
√x +
V 100
π
୮
(2 - λ) φ (λ) αλ
2
y² + (x − x)²
6. In the general motion of a fluid, prove that if F is the
normal acceleration at any point on a closed surface described in a
fluid, the expansion, w the molecular rotation, and Σ the strain
invariant
fg + gh+hf — a² – b² — c³, where ƒ = du/dx, 2a = dw/dy + dv/dz,
dydz.
then
[[ Fas = [[[( +0² + 2w² - 2Σ) dx dy dz.
7. Fluid is moving in a fine tube of variable section «, prove
that the equation of continuity is
d
d
dt
(KP) +
ds
(Kpv) = 0,
where v is the velocity at the point s.
8. If F(x, y, z, t) is the equation of a moving surface the
velocity of the surface normal to itself is
1 dF
R dt'
where R² = (dF/dx)² + (dF/dy)² + (dF/dz)².
Hence deduce equation (19).
9. If x, y and z are given functions of a, b, c and t, where a,
b and c are constants for any particular element of fluid, and if
u, v and w are the values of x, y, z when a, b, c are eliminated,
prove analytically that
d²x du
dť² dt
du du du
+ u +v +w
dx dy dz
10. Liquid which is moving irrotationally in three dimen-
sions is bounded by the ellipsoid (x/a)²+(y/b)² + (z/c)² = 1, where
EXAMPLES.
17
a, b, c are functions of the time, such that the volume of the
ellipsoid remains constant. Prove that if the ellipsoid is rotating
with angular velocities w₁, w, w about its principal axes, and
u, v, w are the component velocities of the liquid parallel to the
principal axes, the equation of continuity and the boundary con-
ditions are satisfied if
น
ȧx w¸ (a² — b³) y¸ w₂ (c² — a³) z
+
α
a²+b²
with similar expressions for v and w.
+
c² + a²
11. If the lines of flow of a fluid lie on the surfaces of coaxial
cones having the same vertex, prove that the equation of con-
tinuity is
dp
p + r
dt
d
d
dr
(up) + 2pu + cosec
аф
(pv) = 0.
12. Show that
x²/(akt²)² + kt² {(y/b)² + (z/c)²} = 1
is a possible form of the bounding surface at time t of a liquid.
13. The position of a point in a plane is determined by the
length r of the tangent from it to a fixed circle of radius a, and
the inclination of the tangent to a fixed line. Show that the
equation of continuity for a liquid moving irrotationally in the
plane will be
1 d² a² ¡d² 1
do
ď² &
1 do
+
+
+
dr²
r dr
r² do²
r² dr²
dp
r dr
α
+
(2
d² 1 do
drde r de
= 0.
2
Hence indicate a method of finding the motion of a liquid
in the developable surface whose edge of regression is a right
helix, pointing out any peculiarities of the motion.
14. If the velocity potential of a liquid is of the form
$=ƒ(®) F(0) x (2), where ∞, 0, z are cylindrical coordinates,
prove that the equation of continuity is satisfied if f, F, x satisfy
the three equations
χ
2
d'f df
d2F
w
+@
do 2 dw
+ (x²∞² — n³) ƒ = 0,
d02
+ n°F = 0, dx − x²'x=0,
dz²
-
where n and κ are constants; and hence show that
x
$ = ΣA cosh × (z — c) cos n (0 — a)
ΣΑ
["
cos (îr sin w — nw) dw.
B.
2
18
HYDROKINEMATICS.
15. In the motion of a liquid in two dimensions, the velocity
at any point is given by two components v, v' along the directions
which pass through two fixed points distant a from one another.
Show that the equation of continuity is
dv dv p² + p²² – a² (dv
+ +
dr dr
-
บ
dv'
V
2rx' dr'+ dr
+ +
r
0,
—
where r, r' are the distances of any point of the liquid from the
fixed points.
CHAPTER II.
ON THE GENERAL EQUATIONS OF MOTION OF A PERFECT
FLUID.
20. It was stated in the preceding chapter, that the pressure
at every point of a perfect fluid is equal in all directions, whether
the fluid be at rest or in motion. It will now be shown that this
property is the consequence of such a fluid being incapable of
offering resistance to a tangential stress.
Let ABCD be a small tetrahedron of fluid, and let p, p' be the
pressures per unit of area upon
the faces ABC and BCD.
By D'Alembert's Principle,
the reversed effective forces and
the impressed forces which act
upon the volume of fluid, together
with the pressures upon its faces,
constitute a system in statical
equilibrium. The first two vary
B

as the volume, and the last vary as the areas of the faces of the
tetrahedron; and therefore if the tetrahedron be made to diminish
indefinitely, the former will vanish in comparison with the latter.
Hence the tetrahedron will ultimately be in equilibrium under the
action of the pressures upon its faces.
Resolve the pressures upon the faces ABC and BCD parallel
to AD. Since the projections of the two faces upon a plane
perpendicular to AD are equal, the conditions of equilibrium
require that p=p', which proves the proposition'.
rest.
1 This proposition is true even in the case of viscous fluids, provided they are at
2-2
20
EQUATIONS OF MOTION.
.
•
The Equations of Motion'.
21. Let X, Y, Z be the components per unit of mass of the
impressed forces which act on the fluid; p its density, and q its
resultant velocity. Describe any imaginary closed surface S in the
fluid, and let e be the angle which the direction of q makes with
the normal to S drawn outwards.
The rate at which momentum flows into S, parallel to x,
together with the rate of increase of the component of momentum
parallel to x, of the fluid contained within S, must be equal to the
component parallel to x of the impressed forces which act on the
fluid within S, together with the component parallel to x of the
pressure upon the boundary of S.
The rate at which momentum flows into S, parallel to x, is
ff pq❜l cos edS = SS pu (lu + mv + nw) dS
by § 7.
- fff (d (pu) + d (pwe) + d (pww) } dædyds
= (pu²), (puv),
dx
dy
dz
The rate of increase of the component of momentum parallel
to a of the fluid contained within S
= [[[ d (pu) dx dy dz.
dt
The component parallel to x of the impressed forces
=
= SSS pXdx dy dz.
The component parallel to x of the pressure upon the boundary
of S, is
Whence
- Splas = - [[[dp dx dy dz.
SpldS=
dx
[[[(pX_dp) dxdydz=[[[{d(pu)+d(pu²)+d(puv)+d(puw)}{dxdydz.
-
dx
which requires that
px
dt
dp_d(pu) d (pu²)d(puv)¸d(puw)
dx dt
+
+
dx dy
+
dz
1 This method of obtaining the equations of motion is due to Prof. Greenhill.
See Encyc. Brit., Art. Hydrodynamics.
EQUATIONS OF MOTION.
21
Taking account of the equation of continuity § 8, (10) the
right hand side of the last equation becomes equal to pôu/dt,
whence
pX
dp ди
dx
P at
Two other symmetrical equations can be obtained, by consider-
ing the rate of change of momentum parallel to the other two
axes, whence the equations of motion are
du du du
+ u +v
+w
dx dy dz
dv dv
1 dp
du
X
ρ
dx
dt
1 dp
dv
Y
+ u +v
p dy
dt
Z
1 dp ___ dw
p dz
dx dy
+w
&&&&
dz
dw dw dw
+u +v +w
dt dx dy dz
(1).
These equations together with the equation of continuity
furnish four relations between the five quantities u, v, w, p, p.
22. If the fluid be an incompressible liquid, p is constant,
and the above mentioned equations together with the boundary
conditions are sufficient to determine the motion; but in the
case of a gas another equation is required, which is furnished
by means of a relation which exists between Ρ and p.
When the motion of the gas is such that the temperature
remains constant, we have by Boyle's Law the equation
where k is a constant.
p = kp
……..(2),
But when the motion is such as to cause a sudden compression
or dilatation, an increase or decrease of temperature will be
produced; and if it is assumed (as is the case with sound waves),
that the compression is so sudden that loss or gain of heat by
radiation may be neglected, the required relation is
p = kp
...(3),
where Y is the ratio of the specific heat at constant pressure to
the specific heat at constant volume¹. This quantity for all
gases has the approximately constant value 1·408.
23. The expressions on the right hand of (1) are the ex-
pressions for the component accelerations of an element of fluid;
it therefore follows that if F and ƒ be the component force and
1 This equation will be proved in the Appendix.
22
EQUATIONS OF MOTION.
acceleration in any direction, and dp/dh be the space variation of
the pressure, the equations of motion are of the form
F_1 dp
p dh
=f.
3
ρ
Hence if the axes instead of being fixed are moving with
angular velocities 0, 0, 0, about themselves, the equations of
motion will be obtained by employing the expressions for the
accelerations given in § 6, (6), and are therefore,
X
1 dp__du
du
p dx
dt
dx
+ U + V + W
du
du
dy
dz
vo₂+ we
2
1 dp
Y
p dy
हाई
dv
dv
dv
dv
+ U
+ V
+ W
dt
dx
dz
dy
1
-ωθι + κθα
3
·(4).
1 dp
dw
dw dw
dw
Z
+ U
+Ꮴ
p dz
+ W
dt
dx
dy
dz
u0₂+ və
2
•
-
24. Let us now suppose that the forces arise from a con-
servative system whose potential is V. Since p is a function of
p, we may put
V.
[dp - V₁
ρ
and the left-hand sides of (1), will be respectively equal to
dQ/dx, dQ/dy, dQ/dz. If therefore we eliminate Q by diffe-
rentiating the second equation with respect to z and the third
with respect to y, we shall obtain
૦૬ du dv
§ τη
Ət dx dx
dw
+ S. Jx
૬૦,
where, n, are the components of molecular rotation and
0 = du/dx + dv/dy+dw/dz. Eliminating ✪ by means of the equa-
tion of continuity Əp/ǝt + p = 0, and taking account of the two
other equations which may be written down from symmetry, we
shall obtain
f
ô (ફ્
§ du
21
a in
at
215
11
a. us! a usl
n dv
+
I
dw
+
p dx
P dx ρ dx
& du
n dvę dw
+
+
p dy
p pdy 'p dy
.(5).
E du
Ət
+
ρ dz
η dv, & dw
+
P dz p dz
25. It was stated in Chapter I., that in many important
problems the motion is such that a velocity potential exists.
MOLECULAR ROTATION.
23
The condition that such should be the case is, that E, n, Č should
each vanish. We shall now prove, that when the fluid is under
the action of a conservative system of forces, a velocity potential
will always exist whenever it exists at any particular instant.
Let us choose the particular instant at which a velocity poten-
tial exists as the origin of the time; then by hypotheses §, n, Š
vanish when t=0; also the coefficients of these quantities in (5),
will not become infinite at any point of the interior of the fluid;
it will therefore be possible to determine a quantity Z, which shall
be a superior limit to the numerical values of these coefficients.
Hence, n, cannot increase faster than if they satisfied the
equations
д
L
(§ + n + (), &c. &c.
P
Ət
But if §+n+= Np, we obtain by adding the above equations
ΘΩ
= 3LQ,
Ət
Q = Ae³Lt
whence
Now 2-0 when t = 0, therefore A=0; and since is the
sum of three quantities each of which is essentially positive, it
follows that §, n, must always remain zero, if they are so at any
particular instant. The above proof is due to Prof. Stokes¹.
26. There is, as was first shown by Prof. Stokes, an important
physical distinction in the character of the motion which takes
place, according as a velocity potential does or does not exist.
Conceive an indefinitely small spherical element of a fluid
in motion to become suddenly solidified, and the fluid about it
to be suddenly destroyed. By the instantaneous solidification
velocities will be suddenly generated or destroyed in the different
portions of the element, and a set of mutual impulsive forces will
be called into action.
Let x, y, z be the coordinates of the centre of inertia G of the
element at the instant of solidification, x + x', y + y', z + z those
of any other point P in it; let u, v, w be the velocities of G along
the three axes just before solidification, u' v', w' the velocities of P
relative to G; also let u, v, w be the velocities of G, u,, v,, w, the
relative velocities of P, and §, n, Y the angular velocities just
1 "On the friction of fluids in motion," Section II. Trans. Camb. Phil. Soc.
vol. VIII.
24
EQUATIONS OF MOTION.
after solidification. Since all the impulsive forces are internal,
we have
ū = U, V V, w=w.
We have also, by the principle of conservation of angular mo-
mentum, Σm {y' (w, - w') — z′ (v, — v')} = 0, &c.
·
m denoting an element of the mass of the element considered.
But u₁ = nz' — ¿y', and u' is ultimately equal to
1
-
du
x² +
dx
dy
du du
y' +
z',
dz
and similar expressions hold good for the other quantities. Sub-
stituting in the above equation, and observing that
12
12
Σmy'z' = Σm`z'x' =Σmx'y' = 0, and Σmx² = Σmy'² = Σmz”,
we have
dw dv
§ =
1-24
(dy dz
&c.
We see then that an indefinitely small spherical element of
the fluid if suddenly solidified and detached from the rest of the
fluid will begin to move with a motion of translation alone, or
a motion of translation combined with one rotation, according as
udx+vdy+wdz is, or is not, an exact differential, and in the latter
case the angular velocities will be determined by the equations
dw dv du dw
2૬ = =
2n
dy dz' dz dx
25=
dv du
dx dy
On account of the physical meaning of the quantities §, n, S,
they are called the components of molecular rotation, and motion
which is such that they do not vanish is called rotational or vortex
motion; when they vanish, the motion is called irrotational.
In the foregoing investigations, it has been assumed that the
pressure is a function of the density and also that the fluid is
under the action of a conservative system of forces; it therefore
follows that vortex motion cannot be produced, and consequently,
if once set up, cannot be destroyed by such a system of forces. We
shall presently show that the theorem is not true if the pressure
is not a function of the density. If therefore by reason of any
chemical action the pressure should cease to be a function of the
density during any interval of time however short, vortex motion
might be produced, or if in existence might be destroyed.
1
LAGRANGE'S EQUATIONS.
Lagrange's Equations.
25
27. In Lagrange's method the initial coordinates a, b, c and
the time are the independent variables, hence the equations of
motion are
dQ
ů,
dx
dQ
dy
dQ
v,
w.
dz
Multiplying the preceding equations by xa, ya, Za, where the
suffixes denote partial differentiation with respect to a, b, c, we obtain
Qa = úxa + vŸ a + wza
Qo
α
lo = tủy t
b
Qe = úx¸ + vÿ¿ + wz.
...(6).
These equations together with the equation of continuity
pJ=p₁, are Lagrange's hydrodynamical equations of motion.
Weber's Transformation.
28. Integrating the right hand side of the first of (6) between
the limits t and 0, the first term becomes.
α
xx dt
rt
úx dt
α
xx dt = = (xx) – [
1 d ft
= = UX a ― U o
2 da 0
u²dt,
where u is the initial value of u.
a
If we treat each of the other
two terms in a similar manner and put
X
" (Q + 1q²) dt,
[*
S
where q is the resultant velocity of the liquid, we obtain
dx
UXa+vYa+Wza - U o da
dx
(7).
Ux z + vYz + wzz — v。
db
dx
Uxc + vyc + wzc — w。
+vYc
dc
*
These equations together with the equation of continuity and
dx/dt = Q+¿q², give five equations for determining x, y, z, p, X ;
p being supposed to have been eliminated by means of (2) or (3).
The above equations may be expressed in a different form, for
multiplying by dJ/dx, dЛ/dx, dJ/dx, and adding, we obtain
dJ
1
dJ
u =
Ио
0
0
dxa
+vo do o
with two similar equations.
α
+ Wo
dJ
de) + dxx
• dic
(8),
26
EQUATIONS OF MOTION.
29. Multiply (7) by da, db, dc and add, and we obtain
udx+vdy + wdz — uda — v¸db — w.dc = dx…........(9).
If at any particular instant which we shall choose for the
origin of the time a velocity potential exists, uda + v¸db+w.dc
will be a perfect differential; hence if p be a function of p, dy will
also be a perfect differential, which proves that if a velocity
potential once exists, it will always exist; but if p is not a function
of p we cannot put Q V-Sp¹ dp, but must write
-1
d
rt
da
['" ($q" - √) di - ["
V) dt
1 dp dt
0
ρ
0
da
for dy/da, in which case the right hand side of (9) becomes
d
ct
dt
['" ($q² - V) at - [" (dp)
0
dt
where d denotes space differentiation. The right hand side of (9)
is no longer a perfect differential; hence uda+vdy + wdz is not a
perfect differential.
If therefore the pressure be not a function of the density, vortex
motion can be generated or destroyed in a perfect fluid moving
under the action of natural forces.
Cauchy's Integrals.
30. Eliminating Q from the last two of (6), we obtain
b
0.
b
Integrate this equation with respect to t, and let u, v, w, be
the initial values of u, v, w; then
But
Uz Xc − UcX z + V ¿Y c − VcY z + WzZc — WcZz
z
dw. dv.
db dc
du
du
du
и
α
dx
Xa+ Ya+
%as
&c. &c.
ar
dz
dy
Substituting these values of u, ur, &c., we obtain the equations
dJ
E dx a
τη
dJ dJ
+5
६,
dy
dz a
us
|
dJ
E doo
દ
dJ
Ax
τη
+ n
dy
dJ
+5
dJ
dJ
Nos
dz
dJ
+5
dy
+ 5 dz.
--
1
CAUCHY'S INTEGRALS.
27
Multiplying these equations by xa, x, x and adding, and
remembering that Jp = p., we obtain
Ροξ
P
Pon
ρ
ροζ
P
= {oxa+noxo +50°
=
Eoya+noyo+Soye
=
α
= {o²a + no²o + 50%
α
..(10).
These equations show that §, n, are always zero, if they are
initially so.
31. The equations of motion can be integrated whenever
a force and a velocity potential exist; for putting
Q
[dp
P
མ,
and multiplying (1) by da, dy, dz respectively and adding, we
obtain
ди
dQ=
=
Ət
მა
dx + ~dy +
Ət
dw
dz.
Ət
Now in the present case
dv dw
+w
dx dx
Ət
22
ди
du
du
+ u
dt
dx
d do ¿q²),
2
dx dt
where q is the resultant velocity. Integrating, we obtain
dp
ρ
аф
dt
+V+ + { q² = F (t) ………….
where F is an arbitrary function.
.(11),
32. DEF. A vortex line is a line whose direction coincides
with the direction of the instantaneous axis of molecular ro-
tation.
The differential equations of a vortex line are thus
dx dy dz
η Š
28
EQUATIONS OF MOTION.
Clebsch's Transformation¹.
33. When a velocity potential does not exist, a first integral
of the general equations of motion can be obtained by means of a
method which depends upon the analytical theorem, that if u, v, w
are any given functions of x, y, z it is always possible to determine
three quantities 4, λ, x, such that
udx +vdy+wdz = dp + λdx
........(12).
In order to prove the theorem, let u', v', w', o be four quantities,
such that
u=u'+$x v=v' + $₂, w = w' + $₂·
These equations involve three relations between the four
quantities u', v', w', p and are therefore insufficient to determine
them as functions of u, v, w; we may therefore assume any relation
between u', v', w' which may be convenient. Let us therefore
suppose that
u' (w'₂ — v'₂) + v′ (u'z — w´x) + w′ (vx' — v'„) = 0.
This is the condition that u'dx + v'dy + w'dz should have an
integrating factor, we may therefore put this quantity equal to
Ady which proves the proposition. It therefore follows that,
аф dx
u = +λ
do dx do
dy'
v = +λ
dx dx' dy
W
dx
+ λ
dz dz
......(13).
The components of molecular rotation are given by the
equations
2§ = λy Xz — λzXv
2η = λ.χα - λ.χ.
25=λxXx − λyXx
(14).
The form of these equations shows that the vortex lines are
the intersections of the surfaces λ = const.,
du d do
dt
dx dt
+ x dx
const., x = const.
dx) + dx xx - dxxx.
dt
dt
Now
dt Xx
Therefore
ди
d/do
dx
du
dv
dw
+λ
+u + v
+ w
Ət
dx dt dt
dx
dx
dx
θλ
дж
+
It Xx
.(15).
Ət
¹ Crelle, vol. LVI. p. 1. See also Hill, Quart. Journ. vol. xvII; Trans. Camb.
Phil. Soc. vol. xiv. p. 1; Phil. Trans. 1884, p. 363; Proc. Lond. Math. Soc.
vol. xvi. p. 171.
!
CLEBSCH'S TRANSFORMATION.
29
Putting
H = − Q+
аф
dx
+ λ
+ 19 . . . . . . . .
.(16),
dt
dt
and substituting the values of du/ot and dQ/dx from (15) and (16)
in (1), we obtain
анал
+
ах
dx
It Xx
Ət
λ. = 0 ...
.(17),
with two similar equations.
Multiplying by §, ʼn, § and adding, we obtain
dH ан dH
τη + m
dx dy dz
=0..
.(18).
If ds be an element of a vortex line, and o be the resultant
molecular rotation, the operator is equal to wd/ds, whence in-
tegrating along a vortex line, we obtain
dp аф dx
P
F (t, λ, x)…………………….(19).
+V+ +λ + {q² = F (t, x, x).
dt dt
Writing for a moment P =əλ/ət, R= əx/ǝt and eliminating H
from (17), we obtain
P₁Xx - R₁λx - P₂Xy + R₂λ₁ = 0
P₂X₂ – R₂λ
Rλ — P₁X² + Rλ = 0
P xX₂ — R¿λ₂ — P₂ Xx + R₂λ = 0.
-
P₂Xx
Multiplying these equations in order by A, A, A, and adding
and taking account of (14), we obtain
¿Px+ŋP₂+ ¿P₁ = 0
………..(20).
If x, y, z be any point on the surface λ=A, where A is an
absolute constant, and if {/w, n/w, c/w be the direction cosines of
the vortex line at this point; equations (14) and (20) show that this
vortex line lies on the surfaces λ= A and λ + dx/dt. dt = A, which
is impossible unless ǝx/dt = 0. Similarly dx/dt = 0; whence the
surfaces λ and x and therefore the vortex lines are always composed
of the same elements of fluid. This important theorem was first
established by Helmholtz¹.
X
Hence it follows from (17) that H¸ Ä, H, are each equal to
zero, and therefore H is a function of the time alone; whence the
pressure is determined by the equation
dp аф
P
dx
+V+ +λ + } q² = F(t) ....
dt dt
1 Crelle, vol. LV. and Phil. Mag. (4) vol. xxxi. p. 485.
.(21).
30
EQUATIONS OF MOTION.
T
34. We can now show that in the case of a liquid, the
integral
SSSS
!!! C didx dyds
+ V) dt dx dy dz..........
.(22),
is a maximum or minimum, where the value of p/p + Vor- Qis
given by (21), and the time remains invariable.
For SQ = usu + v&v + wdw+
αδο, αχ
dt dt
+ dx sx + λ
ddx
dt'
and
Therefore
Su= dsp + dx sx +λ
dx dx
dx
ffffusu dt dx dy dz = ſſſu (84 + λdx) dt dy dz
ddx
+
-「「「「{wxbc-
d
du
δλ
dx
(λκ) δχ
dx
(ra) ôx − a b atdadyde.
Omitting the triple integrals which refer to the boundary we
see that the first three terms of 8Q give rise to the terms
SSSS{(UXx + vXv + wxz) dλ − (uλ„ + vλ, + wλ) dx
which
δλ
= SSS {(@x_dx) or - (3ì
where
Ət
dt
0 =
Ө
-0 (84 +λdx)} dtdx dydz,
dr
dt
dx−0 (8$+λôx)} dtdx dydz,
du dv dw
+ +
dx dy dz
*
Also the last three terms of SQ (omitting triple integrals) give
rise to
Whence
δλ
αλ
dt
[[[[{dx sx - dr sx } dtdx dy dz.
dt
[ [ [ [ 8Q d t d x d y d z = [[]]{x x
θλ
δλ
Ət
Sx − 0 (84+λ8x)} dtdx dydz
+ triple integrals.
In order that the quadruple integral should vanish, we must
have 0 = 0, ǝx/dt = 0, dx/dt = 0, which by virtue of the equation of
continuity and § 33 is obviously the case.
KINETIC ENERGY.
31
On the Application of the Principles of Energy and Least Action.
35. Let S be any imaginary closed surface, which is fixed in
the fluid. The work done during a small interval St upon the
liquid contained within S, by the impressed forces which act
upon its mass, together with the work done by the pressure upon
the boundary of S, must be equal to the increase during the
interval St of the kinetic energy of the liquid contained within S,
together with the kinetic energy which, during the same interval,
flows into S across the boundary.
The work done by the impressed forces
dV dV dV
+ v +w
dx
-- [I]p (u dr to dy
SSSP
dz
St dx dydz.
The work done by the pressure upon the boundary
SS p (lu + mv + nw) St dS
=- SSS (u dp
+ v dy
dp + wdp) St dx dy dz,
by § 7. Hence the total work done
dQ dQ
dz
dQ
= fffp (u de + v de + w 4) &t dx dyds.
=SSSP
dx dy dz
Let T be the kinetic energy per unit of mass, so that
T = {(u² + v² + w²).
The increase in the kinetic energy of the liquid contained
within S
=fffd (TP) St dadyde.
dt
The amount of kinetic energy which flows into S
=
= SS pT (lu + mv + nw) St dS
d
d
= SSS {a e (puT) +
d
dy
(pvT)+ (pwT)} St dx dydz.
dx
dz
Taking account of the equation of continuity §9 (10) the total
increase in the kinetic energy
эт
Ət
= [JJp T St dx dy dz.
32
EQUATIONS OF MOTION.
Whence
dQ
dQ
SSSP (37 - u do + v do
which requires that
Ət
น
dx
dy
dQ
W
St dx dy dz= 0
dz
dQ
ƏT
dQ
dQ
U
+ v
It
+w
dx dy dz
.(23).
If we substitute the values of u, v, w from (13), we find that
dT
dt
=(20
d
dx
d d аф
+ v +w
dy
dz dt
dx
+λ
dt/
αλ
-
- (uλ₂+ vλ + wλ) dx.
dt
+(UXx+vX₂+ WX«) dt
The last two terms vanish by § 33, whence (23) becomes
d d
U + v
dx dy
+w
d
-dz
:) (9-1
do
λ
dt
dt
dx)
= 0.
Now if ds be an elementary arc of a stream line u = qdx/ds, &c.,
and the operator is therefore equal to qd/ds. Integrating along a
stream line, and restoring the values of Q and T, we obtain
p
+ V + 1 q² +
аф
λ
P
dt
+ x d x
dt
= F(t).
36. The equations of motion may be deduced, as Mr Larmor
has shown, by means of the Principle of Least Action combined
with the Lagrangian method.
Let x, y, z be the coordinates at time t of an element of fluid
whose initial coordinates are a, b, c; the Principle of Least Action
requires that
SSSS {\p (x² +ÿ²+¿²) — Vp} dt dx dy dz
should be a maximum or minimum subject to the condition that
d (x, y, z) _ Po
J =
=
Ро
d (a, b, c) P
where the time of the motion is constant.
Hence if represent an undetermined function of x, y, and z,
we must have
©SSSS
− −
© Sƒƒƒ { † (œª² + y² + ¿²) − V − λ
Taking the variation of the first
(εδώ δύ ἐδέ)
+ ÿòÿ + žôê) – (d
SSSS { (àðã
d (x, y, z)
dt dadbdc=0.
d (a, b, c)
two terms, we obtain
dv
dv
Sx +
Sy +
dy
dz
&z) } d
Sz dt da db dc.
LEAST ACTION.
Integrating by parts and omitting the triple integrals, this
dV
dV
33
− − SSSS {(i + dr) 8x + (ÿ + dy ) 8y + (+ d) Se} dit dadb do.
Sx
--
8=
dz
If M₁, M₂, M¸ be the minors of dx/da, dx/db, dx/dc in J
8
d (Sx, y, z)
d (a, b, c)
=
dSx
dSx
M₁
1 da
+ M.
2 db
whence, omitting triple integrals,
SSSS
λ
[SS] x d (da, y, z) dt da db dc = -
b, c)
αλ
dSx
+ M.
8 dc'
1
dM
SSSS [x (dM+dM+ аM
λ
da db
dr
+(M + M. + M₂ do)]
(M, da
2 db
dc
3
Sx dt da db dc.
The first term in brackets vanishes, and the second term is
equal to Jɗ/dx,
whence
-SSSS ~ε
λδ
d (x, y, z)
d (a, b, c)
dt da db dc
dr
Sx
- {dh &z + da by + dr de} Jdidadbdo.
= SSSS
бу Sz
dx dy dz
Hence the conditions of the problem require that
dV
ä +
Ро
αλ
0
dx
P dx
dV
Ро
Po dr
ÿ +
= 0
dy
d V
p dy
.(24).
z +
Ро
dr
0
dz
p dz
Now, ÿ, are the component accelerations of the element
whose coordinates are x, y, z, and are therefore equal to du/dt,
Əvət, and aw/ət respectively; and when we interpret - λp, which
must represent the pressure, equations (24) are the equations of
motion in the ordinary form.
On Steady Motion.
37. When the motion is steady du/dt, dv/dt and dw/dt are
each zero.
In this case the general equations of motion can be
integrated without having recourse to Clebsch's transformation.
It will however be necessary to distinguish between irrotational and
'rotational motion.
B.
3
34
}
EQUATIONS OF MOTION.
The general equations of motion may be written,
ди
du dq²
dQ
ди
R8 EB Ek
+ 1/ · 2v5+ 2wn:
dt
dx
dv
+ 1/2
dq³ _ 2w€ + 2u5=
dQ
dt
dy
&& &&
..(25).
dy
ди аги dq²
dQ
+ 1/ 2un + 2v§ =
dz
Ət dt dz
When the motion is steady and irrotational ú, v, ŵ, §, n, § are
each zero; whence, multiplying by dx, dy, dz, adding and inte-
grating, we obtain
or
Q = 1 q² — C,
dp
+ V + ¿q² = C ....
P
.(26).
In this case the quantity C is evidently an absolute constant.
When the motion is rotational, let ds be an element of a stream
line, then
dx
dy
dz
u = q Ts
v = q
ds
>
w = q Ts
ds
Multiplying the general equations by u, v, w and adding,
we obtain
dQ dq
=
ds ds'
whence
dp
+ V + ¿q² = A.....
P
.(27).
This is Bernoulli's Theorem.
Since we have integrated along a stream line, the quantity A
is not an absolute constant, but a function of the parameter of a
stream line: in other words if y = const., x= const. be two surfaces
whose intersections determine the stream lines, A is a function
of y and
X.
38. Let us now consider the steady motion of a liquid¹ which
is symmetrical with respect to the axis of z. The vortex lines
will evidently be perpendicular to every plane through the axis
of z, hence by § 19 (41) the molecular rotation w will be determined
by the equation
2w
=
du dw
dz do
1 Stokes, "On the steady motion of incompressible fluids," Trans. Camb. Phil.
Soc. vol. vi. p. 439.
STEADY MOTION.
35
Substituting for u and w their values in terms of Stokes'
current function , § 16 (25), we obtain
ď² f
d²
1 dy
+
+ 2ww = 0 ....
dz2
do 2
@ dw
..(28).
dQ
The equations of motion are
du du
d (q²)
= u
+w.
dw
da
dz
1-124
+ 2ww,
do
d Q
dw dw
d (q²)
dz
= U +w
da
1/4
2uw.
dz
dz
Eliminating Q-19², we obtain
dw do /du
И +w +w + dw)
dw dz
dw
The equation of continuity § 9 (13) is
=
= 0.....
dz
.(29).
dot
du dw
+ + = 0,
dz
@
whence (29) becomes
dw
dw
UW
u do
+w
=
0,
dz
®
d
d
or
И +w
da
= 0 ...
dzl w
.(30).
Substituting the values of u, w and w in terms of , (30)
becomes
dy d dy
dz
de do
dz
dy d d) (1 1 d²
do de) { m² (dy + dy
1 dy
dz² do²
a da
=0...(31).
A first integral of this equation is evidently
+
dz²
ď³, ď² 1 dự
da2 w dw
2
@²ƒ (†)
.(32),
whence by (28)
.(33).
2w+of (¥) = 0.......
When the motion takes place in two dimensions, we shall, in
exactly the same way, arrive at the equations
αζ
αζ
W
0,
dx
dy
ď²
and
d
d²v
dx²
+
dy²
+25=0......
…………….(34),
whence
dy d
dy d
dy dx
dx dy
dx dy) \ dx²
:) (
dx² + dy²
3-2
36
EQUATIONS OF MOTION.
"
a first integral of which is
d
ď² +
d
dx² + dy²
=
=ƒ (¥) ………….
..(36),
whence by (34)
(37).
25+ƒ(¥)=0 .....
39. The subject of the steady motion of a liquid has been
treated in the following manner by Clebsch¹.
Let b and c be any functions of x, y, z and t; then if the
suffixes denote differentiation with respect to x, y and 2, we may
evidently put
u=b,c. – b_c, v = b_c – bạc, v= bạc, — b_c......(38),
for these values of u, v and w satisfy the equation of continuity.
From (38) we deduce
ub¸+vb¸ + wb₂ = 0
UCx + VC₂ + Wcz
vCy
= 0
.........
.(39),
hence the stream lines
are the intersections of the surfaces
Putting
2T=u²+v²+w³,
b = const., c = const.
and multiplying equations (25) by dx, dy, dz respectively and
adding, we obtain
dQ − dT = M¸dx + M₂dy + M,dz ....
M₁ = − v (v₂ − u„) + w (u₂ — w₂) - 2v5 + 2wn,
.(40),
From the values of M₁,
1
M¸u+M¸v+M¸w = 0
(41).
where
with similar expressions for M, and M.
M₂, M, it follows that
2
2
Eliminating u, v, w from (39) and (41), we obtain
M₁, b₂, c₂ = 0.
cx
M₂, by, cy
8'
M., b₂, Cz
Hence we may put
T
M₁ = Bb₂+Ccx
1
М M₁ = Bb,+Cc,
2
M₁ = Bb₂+ Cc₂
M¸
3
.(42),
where B and C are quantities whose values we shall hereafter
determine; (40) may now be written
¹ Crelle, vol. LIV. p. 293,
or
CLEBSCH'S TRANSFORMATION.
dQ – dT = B (b₂dx + b₁dy + b₂dz) + C' (c,dx+c,dy +c,dz),
dQ-dT Bdb+ Cdc......
-
.(43).
37
Since the left-hand side of (43) is a perfect differential, the
right-hand side must be so also, whence if F be a function of b and
c, we must have
and therefore
dF
B =
dF
db,
C:
dc
Q− T= F(b, c)
is an integral of the equations of motion.
……..(44),
.(45)
When the motion is irrotational, M₁, M., M, and therefore B
and C are each zero, and therefore F is an absolute constant.
40. We must now find the values of B and C. If we sub-
stitute the values of u, v and w from (38) in the expression for T
and differentiate partially, we shall obtain
dT
dbx
vc₂+ wc,
ข
dT
dby
dT
dbz
d dT
whence
dx db.
x
+
wcx + ucz
uc₁ + vcx
d /dT
dy (db) + 2(d)
dz db₂
Cx (W₂ — v₂) — Cy (Uz — wx) — Cz (Vx — Uy)
=-2 (c+c+c5).
From the first two of equations (42), we obtain
Bw M,c,- M₂c,
=
= 2c, (− v5+ wn) — 2c, ( − w§+u5)
=2w (c‚§+c,n+c₂5)
/dT d/dT
de (db) + d (db) + d (db)
dy
dz
dF
- B =
db
..(46).
by (39). Therefore
d/dT
dx
Similarly
dx
dy
de (de) + d (da) + d (de)
dT
dF
C=
....(47).
dz
dc
•
3-
38
EQUATIONS OF MOTION.
41. By means of the preceding equations it can be shown
that the conditions of steady motion make
SSS (T — F) dx dy dz
a maximum or minimum.
For
and
Ꭲ
dT
ST
=
Sb₂+ &c.,
dbx
[SS
db
[[] dl sb. dx dydz = [] dl sbdydz - [[[ d (d) Sb dx dy dz.
dbx
lab,
db
2
Whence, omitting the double integrals which refer to the
boundary, we obtain
[[[ &T dadyde = -
ST dx dz
-
d/dT ddT
dy db
[[] { 2 (db) + 2(17) +
dx
/dT d/dT
III (d (d²) + d (de) +
-[[]
dx dc
dF dF
dy dc
Sc
/dT
d
dz db
(17)} 86 dadyds
dx dz
d
/dT
(de)} bededyds
dx dz
{d8 86 + dr 8c } dadyds
db
by (46) and (47); whence
dc
fff S (T — F) dx dydz = 0,
which proves the proposition.
Impulsive Motion.
42. Let u, v, w and u', v', w' be the velocities of a fluid, just
before, and just after the impulse; p the impulsive pressure. Then
if S be any closed surface, the change of momentum parallel to
x, of the fluid contained within S, must be equal to the component
parallel to x of the impulsive pressure upon the surface of S.
Hence SSS p (u'-u) dxdy dz=-ff pldS
Therefore
dp
=-SSS
--[[] de dady dz.
dp
dx
p(u' — u) = -
Similarly
p (v' — v) = ·
p (w' — w)
SIS SIS SI
dp
(48).
dy
dz
Multiplying by dx, dy, dz and adding, we obtain
dp
P
=
(u' — u) dx + (v' — v) dy + (w' — w) dz…………….(49).
EXAMPLES AND APPLICATIONS.
39
In the case of a liquid p is constant, whence differentiating
(48) with respect to x, y, z respectively, and taking account of the
equation of continuity, we obtain
V²p = 0.
If the liquid were originally at rest it is clear that the motion
produced by the impulse must be irrotational, whence if o be its
velocity potential, we must have
p=- pp
EXAMPLES AND APPLICATIONS.
.(50).
43. A mass of liquid whose external surface is a sphere of
radius a, and which is subject to a constant pressure II, surrounds a
solid sphere of radius b. The solid sphere is annihilated, it is
required to determine the motion of the liquid.
It is evident that the only possible motion which can take
place is one in which each element of liquid moves towards the
centre, whence the free surfaces will remain spherical. Let R', R
be their external and internal radii at any subsequent time, r the
distance of any point of the liquid from the centre. The
equation of continuity is
d
whence
dr (2² V) = 0,
r²v = F(t).
The equation for the pressure is
whence
1 dp
dv
dv
ข
p dr
dt
dr
F" (t)
dv²
- 1/2
22
dr
Ρ
-
A+
F" (t)
- 1 v²,
Р
r
when r=R', p= II, and when r=R, p=0, whence if V, V' be
the velocities of the internal and external surfaces
also
Π
P
· F' (t) (1
1 1
— § (V²² — V²).
R R
Since the volume of the liquid is constant,
R'³ — R³ = a³ — b³ = c³,
d
F' (t) = 2/4 (RV),
dt
40
EQUATIONS OF MOTION.
whence
II
d
V
P
dR
(R®V)
1
1
(R³ + c³)³ R
- Į V²
R4
(R³ + c³)#
-1}.
Putting z = RV, multiplying by 2R2 and integrating, we
obtain
1
ap
II (R³ – b³) – y²
pR*
(R³ + c³) 3-R)
Ꭱ
>
which determines the velocity of the inner surface.
If the liquid had extended to infinity, we must put c∞, and
we obtain
211
3p
(bº — Rº) = R³ (dR)*,
>
whence if t be the time of filling up the cavity
R* dR
=
0
211 √b³ – R³
г (4)
=b
πρ
V611 г()
611 1 (1)
The preceding example may be solved at once by the Principle
of Energy.
The kinetic energy of the liquid is
r²v²dr = 2πpV2R*
2 TP [H
R'
·R' dr
R
Ꭱ .
Ꭱ
p²
1
= 2πp V²R¹
2πpV³R*
1
R
(R³ + c³)
The work done by the external pressure is
α
4πΠ r²dr = ‡IIπ (a³ – R'³)
Ꭱ
=
= {IIπ (b³ — R³),
1
1
whence
}II (b³ — R³) = V²R*p
R
(R$ + c³)
44. The determination of the motion of a liquid in a vessel of
any given shape is one of great difficulty, and the solution has
been effected in only a comparatively few number of cases. If,
however, liquid is allowed to flow out of a vessel, the inclinations
of whose sides to the vertical are small, an approximate solution
may be obtained by neglecting the horizontal velocity of the
41
I
HYPOTHESIS OF PARALLEL SECTIONS.
liquid. This method of dealing with the problem is called the
hypothesis of parallel sections.
Let us suppose that the vessel is
kept full, and the liquid is allowed to
escape by a small orifice at P. Let h
be the distance of P below the free
surface, and z that of any element of
liquid. Since the motion is steady,
the equation for the pressure will be

p
gz + {v² = C.
P
P
Now if the orifice be small in comparison with the area of the
top of the vessel, the velocity at the free surface will be so small
that it may be neglected; hence if II be the atmospheric pressure,
when z=0, p = II, v=0 and therefore CII/p. At the orifice
p=II, z=h, whence the velocity of efflux is
v=√2gh,
and is therefore the same as that acquired by a body falling from
rest through a height equal to the depth of the orifice below the
upper surface of the liquid. This result is called Torricelli's
Theorem.
45. Let us in the next place suppose that the vessel is a
surface of revolution, which has a finite horizontal aperture, and
which is kept full¹.
Let A be the area of the top AB of the vessel, U the velocity
of the liquid there; let K, u; Z, v be similar quantities for the
aperture CD, and a section ab whose depth below AB is z:
also let h be the depth of CD below AB.
The conditions of continuity require that
AU= Ku= Zv,
and since the horizontal motion is neglected, the
equation for the pressure is
A
B

1 dp
ρ
dv
dv
g
V
dz
dt
dz'
Now U and u are functions of t alone, whilst Z
is a function of z only, whence
dv A dU K du
dt Z dt = Z dt'
1 Besant's Hydromechanics.
a
C
D
42
EQUATIONS OF MOTION.
..
whence
P=F(t)+gz – A
P
when z = 0, p = II, v = U, therefore
dUf² dz
v²,
dt. Z
II
· F (t) − 1 U²,
P
་་
Z-¹ dz,
–
when z=h, p= II, v=u, whence if a =
II
ρ
whence
² = F (t) +gh — AaŮ – žu³,
AaÙ = gh + 1 (U² — u³)
A2
= gh+ U² (1 −
K2
σ(1
Putting (A/K)²-1=2m, 2√ghm = aa, and integrating, we
obtain
gh C-e-at
گا
U
=
m C + εat'
where C is the constant of integration. Now initially U=0 since
the motion is supposed to begin from rest, therefore C=1, and we
obtain
U =
gh
Vỏ tranh fat
m
'gh tanh tghm|a.
m
The velocity of efflux is
u =
√(1+2m)
gh
tanhtghm/a.
m
After a very long time has elapsed tanh t√ghm/a becomes
equal to unity, and if K be very small compared with A, m = ∞,
and we obtain Torricelli's Theorem
u = √2gh.
EXAMPLES.
43
EXAMPLES.
1. A FINE tube whose section k is a function of its length s, in
the form of a closed plane curve of area A, filled with ice is moved
in any manner. When the component angular velocity of the
tube about a normal to its plane is N, the ice melts without change
of volume. Prove that the velocity of the liquid relatively to the
tube at a point where the section is K at any subsequent time
when a is the angular velocity is
2Ac
(2 − w),
K
where 1/c=k¹ds, the integral being taken once round the tube.
2. A centre of force attracting inversely as the square of the
distance is at the centre of a spherical cavity within an infinite
mass of liquid, the pressure on which at an infinite distance is a,
and is such that the work done by this pressure on a unit of area
through a unit of length, is one half the work done by the attractive
force on a unit of volume of the liquid from infinity to the initial
boundary of the cavity; prove that the time of filling up the cavity
will be
πα
♡
2 − (³)*},
a being the initial radius of the cavity, and p the density of the
liquid.
3. In the case of the steady motion of a gas issuing symmet-
rically and subject to no forces, neglecting changes of temperature;
prove the following relation between the velocity v and the
distance r from the centre;
4πvr² = Q exp (v² – u²)/2k,
where Q is the quantity of gas that issues per unit of time, k is
the constant ratio of the pressure to the density, and u is the
velocity at points where the pressure is k.
4. In the steady motion in one plane of a liquid under the
action of natural forces, prove that
vv³u — uv²v = 0,
where
▼² = d²/dx² + d²/dy³.
44
EQUATIONS OF MOTION.
5. Jets of water escape horizontally from orifices along a
generating line of a vertical cylinder kept always full. Show that
(to axes inclined at an angle π to the vertical) the equation of the
lines of equal action for unit mass of water is of the form
x*+y³ = a*.
Show also that the line of equal time for particles of water
issuing simultaneously from the orifices, is the free path of the
water which leaves the vessel by an orifice at a depth below the
surface equal to that time.
6. A cistern discharges water into the atmosphere through a
vertical pipe of uniform section. Show that air would be sucked
in through a small hole in the upper part of the pipe, and explain
how this result is consistent with an atmospheric pressure in the
cistern.
7. A mass of liquid is moving so that the velocity at any point
is proportional to the time, and the pressure is given by
p/p= µxyz — 1ť² (y²z² + z²x² + x²y²);
prove that this motion may have been generated from rest by finite
natural forces independent of the time, with suitable boundary
conditions: and show that if the direction of motion at every point
coincides with the direction of the impressed force, each particle of
liquid describes a curve which is the intersection of two hyperbolic
cylinders.
8. Water is revolving with angular velocity w in a smooth
fine circular tube of radius a which it completely fills, and which
rests on a horizontal plane. If the tube be made to revolve with
uniform angular velocity w' about a pivot O in its plane, show that
the absolute angular velocity of the water round the centre C of
the tube is unaltered. Also if be the average pressure of the
water throughout the tube, show that the mean pressure of the
water for a section through any point P of the tube is a + μaco" cos 0,
and that the resultant pressure on the tube at P per unit of length
is mw/µa+maw² + 2mcw" cos 0, where 0 is the angle between OP
and OC produced, c=OC, m is the mass of water which would
fill a unit length of the tube, and μ that of a unit volume of
water.
12
EXAMPLES.
45
9. Prove that the equations of motion of a fluid referred
to moving axes may be expressed in the form
du du du du
+ u +v + w
dx dy dz
1 dp
p dx
x+
dt
-
2vw,+2ww₂
(w¸² + w¸³) x − (w¸− ∞¸w,) y + (ŵ₂+w¸w,) z = 0,
2
and two similar equations: where u, v, w are the component
velocities of the fluid relative to the moving axes whose component
angular velocities are w₁, w₂, w..
10. A solid sphere of radius a is surrounded by a mass of
liquid whose volume is 4πc³/3, and its centre is a centre of attractive
force varying directly as the square of the distance. If the solid
sphere be suddenly annihilated, show that the velocity of the inner
surface when its radius is x, is given by
x*x³ {(x³ + c³)* − x}
3p +9
211 2
+
не
- 3 µcº) (a³ — a³) (c³ + a³) *‡,
−
where p is the density, II the external pressure and μ the absolute
force.
11. Prove that if a be the impulsive pressure, o, o' the
velocity potentials immediately before and after an impulse acts, V
the potential of the impulses,
@+pV + p (p − p) = const.
12. A mass of compressible fluid is arranged in concentric
spherical layers round a point under its own gravitation. Show
that if radial vibrations be set up, the displacement z of an element
is given by
1 dez
kypr¹
d'z
dr²
dp dz
dp
keypy + de- dr + (2 dr + 2) dr-2 (2-y do + 1) =
ρ dr γ r
the pressure and density being connected by the equation p = kp¼,
and p in the differential equation being the density of the element
when at rest.
13. If p denote the pressure at any point of a liquid moving
irrotationally in two dimensions, under the action of a conservative
system of forces, prove that
2
V² log V² p = 0.
14. The surface of a vessel consists of two equal right cones,
height 2c, with coincident bases; it is fixed with its axis vertical
and filled with water to half way up the axis of the upper cone, the
46
EQUATIONS OF MOTION.
air above this level being initially at atmospheric pressure and the
vessel closed. The water flows out of the vessel from a ring of
apertures on the level of bisection of the axis of the lower cone.
On the hypothesis of parallel sections, obtain a differential equation
for the velocity of efflux, while the free surface is above the
midway point, and show that one equation to find its maximum
value in this stage is
u³ [1 — {c/(2c — x)}*] — 2g (c + x) = 2π [{c/(2c − x)}³ — 1]p¯¹‚
where x =
height of surface above midway point.
15. If the motion of a homogeneous liquid be given by a
single valued velocity potential, prove that the angular momentum
of any spherical portion of the liquid about its centre is always zero.
16. Homogeneous liquid is moving so that
u =yx+ay, v = ßx — yy, w= 0,
and a long cylindrical portion whose section is small and whose axis
is parallel to the axis of z is solidified and the rest of the liquid
destroyed. Prove that the initial angular velocity of the cylinder is
BB - Ax - 2Fy
A+B
where A, B, F are the moments and products of inertia of the
section of the cylinder about the axes.
17. Liquid is contained between two parallel planes; the free
surface is an elliptic cylinder whose axis is perpendicular to the
planes, and the semi-axes of whose section are a, b,. All the
liquid within a confocal elliptic cylinder, the semi-axes of whose
section are a, b, is suddenly annihilated; prove that if II be the
pressure at the outer surface, the initial pressure at any point of
the liquid is
II
log (a+b)—log (a₂+b₂)
log (a+b₁) — log (a+b₂)'
where a and b are the semi-axes of a confocal cylinder through the
point.
18. Fluid is contained within a sphere of small radius; prove
that the momentum of the mass in the direction of the axis of x is
greater than it would be if the whole were moving with the
velocity at the centre by
Ma²
Po
x U x
5p
Pelle + Polly + palla +$pvz
:
EXAMPLES.
47
19. Obtain by means of Clebsch's transformation, § 39, the
equations (33) and (37) of § 38.
20. Prove that when the motion of a liquid is steady, it is
possible to draw a series of surfaces P= const. each of which shall
be covered with a network of stream lines and vortex lines. Prove
also that at every point of such a surface
dP
=
dn qw sin e,
where q and w are the resultant velocity and molecular rotation,
and e is the angle between their directions.
21. A cylindrical vessel of any form which is rotating about its
axis, is filled with liquid which is rotating as a rigid body. Prove
that if the cylinder be reduced to rest, the resulting motion of the
liquid will be steady.
22. If the motion of a liquid be referred to axes moving with
angular velocities 0,, 0, 0, about themselves, prove that the com-
ponents of molecular rotation are determined by the equation
8
dé
dt
8
- no₂+ ço₂ + U
αξ
+ V
dé
+W
=
2
dx
dy
dě du du du
τη +5 √z'
dz dx dy
with two similar equations; where u, v, w are the component
velocities of the liquid parallel to the moving axes, and U, V, W
are its component velocities relative to these axes.
CHAPTER III.
ON SOURCES, DOUBLETS AND IMAGES.
46. WHEN the motion of a liquid is irrotational and symmet-
rical with respect to a fixed point, which we shall choose as the
origin, the value of 4 at any other point P is a function of the
distance alone of P from the origin; and Laplace's equation
becomes
Therefore
and
d², 2 dp
dr²™ r dr
= 0.
m
$ =
аф m
dr
p2
The origin is therefore a singular point, from or to which the
stream lines either diverge or converge, according as m is positive
or negative. In the former case the singular point is called a
source, in the latter case a sink.
The flux across any closed surface surrounding the origin is
I'm cos e
ffade as
as = [["
Im
dr
= 4πη,
p2
ds = m [[an
where do is the solid angle subtended by dS at the origin, and e
is the angle which the direction of motion makes with the normal
to S drawn outwards.
The constant m is called the strength of the source.
VELOCITY POTENTIAL OF A DOUBLET.
49
P
47. A doublet is formed by the coalescence of an equal source
and sink. To find its velocity potential; let there
be a source and sink at S and H respectively, and
let O be the middle point of SH, then

$
m
M
+
SP HP
mSH cos SOP
OP
H
Now let SH diminish and m increase indefinitely, but so that
the product m. SH remains finite and equal to μ, then
&=
μ cos SOP
2.2
με
23
if the axis of z coincides with OS.
Hence the velocity potential due to a doublet is equal to the
magnetic potential of a small magnet whose axis coincides with
the axis of the doublet, and whose negative pole corresponds to
the source end of the doublet.
48. The velocity potential due to a sheet of doublets of
strength m per unit of surface, which is such that the axis of each
doublet coincides with the direction of the normal to the sheet at
the point at which it is situated, is
$ =
- الله
Im cos e
p2
ds
If m be constant,
- ffmdQ.
$ · Ω.
Hence the velocity potential due to a doublet sheet is equal to
the magnetic potential of a simple magnetic shell of strength
m.
49. When the motion is in two dimensions, and is sym-
metrical with respect to the axis of z, Laplace's equation becomes
Therefore
ď²
dr²
+
1 do
r dr
= 0.
& = m log r,
do
m
r
dr
B.
4
50
SOURCES, DOUBLETS, AND IMAGES.
where r is the distance of any point from the axis. This value of
represents a line source of infinite length, whose strength per
unit of length is equal to m.
If be the current function,
Therefore
m
1 dự
r
r do
зва
У
XC
==
тө
= m tan¹
The velocity potential due to a doublet in two dimensional
motion is
$ = m log SP - m log HP
SH
m cos SOP
OP
μ cos SOP
γ
अरे
ᏞᏆ
2.2 ·
2
Theory of Images.
50. Let H₁, H, be any two hydrodynamical systems situated
in an infinite liquid. Since the lines of flow either form closed
curves or have their extremities in the singular points or bound-
aries of the liquid, it will be possible to draw a surface S, which is
not cut by any of the lines of flow, and over which there is there-
fore no flux, such that the two systems H₁, H, are completely shut
off from one another.
2
The surface S may be either a closed surface such as an ellip-
soid, or an infinite surface such as a paraboloid.
If therefore we remove one of the systems (say H₂) and sub-
stitute for it such a surface as S, everything will remain unaltered
on the side of S on which H, is situated; hence the velocity of
the liquid due to the combined effect of H, and H, will be the
same as the velocity due to the system H, in a liquid which is
bounded by the surface S
2
1
1
2
The system H₂ is called the image of H, with respect to the
surface S, and is such that if H, were introduced and S removed,
there would be no flux across S.
2
The method of images was invented by Sir William Thomson,
IMAGE OF A SOURCE IN A SPHERE.
51
and has been developed by Helmholtz, Maxwell and other writers';
it affords a powerful method of solving many important physical
problems.
51. We shall now give some examples.
Let S, S be two sources whose strengths
are m. Through A the middle point of SS'
draw a plane at right angles to SS'. The
normal component of the velocity of the
liquid at any point P on this plane is

m
m
cos PSA +
SP2
cos PS'A=0.
S' P²
A
Hence there is no flux across AP. If therefore
be any
point on the right-hand side of AP, the velocity potential due to
a source at S, in a liquid which is bounded by the fixed plane AP,
is
m
m
φ
SQ S' Q
Hence the image of a source S with respect to a plane is an
equal source, situated at a point H on the other side of the plane,
whose distance from it is equal to that of S.
52. To find the image of a source placed outside a sphere.
P

онм
S
Let S be the source, O the centre of the sphere, a its radius,
OS = ƒ, POS= 0, µ = cos 0.
Then, if ⚫ be the velocity potential due to the source,
m
m
Φ
SP
(r² — 2ƒrµ + ƒ²)} *
1 Helmholtz, Crelle, vol. LV. 1858; Thomson, Reprint of papers on Electricity
and Magnetism, p. 52; Maxwell, Proc. Roy. Soc., 18 Feb. 1872; Electricity and
Magnetism, vol. 11. c. 12.
* W. M. Hicks, "On the Motion of Two Spheres in a Liquid," Phil. Trans. 1880.
4-2
52
SOURCES, DOUBLETS, AND IMAGES.
·
r<f;
Now at all points in the neighbourhood of the sphere, r <ƒ;
hence can be expanded in a convergent series of the form
m m
Φ
Σ
P₁ (u),
ƒ
ƒ
where P₁(u) is the zonal harmonic of degree n.
At all points outside the sphere, the velocity potential of the
image of S can be expanded in a series of the form
1
n
* = 4, (C)* P.
Φ'
Ο
n
Since the sphere is at rest, the surface condition is
do
+
dr
do
0,
dr
when
Therefore
भ
whence A= 0,
nan-
r = a.
mΣ² 7 P₁ + Σ, 4, (+1) P₁ = 0;
1
fn+1
P₂+Σ A
n
(n
a²
n
mn
A.
n+1
therefore
a
n+1
n a2n+1
α
Q' = − mΣ, n + 1 (fr)" + 1 P₂
ma
n + 1
=-72
Ch
f
+ 1
ma
n
c"
P.
n
P+7 +1
where ca²lf. Now if c<r,
0
αλ
(r² − 2λrµ + x²)
f
pr+1
p² + 1 n + 1
n+1
P
n
n + 1°
.(1),
Hence, adding and subtracting ma/fr from (1), the value of '
may be written
ma
Φ
1
+
ƒ' (r² — 2rcµ + c²) $
f
-
m
dr
α (r² − 2λrµ + λ²)³ °
The first term represents a source of strength ma/f, situated at
a point H such that OH=calf, and which therefore coincides
with the electrical image of S with respect to the sphere: the
IMAGE OF A DOUBLET IN A SPHERE.
53
second term represents a line sink of strength m/a per unit of
length, extending from the inverse point H to the centre of the
sphere.
53. To find the image in a sphere of a doublet whose axis
passes through the centre of the sphere.

H H'
S' S
Let O be the centre of the sphere, a its radius, S a source of
strength µ, S′an equal sink, and let H, H' be the inverse points
of S, S'; also let OS=f, HP=r, PHS=0. Then, if o be the
velocity potential of the image,
$=-
μα
1
µHH'
aHP'
μα 1
+
ƒ'HP¹‍ƒ— SS''H'P
But OH. OS OH'. OS" = a², therefore
also
Therefore
Ф=
μα
=
HH' = OH' — OH = α²
a²SS"
f²
1
1
2
OS
OS
H'P=HP-HH' cos 0.
80).
+55) (1+ 235 cose) - 493
fr + Ha (1 + SS)
μα
fr
μSS'a³
cos 0.
ƒ³µ²
a²SS"
for
μας
ƒ³r
Now μSS' =m, where m is the strength of the original doublet,
hence
a
ф=т
(3) cos e
p2
This is the velocity potential of a doublet situated at the
inverse point H, whose strength =- m (a/f)³.
54
SOURCES, DOUBLETS, AND IMAGES.
- J
54. To find the image of a doublet whose axis is perpendicular
to the line joining it with the centre of the sphere.

R' H'
RH
Let S be a source, S' an equal sink; H, H' the inverse points
S, S'. The image of S is a source of strength µa/ƒ at H, and a
line sink of strength μ/a per unit of length from 0 to H.
Now
HH'
=
SS'a 2
f²
whence the source and sink at H, H' coalesce into a doublet at H
of strength
3
μSS'a mas
ƒ³
>
where m is the strength of the original doublet.
Let R, R' be any points on OH, OH', such that
OR. OS OR'. OS";
then
SS'. OR'
RR'
f
and the two sink and source elements at R, R' coalesce into a
doublet of strength
μ SS'. OR m OR.
α f
af
Hence the image of S is a positive doublet at H of strength
ma³/ƒ³, together with a negative line doublet of strength —m OR/af
per unit of length, extending from 0 to H.
55. In the next place, let there be a source of strength m at
a point P outside a sphere whose centre is 0 and radius a; and a
line sink from P to Q, (where Q is a point on OP which lies on
the side of P remote from 0), of strength-m/PQ per unit of
length'. Let R be any point between P and Q; P', R', Q' the
inverse points of P, R, Q. Also let OR = x, OR = y.
1 W. M. Hicks, "On the Problem of Two Pulsating Spheres in a Fluid," Proc.
Camb. Phil. Soc. vol. III. p. 276.
IMAGE OF A SOURCE AND LINE SINK.
The image of P consists of
(1) a source at P' of strength ma/OP,
55
(2) a line sink from 0 to P' of strength-m/a per unit of
length.
The image of the line sink element de at R produces
(3) a sink at R', of strength
madx
PQ.x
mady
PQ.y
and
(4) a line source from 0 to R' of strength mdx/PQ.a per
unit of length.
In order to calculate the image of the line sink between P and
Q, it will be convenient to consider first the portion of the image
between 0 and Q', and secondly the portion between Q' and P'.
Since every element of PQ produces an elementary line source
of strength mdx/PQ.a between 0 and Q', the resultant is a line
source between O and Q' whose strength per unit of length is
m
coe mdx
PQ. a α
ОР
This line source cancels the portion of (2) which lies between
O and Q'.
Only those elements of PQ which lie between P and R con-
tribute anything to the density at R', hence, adding the effects of
(2), (3) and (4), the total strength at R' is
m
m
+
mOP) dy.
- (2
α
mOQ
a PQ
a. PQ
pdy
mdy
ɑ
mady
PQ.OR'
+
mPRdy
a PQ
p =
a
(1+
OP
PQ
=
a²P'Q'
OP.OQ
P'Q. OP. OQ
a²
Therefore
But PQ = OQ – OP
Therefore
ma
p=
OP.P'Q' '
Hence the image consists of a single source at P' of strength
ma/OP, and a line sink from P' to Q of strength - ma/OP. P'Q'
per unit of length; that is, the image is an arrangement of the
same form as the object.
56
SOURCES, DOUBLETS, AND IMAGES.
56. To find the image of a source in a cylinder, the motion
being in two dimensions.

P
O H A
S
Let S be a source situated outside a cylinder, H the inverse
point. Then, if an equal source be placed at H, the normal velocity
due to the combined effect of both is
q
m
m
q
cos OPY+
SP
cos OPH
HP
But since OH. OS=OP², the triangles OSP and OPH are
similar, therefore
m
q=
cos OPY +
SP
m
OP'
m
HP. Os (SP+ OP cos OPY)
Hence the image of a source at S is an equal source at the
inverse point H, together with an equal sink at O the centre of
the cylinder.
Similarly the image of a doublet is an equal doublet, but of
opposite strength, situated at H.
57. We shall conclude this chapter by applying the method
of images to determine the velocity potential due to a source
situated between two infinite parallel planes¹.
0 P
P₁
2
A
B
1 W. M. Hicks, Quarterly Journal, vol. xv. p. 274.
7
1
SOURCE BETWEEN TWO PARALLEL PLANES.
57
Let P be the source, and let the origin be the middle point 0,
of a line through P perpendicular to the two planes.
2
2
1
The image of P in the plane B will be another source P₁, such
that BP₁= BP; the image of P, in the plane A will be another
source P, such that AP, AP,, and so on for an infinite series.
Similarly the image of P in the plane A will be a source P',, such
that AP'₁ = AP, and the image of P', in B will be a source P',
such that BP', = BP',, and so on. The velocity potential of the
motion of the liquid contained between the two planes due to the
source P, will be equal to the velocity potential of P together
with the velocity potential of the two infinite trails of images.
Let
then
1
1
AB=2a, OP= §,
1
OP₁ = a + BP = 2a - §,
1
OP₁ = a + AP₁ = 4a — §,
2
1
2)
Similarly
OP₁ = a+ BP, 6a - §,
=
ξ,
3
OP₁ = 2na — §.
n
OP' = 2na + §.
n
(i) Let the motion be in three dimensions, and let z, a be
the coordinates of any point Q of the liquid; then
$ =
1
√ {(≈ − §)² +∞²}
8
+Σ00
1
Therefore
+ Σ
[ve
1
0
1
—
√ {(≈ + § − 2a − 4na)² +∞²}
+
+
1
~ √{(≈ + § + 2a + 4na)² + w²
1
.√ {(z − § + 4na)² + ∞²} ⋅ √ {(≈ — § — 4na)² +z '}
1
+ = = = [√ICE + E-
∞
+
1
ปี
1
(z + § − 2a + 4na)² + w²} ⋅ √{(≈ − § + 4na)² + :
Each of these expressions is of the form F (z, w), where
1
F (z, w) = Σ_∞ √/{(z+4na)² + w*}
'
We proceed to find a finite expression for this series. If a is
positive,
1
2a (tr
√ (a² + b²)
π
ᏧᎾ
a² + b² cos² 0 ;
0
A
58
SOURCES, DOUBLETS, AND IMAGES.
therefore
1
√{(z+4na)² + ∞²}
Also
T
1
ㅠ
​2w fπ
do
π (2 + 4na) * cos² + w²
0
1
(z+4na) cos 0 - wi
πφ πφ
1
(≈ + 4na) coș 0 + 50
sin TO-TO (1-2)... (1-4).
C
C
n²c²
therefore, taking logarithms and differentiating, we obtain
7 cot
πό 1
+ Σ
1
C
C
20
p² - n²c²
1
Σ
$+nc
do.
။
T
Therefore F (z, ∞)
1 ft.
1
:S
{
0
4αι
0
1
π
1
z cos 0 — wɩ + 4na cos 0
π
cot (2 - wi
4a
wɩ sec ) — cot
π
4a
sec О sinh (π sec 0/2a) d0
1
z cos 0 + wɩ + 4na cos
0)}
(z+wɩ sec )} sec 0 do
πι
2a cosh (π sec 0/2α) — ços (πz/2α)
0
1 d fπ
—
= log (cosh (7 sec 0/2a) — cos (πz/2a)} d0.
παπ
0
do
The first integral becomes infinite at the upper limit, but since
the variable part of potential functions is the only part which it is
necessary to consider, we may subtract
1 π
2a
make the integral finite, and we shall obtain
F(x, w) = —
1
π
sec e de, which will
**” exp ( — π™ sec 0/2a) — cos (πz/2a)
2a. cosh (π sec 0/2a) — cos (πz/2α)
0
sec e do.
And since = F {(≈ + § − 2α), ∞} + F {z − §, w}, we finally obtain
1
Επ
$ =
2a
exp (− πш sеc 0/2a) + cos π (≈ + §)/2a
cosh (Ta sec 0/2a) + cos π (z + §)/2a
+
Π
exp (− π sеc 0/2a) cos π (z — §)/2a
cosh (π sec 0/2a) — cos π (≈ — §)/2a
sec o do.
(ii) Let the motion be in two dimensions: writing x and y
for z and a respectively, we obtain
=Σlog {(x + § − 2a + 4na)² + y²}
+ }Σlog {(x − § + 4na)² + y²}.
-
Let
EXAMPLES.
ƒ (x, y) = log II°∞∞ {(x+4na)³ + y²},
59
where the symbol II denotes the infinite product for all positive
and negative integral values of n including zero.
пв πθ
02
Now
sin
1
C
C
.(1-
62
2.2
nc
πθ
II'
C
(1 + 1)
where II' denotes that the value n = 0 is excluded. Now
П1° {(x+na)² — y²} = II°。(x+y+na) (x −y+na)
therefore
x
x-
= (x + y) (x − y) II'° ¸n²a² ( 1 + ªty) (1 + ¤ y),
(x y)
Π
na
x vy
ƒ (x, y) = log II´¨¸„4*n*a* + log (* +13) II" (1 + 2 + 1 )
4a
x - by
4na
x
+log II (1+1)
。
4a
4na
The first term which is constant may be omitted; we therefore
obtain
π
ƒ (x, y) = log sin
π
4a (x + y) sin (x — iy)
4α
= log(cosh Ty/2a - cos πx/2α) - log 2;
Пх
whence, omitting constant terms, the value of & may be written
$ = log {cosh πу/2α
-
C cos π (x — §)/2a}
+ ½ log {cosh πy/2a + cos π (x + §)/2a}.
EXAMPLES.
1. Prove that when the motion of a liquid is irrotational
and symmetrical with respect to an axis, Stokes' current function
satisfies the equation
d² sin o d
+
cosec
0
dr
dib) = 0,
and thence show that the current function due to a source of
strength m is
s✪
¥ = — m cos
m cos 0 + const.
60
SOURCES, DOUBLETS, AND IMAGES.
2. When the motion is in two dimensions, prove that the
current function due to a source is me, and apply this result to
find the image of a source in a circular cylinder.
3. The motion of a liquid is in two dimensions, and there is
a constant source at one point A in the liquid and an equal sink
at another point B; find the form of the stream lines, and prove
that the velocity at a point P varies as (AP.BP)˜¹, the plane
of the motion being unlimited.
If the liquid is bounded by the planes = 0, x = a, y = 0, y = a,
x
and if the source is at the point (0, a) and the sink at (a, 0), find
an expression for the velocity potential.
4. The motion of a liquid in two dimensions is steady, and is
due to the presence of any number of sources and sinks. If the
mass of any source or sink be supposed equal to that of the liquid
which it would generate per unit of time (the masses of the sinks
being negative), show that any source has a tendency to move
with an acceleration made up of accelerations from every other
source and towards every sink, and proportional in each case to
the numerical strength of the source and sink, and the inverse of
its distance.
5. Liquid is bounded by two perpendicular planes OX, OY.
A source is situated at a point whose distances from the planes
are a and b respectively. Find the pressure at any point of either
of the planes, (i) when the motion is in two dimensions, and (ii)
when in three dimensions.
6. The boundary of a liquid consists of an infinite plane
having a hemispherical boss, whose radius is a and centre 0. A
doublet of unit strength is situated at a point S, whose axis
coincides with OS, where OS is perpendicular to the plane. P is
any point on the plane, OP = ∞, OS=f. Prove that the velocity
of the liquid at P is
7. Prove that
6f w
αδ
· (a²+ ƒ²∞²)
1
(f² + @
$ = ƒ (t){(r² + a³ − 2az)*+ (r² + a² + 2az)* − r¯`¹} + if (t)
is the velocity potential of a liquid, and interpret it. Find the
surfaces of equal pressure if gravity in the negative direction of
the axis of z be the only force acting.
•
EXAMPLES.
61
8. Liquid enters a right circular cylindrical vessel by a supply
pipe at the centre O and escapes by a pipe at a point A in the
circumference; show that the velocity at any point P is pro-
portional to PB/PA. PO, where B is the other end of the diameter
AO. The vessel is supposed so shallow that the motion is in two
dimensions.
9. A source is placed midway between two planes whose
distance from one another is 2a. Find the equation of the stream
lines when the motion is in two dimensions; and show that those
particles which at an infinite distance are distant a from one of
the boundaries, issued from the source in a direction making an
angle π/4 with it.
10. The boundaries of a liquid are given by 0 = ± π/2n, and
a source of strength m exists at the point = 0, r = a. Prove
that the current function for two dimensional motion is
M
tan-1
2π
r²n sin 2n0
p²n cos 2n0 — a²n
11. A quantity of liquid moves in that quadrant of the plane
of xy in which x and y are both positive, and which is bounded by
the planes yz, zx: at the point (a, 0) is a semicircular source of
liquid, and at the origin a quadrantal sink. Assuming that the
amount of liquid flowing out of the source per unit of time is equal
to the amount which flows into the sink, and that the motion is in
two dimensions; find the velocity potential, and prove that the
general equation of the stream lines is
(x² + y²)² — a³ (x² — y²) = λa³xy.
+
CHAPTER IV.
!
i
VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
1
58. THE most general kind of motion of which a fluid is
capable is one which is a combination of rotational and irrotational
motion; that is to say, the component velocities may be regarded
as consisting of two parts, u₁, v₁, w₁ and u, v,, w„, where the former
quantities are derivable from a velocity potential, whilst the latter,
which depend upon the molecular rotation, cannot be so derived.
The peculiarities of the motion specified by the latter quantities,
and which depend upon the molecular rotation, were first investi-
gated by Helmholtz' and will now be considered.
59. We have defined a vortex line to be a line whose direction
coincides with the direction of the instantaneous axis of molecular
rotation. If through every point of a small closed curve a series of
vortex lines be drawn, they will enclose a mass of fluid which
may be called a vortex filament, or shortly a vortex.
We have shown that if the forces which act on the fluid have a
potential, and the density is a function of the pressure, the motion
of the fluid constituting the vortex can never become irrotational.
It will now be shown that every vortex possesses the following
three fundamental properties:
(i) Every vortex is always composed of the same elements of
fluid.
(ii) The product of the angular velocity of any vortex into its
cross section is constant with respect to the time, and is the same
throughout its length.
1 Crelle, vol. LV. p. 25; translated by Tait, Phil. Mag. (4) xxxiii. p. 485.
!
PROPERTIES OF VORTEX FILAMENTS.
63
(iii) Every vortex must either form a closed curve or have its
extremities in the boundaries of the fluid.
Let a, b, c be the initial coordinates of an element of fluid
whose coordinates at time t are x, y, z. Then
da db dc ds.
= λ.
مدا
no 50 ωο
dx
dx
dx
But
dx
da +
db +
dc
da
db
dc
dx
dx dx
=
-λιξ
To da
+ No db
+ So dc
+5%
ροξάς.
ρωο
by § 30 (10); hence
Pods。_pds
@
= €
W
.(1).
Let u, v, w be the component velocities at x, y, z; and let
u+du, v+dv, w+dw be the velocities at a neighbouring point
x+dx, y+dy, z+dz on the same vortex line. Since
dx dy dz ds €
ૐ η Š W
du
du du
τη + 5 de
dy dz
€
therefore
du
ρ
dx
Ε
du
ρ
a
امد
= €
Ət
(§ 12 + 1
ૐ
dx
.*
dv dw
+5
dx
dx
by § 24 (5).
The quantity du is the rate at which the projection of the
element ds on the axis of x is increasing in length; and since this
projection is equal to ed(p¯¹¿)/dt, the line ds still forms part of a
vortex line.
This proves the first theorem.
To prove (ii) let σ be the area of the cross section at time t,
then, since the mass of the element remains unchanged,
Pods = pods.
Therefore by (1)
σοπο = σω,
which proves that ow is independent of the time.
σω
64 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
1
..
Also, by § 7 and § 17 (26),
(d€
15
SS (IE + mn + n5) dS = [[] (de + d + d) dxdydz = 0,
(15+mn
dy dz
or ſfw cos edS′ = 0,
where e is the angle between the axis of rotation and the normal
to S drawn outwards.
Now if we choose S so as to coincide with the surface of any
finite portion of a vortex of small section, together with its two
ends, cos e vanishes except at the two ends; and is equal to +1 at
one end, and 1 at the other; hence
w‚dS¸ — w‚dS₂ = 0,
1
which proves the second part of (ii).
2
To prove the third theorem, we observe that if a vortex did
not form a closed curve or have its extremities in the boundary, it
would be possible to draw a closed surface cutting the vortex once
only, and the surface integral would not vanish.
The first theorem and the first part of the second theorem
depend on dynamical considerations; the second part of this
theorem and the third theorem are kinematical.
60. Since every kind of motion may be regarded as a combi-
nation of rotational and irrotational motion, we may put
u =
do dN dM
dx
+
dy
dz
dø
dL
dN
บ
+
dy
dz dx
where
do dM dL
W = +
dz dx
dy'
is the velocity potential of that part of the motion which
does not depend on the molecular rotation.
Hence in the case of a gas,
du dv dw
1 др
V= + +
dx dy ¹ dz
P at
but in the case of a liquid V2=0: in addition to the above
equations which must satisfy at every point of the fluid, & must
also be determined so as to satisfy the boundary conditions.
If we put
J = dLdM
dM
+
+
dx
dN
dy dz'
we obtain
INTEGRATION OF THE EQUATIONS.
23=
dw dv dJ
dy dz dx
V³L,
65
with two similar equations. Hence if
we have
J = 0 or a constant
▼²L+2§=0, V³M+2ŋ=0, V³N +25=0………......(2).
It follows from (2) that if J=0 or a constant, the quantities
L, M, N are the potentials of distributions of matter whose densities.
are respectively equal to έ/2π, n/2π, /2π; hence if x, y, z be any
point where molecular rotation exists, x, y, z any other point, and ƒ
the reciprocal of the distance between these two points, then
1
L =
2π
1
[√]
E'ƒdx'dy'dz'
M = [[["ƒda'dy'de'
N
==
2π
1
=[[[
2π
'fdx'dy'o
'dz'
(3),
where', n', ' are the values of the components of molecular
rotation at (x, y, z) and the integrations extend throughout those
portions of fluid where there is vortex motion.
We have now to prove that the above values of L, M, N make
J=0 or a constant.
Since
df
df
dx
dx
we have
J
27
(& af
df
dx'
+
1
-2-(
2π
df df
dz
dz
++ d) da'dy' de
21747 [[ƒ (VE² + mn' + n5)
2π
1 dě dn
de
+
dn' d5)
dS
2π dx dy+dz pdx'dy'dz'.
The volume integral vanishes by § 17 (26), and if the vortices
form closed rings the surface integral vanishes, since at the
surface of each vortex l§'+mn' + n5' = 0.
Also, if the fluid extends to infinity and is at rest there, the
surface integral will either vanish or be equal to a constant, since
§', n', ' and ƒ all vanish at infinity. But if the fluid is bounded
either externally or internally, and some of the vortices extend to
this boundary and then break off, we must suppose the boundaries
4.
B.
5
66 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
removed and a hydrodynamical system substituted for them, such
that the velocity at points occupied by the boundary remains un-
changed. This hydrodynamical system will necessarily consist, in
part, of the continuations of these vortices, which must either
extend to infinity or form closed curves, and in either case the
surface integral taken throughout the vortices included in this
larger region, as well as throughout those included in the original
region, will vanish or be constant.
61. If Su, Sv, Sw be the component velocities at a point x, y, z
of a fluid due to an element da'dy'dz' whose rotations are ', n', 5';
then
би:
2π
2TT (5
dy
df
15)
df
dz
dx'dy' dz',
-1
whence if r¹=ƒ, we obtain
1
2πrs
1
би
=
Sv
1
δω
2πrs
2πp³
{n' (z — z′) — 5' (y — y')} dx'dy´dz',
{S' (x — x' ) — §' (z — z′)} dx'dy'dz',
……..(4).
{§' (y — y') — n' (x — x')} dx'dy'dz',
Hence, if q is the resultant velocity due to the element,
q
=
w sin e
2πr²
dx'dy'dz'……..
.(5),
where e is the angle which r makes with the direction of the axis of
rotation of the vortex element. It also appears from (4), that this
velocity is perpendicular to the plane containing the direction of
r and the vortex element, and that its direction is that in which
the point (x, y, z) would move if it were rigidly attached to a body
moving with the vortex element.
62. At all points external to a vortex the motion is irrota-
tional, and a velocity potential consequently exists. We shall
now show that the velocity potential at any point, due to a vortex
of small cross section, is proportional to the solid angle subtended
by the vortex at that point.
Let x, y, z be the given point, x', y', z' any point on the vortex,
r the distance between (x, y, z) and (x', y', z'). Using polar co-
ordinates r, 0, x referred to (x', y', z') as origin, we have
x − x′ = r sin 0 cos x, y-y' = r sin 0 sin x, z-z' = r cos 0.
VELOCITY POTENTIAL OF A VORTEX.
67
Now if be the solid angle subtended at (x, y, z) by the
vortex,
Q = ff sin Oded x
= ƒ (1 − cos 0) dx
= [ dx - [cos e dx ds.
where the integration with respect to s extends once round the
vortex.
X
x
Now
- cot X•
y-y
dx'
dy
Therefore
(y - y')
· (x − x')
r² sin² 0
dx
ds
ds
ds
Therefore
Ω
= f.dx - f²²
γ
{
dx'
dy'
ds
-
(y — y')
- (x − x')
ds
ds) (x − x')² + (y — y')² '
The first term is equal to 2π or zero according as the vortex
does or does not embrace the axis of z; also
ΦΩ
dz
dx
ds
(y-y') (x-x')
dy') ds
ds ps
Now by (4) if w be the z-component of the velocity due to a
vortex of small cross section σ,
or
аф
dz
ωσ
dx'
w=
2π
(y — y')
ds
-(x-x') dy ds.
Hence
аф
wo do
dz
2π dz
ωσ
φ
Ω ..
2πT
.(6).
If the section of the vortex be of finite area, the velocity
potential will be
$=
1
= SS
[[ wndo
Ωσ
2π
where the double integral extends over the cross section.
·(7),
Since the solid angle & diminishes by 4π, whenever the point
x, y, z describes a closed curve in the positive direction, which
embraces the vortex once, & is a many valued or cyclic function.
The product of the angular velocity and the cross section of a
vortex filament, is called the strength of the filament.
5-2
68 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
Vortex Sheets.
63. If we have a sheet of thickness h, consisting of rotation-
ally moving liquid, and if o increase and h diminish indefinitely
but so that the product wh remains finite, equal to w', we ultimately
obtain an indefinitely thin film of rotationally moving liquid
whose molecular rotation is w'. Such a film is called a Vortex
Sheet.
By (3), if §', n', y' be the components of w', the quantities L, M, N
which determine the velocities are given by the equations
1
1
1
=
N
// ½ 27 // $ds... (8),
L-as, M = 2 as,
Ꭱ
2π
2π
where R is the distance between any point on the vortex sheet
and the point (x, y, z), and the integration extends over the
vortex sheet.
64. It was first pointed out by Helmholtz', that the equations
of motion and the equation of continuity of a perfect fluid do not
exclude the possibility of slipping taking place along a surface;
for the only conditions to which the velocity must be subject are,
that it must be finite at all points of the fluid, other than points
where sources or sinks exist, and also that its normal component
at all points of any surface drawn in the fluid must be continuous.
The above conditions obviously do not require that the tangential
component should be the same on both sides of such a surface,
and hence the conditions to which the velocity must be subject
will not be violated if slipping takes place.
65. We shall now show that every surface of discontinuity
over which slipping takes place has the properties of a vortex
sheet.
Let l, m, n be the direction cosines of the normal at any point
P of such a surface of discontinuity; u, v, w; u, v, w' the compo-
nent velocities on the positive and negative sides of the surface.
It is evident that it will be possible to draw a line in the tangent
plane at P such that the tangential components along this line of
the velocities on both sides of the surface shall be equal. Let
X', ', ' be the direction cosines of this line; and let λ, u, v be those
¹ Phil. Mag. Nov. 1868.
Then
VORTEX SHEETS.
69
of the line through P perpendicular to l, m, n and X', µ', v', and
which is therefore the line along which slipping must take place.
I (u — u') + m (v − v') + n (w − w') = 0,
x' (u — u') + µ' (v — v') + v' (w — w') = 0 ;
λ (u−u')+µ (v − v') + v (w — w') = o.
also let
From these equations we easily obtain
Let
น
v-v'
พ w'
=σ
λ
……..(9).
μ
V
1
L
4πT
[Jon' as
.(10),
the integration extending over the positive side of the sheet only;
then
1
Ꮮ
47
1
4π
— wa) []
(vm - μn) + ds
Ads
R
(0-0)}
ds
R
· []{m — —
m (w — w´) — n (v –
Now the surface S may be regarded as the limit of the surface
of a solid bounded by S and another surface indefinitely near S
whose distance from it is h; we may therefore write
1
L =
417T
[f(mw — nv)
1
nv) ds-
4π
= SSSC
พ
d 1
dy R
d 1
ข
1)
dz R dx dy dz
where the surface integral extends over the surface S and the
surface indefinitely near it, and the volume integral extends
throughout the volume enclosed by the two surfaces. The latter
integral evidently vanishes in the limit. Integrating by parts we
obtain
ultimately.
L
=
1
4π
1
元
​dw dv
R (dy-de) dady dz
as
- 2 - // dh ds
2π
1
2π
:SS
ff as
Comparing (10) and (11), we obtain
ξ = σλ, ή = εσμ, ζ' = εσύ, ώ = ησ
..(11),
It therefore follows that the effect of the surface of discon-
tinuity is the same as that of a vortex sheet whose molecular
rotation is to, and that the direction of the vortex lines is perpen-
dicular to that of slipping.
70 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
•
Circulation.
66. We have shown that the motion of a fluid may be
separated into two kinds, rotational and irrotational motion; and
it appears from § 62 that irrotational motion may be subdivided
into two classes according as 4 is a single valued or a many valued
function. In the former case the motion is called acyclic, and in
the latter case cyclic irrotational motion.
67. The line integral (udx + vdy+wdz) taken along any
curve joining a fixed point A, with a variable point P, is called
the flow from A to P.
If the points A and P coincide, so that the curve along which
the integration takes place is a closed curve, this line integral
is called the circulation round the closed curve.
If
any surface which is bounded by a closed curve be divided
into elementary areas by a series of
lines drawn upon it, the circulation
round the bounding curve is equal to
the sum of the circulations round each
of the elementary areas; for the flow
along the sides of all the elements,
except those sides which form part of
the boundary, is taken twice over and
in opposite directions.
In the same way it can be shown
that the circulation round any closed
curve is equal to the sum of the circulations round its projections
on the coordinate planes.

D
68. Let us now determine the circulation for an elementary
0
A
Z
C
B
rectangle ABCD, whose sides are dy,
dz, the positive direction being from
the axis of y to that of z.
Let x, y, z be the coordinates of 0,
the centre of inertia of the rectangle ;
u, v, w the velocities at 0.

The portion of the circulation due to the two sides B and D is
(w + {w,dy) dz − (w - w, dy) dz=w, dy dz
and that due to the two sides C and A=- v, dy dz.
STOKES' THEOREM.
Hence the circulation
=
= (w, - v₁) dydz.
71
Hence, if dS be an element of a surface S whose projection on
the plane yz is the rectangle ABCD, the circulation round the
boundary of S
=
= SS[(w, — v„) dydz + (u„ — w₂) dz dx + (v% − u¸) dx dy].
Hence we obtain the following important analytical theorem,
which is due to Prof. Stokes'¹, viz.
dv
[[{1 (dw-de) + m (du - dw) + n (dr_du) } ds
dy dz
= f(udx + vdy + wdz)
=
dx dy
dy)}
..(12),
where the surface integral is taken over any surface bounded by a
given curve, and the line integral is taken once round the curve.
Substituting the quantities §, n, §, we obtain
2 ƒƒ (l§ + mn +ng) dS = f(udx + vdy + wdz) …………..(13).
69. Several important consequences can be deduced from this
theorem.
If there are no vortices in the fluid, §, n, are everywhere zero,
and the circulation vanishes. Hence in this case & must be a
single valued function.
Since every vortex must either form a closed curve, or have its
extremities in the boundaries of the fluid, it follows that if the cir-
culation be taken round a closed curve which embraces a vortex
once only, the surface S must cut the vortex an odd number of
times. Hence in this case the circulation will not vanish, but will
be equal to twice the surface integral on the left-hand side of (13).
Since §, n, are zero at all points of S, except those which lie
in the vortex, the value of the circulation is 2ffw cos edo where do
is an element of that portion of S which is cut off by the vortex, w
the molecular rotation, and e the angle which its direction makes
with the normal to σ drawn outwards.
Hence the value of at any point P of a closed curve which
embraces a vortex experiences a constant augmentation every time
P travels round the curve to its original position, which is equal to
twice the above-mentioned surface integral. This constant aug-
mentation is called the cyclic constant of p.
1 Smith's Prize Examination, 1854.
72 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
If the line integral is taken round a closed curve which does not
embrace a vortex, S can be drawn so as not to cut any of the
vortices; if S cut any vortex once, it must cut it again, and by § 59
the two portions of the surface integral cancel one another; hence
the surface integral and therefore the circulation round such a
curve will be zero.
Since the circulation taken round any indefinitely thin vortex
filament is equal to twice the product of its angular velocity and its
cross section, it follows from § 59 that the circulation is inde-
pendent of the time; and since every vortex of finite section can
be divided into indefinitely thin vortex filaments, it follows that
the circulation round a vortex of finite section is also independent
of the time.
70. It thus appears that whenever there is circulation the
velocity potential is such as would be due to some distribution of
vortices in the fluid. These vortices need not however have an
actual existence, since in the case of a liquid it is possible for hollow
spaces to exist in the liquid round which circulation takes place ;
or the vortices of which is the velocity potential may lie beyond
the boundaries of the fluid. For example, if tan¹ y/x= 0,
=
is a two dimensional many valued velocity potential whose cyclic
constant is 2π for all circuits which embrace the origin, and zero
for all other circuits: and it will be shown in the second volume,
that if the pressure at a distance from the origin be properly adjusted
by means of suitable boundary conditions, it is possible for the
cylinder ra to be a free surface, which forms the inner boundary
of a liquid, and the space within which is devoid of liquid. It is
also possible to have circulation round a fixed rigid cylinder, in
which case will be the velocity potential of one of the possible
motions of the liquid which may take place.
71. Since a fluid always flows from places of lower to places of
higher velocity potential, it follows that when the motion is acyclic
the stream lines cannot form closed curves but must begin or end
in the boundaries or singular points of the fluid; but when the
motion is cyclic some of the stream lines may be closed curves,
whilst others begin and end in the boundaries of the fluid.
72. The circulation round any closed circuit may be shown
not to alter with the time as follows¹.
1 Sir W. Thomson, "Vortex Motion," Trans. Roy. Soc. Edin., vol. xxv.
FLOW AND CIRCULATION.
73
Let AB be a curve joining two points A and B of a fluid which
always passes through the same elements of fluid; also let ƒ be
the tangential velocity of the fluid at any point P of AB; then
fds=udx+vdy + wdz;
therefore
a
ди
Ət
(fds)
-
dx + u
d (dx)
+ &c.
Ət
dt
Let pq be the projection of ds on the axis of x; u, u + du the
component velocities of p and q parallel to x; then
u = dx/dt, u+du = d (x+dx)/dt;
hence du = d. dx/dt, therefore
a
ди
av
ди
Ət
(udx+vdy+wdz)
dx +
dy +
dz +udu+vdv+ wdw,
It
at
Əz
d (Q + {q²).
гв
Therefore
9]*$ (udx+vdy + wdz) = [Q+49°]z − [Q + $9®]4.
Ət A
B
Since Q and q are always single valued functions, the right-
hand side vanishes when the integration is taken round a closed
curve, which proves the proposition.
73. If at some particular instant, which we shall choose as the
origin of the time, the motion is irrotational and acyclic, the cir-
culation will be zero round every closed circuit, and the preceding
proposition shows that it will always remain zero.
Hence we obtain another proof of the proposition that motion
which is once irrotational is always so; and also that irrotational
motion which at any particular instant is acyclic, always remains so.
Simply and Multiply-Connected Regions.
74. Whenever the motion is cyclic, the flow between two
points will not have the same value for two different lines joining
them, unless the lines are such as are capable of being made
to coincide, without cutting through any of the vortices or passing
through the boundaries of the fluid. The latter class of lines are
called reconcileable lines, the former irreconcileable lines.
75. We are thus led to consider the properties of simply and
multiply-connected regions, which are defined as follows.
•
}
74 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
A simply-connected region, is one in which any two lines join-
ing two given points, may be made to coincide with one another,
without passing out of the region in question.
The spaces inside or outside an ellipsoid or paraboloid are simply
connected regions.
A multiply-connected region, is one in which two or more lines
can be drawn connecting two points, which cannot be made to
coincide with each other without passing out of the region in
question.
The space inside or outside an anchor ring, is an example of a
doubly-connected region.
A region in which n irreconcileable lines can be drawn, is called
an n-ply connected region.
Hence in a simply-connected region, every closed circuit is
capable of being contracted to a point without passing out of
the region. In an n-ply connected region, it is possible to draw
n-1 different circuits, which cannot be contracted to a point
or be made to coincide with one another without passing out of
the region.
Any circuit drawn in a multiply-connected region, which is
capable of being contracted to a point without passing out of the
region, is called an evanescible circuit; and any two circuits which
can be made to coincide with each other without passing out of the
region, are called mutually reconcileable.
76. Every n-ply connected region, may be reduced to a simply
connected region, by drawing n-1 barriers or diaphragms, such that
each diaphragm meets every simple non-evanescible circuit once
only. For example, the space outside two circles which do not cut
one another, is a triply-connected region in two dimensions; but
if from a point on each of the circles, we draw two lines to infinity
which do not cut one another, the region becomes simply-con-
nected.
77. If o be a polycyclic velocity potential, the circulation round
any closed curve, which does not cut any of the barriers is conse-
quently zero: if the circuit cuts all of the barriers once only, the
circulation is ₁ + K₂+ &c. where x,, K, are the cyclic constants
corresponding to each barrier. The number of barriers which
1
2
2
VORTICITY.
75
must be drawn, in order to make the circulation round every
closed curve vanish, is equal to the number of cyclic constants
of p.
78. Every polycyclic function may be expressed as the sum
of the same number of monocyclic functions, as the function has
cyclic constants. For if the number of cyclic constants be n there
will be n simple non-evanescible circuits round which the circulation
does not vanish; hence if
where N₁, N₂……….
2
=
n 'n'
...... are monocyclic functions, whose cyclic constants
are unity; and which are such that the line integral
n
´d Q, dx, dîn dy, dî, dz\
dx ds dy ds dz ds
+
n
+
ds,
taken round any closed circuit is zero, except when the circuit cuts
the barrier corresponding to ₂, it follows that the circulations
round each of the simple n non-evanescible circuits, are respectively
equal to K₁, K₂......, hence the circulation round a circuit which
cuts each barrier once only is equal to к¸ + ê₂+ …..... + K‚·
19 2
Vorticity.
1
n
79. Let a mass of rotationally moving fluid be divided up into
elementary vortex filaments; and let P be any point on the axis
of one of these filaments, dm the mass of the filament which
contains P, w and dS the molecular rotation and cross section of
the filament at P at time t. Then the quantity wdS/dm is called
the vorticity of the fluid at the point P.
This quantity has the same value at all points of the filament
which contains P, and is constant with respect to the time, for if
the suffixes denote the initial values of the quantities (or their
values at some given epoch) and ds is an element of the axis
of the vortex element, the vorticity
wds
dm
wds.
lopodis - lopo'
by § 59, (1); where 7, is the initial length of the filament.
The aggregate vorticity of a mass M of rotationally moving
fluid is equal to the sum of the vorticities of every filament, and
therefore
1
Σωλ
ΣpdSds M
G ff w c
w cos ed S,
'
76 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
where dS is an element of any surface which cuts all the vortex
filaments once only, and e is the angle between the direction of w
and the normal to S drawn outwards.
If the rotationally moving fluid is surrounded by irrotationally
moving fluid, and consists of an arrangement such as a circular
vortex ring, which is resolvable into elementary circular filaments
which are perpendicular to the meridian sections of the ring, the
aggregate vorticity is equal to /M, where x is the circulation
round any closed circuit which embraces the ring once. But if
the rotationally moving fluid consisted of the arrangement above
described, together with an outer sheet which is resolvable into
filaments lying in planes passing through the meridian sections of
the ring, the circulation will remain unaltered, but the aggregate
vorticity will be
1
K
1
ods,
* + + [[wds.
2M M
1
2
2
where M, is the mass of the inner ring, M₂ that of the sheet, and
w, dS are the molecular rotation and cross section at any point of
one of the elementary filaments of the sheet. Hence the aggregate
vorticity is not necessarily proportional to the circulation.
Green's Theorem.
80. The following theorem, which is of great importance in
Electricity and various branches of physics, is due to Green¹.
Let and be any two functions, which throughout the interior
of a closed surface S are single valued, and which together with
their first and second derivatives are finite and continuous at every
point within S; then
/// (dd dyb + do dy do dr
dx dx dy dy+dz dz) dxdy dz
df
dS
dn
ds - [[[ÞV³µ dædydz ...(14),
=SS
зва
аф
dn
pdædydz ……….(15),
do ds – [√] Þv³ da dy dz .... (15),
where the triple integrals extend throughout the volume of S, and the
surface integrals over the surface of S, and dn denotes an element of
the normal to 8 drawn outwards.
1 Mathematical Papers, p. 24.
GREEN'S THEOREM.
Integrating the left-hand side by parts, we obtain
d
[[[dd dyr dødyds = []]+
dx dz $d dyde] - Jay dady dz.(16),
do d¥
dx dx
dx
dx²
77
where the brackets denote that the double integral is to be taken
within proper limits. Now since the surface is a closed surface, any
line parallel to x, which enters the surface a given number of times,
must issue from it the same number of times; also the x-direction
cosine of the normal at the point of entrance, will be of contrary
sign to the same direction cosine at the corresponding point of
exit; hence the surface integral
-Idas.
dx
ld S.
Treating each of the other terms in a similar manner, we find
that the left-hand side of (16)
= [[ $ dv ds - [[[ þv³y dady dz.
φ as
dn
The second equation (15) is obtained by interchanging
and y.
81. We may deduce several important corollaries.
φ
(i) Let & be the velocity potential of a liquid, and let √ = 1;
then ² = 0, and we obtain
0 =
2
· [[] v'p dx dy dz = [[dp as
dn
…………..(17).
The right-hand side is the analytical expression for the fact
that the total flux across the closed surface is zero; in other words
as much liquid enters the surface as issues from it.
be both velocity potentials, then
(ii) Let and
φ
dv ds =
аф
ป
зва
ds
dn
dn
……..(18).
(iii) Let =√, where is the velocity potential of a
liquid; then
2
2
dx dz o do ds .. (19).
SIS {(de) + (de)" + (de) } dadyde -
dx dy dz
dn
If we multiply both sides of (19) by tp, the left-hand side is
equal to the kinetic energy of a liquid, and the equation shows
that the kinetic energy of a liquid whose motion is acyclic and
irrotational, which is contained within a closed surface, depends
solely upon the motion of the surface.
78 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
82. Let us now suppose that liquid contained within such a
surface is originally at rest, and let the liquid be set in motion by
means of an impulsive pressure p applied to every point of the
surface. The motion produced must be necessarily irrotational, and
acyclic; also if be its velocity potential, it follows from § 42 (50)
that p-pp. Now the work done by an impulse, is equal to the
product of the impulse into half the components in the direction
of the impulse, of the initial and final velocities of the point to
which it is applied; hence the work done,
124 p
аф
dn
аф
dS = p
$P //
ds,
dn
and equation (19) asserts that the work done by the impulse is
equal to the kinetic energy of the motion produced by it, which is
a particular case of the Principle of Energy.
83. Let us in the next place suppose that liquid is contained
within a closed surface which is in motion; and let the motion of
the liquid be irrotational and acyclic; also let the surface be
suddenly reduced to rest. Then if o be the new velocity potential,
do/dn = 0, and therefore
SSS {(
2
аф αφ
+
dx dy
+
(dd)} dx dy dz = 0,
whence dp/dx, dø/dy, and dø/dz are each zero, and therefore the
liquid is reduced to rest.
84. In proving Green's Theorem, we have supposed that the
region through which we integrate, is contained within a single
closed surface, but if the region were bounded externally and
internally by two or more closed surfaces, the theorem would still
be true, provided we take the surface integral with the positive
sign over the external boundary, and with the negative sign over
each of the internal boundaries.
85. Let us suppose that the liquid extends to infinity and is
at rest there, and is bounded internally by one or more closed
surfaces S,, S, &c., and let us calculate the value of T for the space
bounded by S₁, S, &c., and a very large sphere S whose centre is
the origin. Then
2
аф
T= p[[dds - ip[as]
аф
dn
ερ
dS
dn
where the square brackets indicate that the integral is to be taken
over each of the internal boundaries.
.
GREEN'S THEOREM.
79
Now at the surface of S, & will be of the order m/r, where m
is a constant, and do/dn = do/dr=-m/r²; also if d be the solid
angle subtended by dS at the origin, dS=r'd; therefore
аф
2
[[ $ do as = - m² [fan-
φ dn
p
4πm²
r
which vanishes when r= ∞. Hence the kinetic energy of an
infinite liquid which is at rest at infinity, and which is bounded
internally by closed surfaces is
аф
T= -1 p
3 • [/l & do as]
dn
..(20),
where the surface integral is to be taken over each of the internal
boundaries.
The preceding expressions for the kinetic energy show that
if the motion is acyclic and the internal boundaries of the liquid
be suddenly reduced to rest, the whole liquid will be reduced to
rest.
86. When the motion takes place in two dimensions, Green's
Theorem may be established in a similar manner. Let the liquid
be bounded externally by a closed surface S, and internally by one
or more surfaces S₁, S.... Then
đẹp đẹ
[[(dib dé + dy do) dedy = [ v (do
da dy
dx
-[S
do аф
ป
зв
аф
dx dy
аф
dy + do dx)
↓³
2
dxdy,
dy + do dx)] − SS ↓
– $
dx dy
2
where y² = d²/dx² + d²/dy² and the square brackets denote that the
line integral is to be taken once round the circumferences of each
of the internal boundaries. Now if we integrate round the
boundary of the liquid in the contrary directions of the hands of
a watch, the integration with respect to y will be in the same
direction and that with respect to x in the opposite direction
to s, whence the first integral becomes
ป
dx
(do do do dx) ds,
da ds dy ds
also if dn be an element of the normal drawn outwards,
dx/ds=-dy/dn, dy/ds
dy/dn, dy/ds= dx/dn,
80 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
whence
ff(d
đẹp đẹp
dx dx
+
đẹp đẹp
dy dy
аф
- [y do ds - [[
dn
зв
d ds] - [] v vp dødy …..(21),
dn
- dy ds - [4 dy de] - [[ v³y dady ...(22).
=/¢
dn
dn
ds
This is Green's Theorem for two-dimensional space.
Hence the kinetic energy of the liquid is
аф
T = bp √ 4 do ds - kp [[ $
4p [4 do ds]
аф
dn
dn
………….(23).
In this equation may be either the velocity potential or the
current function.
If the liquid extends to infinity and is at rest there, the value
if single valued, at a great distance from the origin, must
be of the form
of
A log r+r¹ (B cos 0 + C sin 0),
and therefore when r is very large the first integral becomes equal
to 2πp 42 logr which becomes infinite when r∞ unless A = 0;
when this is the case, since all the other terms vanish, we obtain
T = − 1 p
# [
аф
$
ds..
dn
...
.(24),
the integrations being taken round the internal boundaries only.
87. All the results of the last article may be also obtained by
means of Stokes' theorem § 68 (12), and they may be extended
to the case of polycyclic velocity potentials in the same way as in
the next article. It should however be noticed that if o be a
polycyclic function, it will contain terms of the form 40, and
hence will contain terms of the form A log r and will therefore
be single valued. We may therefore, in the case of cyclic motion,
employ the single valued current function, instead of the velocity
potential; but when there is circulation it follows from the
last article that the kinetic energy will be infinite if the liquid
extends to infinity. We shall show how the difficulty thus intro-
duced may be evaded in Chapter VIII.
THOMSON'S THEOREM.
81
Thomson's Extension of Green's Theorem.
88. The proof of Green's Theorem given in § 80 holds good
only when & and ↓ are single valued functions. If they are poly-
cyclic functions, the surface and volume integrals on the right
hand side of (14) and (15) become indeterminate. The extension
of this theorem when and are polycyclic functions is due to
Sir W. Thomson¹.
2
Let us suppose that the region is multiply-connected, and that
4 is a polycyclic function whose cyclic constants are k₁, K... Let
the region be made simply connected by drawing the requisite
number of barriers. Since we are not allowed to cross any barrier
during the integration, we must include the surface on both sides
of the barrier in the surface integrals. Hence if do,, do,...be
elements of the different barriers corresponding to the quantities
K₁, K₂…… respectively
2
...
[] + dv ds = [[ $ dv ds + [[ $ dy do........
φ dn
dn
dn 1
where on the right-hand side, the integration with respect to S
extends over the boundaries, and that with respect to σ, over both
sides of the barrier σ₁.
Now the values of dy/dn are equal in magnitude and of
contrary sign at two contiguous points situated on different sides
of the barrier, also the value of on the negative side exceeds
that on the positive side by ₁, therefore
K
K
dr
do
dn
[[ & dve do₁ = [[ $ dv do, - [] ($+*,)
dn 1
φ
dn
K
SS
dif
do,,
dn
where the integration on the left-hand side extends over both
sides of the barriers, and that on the right over the positive side
only.
Hence instead of ſſødy/dn. dS, we must write
SS
d¥
φ
dn
dv ds - K
K₂
dn
ffdy do, - kdy do......
dn
1 “On Vortex Motion," Trans. Roy. Soc. Edin., vol. xxv. p. 217.
B.
6
82 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
Similarly if were a polycyclic function whose cyclic constants
are κ', '………….. we must write instead of ffydd/dn.dS,
аф
ffy do as - K
K'
ffan do, - K', ffan
2
[]
do₂-
dn
dn
dn
Hence when and are polycyclic functions and the region
is a multiply-connected one, Green's Theorem becomes
do dv
+
dx dx dy dy
SSS (d
= [[ & dv ds = [[[ $ dv ds ] + x, dy do + x. [da do +.....
dn
ESS
φ
dn
dn
đẹp đẹp đẹp đ
+
dz dz
dx dy dz
K
1
dn
do₁ K K2 do₂+
(25),
K'
2
Jan do₂+
(26),
- Sp▼²↓ dx dydz.
- [4 db as - [[fy do ds] + x', [de do +x [do do +.....
аф
dn
аф
dn
K
1
1
dn
2
– SSS↓▼²pdx dy dz.
where the first integrals on the right hand side are to be taken
over the outer boundary, and the square brackets denote that the
second integrals are to be taken over each of the internal bound-
aries.
89. Putting =, it follows that if the liquid extends to
infinity and is at rest there,
аф
- ip[//dd as]-
T = − 1 p
гаф
аф
φ as] + xp [[ando, + bx, [[do do₂+...(27).
dn
dn
The first term represents the work done by the impulsive
pressure which must be applied to the boundaries S in order to
produce the actual motion from rest. The second term represents
the work done by a uniform impulsive pressure «、p, applied in
the positive direction to every point of the barrier corresponding
to . Hence cyclic irrotational motion may be artificially gene-
K₁.
rated by means of a proper impulsive pressure applied to every
point of the boundaries, together with uniform impulsive pressures
K、p, k‚ρ………….., applied respectively to every point of the barriers,
which must be drawn in order to make the region occupied by
the liquid simply connected. We may therefore generalise the
theorem of § 85, by asserting that if irrotationally moving liquid
occupying a multiply-connected space, is bounded by moving surfaces,
which are suddenly brought to rest, the whole liquid will be reduced
KINETIC ENERGY.
83
to rest unless its motion is cyclic; and that in the latter case, the
cyclic motion which could have been generated in the manner above
described will not be destroyed.
90. The foregoing arguments show that if the bounding surface
of a liquid which was originally at rest, be made to vary in a given
arbitrary manner, the kinetic energy of the liquid at each instant,
will be less than it would be if the liquid had any other motion con-
sistent with the given motion of the bounding surface.
Since the liquid is originally at rest, the motion which is
caused by the variation of the bounding surface will be acyclic
irrotational motion. But the most general kind of motion which
is possible within the surface is a combination of acyclic, cyclic
irrotational motion, and vortex motion. The first can be destroyed
by means of a suitable impulsive pressure applied to every point
of the boundary, but the two latter cannot be destroyed by any
operations performed on the boundary alone. Hence the kinetic
energy of the acyclic motion alone, must always be less than the
kinetic energy of the most general possible motion.
This theorem is due to Sir W. Thomson¹.
91. When the motion is rotational the kinetic energy cannot
be obtained by Green's Theorem, since within a vortex there is no
velocity potential. In this case
T = {p SSS (u² + v² + w²) dxdy dz,
dø, dN
= $p SSS {u (d/2x + dy
部
​dM
+
dz
аф
do dM
+w
+
dz dx
do dL dN
+
dy™ dz dx
dL
_dL
d)} dady dz,
by § 60. Integrating by parts, the terms involving
= ≤ p √S $ (lu + mv + nw) dS,
since the volume integral vanishes by virtue of the equation of
continuity. The other terms
-
= {p SS {L (nv — mw) + M (lw − nu) + N (mu − lv)} dS,
{pSS {L(nv
►
έρ
+ SSS I
+ Pfff { 1 (dy - de
dv
du
dw
dv
du
+ M
M
dz
dz
+ N
da
dx
dy
dy)} dx dy dz.
1 "Notes on Hydrodynamics," Camb. and Dubl. Math. Journ., vol. iv. p. 90.
6—2
r
!
84 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
If the liquid extends to infinity and is at rest there, and all the
vortices are within a finite distance of the origin, the surface
integrals will vanish and we obtain
T = p SSS (L§ + Mn + NG) dxdydz…………………………….(28).
92. Let us now suppose that we have two closed vortices of
small cross sections σ,, σ. Let ds,, ds, be elements of their lengths;
к₁, к the circulations due to them; then
1' 2
2
dx
T = tx₁p √(Laz
dx
ds₁
dz
dy
+ M +N ds,
ds, ds,
dy dz
+ M as₂
+ N dsq
ds.
where the line integrals extend round each respective vortex. Now
K1
1 dx
K2
dx
L=
L = = f. do, de' + 2 ft da, de ; &c. &c.
4πrds, 4π Jr ds2
1
Therefore
- If
T= L
8π
ρ
2
(Ak¸² +2Bk‚k₂+ Ck₂²)
dy dy
dz dz
dx
+
· (da, da, ds, ds,+ da, ds,)
where
A :
=
B = []
ds,'
+
+
dx dx dy dy dz dz
r ds,
\ds d³¸* ds, ds* ds, ds
1
24
ds, ds,
1
dsds,
and C is obtained from A by changing s₁, s' into s„, s'. If e be the
angle between the two elements ds₁, ds, these expressions may be
written
COS
A = [[cos € ds.ds",
B =
ds₁ds₂, C= foose ds.ds'.
[[cos e do do
2
γ
The quantities A and C are evidently the coefficients of
self-induction of two electric currents of unit strengths which
coincide with the vortices, and κ, respectively, and the quantity
B is the coefficient of mutual induction of two such currents.
Hence the kinetic energy of the hydrodynamical system is equal
to the electro-kinetic energy of two currents of strengths, (0/π)±
and (p/π) respectively, which occupy the positions of the
vortices. This proposition may easily be extended to any number
of vortices.
!
KINETIC ENERGY.
93. Another expression for T may be obtained in the form
T=2pSSS{u (y5 — zn)+v (z§ − x5)+w (xn− y§)}dxdy dz…….(29).
For the first term
- pfffu {y (do - dy
dx
dx
dv du
du dw
(dz-d)} dady dz
- P [[]{vy ·
du
(vy+wz)
v
dx dy dz,
dy
85
since the surface integral vanishes. Transforming the other terms
in the same way, adding, and making use of the equation of
continuity, we obtain
du
dv
dw\
P
SSS(~²
u² + v² + w² + xu
dx
+yv dy
+ zw
dz
du) dx dy dz.
Integrating the last three terms by parts, the right hand side
of (29)
= } p SSS (u² + v² + w²) dxdydz.
94. When the motion is symmetrical with respect to the axis
of z, an expression for T may be obtained in terms of Stokes'
current function; for
2
T=wp[] = {(d) + (de) } dwds.
Τ
Therefore
T
=πρ
w
dz
I _ [v (dy de + dv de) - [[(dy de + dy d=)]
πρ w
dz
da
dz
- // * (dy -
dz
đỉnh
´ď³↓ 1 d¥, ď²
+ d) dadz,
2
@\dw?
παπ
dz²
where the first integral refers to the external, and the second
integral to the internal boundaries of the liquid.
Now in order that this kind of motion may be possible, it is
necessary that the boundaries should be surfaces of revolution
whose axes coincide with the axis of z. Let s be an element of
the meridian curve of one of the boundaries, and let the inte-
gration with respect to s be taken from z to w. Since the
integration with respect to a will be in the same direction, and
that with respect to z in the opposite direction to s, the first
integral becomes
[ 14 (dry do _ dry de) ds =
w dz ds da ds
↓ dy ds,
fb
w dn
86 VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.
where dn is an element of the normal drawn outwards. The
volume integral is equal to
- 2fffwdw dz,
where w is the molecular rotation: whence
ψ αψ
↓ dy ds-πp
T=TP πp [[ V dv ds ] + 2[[yo dads...(30).
@ dn
@ dn
If the motion is irrotational and the liquid extends to infinity,
and is at rest there,
ψ
T=
τρ
#p[[V dv de].
dn
(31),
where the integration is taken once round the meridian curves of
each of the internal boundaries.
On the Connection between Vortex Motion and Electromagnetism.
95. In § 60, we have shown that the velocity potential at P
due to a single closed vortex filament of strength m, is
$= − mQ/2π,
where is the solid angle subtended by the vortex at P.
This is the magnetic potential of an electric current of strength
- m/2π, which flows round a closed circuit coinciding with the
vortex (Maxwell, Electricity and Magnetism, vol. II. §§ 410 and
484). Now the magnetic potential due to such a current is the
same as that due to a simple magnetic shell of strength-m/2π
which is bounded by the current; also by § 48, 4 is the velocity
potential due to a doublet sheet of strength m/2π bounded by the
vortex. Hence a vortex filament and a doublet sheet respectively
correspond to an electric current and a magnetic shell, and a
vortex sheet may be replaced by a doublet sheet in the same
manner as an electric current may be replaced by a magnetic
shell.
The action of a vortex filament upon the surrounding liquid is
determined by the quantities L, M, N, whence it follows from (3)
that the molecular relation corresponds to an electric current: the
quantities L, M, N to the components F, G, H of electromagnetic
momentum; and the velocities u, v, w to the components a, ß,
of magnetic force (see Maxwell, § 616).
VORTEX MOTION AND ELECTROMAGNETISM.
87
Also the magnetic potential of a magnetic shell, and the
velocity potential due to a doublet sheet are essentially single
valued functions, since the line integral of magnetic force and
the circulation are zero for all circuits which do not cut the shell
or doublet sheet, and which it is not permissible to cross; on the
other hand the magnetic potential due to an electric current, and
the velocity potential due to a vortex, although represented by
the same quantities, are cyclic functions, the cyclic constant being
equal to 2m, where m is the strength of the vortex. This cyclic
constant is equal to the line integral fdp/ds. ds taken once round
a closed circuit embracing the vortex or current once; and in the
former case it represents the circulation, and in the latter case
the work which would have to be done in moving a magnetic
pole once round the current in opposition to the magnetic force
exercised by the current (Maxwell, § 480).
The potential energy of a magnetic shell of strength -1,
placed in a magnetic field, the components of whose vector
potential are F, G, H is (Maxwell, § 423)
F
|(r da
ds
+ G
dz
dy + H de)
ds ds
c) ds.
The flux through a closed vortex ring is,
SS (lu+mv + nw) dS
=[[{1 (dN_dM).
L
"
dy
dx
dy
dL dN
dM
dN
+ m
+ n
dS
dx
dx dy
(
dz
dy + Nde) ds,
= [(1 de + May
ds
ds
dz
ds
and this corresponds to the potential energy of the magnetic
shell.
The following table shows the connection between the two
subjects:
+
88
VORTEX MOTION AND CYCLIC IRROTATIONAL MOTION.

譬
​Hydrodynamical Quantities
Electromagnetic Quantities
Name
Symbol
Name
Symbol
Velocity of Liquid
И, о, го
Magnetic Force
a, B, y
Molecular Rotation
έ, η, ζ
Electric Current
u, v, w
L, M, N
Electromagnetic
F, G, H
Velocity Potential
due to Vortex
Vortex Filament
Doublet Sheet
Circulation
Flux through Vortex
Ө
K
Momentum
Magnetic Potential
of Current
Electric Current
Magnetic Shell
Work done in
moving a Magnetic
Pole once round
Current
Potential Energy
of Magnetic Shell
Ω
In addition to the papers cited in the preceding chapter, we
may refer to the following by Sir W. Thomson: "Vortex Atoms,"
Phil. Mag. (4) XXXIV.; "Vortex Statics," Proc. Roy. Soc. Edin.
1876; "On Maximum and Minimum Energy in Vortex Motion,"
Phil. Mag. (5) XXIII. p. 529.
The theory of rectilinear and circular vortices will be discussed
in the second volume.
EXAMPLES.
89
EXAMPLES.
w
1. Liquid is contained in a simply-connected surface S; if
is the impulsive pressure at any point of the liquid due to any
arbitrary deformation of S subject to the condition that the
enclosed volume is not changed, and the impulsive pressure for
a different deformation, show that
I do' d8 - ff de ds.
dn
dn
2. If a sphere be immersed in a liquid, prove that the
kinetic energy of the liquid due to a given deformation of its sur-
face, will be greater when the sphere is fixed than when it is free.
3. If V be the attraction potential of a uniform circular
lamina of unit density, in the plane of xy, prove that wdV/dz will
be the velocity potential of a circular vortex filament coinciding
with the boundary of the lamina.
4. The boundaries of a liquid are two fixed concentric cylinders
of radii a and c. Prove that if the motion of the liquid is irrotational
and in two dimensions, the velocity potential must be equal to
к0/2π, where к is the circulation round any closed circuit which
embraces the inner cylinder once only; and that the kinetic
energy is equal to x² (4π)-¹ log a/c.
K
5. Apply the equations of impulsive motion, to show that if
liquid be contained within a closed surface, the circulation and the
molecular rotation cannot be altered by any impulse applied to
the boundary.
6. A mass of ice is contained within an ellipsoidal case which
is rotating in any manner about its centre: prove that if the ice
be melted and the boundary be deformed in such a manner that
it remains ellipsoidal, the resultant molecular rotation at any
point is proportional to the diameter of the ellipsoid which is
parallel to the tangent to the vortex line at that point.
.
*
CHAPTER V.
ON THE MOTION OF A LIQUID IN TWO DIMENSIONS.
96. THE solution of questions relating to the motion of a
liquid in two dimensions, can be most conveniently effected by
means of Earnshaw's current function . This function when the
motion is irrotational, which will be the case in most of the
problems discussed in the present chapter, satisfies the equation
+
=
= 0 .....
dx² dy²
.(1),
the solution of which is
¥ =ƒ (x+cy) + F (x − ɩy) ..
.(2).
Also
u =
dy
dy'
dy
ข
dx
.(3).
If the liquid is bounded by fixed surfaces, the normal component
of the velocity must vanish at the boundaries. This condition
requires that const. at all points of boundaries which are fixed.
97. When the cylindrical boundary is in motion, the following
conditions must be satisfied at its surface.
(i) Let the cylinder be moving with velocity U parallel to the
axis of x, and let be the angle which the normal to the cylinder
makes with this axis; then at the surface
u cos + v sin 0 = U cos 0.
0
Now cos 0 = dy/ds; sin 0 =— dx/ds; therefore by (3)
dự
ds
=
U
dy
ds
:
CONJUGATE FUNCTIONS.
Integrating along the boundary, we obtain
where A is a constant.
¥ = Uy + A .
(4),
91
(ii) If the cylinder be moving with velocity V parallel to the
axis of y, the surface condition in the same manner can be shewn
to be
зв
Vx+B......
(5).
(iii) Let the cylinder be rotating with angular velocity w;
then at the surface
wy cose + wx sin
= — wy
u cos ✪ + v sin 0 =
αψ
dr
or
wr
ds
ds'
Therefore
зв
{{wr² + C ..
where r = √x² + y³.
………….(6),
When there are any number of moving cylinders in the liquid,
conditions (4), (5) and (6) must be satisfied at the surfaces of each
of the moving cylinders.
In addition to the surface conditions, must satisfy the
following conditions at every point of space occupied by the
liquid; viz. ↓ must be a function which is a solution of Laplace's
Equation (1), and which together with its first derivatives must be
finite and continuous at every point of the liquid. If the liquid
extends to infinity and is at rest there, the first derivatives must
vanish at infinity.
Conjugate Functions.
98. DEF. If § and ŋ are functions of x and y such that
§+in=ƒ (x+iy)…………..
then & and n are called conjugate functions of x and y.
(7)
Differentiate (7) with respect to x and y respectively, eliminate
the function f, and equate the real and imaginary parts in the
resulting equation, and we shall obtain
dę _dn dę
de
dn
dx dy' dy
;
dx
(8).
92
MOTION IN TWO DIMENSIONS.
[
Now, if and be the velocity potential and current function
of a liquid, it follows that if and are written for § and ʼn
respectively, equations (8) are satisfied; hence and ✈ are
conjugate functions of x and y.
99. Again
dę dn, d§ dn
dx dx dy dy
2
+
2
0 ..
.(9),
αξ
de dn dn
2
+
+
= J²
……………….(10),
dx
dy
dx dy
V² = 0, V²n = 0
………….(11),
where
▼² = d²/dx² + d²/dy³.
Equation (9) shows that the curves = const., n = const. form
an orthogonal system. Equations (2), (7) and (11) show that
whence
2§ = ƒ (x + ɩy) + F (x − iy)\
2ın = f (x + ɩy) − F (x — iy)S
§ — in = F (x − iy).
(12),
Hence if (x, y, c) = 0 be the equation of any family of
curves which can be expressed in the form
2x (c) = 2§ = ƒ (x + ɩy) + F′ (x − ɩy)
the equation of the orthogonal system of curves will be
2ın = f (x+ iy) − F (x — ɩy),
where n is constant along each curve of the orthogonal system.
η
Again we have
m
αξ
de
dx +
dy,
dé
=
dx
dn
dn=
dx +
an dy.
dx
dy
Therefore if ds be the distance between two adjacent points,
J³ds² = d¿² + dn².
Hence if ds, ds, be small arcs of the curves § and ʼn respec-
tively
Jdst = dn\
Jds₁ = des
dé)
.(13).
100. If and are conjugate functions of § and ŋ, then & and
are conjugate functions of x and y.
CONJUGATE FUNCTIONS.
93
For
and
therefore
$+ vf = F (§ + in)
§ + in=ƒ (x + ɩy),
$+cf = x(x + ɩy).
101. Let p and q be the velocities perpendicular to § and ŋ
in the directions in which these quantities increase, then
df
=J
p
dsn
αξ
dn
аф
do
q
J
Jdy
ds&
dn
αξ
(14).
If we consider a small curvilinear rectangle bounded by the
curves §, n; §+d§, n + dn, the difference between the fluxes over
the faces +§§ and ʼn + dŋ, and those over the faces § and η is
dp αφ
+ dźdn = Jª (db
de dn²
ď &
+ dady,
d§² ' dn².
2
η
but if we choose the two tangents to the curves § and ʼn at their
point of intersection as the axes of x and y, the difference between
these fluxes will be
Hence
V²p = J²
V²pdxdy.
độ đ
dE dn
+
2
(15).
In the case of an irrotationally moving liquid, both sides of
this equation must be zero; hence Laplace's equation when trans-
formed into any variables §, which are conjugate functions of
x and
y, becomes
Ꮠ
+
dr²' dn²
ď &
=
0......
…………….(16).
If we assume as the value of any solution of (1) or (16)
and substitute this value in any of the three equations (4), (5)
or (6), we shall obtain a system of curves, any one of which would,
by its motion in the prescribed manner, produce lines of flow
determined by the equation = const.
102. We shall now give some examples.
1
1
(i) Let
y = - & Va²
+
x + ly X
Ly
Vax
дов
..(17).
94
MOTION IN TWO DIMENSIONS.
:
1
f
When ra, - Va; also the velocity is finite and con-
tinuous at all points outside the cylinder r=a, and vanishes at
infinity; hence is the current function when a circular cylinder
of radius a is moving in an infinite liquid with velocity V parallel
to y.
The velocity potential is
1
or
1
210 = Va²
x + iy
x — iy
φ
Vary
p²
(18).
The paths of individual particles of liquid due to the motion
of a cylinder along a straight line, have been calculated and traced
by Clerk-Maxwell'.
(ii) If the liquid instead of extending to infinity is bounded
by a fixed concentric cylinder of radius c, the initial motion of the
liquid can be obtained as follows.
Since (x + y)" is a solution of Laplace's equation, it follows
that " (A cos no + B sin no) is also a solution, where n is any
quantity positive, negative, real or complex.
Ꮳ
Hence if the inner cylinder be moved along the axis of a with
initial velocity U, we may put
B
$ = (Ar + 2) 008 0.
cos
When r=a, dø/dr = U cos 0, whence
B
A
U.
a²
When r=c, do/dr = 0, whence
B
А A-2=0.
Therefore
Ua2
Φ
r
c² - a²
(~ + = 9)
cos 0.
r
.
This is the expression for the initial value of the velocity
potential. The motion at any subsequent time after the cylinders
have ceased to be concentric will be determined in § 122.
1 "On the Displacement in a case of Fluid Motion," Proc. Lond, Math. Soc.
vol. III. p. 82,
EXAMPLES OF CONJUGATE FUNCTIONS.
(iii) Let
¥=†A {(x+vy)³ + (x — ıy)³}
= A(x³-3xу³) = Ar³ cos 30.
Substituting in (6) the equation of the boundary becomes,
▲ (x³-3xу³) + {w (x² + y²) = C ... (19).
95
If we choose the constants so that the straight line x = a, may
form part of the boundary, we find
W
2wa²
A ; C =
6a
3
Hence (19) splits up into the factors
(x − a) ; x+y√3 + 2a ; x − y√3 + 2a.
The boundary therefore consists of three straight lines forming
an equilateral triangle, whose centre is the origin.
Hence is the current function due to liquid contained in an
equilateral prism, which is rotating with angular velocity w about
an axis through the centre of inertia of its cross section. The values
of ↓ and 4, when cleared of imaginaries, are
W
за
p³ cos 30,
6a
(iv) Let
W
Ф p³ sin 30.
6a
¥ = {A {(x + ɩy)² + (x − ıy)²}
= A (x² — y³).·
Substituting in (6) we find
Putting
▲ (x² − y³) + {w (x² + y³) = C.
w+2A
1
W
- 2A
1
20
α
a²;
2
20
b2
>
.(20).
the equation of the boundary becomes
x²
a²
+
and
= 1 ....
α
b
(21),
w a² — b²
ป
2 a² + b²
(x² — y²) ………….
(22),
is therefore the current function due to the motion of liquid
contained in an elliptic cylinder, which is rotating about its axis.
The preceding value of is also the current function, when
the liquid is bounded by two concentric, similar and similarly
situated elliptic cylinders,
96
MOTION IN TWO DIMENSIONS.
103. To find the current function when liquid is contained in a
rectangular prism which is rotating with angular velocity w about
its axis¹.
If 2a, 2c be the sides of the cross section of the prism, the
boundary conditions are
df
น
wy,
when x = ±a,
dy
df
ข
= wx,
when y = c.
dx
Also
Let
√² = 0.
x − 1 w (x² + y²) = ↓ ;
then
dx = 0,
dy
x = ± α
(23),
and
dx = 0,
dx
▼²x - 2w = 0
y = ± c
.(24).
Let
x= Σ (0 cos λx + y sin λx),
where
and
are functions of y alone.
Substituting in the first
of (23) we obtain,
do
de
Σ
cos λa +
λα
dy
dy
sin λa) = 0,
therefore
<= 0,
Hence
λ = (2n+1)
π
2a
x=Σθ cos (2n + 1)
2n+1
TX
2a
Substituting this value of x in (24), we obtain
π-2
(2n+1)² Ꮎ cos (2n + 1)
d²
Σ
dy²
Now
S
ra
and
-a
-
·a
4a2
2n+1
cos (2n + 1)
cos (2m + 1)
2a
...
(25).
πTOC
2w = 0 ...(26).
TX
dx
2a
(−)" 4α
(2n + 1) π
πTOC
πTOC
cos (2n + 1)
dx = 0 or a,
2a
2a
according as m is unequal or equal to n.
1 Stokes, "On some cases of Fluid Motion," Trans. Camb. Phil. Soc. vol. vIII.
p. 105. Ferrers, "Solution of certain questions in Potentials and Motion of
Liquids," Quart. Journ. vol. xv. p. 83. For the expressions for the component
velocities of the liquid in terms of elliptic functions, see Greenhill, Quart. Journ.
vol. xv. p. 144,
A
RECTANGULAR PRISM.
97
Multiplying (26) by cos (2n + 1) π/2α, and integrating between
the limits a and — a, we obtain
Ꮎ.
dy³
2n+1
(2n+1)³π³
4a2
Ꮎ .
2n+1
(−)¹8w
(2n+1) π
=
= 0,
therefore
02n+1=A2n+1 cosh (2n + 1)
пу
2a
2n+1
+ B₂+1 sinh (2n + 1)
TY
( − )"32a³w
2a
(2n+1) 773
1)³π³ •
If we substitute this value of 0n+1 in (25), and then substitute
the resulting value of x in the second of (23), we obtain
B
2n+1
=
= 0,
χ
(-)"32a³w
A 2n+1
(2n + 1)³π³ cosh (2n + 1) πс/2α'
whence
8
x=
77-3
32 a²∞ ( - )" cos (2n+1) πx/2a. cosh (2n + 1) πу/2α
(2n + 1)³ cosh (2n + 1) πc/2α
32a²w
77-3
Στο
(− )" cos (2n + 1) πж/2α
(2n+1)³
Now if z lies between 7 and – ½π,
1
1
8
Π
COS Z
cos 3z +
33
53
cos 52 –
32
8
hence the second series is equal to a (a² - x²), and the value of
is therefore
¥ = − wa² — {w (x² — y³)
зва
32a²w Σ (−)" cos (2n + 1) πx/2a cosh (2n+1) πy/2α
+ 3
π
(2n+1)³ cosh (2n + 1) πс/2α
A more symmetrical expression may be obtained from the
consideration that must be unaltered when a and c are written
for c and y; making these changes and adding the results we
obtain,
↓ = - {w (a² + c²)
16a²w
+
Σ
πT-³
3
16c²w
+
ΣΠ
7-8
(−)” cos (2n + 1) π/2а cosh (2n + 1) πy/2α
(2n+1)³ cosh (2n + 1) πс/2α
-)" cos (2n+1) Ty/2c cosh (2n+1) π/2c
(2n+1)³ cosh (2n + 1) πα/2c
B.
7
98
MOTION IN TWO DIMENSIONS.
}
104. To find the velocity potential when liquid is contained in
a cylinder whose cross section is the sector of a circle, which is
rotating about an axis through the centre of the circle¹.
Let a be the angle of the sector, a the radius of the cylinder,
w its angular velocity, then
đợi
+
dr²
dp 1 dp, 1 do
+ m² do
r dr
0 .....
(27),
and the surface conditions are
1 do
= wr, when 0 = 0 or a
(28),
r de
аф
0,
when r=a
……………….(29),
dr
Let
also must not become infinite when r = 0.
$ = R cos λ (0 +B),
where R is a function of r alone.
Substituting in (27) we obtain
d² R
1 dR
Rx²
2
+
0,
dr²
r dr
22
the solution of which is
R = Ar² + Br-λ.
Hence since has not at present been determined, the value of
may be written
= A¸ r² cos 2 (0 + ßo) + Σ (Ar² + Br−λ) cos λ (0 + B).
Substituting in (28) we obtain
24¸ r² sin 2 (0 + B。) + Σ λ (Ar² + Br¯λ) sin λ (0 + ß) = — wr².
0
This equation is satisfied, provided
= ω, when a or 0,
2A¸ sin (20 + 2,6.) — — w
Σλ (Ar² + Br−λ) sin λ (0 + ß) = 0,
which requires that
=
2B = ½π — α,
B=0,
24¸cos
24¸ cos a = — @,
λ = (2n+1) π/x.
1 Stokes, "On the critical values of the sums of periodic series," Trans. Camb.
Phil. Soc. vol. VIII. p. 533. Greenhill, "Fluid motion in a rotating semi-circular
cylinder," Mess. Math. vol. vIII. p. 42; "Fluid motion in a rotating quadrantal
cylinder," Ibid. p. 89; "Fluid motion in a rotating rectangle formed by two
concentric circular arcs and two radii," Ibid. vol. ix. p. 35; "On the motion of a
frictionless liquid in a rotating sector," Ibid. vol. x. p. 83.
ROTATING SECTOR.
99
Therefore
φ
@p2
2 cos a
sin (20 — 1)
+ Σo {A¸r(2n+1)π/a + B₂r-(2n+1)/a}cos (2n + 1) π0/α.
0
n
Since must not be infinite when r=0, B = 0; substituting
in (29), we find that for all values of ✪ between a and 0,
π
wa sec a sin (20 − a) + 4 (2n+1) a(2n+1)/a cos (2n +1) π0/x = 0,
α
n
whence by Fourier's theorem
‡πA„ (2n+1)a (2n+1)=/a = wa² sec a sin (20-a) cos (2n+1)π0/a de
n
therefore A,
n
W
and o
2 cos a
0
4wa²a²
2)
4a² — (2n + 1)² π² ›
8waⓇa
2
π (2n + 1) {(2n + 1)² π² — 4x²}
r² sin (20 — a)
a¯(2n+1)π/a¸
∞ /~\ (2n+1)π/α
+ Swa²a Σo (1)
0
cos (2n + 1) π0/α
π (2n + 1) {(2n + 1)² π² — 4a³} *
105. The interpretation of this expression presents no difficulty
so long as a<π, but when a> the velocity becomes infinite at
the origin. The following explanation of the motion which takes
place when this is the case, is given by Prof. Stokes :
A
"Let OAB be a section of the sector made by a plane
perpendicular to the axis, and cutting it
in O. Suppose the cylinder turning round
O in the direction indicated by the arrow.
Then the liquid in contact with OA and
near O, will be flowing relatively to OA,
towards O, as indicated by the arrow at O.
When it gets to O, it will shoot past the
face OB; so that there will be formed a

B
surface of discontinuity indicated by the dotted line, extending
some way into the liquid, the liquid underneath this line and near
O flowing in the direction AO, while the liquid above is neatly
at rest.”
7
100
MOTION IN TWO DIMENSIONS.
*
Whenever a liquid is flowing past a sharp edge, the analytical
expression for the velocity, calculated on the assumption that the
liquid is perfect and flows according to the electrical law of flow,
always becomes infinite at the edge; a result analogous to that
which occurs in the theory of the distribution of electricity on
conductors, where it is found that the analytical expression for the
density upon a conductor having a sharp edge becomes infinite at
the edge.
The mathematical investigation of the discontinuous motion
which takes place in such cases is one of great difficulty, but
certain special cases will be considered in the next Chapter.
霉
​106. The problem of finding the velocity potential and current
function, when a cylinder whose cross section is a given curve, is
moving in an infinite liquid, has been solved in comparatively
few cases. The theory of conjugate functions affords a powerful
method of attacking such problems, but the principal difficulty
consists in finding a relation between the complexes + in and
x+ɩy, such that the given boundary shall be represented by some
particular value of one of the functions § or 7.
The principal solutions of this problem, which have hitherto
been obtained, will be given in the following articles.
107. Let
then
x + iy = c cos (§ — in)
.
x = c cos & cosh η,
y:
= c sin ‡ sinh
η,
(30),
η
and the curves = const., §= const. are a family of confocal ellipses
and hyperbolas.
If a and b be the semi-axes of the cross section of the ellipse
n=B, then
a = c cosh ß,
Also
b = c sinh B,
a² — b² = c².
2
J2
c² (cosh 2n
cos 2€)
...
(31).
Here ʼn may have any positive value, and § may have any real
value whatever; when ŋ=0, the ellipse becomes a double line
joining the foci; and when ŋ= ∞ the curves become circles; also.
J vanishes at infinity.
ELLIPTIC CYLINDER.
Now satisfies the equation
101
ď³, d³
+
0......
dε² + dn²
(32),
and this equation is satisfied by the expression
An +Σ¸ª e¯nn (A„ cos n§ + B₂ sin n§) ………………….. (33),
which is the proper form of a potential function outside an elliptic
cylinder, since by (14) and (31), it makes the velocity vanish at
infinity.
To find the form of inside the cylinder, we observe that (32)
is also satisfied by the series
1
(A, cosh nn cos n§+B, sinh nn sin n§+C, sinh nn cos ng
+ D„ cosh nŋ sin n§) …………… (34).
Now if we examine the components of the velocity in the
neighbourhood of the line joining the foci, it will be found that
they will be discontinuous, unless dy/dn and dy/dę either vanish
or change sign in passing from one side of this line to the other;
the last two terms of (34) are therefore inadmissible. Hence every
potential function, which together with its first derivatives is finite
and continuous inside an elliptic cylinder, must be of the form
Σ₁ (4„ cosh nŋ cos n§‍+ B„ sinh nŋ sin n§)……………………(35).
This value also makes the component velocities finite at the
foci; for in the neighbourhood of these points Jc = (dn² + SE²) −³,
and from (35) both dy/de and dy/dn are infinitesimals of the first
order.
y
Hence, by (4) and (5) if ¥., ↓ be the current functions when
the cylinder n =ẞ is moving parallel to x and y with velocities
U and V respectively,
η
20
| = Uce-n+® sinh 6 sin f
(36).
Again,
Y₁ = − Vce-n+³ cosh ß cos §
ม
-
2
r² = x² + y² = {c² (cosh 2n + cosh 2§).
Hence, if, be the current function when the cylinder is
surrounded with liquid and is rotating with angular velocity w, we
must put
√3 = A€˜2(n−ß) cos 2§.
102
MOTION IN TWO DIMENSIONS.
and
Substituting in (6) and putting ʼn = ß, we obtain
Hence
A cos 25+ c²w (cosh 28+ cos 2§) = C.
C = 4c²∞ cosh 26,
W
A =-1c²w,
Y₁ = {c² wε 2n+28 cos 2.
3
(37).
The value of the velocity potential may be deduced from the
preceding values of or from the corresponding expressions for an
ellipsoid, which will be given in Chapter VII. and which were first
obtained by Green' and Clebsch". The expressions in the text are
due to Prof. Lamb³.
The motion of a liquid in a rotating cylinder, whose cross
section is formed (i) by the arcs of confocal ellipses and hyperbolas,
(ii) by arcs of confocal parabolas, has been investigated by Dr
Ferrers*.
108. We shall now solve the same problem for a cylinder
whose cross section is the inverse of an ellipse with respect to its
centre".
Let
.(38),
then
x + by
= c sec (§+ in)…………..
ос
p² = c²
y²
cosh2n
+
sinh2
>
x²
y
p² = c² 2
cos²
2
sin² E
whence the curves
a, n =ẞ are the inverses of a family of
confocal hyperbolas and ellipses with respect to their common
centre.
CX
Also
p2
cosh n cos §,
cy
p2
Y = sinh n sin
2c²
22
cosh 2n + cos 2§,
(cosh 2n + cos 2§)²
J
2c² (cosh 2n — cos 2§)
1 Trans. Roy. Soc. Edin. vol. xiii. p. 54.
2 Crelle, vol. LII. p. 119.
3 "Some hydrodynamical solutions," Quart. Journ. xiv. p. 40.
4 Quart. Journ. XVII. p. 227.
5 Ibid. vol. XIX. p. 190, and vol. xxi. p. 336.
(39).
CROSS SECTION THE INVERSE OF AN ELLIPSE. 103
Here ŋ may have any positive value, and § any value positive or
negative, but as the values of x and y are periodic with respect to §,
it is only necessary to consider values of § lying between 0 and 2π.
ๆ
When is large the curves n=const. consist of small oval
curves about the origin, with which they ultimately coincide when
n=∞; and when 70 they become two double lines extending
from the points x=±c to infinity in the positive and negative
directions respectively.
η
Also when ʼn is large
η
J=1€/c.
Hence, within the cylinder, every potential function must be of
the form
form
Σ₁ e¬nn (A„ cos n§ + B sin n§) …………………………………….(40).
N
Outside the cylinder, every potential function must be of the
Σ₁ (4„ cosh nn cos n§ + B„ sinh nŋ sin n§)…………………….(41),
for the velocities will be discontinuous along the two double
lines, unless dy/d§ and dy/dn either vanish or change sign in
crossing from one side of these lines to the other, and (41) is the
only form which satisfies these conditions. This form also makes
the velocity at the points = ±c finite.
Now
x + vy
= c sec (§ + in)
2ce-n+is
1+€¬2n+2ı§
= 2c (e-n+1 — €−3n+3¢§ + € −5n+5c§ — &c.);
therefore
x = 2c Σ (−)" e−(2n+1)n cos (2n + 1) §,
y = 2c Σ (−)" e−(2n+1) ʼn sin (2n + 1) §.
ย
Hence, if √ √, be the current functions when the cylinder is
moving with velocities U and V parallel to x and y respectively,
Ꮳ
¥x= QUcΣ*(-)”
€−(2n+1)ß sinh (2n+ 1) ŋ sin (2n + 1)§
...(42),
sinh (2n+1) B
Y
↓₂ = − 2Vc Σ (−)n
e-(2n+1)ẞ cosh (2n + 1) n cos (2n+1)§
cosh (2n+1) B
η
..(43),
where ẞ is the value of ʼn at the boundary.
n
104
MOTION IN TWO DIMENSIONS.
109. The two series (42) and (43) constitute the complete
solution of the problem when the motion of the cylinder is one of
translation. The results can however be put into a more compact
form by means of elliptic functions. To do this, let
§ + in = u, § — in = v; K/π = K'/28, so that q=e-28; then
cos (2n + 1) u — cos (2n + 1) v}
↓ x = 2 Uci Σo 1 = q²n +1
(−)nq2n+1
1
UK
{cosecam
cosecam (2K
+.
K)-
cosecam
π
π
2Kv
π
+ K
K)}
therefore
- 1 Ucı (sec u sec v);
1
UKci
20
π
{
2Ku
cosecam
π
+K)
2 Kv
-
cosecam
2
π
+ K + Uy.
Putting 2K/T + K=0, 2Kn/π= p, dn² 0 = a, sn² (p, k') = ß,
and clearing of imaginaries, the term in brackets becomes
S=
2ɩ (1 — aß) k² sn & cn & cn 6 dn 0
(1 − a) (1 — k´²ß) + aß (1 − ß) (α — k¹²)
the functions of being to modulus k'.
The denominator of S
'
therefore
where
= (1 − aß) (1 + aß — a — k´²ß),
S=
2
2ık² sn & cn & cn 0 dn 0
12
2
2
dn* ¢ – dn 0 cn
2ık" sn & cn & sn x
1 - sn² Χ dn²
x=2Kg/π.
Hence we finally obtain
2
Xx
π
UKk'c.
sn (2K§/π) sn (K´ŋ/ß)cn (K'n/B) + Uy...(44),
1 — sn² (2K§/π) dn² (K'n/ß)
the functions of being to modulus k, and those of n to modulus k'.
Similarly
ป
1
2 Ku
2Kv
VKk'c secam
+ secam
- Vx.
π
π
CROSS SECTION THE INVERSE OF AN ELLIPSE.
105
Putting 2K§/π= x, 2Kn/π = 4, dn³ x=a, sn² (4, k') = ß, and
clearing of imaginaries, the term in brackets becomes
S=
2k² (1-aß) cn 0 cn
(a − k²²) (1 − B) + aß (1 − a) (1 − k’³ß) '
The denominator
= (1 — aẞ) {a — k'² (1 + aß − ß)},
2 cn x cn &
therefore
S
1- sn² x dn³
Hence we finally obtain
2
ป
VKkc.
π
- Vx...(45).
110. When the cylinder is rotating about its axis with angular
velocity w, the surface condition is
cn (2K§/π) cn (K'n/B)
1-sn²(2K/π)dn³ (K'n/B)
3
- {wr² + C.
Now 1+
sinh 28
cosh 26 + cos 2
26
€¹³ + cos 2
є
cosh 26+ cos 2§
1
1
1 + c−2(B−ı§)
1+€
+
1+€
S
2 (B +ı§)
= 2 + 2Σ, (−)″ €
n
-2nß
cos 2né,
therefore
1 22 2
c²
cosh 28+ cos 2§
= c² cosech 28 + 2c² cosech 28 Σ (−)" e−2nß cos 2n§.
Therefore
¥3
- wc² cosech 2ß
- 2ẞ € 2nß
- 2 wc³ cosech 28Σ (-)" -2ns cosh 2nn cos 2n
cosh 2nß
..(46).
111. If liquid is contained in a cylindrical cavity bounded by
the curve n = B,
↓ = - wc² cosech 28 - 2wc² cosech 2ẞ Σ, (−)" €
1
-2nn
cos 2n
2,8 (co
sinh 2n
-
cosh 28+ cos 2
-1)
=-wc² cosech 28- wc cosech 28
c²w cosech 2ẞ sinh 2n
cosh 2n + cos 2§
…..(47).
...
106
MOTION IN TWO DIMENSIONS.
112. The results of § 109 admit of various interpretations, by
means of which we can obtain the solutions of several problems in
other branches of physics. Thus the function Y, is
ย
(i) The potential without the cylinder, of the induced
charge, when the cylinder is placed in a field of uniform electric
force parallel to x.
If we invert with respect to the origin, which is equivalent to
putting c²x/r² for x, and x + y = c cos (§ — în), ¥, is
(ii) The potential of the induced charge within an elliptic
cylinder which encloses an electric system whose potential is
Vc²x/r².
(iii), is the temperature within a solid elliptic cylinder
whose boundary is maintained at a temperature - Vc²x/r².
113. The equation
x + y = 2c sec³ 1 (§ + in)
represents a family of confocal limaçons. The curves ŋ= const. are
the inverses with respect to a focus of a family of confocal ellipses,
whilst the curves = const. are the inverses with respect to the
same focus of the orthogonal family of confocal hyperbolas. The
current functions due to the motion in an infinite liquid of a
cylinder whose cross section is the curve n = ẞ, and also of liquid
contained in a rotating cylindrical cavity of this form, may be
obtained in a similar manner to that employed in §§ 109-111 (see
Quarterly Journal, Vol. xx. p. 234).
114. Let us now consider the system of curves given by the
equation
§ + en = } log (œ + sy)* — c°
in
This is equivalent to the system
-
c²
(x² — y² — c²)² + 4x²y² = c²e¹¢
x²- y² - c² = 2xy cot 2n...
— —
………….(48),
...
(49).
(48) is the equation of a family of confocal lemniscates, the
distance between whose foci is 2c; and (49) is the equation of
a family of rectangular hyperbolas, each of which passes through
the foci of the lemniscates and cuts them orthogonally.
CROSS SECTION A LEMNISCATE.
107
It is easily seen that
2x =
= c {1 + €²(E+cn)}³ +c {1 + €²(§−in)}³,
2ty = c {1 + c2(E+en)}} − c {1 + €2(E−in)}},
r² = c² (1 + 2€² cos 2n + e¹5)³,
re-25
J
c²
€
(1 + 2e-2 cos 2n + €−4)³.
C
ʼn
=
and may have any values whatever. At infinity, § = ∞,
J=0; at either of the foci §= ∞ and J = e−2/coo. When
=0 the curve becomes the lemniscate of Bernoulli (r² = 2c² cos 20);
ʼn andπ +ʼn are the angles which the asymptotes of the hyperbola
make with the axis of x, and in the first quadrant ŋ varies from
0 to π.
η
Hence, for motion parallel to x,
4 x
n
− §Ucı [{1 + e−2(§−2a−ın)}³ — {1 + e−2(§−2a+in)}³]…..(50),
and for motion parallel to y
- † Vc [{1 + c−2(§−2a−in)}³ + {1 + c−2(§−2a+in)}³]…….(51),
where a is the value of § at the surface.
115. Before dealing with the rotation of the cylinders, we
shall make a short digression for the purpose of considering the
coefficients of cos ne in the expansion of (1 + 2c cos 0 + c²)³, which
we shall denote by L,, where c< 1.
Now
(1 + 2c cos 0 + c²)² = (1+ceïº)³ (1 + ce¯ïº) ś
where
S
n
= (1 + 1½ ceio + √₂c² €²ie + ... S₂ce-nio +…..)
+ ... Snc”€˜nio
× (1+ ce-i + S₂c²e-2i0 +
1
(−)”-¹1.3. 5 ... (2n − 3)
2".n!
S₂c"ε-nie + ...),
;
therefore L₂ = 2c" {Sn + & Sn+1 C² + Sn+2 S₂c² + Sn+з S₂c® + ...}.
n
The value of L„, however, may be put into a more convenient
form for calculation, for
c²j¹
+1₂+1 = [ " (1 + 2c cos 0 + c³)³ cos (n + 1) OdO
C
π
cos no
n✪ − cos (n + 2)0
2 (n + 1)√ 。 (1 + 2c cos 0 + c²)³
do.
เ
108
MOTION IN TWO DIMENSIONS.
Also
T
bπ (L₂ — Ln+2) = ["
0
(1+2c cos+c) {cos no-cos (n+2) 0}
π (1+ c²) (n + 1) L. +c|
C
π (1 + c²) (n + 1)
therefore
C
(1 + 2c cos 0 + c²)
Π
0
cos (n-1) - cos (n+3) 0
(1 + 2c cos 0 + c²)³
Ln+1+π {nL₂+ (n + 2) In + ₂} ;
(2n +5) Ln+2 + (2n − 1) L₂+
Also
ኬ+2
2 (1 + c²) (n + 1)
C
+πL₂ = √" (1 + 2c cos 0 + c²)³ do
0
= (1 + c) E (k, žπ), where k
E (k) = k'² ( F + k
dF
dk
F (k) = (1 + c) F (c);
Now
Also
therefore
dF (b) = {F (c) + (1 + c)
dk
L
do
ᏧᎾ
= 0... (52).
n+1
2/c
1+c
dF (c)) (1 + c)2 √//c
dc
= {E (c) − (1 − c ) F (c)
1-c
(1 + c) 2
(1 − c)² √c ³
= {
− (1 − c) F (c) ;
therefore
E (k)
=
2E (c)
1+ c
L。 = 2
{2E − (1 − c²) F'}………..
…………….(53).
therefore
Again,
+πL₁ = √" (1 + 2c cos 0 +
(1 + 2c cos 0 + c²)³ cos de
0
π
c sin² Ode
0
(1 + 2c cos 0 + c²)
Π
cos ede
CF (c) − + πL, + ↓ (1 + c²) ["。 (1 + 20 cos @ + c)
cF — 1πL₁+
c²)
1+c²
therefore
T.
‡πL₁ = cF+
{½πL。 − (1 + c²) F};
4c
therefore
L₁
1
2
3пс
{(1 + c²) E − (1 − c²) F}.. (54).
..
.:
CROSS SECTION A LEMNISCATE.
109
Having obtained the values of L and L,, the values of the
successive functions can be calculated by means of the sequence
equation (52).
116. To find the current function due to the rotation of the
cylinder in an infinite liquid.
(i) Let & be positive at the surface of the cylinder and equal
to a, then
r² = c²€²ª (1 + 2e−2ª cos 2n +€−4α)}
c'e²a L (a) cos 2nn,
n
where L (a) is put for L₁ (e-2a).
n
n
Hence ¥₁ = − {wc²€²α ΣL, (a) e−2n (-a) cos 2nŋ.........(55).
3
n
(ii) When § is negative at the surface, the cylinder consists
of two portions, which we must suppose to be rigidly connected
together; in this case let = -x at the surface, where a is
positive; then
3
- {wc² Σ° L₂ (a) e−2n(§+a) cos 2nn
(56).
In the case of a cylindrical cavity filled with liquid, the values
of y are
2a
2n
— † wc² 2ª Σ00 €²n (§-a) L (a) cos 2nn - wc²e2ª L。 (a)…………..(57),
∞ 2n (a+§)
n
Ε L₁ (a) cos 2nn – ½wc³L。 (x)…………………… (58).
and
- ½ wc² Σ₁ €
1
n
117. When a = 0, and the cross section becomes a lemniscate
of Bernoulli, the preceding formulae become much simplified.
Putting u = x + ɩy, v = x − ɩy, we obtain
¥ x = − & Ucı
Vc
ข
U
/v² - c² √u².
V
น
.(59),
.(60).
+
'v² - c² ' √u² -
118. The values of when the cylinder is rotating about its
axis may be obtained in this case without having recourse to the
general formulae of § 116, for the value of ² at the boundary is
2c² cos n, whence †, wc² e- cos n. This may be expressed in
the form
3
- twcs
u²
2
1
C
+
1
√o² -
.....
...(61).
110
MOTION IN TWO DIMENSIONS.
119. To find when the liquid is contained in a cylindrical
cavity formed by one of the loops of the curve, we observe that
cannot contain any lower power of ethan 2 (§ being of course
negative), otherwise the velocities would be infinite at the foci,
where Je-2/c. Now
r² = 2c² cos n i
also for all values of ʼn between π and —
Therefore
n=∞
cos η = π + Σ
n=1
n-1
both exclusive,
(−)"−¹ cos (2n + 1) n
2n+1
·1 §
¥ = − wc² { ± π + Σ°° (−)"−¹ e(2n+1) ŝ cos (2n + 1) n
2n+1
wc² (1π + €³ cos n + 1 tan¬¹
+ 1) ₂ }
cos n
(62).
sinh
120. Lastly, let us consider the equation
Then
x + y = c tan 1 (§+ in).
tan § = tan 1 (§ + in + § − in)
2cx
c² — x² — y² ·
.(63).
Therefore
Also
x² + y²+2cx cot
c² 0...............(64).
ηπ
+ tanh ? = tan } (+ in - §tin)
+
2cıy
c² + x² + y² ·
Therefore
x² + y² - 2cy coth n + c² = 0
n+c²
……....
..(65).
Again,
x + iy = c
=c
=
Y
1
J =
C
sin † (§ + in) cos † (§ — in)
cos § (§+ in) cos § (§ — in)
sin § + sinh n
cosh n + cos &
c sin
cosh n + cos &
c sinh n
cosh n + cos
= = (cosh n+cos §).
Therefore
X =
..(66),
DIPOLAR COORDINATES.
111
121. Equation (64) represents a family of circles whose cen-
tres lie on the axis of x, at a distance - c cot & from the origin 0,
and whose radii are equal to c cosec . Each circle passes through
two fixed points A and B on the axis of y, whose distances from 0
are c and — c.

A
0
Χ
B
The angle & is half the angle subtended by AB at the centre
of the circle. Hence the curve = 0 represents the portion AB of
the axis of y. When & has any positive value between 0 and π the
curve consists of that segment of a circle passing through A and B
which lies on the positive side of the axis of y; and when §=π
the curve becomes the whole of the axis y except the portion AB.
When έ has any negative value between 0 and =π the curves
consist of segments of circles described on AB, and which lie on the
negative side of the axis of y.
Equation (65) represents two families of circles whose centres
lie on the axis of y, at distances + c coth n from 0, and whose radii
are equal to c cosech n. These circles do not cut the axis of x.
η
ๆ
When = ∞ the curve reduces to the point A; when ŋ has
any positive value the curve represents a circle surrounding this
point; and when
and when ŋ= 0 the curve becomes the axis of x. When
η
η
n has any negative value the curve represents a circle surrounding
the point B, with which it ultimately coincides, when ŋ= — ∞ .
n= oc.
112
MOTION IN TWO DIMENSIONS.
Let P be any point on one of the circles A, then
AP² = x² + (y − c)²
2cy (coth n-1),
BP = 2cy (coth n + 1);
AP/BP = €¯”.
η
Whence every circle of the system is such that the ratio
AP/BP is constant along each circle; therefore A and B are the
common inverse points of each circle of this system. In con-
sequence of this property the coordinates § and ʼn are called di-
polar coordinates.
122. We can now find the current function when two circular
cylinders are moving in any manner in an infinite liquid¹.
25
ปูน
Let n=a, n=-ẞ be the equations of the two cylinders sur-
rounding the points A and B respectively; and let x₁, Y₁; x₂, — Y₂
be coordinates of any point on the cylinders A and B respectively,
then
x₁₂+ iy₁ = ctan
1
therefore
Again,
= CL
(§ + ca)
1 - €
ιξ α
1 + €¹Ê-a
∞
= cɩ {1 + 2Σ1° ( − )n ε−na (cos n§ +ɩ sin n§)};
201
n§)
2c, (-)" e-na sin ng
γα
Y₁ = c + 2c Σ₁ ( − )n e−na cos n§)
1
x₁₂- iy₂ = c tan 1 (§ — iẞ)
2 2
c 1 — €¯¹Ê—ß
し
​therefore
—
n§
1+2Σ,1° ( − )” e−nß (cos ng — ɩ sin n§)
x₂ = − 2c Σ¸° ( − )n ε−n³ sin n§
E – e-n® sin
X2
Y₂ = c + 2c Σ₁ (− )n €¯nß cos n§)
૧)
…..(67).
..(68).
Let u, v be the component velocities parallel to x and y of
the cylinder A, and u', v' those of B; then
↓ = uy₁ − vx₁ + const. at A
1
..(69).
зв
u'y₂ — v'x, + const. at B
¹ Greenhill, "Functional Images in Cartesians," Quart. Journ., vol. xv. pp.
356-362. See also Hicks, Ibid. vol. xvi. pp. 113 and 193.
TWO CIRCULAR CYLINDERS.
113
Hence
↓ = 2c Σ₁ (− )n e-na
Σ1º
sinh n(n+B)
sinh n (a + B)
(u cos ng + v sin n§)
— 2c Σ1 (− )n e−nß
sinh n (a − n)
Ε
sinh n(a+B)
(u' cos n§ – v' sin n§)…..(70).
If the cylinder a were moving inside the cylinder ß, we should
obtain in the same manner
1
↓ = 2c Σ₁ ( − )n e¬na
sinh (n − B)
sinh (α — ß)
(u cos n§+ v sin n§)
sinh (a − n)
+2cΣ₁ ( − )n e−nß
(u'cos n§+v' sin n§) ...(71).
sinh (a – B)
123. We shall hereafter require an expression for the kinetic
energy T of an infinite liquid in which two cylinders are moving.
By Green's theorem,
27
P
[
Now
-π
df
ป de
dn
dy
зв
de
n=a
π
dn
n=ß
Ya = 2cΣ₁ (−)n e-na (u cos ng + v sin n§),
df
dn
α
α
1
= 2c Σ₁ (−)" ne-na coth n (a + B) (u cos n§ + v sin n§)
+2c Σ₁ (−)n ne¯n³ cosech n (a + B) (u´ cos n§ – v' sin n§).
Hence the first integral
Σ¸¸ne
(a+B)
= 4πc² (u² + v²) Σ, ne-2na coth n (a + B)
+4πc² (uu′ — vv) Σ, ne¬n (a+ß) cosech n (a + B).
1
Similarly the second integral is equal to
4πc² (u´² + v´'²) Σ, ne-2nß coth n (a + B)
Hence
Σ*°
— 4πc² (uu′ — vv) Σ, ne¬na+B) cosech n (a + B).
2T = P (u² + v²) + Q (u'² + v²²) + 2L (uu' — vv') …………….(72),
P=4πρc² Σ₁₂ne-2na coth n (a+B)
4προΣ, ε
where
Q
=
4πрc² Σ₁ ne-2m³ coth n (a + B)
………..(73).
L = 4πρc² Σ₁₂ ne-n(a+ß) cosech n (a + B)
X
1
B.
8
114
MOTION IN TWO DIMENSIONS.
124.
Before we can make use of the foregoing values of P, Q
and L, it will be necessary to express them in terms of the radii
a and b of the two circles and their coordinates. To do this, let
0₁ = ε¯ª, 0₂ = ε˜³, q=e-a-8; then
0,
1
2
P = 4πο Στηθ
2n
1+9²n
1
1
1 – q²n
2n
m=∞
Now
n=00
= 4πc² Σ no 2" + 2n Σ n0²n q
Σηθ Σταθε
n=1
1
m=1
q+2q²+3q³ +...... =
9
2mn
}
(1 − q)² ;
therefore, inverting the order of summation,
m=∞
Ρ = Απο Σ
m=1
0,
Ꮎ
2
+
1
20,2q2m
1
2 2m\2
(1 — 0,²)² † (1
(1 − 0,² q²m)²)
1
a = c cosech a = 2c0¸/(1 − 0¸³)
Now
therefore
2
P = πа² 1+ 2Σ
Similarly
Again
1
(1-0,2) 3q2m)
(1 − 0, "q*¹)³)
Q = Σ,
πb² 1 + 2 x∞ ( 1
{1
L = 8πc² Σ,
2
nq²n
1 - qin
N=∞ m=∞
- 8πο Σ
=
Σng"
n=1 m=1
m=∞
1
2mn
q²m
2m\2
8πε Σ
:1 (1 − q²m)³
m=1
m=00
1
2 2ın\2
2
− 0¸²)² q
(1 −
2
2m
2 2m\2
0,² q²m)²
=
- 2παό Σ
1
(1 − 0¸²) (1 − 0¸²) q²m-1
2
2
m=1
(1 - 92m) 2
(74).
.(75.)

.(76).
2
Since the quantities 0, 0, are functions of the respective
distances of the circles a and B from the axis of y, these values
of P, Q and L are of the required form. The coordinate a does
not enter into the expressions for the coefficients.
The kinetic energy of a liquid in which two cylinders are
moving, was first obtained by Mr W. M. Hicks¹: the investigation
given in the text is due to Prof. Greenhill".
1 "On the motion of two cylinders in a fluid," Quart. Journ. vol. xvi. pp. 113
and 193.
2 "Functional Images in Cartesians," Quart. Journ. vol. xvII. pp. 356–362.
+
EXAMPLES.
115
EXAMPLES.
1. An elliptic cylinder is filled with liquid which has molecular
rotation at every point, and whose particles move in planes
perpendicular to the axis; prove that the lines of flow are similar
ellipses described in periodic time
π (a² + b²)
abs
2. A fixed cylinder whose cross section is any one of the
lemniscates rr' = c, where c is any constant and 2a is the distance
between the points from which r, r' are measured, is surrounded
by an infinite mass of water in steady cyclic irrotational motion;
show that the stream lines are all lemniscates of the same system,
and that the velocity along a stream line at any point varies as the
distance from the centre.
Prove also that the polar coordinates (referred to the centre) of
a liquid particle in terms of the time t are given by
r² = a²cnµt ± c³dnµt,
20=amut, k = a/c.
3. The cross section of a cylinder is a sector formed by the
circle r = a, and the lines = ±a. Prove that if the cylinder be
rotating with angular velocity w,
↓ = −1 wr²
cos 20
cos 2x
Swa² aΣ®® (-)+1
(r/a) (2n+1) π/2a
cos (2n+1)π0/2x
(2n + 1) π {(2n + 1)² π² — 16a²}'
4. The transverse section of a uniform prismatic vessel is of
the form bounded by the two intersecting hyperbolas represented
by the equations
√/2 (x³ — y³) + x² + y² = a³, √/2 (y² − x²) + x² + y² = b².
If the vessel be filled with water and made to rotate with
angular velocity w about its axis, prove that the initial component
velocities at any point (x, y) of the water will be
W {2y³ − 6x³y + √/2 (a² — b²) y}
a² + b²
W
a²+b²
respectively.
{2x³ — 6xy² + √√/2 (b² — a²) x}
8-2
116
MOTION IN TWO DIMENSIONS.
5. A cylinder whose cross section is the limaçon
γ
cos² 10 sech² 1 + sin² 10 cosech² 1ß,
2c
is in motion in an infinite liquid with velocities U, V parallel to
respectively; prove that
the lines = 0,
0 0
↓ = 8 UcΣ₁ (−)″-¹ ne-ns cosech nẞ sinh nŋ sin ng
n-1
— 8VcΣ¶® (−)”−¹ ne¯n³ sech nẞ cosh nŋ cos n§,
1
where § and ʼn are conjugate functions such that
x + y = 2c sec² † (§ + in).
6. Prove that if the cylinder in the last example be rotating
in an infinite liquid with angular velocity w,
1
↓=-8wc³ cosech³ß {cosh ẞ+2coshẞΣ (−)" e-n³ sechnßcoshnn cosnę
+ 2 sinh ẞΣ, n (−)” €¯n² sech nẞ cosh nn cos n§},
and that if a cylindrical cavity of this form be filled with liquid
and made to rotate,
зва
η
8c* [cosh 8 sinh
sinh® 3 ]cosh n+cos $
sinh ẞ (1 + cosh n cos έ))
(cosh ŋ + cos §)²
2
7. A circular cylinder is moving parallel to the axis of x;
prove that if there is cyclic irrotational motion about the cylinder
the velocity potential is
K
кө a²x
$
2π
where x is the circulation round any closed circuit embracing the
cylinder once.
8. A hollow cylinder of radius a, closed at both ends, is
divided into two parts by a plane diaphragm through its axis, and
filled with liquid. If the vessel be made to rotate about its axis
with angular velocity w, prove that the motion of the liquid
relative to the vessel will be such that its velocity potential is
r²+2ar cos 0+a²
cos 20-2 logar cos+a²
wa2
p=C+{wr²sin 20+
8π
[G
+
a²
2ar sin
- 2
2
γ
2.2
sin 20 tan
sin 20
-1
4
a² — p²
a
(2 + 2) cos 0],
where r, ◊ are polar coordinates of any point of the liquid.
9. Prove that
EXAMPLES.
117
* = log (a+a)*+
y²
(x − a)² + y³
gives a possible motion in two dimensions. Find the form of the
stream lines, and prove that the curves of equal velocity are
lemniscates.
10. In the irrotational motion of a liquid, prove that the
motion derived from it by turning the direction of motion at each
point in one direction through 90° without changing the velocity,
will also be a possible irrotational motion, the conditions at the
boundaries being altered so as to suit the new motion.
Discuss the motion obtained in this way from the preceding
example.
11.
Liquid is moving irrotationally in two dimensions, be-
tween the space bounded by the two lines =+ and the
0
curve r³ cos 30a³. The bounding curves being at rest, prove
that the velocity potential is of the form
=r³ sin 30.
12. The space between the elliptic cylinder (x/a)² + (y/b)² = 1,
and a similarly situated and coaxial cylinder bounded by planes
perpendicular to the axis is filled with liquid, and made to rotate
with angular velocity w about a fixed axis. Prove that the
velocity potential with reference to the principal axes of the
cylinder is w (a² — b²) xy/(a² + b²), and that the surfaces of equal
pressure when the angular velocity is constant, are the hyperbolic
cylinders
x²
2
y²
C.
3a²+b² 362+ a²
13. If & = ƒ (x, y), ¥ = F(x, y) are the velocity potential and
current function of a liquid, and if we write
x =ƒ(Þ, ¥), y= F(4, ¥)
and from these expressions find and ; prove that the new
values of and will be the velocity potential and current
function of some other motion of a liquid.
=
Hence prove that if = x²-y, 2xy, the transformation
gives the motion of a liquid in the space bounded by two confocal
and coaxial parabolic cylinders.
*
118
MOTION IN TWO DIMENSIONS.
14. In example 12 prove that the paths of the particles relative
to the cylinder are similar ellipses, and that the paths in space are
similar to the pericycloid
x = (a+b) cos 0+ (a - b) cos
α
-
a+b`
b
2
2
y = (a + b) sin 0 + (a − b) sin (a+b).
−
α
Ꮎ .
15. Water is enclosed in a vessel bounded by the axis of y
and the hyperbola 2 (x² - 3y²)+x+ my=0, and the vessel is set
2(∞²
rotating about the axis of z. Prove that
$ = 2 (3x²y − y³) + xy — ½m (x² — y³),
¥
=2(x³ — 3xy³) + ½ (x² − y³) + mxy.
16. When the stream lines for steady motion are similar
concentric and similarly situated ellipses, the motion of a particle
is the same as if it were acted upon by a central force to the
centre; and if the potential of the impressed forces is a function
of the distance from the centre, the lines of equal pressure are
circles.
17. The coordinates (x, y) of a particle at time t are given by
x = a + A cos 2nπt + В sin 2nπt,
y=b+λA sin 2nπt - AB cos 2nπt,
where A, B, λ and n are constants with regard to x and y, but A
and B functions of a and b. Prove that if the different particles
corresponding to different values of a and b are the particles of a
liquid, A and B must be conjugate functions of the complex
a+bλ. Under what conditions is a free surface possible?
18. The space between two confocal coaxial elliptic cylinders
is filled with liquid which is at rest. Prove that if the outer
cylinder be moved with velocity U parallel to the major axis, and
the inner with relative velocity V in the same direction, the
velocity potential of the initial motion will be
cosh (B — n)
$ = Uc cosh n cos § – Vc
sinh a cos %,
cosh (Ba)
where n =ẞ, n=a are the equations of the outer and inner
cylinders respectively, and 2c the distance between their foci.
EXAMPLES.
119
19. If in the last example the outer cylinder were to rotate
with angular velocity, and the inner with angular velocity w,
prove that initially
cosh 2 (-a)
$ = +Nc²
sinh 2 (3 – a)
sin 2§ – 1wc²
cosh 2 (B — n)
sin 28.
sinh 2 (B − a)
->
20. If u = x + iy, v = x y, and n be any positive real
quantity, prove that when a cylinder whose cross section is the
curver" = 2c" cos no is moving with component velocities U, V
parallel to the axes, in an infinite liquid, the current function is
¥ = U¥₂+ V¥vi
where
+x
Y
Hence prove
зва
— § cı {v (v″ — c″)¯* — u (u" — c″)˜}},
_
† c {v (v” — c”)˜¯* + u (u” — c″)˜}}.
·Zc
"
that if the cross section is the cardioid
r = 2c (1 + cos 0),
↓₂ = 2rcª sin ¿0 (√r − √c cos †0) (r + c − 2 √rc cos ¿0)−²,
20
↓₁ = − rc (r + c cos 0-2 √rc cos 0) (r + c − 2 √rc cos 10)˜º.
ย
CHAPTER VI.
ON DISCONTINUOUS MOTION.
125. IN the preceding chapter, we obtained expressions for
the velocity potential and the current function of a liquid which
is flowing past an elliptic cylinder, and it might be thought that
by making the minor axis of the cross section vanish, we could
obtain the solution for a stream which is flowing past a rect-
angular plate. This however is not the case; for if the minor
axis be made to vanish, it will be found that the velocity of
the liquid becomes infinite at the edges, and therefore the pressure
becomes equal to -, which indicates that a hollow would be
formed in the neighbourhood of the edges. In order that the
motion represented by the formulae should be possible, it would
be necessary that at every point of the liquid boundary of
the hollow, the pressure should be constant, and therefore the
liquid boundary would have to be a line of constant pressure
as well as a stream line; but it is not difficult to show from
the formulae that it is not possible for a line of constant pressure
to coincide with a stream line, and hence the formulae fail when
the cylinder degenerates into a rectangular plate.
126. The problem of determining the steady motion of heat
and electricity, is precisely the same as that of determining the
motion of an irrotationally moving liquid subjected to the same
boundary conditions, so far as the velocity potential is concerned;
but there is an important distinction between the two problems,
for in the former the pressure condition does not exist. Hence the
solution of problems in the conduction of heat or electricity cannot
COMPLEX VARIABLES.
121
receive a hydrodynamical interpretation, unless the value of the
pressure given by that solution never becomes negative at any
point occupied by the liquid;-in other words, whenever it is
possible for the liquid to flow according to the electrical law of
flow; but when this is not the case, the hydrodynamical applica-
tion of such formulae would give results, which although in many
cases approximately representative of the motion at a considerable
distance from the region of negative pressure, certainly do not
give correct results in the neighbourhood of this region.
127. We have noticed in Chapter IV, that there is nothing
in the nature of a perfect fluid to prevent slipping taking place
between two contiguous layers, and we have shown that a surface
along which slipping takes place is a surface of discontinuity,
which possesses the properties of a vortex sheet; but the possibility
of such slipping is not taken into account in the ordinary theory,
which assumes that the liquid flows according to the electrical
law. But in order to solve problems in which liquid is flowing past
a sharp edge, it will be necessary to take into consideration the
possibility of slipping; and we must therefore endeavour to obtain
a solution, such that a certain surface of no flux which passes
through the sharp edge shall also be a surface of constant pressure.
This surface of no flux will either form the free boundary of the
liquid, or will constitute a surface of separation between the moving
liquid and a region of liquid at rest, and in the latter case will be
a surface of discontinuity along which slipping must take place.
The only problems of this class which have yet been solved are
problems of two dimensional motion, and the method of solution
is due to Kirchhoff¹ and depends on the properties of complex
variables.
128. Any complex variable x+y, may be represented geo-
metrically by means of a vector drawn from the origin to the
point whose rectangular coordinates are (x, y).
If we put x = r cos 0, y =r sin 0, the length of the vector will
ber, and will be the angle which its direction makes with the
axis of x.
The quantities r and are respectively called the
modulus and amplitude of the complex x+ɩy.
The sum of two vectors x + y and a + b is x+a+ (y + b),
which represents a vector drawn from the origin to the point
(x+a, y+b). Hence the sum of two vectors is represented by
¹ Crelle, vol. LXX.; and Vorles. über Math. Phy. Chapters XXI., XXII.
122
DISCONTINUOUS MOTION.
!
the diagonal of the parallelogram of which the two vectors are
adjacent sides.
a
Similarly the difference between two vectors is represented by
a line drawn from the origin, which is equal and parallel to the
line joining the opposite extremities of the two vectors.
The product of the two vectors is
where
Hence
(x+ iy) (a + b) = ax − by + ɩ (bx + ay)
= R(cos + sin o),
し
​R cos = ax – by, R sin & = bx +ay.
R² = (a² + b²) (x² + y²),
b/a+y/x
tan o
1- by/ax
Whence the product of two vectors is a vector whose length is
equal to the square root of the product of the two vectors, and
whose direction is inclined to the axis of x, at an angle which is
equal to the sum of the inclinations of its factors.
Similarly the quotient of two vectors is a vector whose length
is equal to the square root of the quotient of the two vectors, and
whose direction is inclined to the axis of x, at an angle which is equal
to the difference of the inclinations of the dividend and divisor.
X
129. Let z and w denote the two complexes x + ɩy and +i;
and let x and y be rectangular coordinates of a point P in a plane,
which we shall call the plane of z; and let ☀ and be rectangular
coordinates of a point P' in another plane which we shall call the
plane of w. Then if w and z be connected by any relation
w = f(z), it follows that if P trace out any curve in the plane of
z, P' will trace out a corresponding curve in the plane of w.
for
130. Every function of a complex has a differential coefficient,
dw_do+id¥
dx + idy
f' (z)
dz
(dp/dx + id¥/dx) dx + (dø/dy +id/dy) dy
dx + idy
do dv
do
dy
And since
+
+
dx
dx
dy dy
this ratio is independent of the ratio dy/dx.
KIRCHHOFF'S METHOD.
123
If and be the velocity potential and current function of
a liquid,
dw_do d↓
dz
dz
1
Therefore
dw
и
-
+
dx dx
ιυ
1
= U w.
(u + iv) = 5 (say),
where q is the resultant velocity of the liquid; hence the vector
represents the reciprocal of the velocity of the liquid.
131. In the class of problems which we are about to consider,
the boundaries of the liquid consist partly of straight lines which
constitute the fixed boundaries of the liquid, and along which the
direction of the velocity is necessarily constant; and partly of the
free surface of the liquid or of surfaces of discontinuity, which
divide the moving liquid from the region of liquid at rest, and along
which the pressure and consequently the magnitude of the velocity
must be constant. Hence, if we choose the scale of measurement
such that g=1 along the latter surfaces, the boundaries will
become transformed in the plane of into an arc of a circle
of unit radius, which corresponds to the free surface, or surfaces of
discontinuity; and into the radii of this circle, which correspond
to the fixed boundaries. The points where the radii meet the
circle correspond to the points where the fixed and free boundaries
intersect; also since the velocity must not become infinite, can
never vanish, and therefore the portion of the plane of external
to this circle and included between the two radii, corresponds
to the portion of the plane of w occupied by the moving
liquid.
Along the boundaries fixed and free, of the liquid in the plane
of %, we must have a, and y = ß, where a and ẞ are constants;
hence the corresponding portion of the plane of w consists of the
space included between the two parallel straight lines y=a,
y = B.
We must therefore endeavour to connect and w by a relation,
such that the above mentioned portions of the two planes of
and w shall correspond; and also that certain points in these
two planes shall correspond to certain points in the plane of z.
When this has been effected, the relation between z and w, which
determines and in terms of x and y, must be obtained by
integration.
124
DISCONTINUOUS MOTION.
.'
132. We shall define a lune as the space which is included
between two circular arcs which meet but do not cross.
The angle of a lune is the angle at which the arcs meet.
Let z=x+iy, z=x+y', where (x, y), (x', y') are the rect-
angular coordinates of two points P, P' in the planes of z, z′
respectively. We shall now show that if P trace out any lune of
angle a in the plane of z, and P' trace out another lune of angle a'
in the plane of z', it is possible to connect z and z by a relation,
such that the angular points of the two lunes shall correspond;
and also that any third point on the perimeter of one lune
shall correspond to any third point on the perimeter of the
other.
!
The equation
Z
AZ + B
CZ + D
· DZ + B
or Z
CZ' + A
……..(1),
where A, B, C, D are complex constants, transforms any circle in
the plane of Z into another circle in the plane of Z'. For if the
point P describe a circle about the point c = a + ib as centre, we
must have
mod (Z-c) = const.
.(2)
or
(x − a)² + (y — b)²= const.
Substituting the value of Z in terms of Z from (1), (2)
becomes
mod K
Z C
Z' - C
= const.
………….(3),
2
P2
where K, C₁, C, are new complex constants. Now if k, P₁, P₂ are
the moduli of K, Z' - C₁₂, Z' - C₂, (3) may be written
kp₁
1
P₂
const.,
whence P' moves so that the ratio of its distances from the
two fixed points C₁, C₂ is constant, and therefore describes a
circle.
Since (1) contains three disposable constants, viz. the ratios of
the three quantities A, B, C, to D, it follows that these ratios may
be chosen, so that a circle passing through three given points in
the plane of Z shall correspond to a circle passing through three
given points in the plane of Z'.
TRANSFORMATION BY COMPLEX VARIABLES.
133. Let
X+iY=Z=
2 C2
where
c₁ = a+ib, c,= a + iß.
125
.(4)
1
Let A and B be the points c, and c₂. The vector Z being
the quotient of the two vectors AP and BP, is represented in the
plane of by a straight line whose inclination to the axis of X
is equal to APB. Now if P describe a circle passing through A
and B, the angle APB is constant, hence every circle passing
through the points A and B in the plane of z, corresponds to a
straight line passing through the origin in the plane of Z.
Also if P and Q are any two points on two different circles passing
through A and B, the inclination of the two corresponding lines
in the plane of Z is equal to BQA – BPA, that is to the angle of
the lune AQBPA. Hence (4) transforms any lune in the plane
of z into two straight lines in the plane of whose inclination is
equal to the angle of the lune.
If we put
Z=Z",
the two straight lines in the plane of Z become transformed into
two straight lines in the plane of Z inclined at an angle n times
as great; hence if a be the angle of the lune and n =π/a, the
equation
π
Z:
(==)
a
(5)
transforms a lune in the plane of z whose angle is a and whose
angular points are c₁, c₂ into a single straight line in the plane
of Z.
Similarly if z' be any other plane, the equation
Z
π
..(6)
transforms a lune in the plane of z whose angle is a' and whose
angular points are c', c' into a single straight line in the plane
of Z'.
If therefore we substitute the values of Z, Z' from (5) and (6)
in (1), the resulting equation transforms any lune of angle a in the
plane of z into a lune of angle a' in the plane of z; and by
suitably choosing the ratios A: B: C: D, we may make any three
points on the perimeter of one lune correspond to any three points
on the perimeter of the other.
•
126
DISCONTINUOUS MOTION.
ག
134.
We must now notice some particular cases.
(i) Let
whence
$+1
z=εw or x + vy
x=eº cos ↓, y=eº sin y.
(7),
When = 0 or π, y=0; hence (7) transforms the two parallel
straight lines =0, π in the plane of w into the single
y y
straight line y = 0 in the plane of z.
=
(ii) Let
z=√/w or x+iy=√&+1&
Putting
φ
4 = R cos x, y = R sin
X,
we obtain
x
= √R cos x,
y=√R sin x-
(8).
When √R sin x = const. = c, y = c; hence (8) transforms the
confocal parabolas R sin xc in the plane of w into the
parallel straight lines y=c in the plane of z. Now if c = 0 the
parabolas degenerate into a double line extending from the focus
to ∞. Hence (8) transforms a straight line in the plane of w ex-
tending from a fixed point to infinity, into the whole of the axis
of x, in the plane of z.
(iii) Let us now consider the portion of space bounded by
the straight lines OA, OB in the plane of ,
which is external to the circular arc AB.

B
γ
A
is
If y is the inclination of OA, OB, the
equation ' = ç™/ transforms the two straight
lines OA, OB in the plane of into a single
straight line in the plane of
arc AB into the semicircle ab.
'; and the
Hence the
transformed region in the plane of y', is the portion of space lying
р

b
0
α
a'
on the upper side of a'b', and which is bounded by the semicircle
and the infinite straight lines aa', bb'. This region may be regarded
as a lune of angle π, one of whose arcs is the semicircle apb;
and whose other arc consists of the infinite lines aa', bb', which
MOTION OF A JET.
127
may be regarded as an arc of a circle whose centre is at infinity.
By (5), the equation
-1
2
2
'+1
transforms this lune into a single straight line in the plane of z,
hence the required transformation is
2
Σπιν
Gπ/Y + 1,
1
(9).
135. We shall now apply the preceding method to the solu-
tions of some special problems.
A jet of liquid escapes by a slit AB from a large cistern of
which the side is x'x; required the motion, which is supposed to
be in two dimensions.



Χ
A
B
Χ
X
B
B
P
x
W
A
P
The figures show the corresponding lines in the planes of z,
and w; corresponding points being represented by the same letters
in each of the three planes, and the fixed and free boundaries and
their corresponding lines by thick and thin lines respectively.
The lines A, Bx along which the direction of the velocity is
invariable, are represented in the plane of by the straight lines
x'A, Bx; and the free surface of the jet along which the magnitude
of the velocity is invariable and equal to unity, by the semicircle
APQB. The portion of the plane of lying above the line
xBQPAx', corresponds to the space occupied by the liquid. In
the plane of w this space corresponds to the region contained
+
128
DISCONTINUOUS MOTION.
between the parallel straight lines d'AP and xBQ. Let &=0,
=π be the stream lines x'AP, and xBQ: also let & = 0 be the
equipotential surface passing through A and B.
In order to transform the region in the plane of to that
in the plane of w, we must put y = π in (9) and we obtain from (7)
and (1),
१-
2
(3-1) = 4e" + B
Cεw + D'
Since a liquid flows from places of lower to places of higher
velocity potential, the following conditions must be satisfied:
(i) ☀=-∞, Y=∞, (ii) µ = ∞, ( = − 1,
φ
$
(iii) w = 0, {=1, (iv) w = IT,
iπ, $= — 1.
Of these (i) gives B=D; (ii) gives A =- C; and (iii) and (iv)
both give A =- B; whence
or
5
(1)
dz
2 1 — εw
1 + εw'
พ
€
+
2w — 1.
-
dw
Let be the angle which the tangent to AP makes with AB;
0, and p is positive; hence
along AP q=1,
whence
Also
cos 0 +ɩ sin 0 = e−º + i 1 − e−26,
cos 0 = €˜º,
sin 0=√1-e-26.
do
ds
=
1,
therefore measuring s from A, we obtain
and
therefore
s = $
dx
= cos 0 = €➖s,
ds
x = 1— € 8.
∞
(10),
A being the origin. When s∞, = 1; also since the final
width of the jet is π, the width of the slit is π +2.
The ratio of the final width of the jet to the width of the slit,
is called the coefficient of contraction of the jet, which is there-
fore equal to π/(π + 2) or '611.
Again
dy
ds
MOTION OF A JET.
129
= sin 0 = √1 — €−28,
1 + √1 − e−28
(11).
1 − √1 − e−2s
y = √1 − e−28 — log
Eliminating s between (10) and (11), the equation of the
free surface of the jet is
y = √ 2x − x² – † log
1
1+√2x
- √2x - x²
―
x²
-
Also the radius of curvature is tan 0, which vanishes at the
origin.
If we put /a for we obtain the solution when the boundaries
xB, x'A are inclined at an angle a.
136. Let us now suppose that the conditions of the last
example are varied by introducing a tube projecting inwards'.


A'
Β'
Α'
A
A'
B
B'
σ
A
Β'
ω
B
The containing vessel is supposed to be so large that we may
disregard what takes place at the sides. The motion will then be
as follows. The liquid will flow along the side B'B, and at B the
direction of its velocity will begin to change, and the liquid will
finally flow out in a stream whose section will be less than that of
the tube.
¹ Helmholtz, Phil. Mag. Nov. 1868.
B.
9
130
DISCONTINUOUS MOTION.
1:
Since the aperture of the tube is supposed to be small, the
curve in the plane of which corresponds to the free boundaries
may be approximately regarded as a circle, and if we put ¿' =N'S
the space bounded internally by this circle and the lines AA',
BB', will be transformed into the region in the plane of in
the last example. The solution in this case may be obtained from
the last example by writing √ for §, and we obtain
dz
dw
· 5 = (e¬w + √2e−2w — 1)²,
2€˜2w — 1 + 2€˜w √€˜2w — 1.
Along the free surface of the jet, we have
S =
φ
cos 0 + ɩ sin 0 = 2e−2 − 1 + 2ɩe˜º √ 1 − e−26,
dx
し
​therefore
= cos 0 = 2e−2s — 1,
ds
x = 1 − s − e−2s,
dy
ds
= sin 0 = 2€˜³ √1 − e−2s,
y = e
e-8 √1-e-28 + sin-¹ e-s+y',
the middle point of AB being the origin. When s=∞,
y=y',
so that 2y' is the final breadth of the stream and is therefore equal
to π; when s=
y'
r; when s=0, y = ½π + у' =π, whence AB=2″, and the co-
efficient of contraction.
137. Lord Rayleigh' has shown that if the vessel were of
finite dimensions, the coefficient of contraction must always be
greater than; for let o" be the area of a section of the vessel so
far removed from the orifice that the velocity over it is sensibly
constant and equal to v". Let v', o' be the ultimate velocity and
section of the jet, σ the section of the tube. The equation of
continuity gives
By the principle of energy
v'o' = v"o".
p = { (v²² — v′¹²),
and by the principle of momentum
12
1/2
po = σ'v² — σ"v"2.
The Contracted Vein," Phil. Mag. Dec. 1876.
STREAM FLOWING PAST A LAMINA.
131
From these equations we obtain
2
1
1
+
σ
which shows that the section of the tube is an harmonic mean
between the sections of the cylinder and jet. When o″ = ∞,
o'/σas before.
=
138. When a rectangular lamina is held fixed in a stream
which meets it obliquely, there will be a region of dead water
behind the lamina, which will be at rest, and the total pressure on
the lamina will be due to the difference of pressures upon its
anterior and posterior faces.
The stream line
0 meets the lamina at some point 0 and
then divides, each branch following the lamina to its edges, and
afterwards forming the boundary between the moving liquid and
the dead water behind the lamina.

A'
Z
78
B
Β'
A
B
α
A
A
B

Α
B
W
९
is
'B'
The portion of the plane of
corresponding to the moving
liquid is that which lies below the semicircle AA'B'B and the two
infinite lines Bb, Aa; and the points +∞ correspond to O. The
whole of the plane of w corresponds to the portion occupied by the
moving liquid, with the exception of the double line shown in the
figure, which may be regarded as the limiting form of a parabola.
Let a be the angle at which the stream meets the lamina;
since the equation w/w converts the double line in the plane
9-2
132
DISCONTINUOUS MOTION.
"
of w, into a single straight line in the plane of w', we must
put
1
A √w + B
8+ 1,
C Nw + D'
The conditions to be satisfied are
(i)
=±∞,
= cos a — i sin a,
(ii)
w=0,
<=∞0.
From (i) we obtain
and from (ii) B= D, whence
'- 1)² ___ - С√/w tan³½ï+B
AC tanta,
C
18+ 1,
C √w + B
C
Let
B
= √K (1 + cos a), w =
› = Kw,
2
and we obtain
-1 1 − (1 − cos a) N/
+1
or
1+ (1 + cos a) √/w'
§ = 2 + √2³ − 1 ....
where
1
Ω
= cos a +
Væ
·
(12),
When the velocity of the stream at infinity is equal to V, which
will be supposed to be the case in what follows, we must change
into V, and (12) becomes
¿V = Q + √Q² − 1………..
(13).
In the plane of z let O be the origin, OB the axis of x; along
AB must be real and equal to u, and at A and B = V¬¹.
Hence at all points of the lamina we must have > 1, and at A
and B, 2-1 and + 1 respectively.
Let be the breadth of the lamina, then since along AB
K$=w and dø/dx
= u,
=
· [db = √(√2 + √12² − 1)
-
the limits of integration being determined by
do
(14),
VK
1
Ω
Q = cos a +
±1. .
Nw
MINA.
133
PRESSURE ON THE LAMINA.
If ẞ be a new variable such that
B = √w sin² a
cos a,
the limits of ẞ will be ± 1, and we obtain
ᏙᏦᏓ =
1
{(B+cos a) cos x + sin³z+ √1 − ß² sin a} cosec¹ a dß.
-
}VKI =
| −
Whence
-1
4 + π sin a
K =
Vl sin¹ a
Along the lines AA', BB' the pressure p′ = p (C – † V²), which
must be equal to the hydrostatic pressure of the dead water. At
the surface of the lamina,
P = C – žu²
P
p
+
+ } ( V² — u²).
ρ
Hence the total pressure on the lamina is,
u³)
do
· = [( p − p') dx = $pf (V² — w²) dẹ
= pV ƒ(N² − 1)* d¢
2 Vp
K sin³ a
|` √ I - B² dB =
π V² lp sin a
4 + π sin a
=
И
πVρ
K sin³ a
.(15),
which determines the resistance which the lamina offers to the
stream, and shows that it depends partly upon the square of the
velocity and partly upon the angle which the stream makes with
the lamina.
The moment of the pressure is
1
2VP
G
K sin³ a
-1
Now by (14),
VKx=f{cos a (B+ cos a) + sin² a + √1-B sin a} cosec* adß.
Hence, if the origin be suitably chosen, the value of x will be
30=
B² cos a + 2B + {B √1 − ß² + sin¯¹ ß} sin a
-
VK sin* a
134
DISCONTINUOUS MOTION.
The odd terms in ẞ contribute nothing to the integral, and
therefore
2p
G
2
K* sin' a
--1
πρ cos a
ẞ² √1 - ß² cos adß
@ cos a
4K² sin' a
4KV sin* a
The distance of the middle
point of the lamina from the origin
is cos a/VK sin* a; hence the distance of the centre of pressure
from line middle point is
3 cos a
ᏎᏦᏙ sin* a
a
31 cos a:
4 (4+π sin a)
If π >a>0, the negative sign shows that the centre of pressure
is on the upstream side of the middle point; hence if the lamina
be free to turn about an axis parallel to its edges whose distance
from the middle point is
x=
31 cos a
4 (4+π sin a)
=
.(16);
it will be in equilibrium. If a π, x=0; and the lamina will
set itself transversely to the stream. When a = 0, x is a maximum
and is equal to 37/16, in which case the axis divides the lamina in
the ratio 11: 5.
139. The results of equations (15 and 16), which are due to
Lord Rayleigh', may be stated in another form as follows. "If the
axis of suspension divide the width in a more extreme ratio than
11 5, there is but one position of stable equilibrium, that namely
in which the lamina is parallel to the stream with the narrower
portion directed upwards. If the axis be situated exactly at the
point which divides the width in the ratio 11: 5, this position
becomes neutral, in the sense that for small displacements the
force of restitution is of the second order, but the equilibrium is in
reality stable. When the axis is still nearer the centre of figure,
the position parallel to the stream becomes unstable, and is
replaced by two inclined positions making with the stream equal
angles, which increase from zero to a right angle as the axis moves
towards the centre. With the centre line itself for axis, the lamina
can only remain at rest when transverse to the stream although of
course with either face turned upwards"."
1 "On the resistance of fluids," Phil. Mag. Dec. 1876.
2 Ibid.
135
$
EXAMPLES.
140. In order to obtain the intrinsic equation of the surface
of separation, we have along this surface
¿V=Q+WI − Q³.
1
น
Therefore
Q = cos a +
V
√ K&
аф
Now
ds
therefore
Φ
=
V,
$ = V (s+c),
dx
1
and therefore
= cos 0 = cos a ±
ds
V{VK (s+c)} °
The constant c is to be determined from the fact that when s=0,
cos = ±1. In the case of perpendicular incidence, we have
e
c=1/VK, whence
dx
ds
C
s+c
or
X = 2 (cs+c²)* + constant,
from which it appears that x does not approach a finite limit as s
increases indefinitely.
The methods of this chapter only apply when the motion is in
two dimensions; so far as I am aware, no problem of this class has
been solved when the motion is in three dimensions.
MISCELLANEOUS EXAMPLES.
1. If u, v, w, o are any functions of x, y, z, prove that
uda + vdy+wdz - do has an integrating factor; hence show that
údx
if u, v, w be the velocities of a fluid, then along any vortex line
udx+vdy+wdz =
do.
2. If in an infinite mass of homogeneous incompressible fluid
in equilibrium under finite fluid pressure only, an indefinitely long
cylindrical column be suddenly annihilated, prove that no motion
will take place.
3. Prove that the velocity potential due to a unit source
placed outside a sphere of radius a, and at a distance ƒ from its
centre is
$ = − (r² – 2ƒr cos 0 + ƒ²)¯* — aƒ˜¹ (r² – 2c r cos 0 + c²)−¹
+a¹ {log [c − r cos 0 + (r² − 2c r cos 0 + c²)*] — log r (1 − cos ()},
where (",) are polar coordinates referred to the centre of the
sphere as origin, and c = a²/f.
2
136
DISCONTINUOUS MOTION.
4. Prove that the rate at which the energy of a mass of
liquid, contained within an imaginary closed surface described in
the liquid is increasing, is equal to
[[(p+ p V) q cos edS,
where p
is the pressure, V the potential of the impressed forces,
q the resultant velocity at any point of S, and e is the angle
between the direction of q and the normal to S drawn out-
wards.
5. If a, b, c be curvilinear coordinates of any point (x, y, z) of
a liquid, such that the lines of flow are the intersections of the
surfaces b= const., c = const.; apply § 39 to prove that when the
motion of the liquid is not steady, a first integral of the general
equations of motion is
p
P
where
$f.
q dq
+ V + b q² + 1 3 4 da = F (b, c, t),
J dt
J
d (a, b, c)
d(x, y, z)
-1
6. If the molecular rotation of a mass of liquid which com-
pletely fills a rigid circular cylinder be equal to r¹F" (r), where
r¹F" (r) is any function of r which does not become infinite
within the cylinder; prove that the paths of individual particles
of liquid are circles described in periodic time
2πr³/F (r).
7. In § 135, if v be the velocity at any point on the middle
line of the jet, whose distance from the orifice is y, prove that
1
1+ v
y
log
V
1-v'
the ultimate velocity of the jet being unity, and the scale of
measurement being such that π+2 is the width of the orifice.
CHAPTER VII.
ON THE KINEMATICS OF SOLID BODIES MOVING IN A
LIQUID.
141. In the present chapter we shall obtain expressions for
the velocity potential, in a variety of cases in which a liquid
is bounded externally or internally by moving solids, when the
motion is in three dimensions. We shall suppose that the motion
of the liquid is irrotational and acyclic, and consequently the
motion will be completely determined by means of a velocity
potential which must satisfy the following conditions;
(i)
must be a single valued function, which at all points of
the liquid satisfies the equation V³ = 0 ;
(ii) and its first derivatives must be finite and continuous
at all points of the liquid, and must vanish at infinity if any
portion of the liquid extends to infinity;
(iii) At all points of the liquid which are in contact with a
moving solid, dø/dn must be equal to the normal velocity of the
solid, where dn is an element of the normal to the solid drawn
outwards; if any portion of the liquid is in contact with fixed
boundaries, do/dn must be zero at every point of these fixed
boundaries.
142. Let us now suppose that a single solid is in motion in an
infinite liquid.
Let Ox, Oy, Oz be three rectangular axes fixed in the solid, and
let, be the velocity potential when the solid is moving with unit
velocity parallel to Ox, and let x, be the velocity potential when
the solid is rotating with unit angular velocity about Ox. Let „
P X2 X3 be similar quantities with respect to Oy and Oz. Also
let u, v, w be the linear velocities of the solid parallel to, and
w₁, w,, w₂ be its angular velocities about the axes.
W
2)
138
KINEMATICS OF MOVING SOLIDS.
The velocity potential of the whole motion will be
$ = u$, + v$₂ + w¢¸ + w₁X₁ + @₂X½ + w3X3…………………..(1).
2
3
For if λ, μ, v be the direction cosines of the normal at any
point x, y, z on the surface of the solid, we must have at the
surface
do₁
dn
3
1
λ,
d&2
dn
dos
μ,
=V,
dn
dx₁
dx2
dxⓇ
=== λε
vx,
vy - με,
dn
dn
dn
μα - λψ.
аф
Hence
dn
=(u−yw¸+zw¸)λ+(v−zw¸+xw¸)µ+(w− xw₂+yw₁) v
= normal velocity of the solid.
143. To find the velocity potential when a sphere of radius a is
moving parallel to the axis of x¹.
Let u be the velocity of the sphere, a its radius, the angle
which the radius to any point on its surface makes with Ox, then
at the surface,
or
when r = a.
аф
= u cos 6,
dn
αφ
= u cos 0………..
dr
(2),
Since the motion is symmetrical with respect to Ox, and the
velocity must vanish at infinity, & must be of the form
A‚Ð₂
φ
+
A。 A,P,
1 1
+
2 2
+......
ņ
p2
p3
where P is the zonal harmonic of degree n. Substituting in (2),
we obtain
n
Ꮎ
A 2A, cos 0
-48
0
1
a³
α
&c. = u cos 0,
A₁ = A
A₂ = &c. = 0,
whence
and
A₁
therefore
1
2
- Lua³,
$ = − 1ua³
cos
22
OC
- Lua³
23
(3).
1 Poisson, "Mémoire sur les mouvements simultanés d'un pendule et de l'air
environnant," Mém. de l'Acad. des Sciences, Paris, vol. ix. p. 521.
TWO SPHERES CUTTING ORTHOGONALLY.
139
144. If the sphere were moving with component velocities
u, v, w, parallel to the axes, the velocity potential would be
$
a³
27.3
(ux + vy+wz).
This expression is the velocity potential of a doublet situated
at the centre of the sphere, whose axis coincides with and whose
source end is turned towards the direction of motion of the sphere.
145. The velocity potential may be determined by the method
of images, when the solid, which is formed by the revolution about
the line joining their centres, of two spheres which intersect at
right angles, is moving parallel to its axis¹.
Let A and B be the centres of
the two spheres, Ca point on their
circle of intersection; then if CS
is perpendicular to AB, S is the
common image of B and A with
respect to the spheres A and B.
Let AC = a, BC =b,
AB=c=√a +b²,

A
S
and let u be the velocity of the solid along AB; also let (r, 0),
(r₁, 0), (r₂, 0₂) be the polar coordinates of any point P referred
to B, S and A respectively as origin.
The velocity potential due to the motion of B alone is
ubs
Φι
cos 0,
2p²
which is the same as that due to a doublet of strength {ub³ at B.
The image of this in A is a doublet at S of strength
3
α
3
AB
- Lub³ (1)",
and the image of this in B is a doublet at A of strength
ab 3
ub³
AB. BS
= {ua³.
This is precisely what is required to give the requisite normal
velocity over A and B, whence
$ = − 1 u
b³ cos 0
p
a³b³ cos 0
+
c³r,*
a³ cos 0.
r,
2
1 Stokes, Math. and Phys. Papers, vol. 1. p. 230.
140
KINEMATICS OF MOVING SOLIDS.
146. The motion of two spheres will be discussed in Chapter
XI., but when the space between two concentric spheres is filled
with liquid, and the spheres are moved in any manner, the velocity
potential of the initial motion can be obtained as follows¹.
Let a and b be the radii of the outer and inner spheres respec-
tively, O their common centre; and let the outer sphere be moved
with velocity u along any direction OA, also let the inner sphere
be moved with velocity v along a direction OB which is perpendi-
cular to OA. Let be the angle which the radius to any point P
makes with OA, X the angle which the plane OAP makes with
the plane OAB.
The surface conditions are
αφ
dr
do
= u cos 0,
dr
arlo
= v sin cos x·
.....(4).
α
The function
B
• = (Ar +232) cos 0 + (Cr+
+(cr 3) sin 6
satisfies Laplace's equation. Substituting in the first of (4) we
sin e cos x
X
must have
2B
2D
A
αγ
3
= U,
C
= 0,
α
3
a³
and from the second of (4)
2B
2D
A
0, C
b³
b3
= v,
whence
A = ua³ /(a³ — b³),
C = — vb³ /(a³ — b³),
B = {ua³b³/ (a³ — b³),
D= { va³b³ / (a³ — b³)
-
ua³
and
$
as bs
じ
​r+
b³
27.2
vb³
3
8
α
cos -
Ꮎ .
-
r+
α
a³ - b³
2r²
sin e cos X.
147. The velocity potential due to the motion of an ellipsoid
in an infinite liquid was first obtained by Green in 1833, for the
case of translation only; the solution was completed for the case
of rotation by Clebsch in 18569.
(i) Let the ellipsoid move parallel to the axis of x with unit
velocity.
¹ Stokes, "On some cases of fluid motion," Trans. Camb. Phil. Soc., vIII. p. 105.
2 "Researches on the vibration of pendulums in fluid media," Trans. Roy. Soc.
Edin., 1833.
Ueber die Bewegung einer Ellipsoids in einer tropfbaren Flüssigkeit," Crelle,
LII. p. 103.
".
AN ELLIPSOID.
141
If V be the potential at an external point of a homogeneous
ellipsoid of attracting matter of unit density, the equation of
whose bounding surface is
(x/a)² + (y/b)² + (z/c)³ = 1,
V=Tabс
bcf" (a²
+
y³
+
شمع
+ y + b² + y c² + f
-1)
dy
{(a²+¥)(b²+¥)(c²+¥)}³ ³
همچ
+
1 ...
(5).
where X is the positive root of the equation
ха
+
y²
a² + λ ˜ b² + λ c² + λ
The potential at an internal point is obtained by putting
x = 0 in the definite integral. We shall write this expression in
the form
V = { (Axx² + Bλу² + С₁²²) — H、…………..
where
Αλ 2πabc
dự
λ (a² +¥)P
&c.
Hλ = παbc
df
P
λ
...(6),
(7),
P = {(a² + y) (b² + ¥) (c² +¥)}³,
and we shall drop the suffix λ, when these quantities refer to an
internal point.
If p is the perpendicular from the centre on to the tangent
plane at x, y, z; the surface condition is,
or
do 1
1
px
dn
α
a²
x do₁, y dó₁
z do₁
OC
+
+
a² dx
b² dy
c² dz α
2
(8).
Since Ax is the x-component of the attraction of the
ellipsoid, this quantity obviously satisfies conditions (i) and (ii) of
§ 142; we may therefore assume that
₁=aAx.
Hence at the surface
афг
= A
dx
2 (4.
2πα αλ
a² dx
do₁
2παχ αλ
dy
do₁
dz
a² dy'
2παχ αλ
a² dz
鸞
​142
KINEMATICS OF MOVING SOLIDS.
Differentiating (5) with respect to x, and then putting λ=0,
we obtain
dλ 2xp²
dx
a²
dλ__2yp²
dλ 2zp²
dy
b2
dz
C²
hence the left-hand side of (8) becomes.
ax (A – 4π)/α²,
2
whence
and
a = (A — 4π)¯¹,
Φι
А xx
A - 4π
It therefore follows that if the ellipsoid is moving with
velocities, u, v, w parallel to the axes.
$
Аxu x Bxvy C₁wz
+
+
A - 4π B-4π C-4π
·(9).
(ii) Let the ellipsoid be rotating with unit angular velocity
about Ox; then the surface condition is
dx₁
dn
= w₁ (ny — mx)
=
w₁pyz (b² — c²)
b2c²
.(10).
Writing for a moment Y and Z for Bay and Cz, it can easily
be shown that the function z Y - yZ satisfies Laplace's equation,
for
dY dZ
▼² (≈ Y − yZ) = 2
dz
dy
d² V
d² V
= 2
= 0,
dzdy
dy dz
2
also at great distance from the origin Y and Z are at least of
the order r², and therefore X, is at least of the order r¹ and
therefore vanishes at infinity.
Let us therefore assume
X₁ = a' (z Y − yZ) = a’yz (Bx — Cλ),
X1
then at the surface
dx₁ = a' (ydz - Z
dn
-zdy
+2
dY
dz
dn dn dn y dn
_ pú′yz {(B − C) (b² + c²) + 4π (b² — c²)}.
b2c
Substituting in (10) we obtain
ď
(b² — c²)
(B − C) (b² + c²) + 4π (b² — c²) ›

therefore
(b² — c³) (Bx — Cλ) yz
Χι
(B − C) (b² + c²) + 4π (b² — c²)
.(11).
The functions X2, X, can be written down from symmetry.
ELLIPSOIDS OF REVOLUTION.
143
148. The quantities A, B and C may be expressed in terms
of elliptic functions of the first and second kinds; but the most
important case is when the ellipsoid is one of revolution.
and
(i) If we put b=c<a, the surface becomes an ovary ellipsoid
Ax
=
2παρ
Απαρ
d¥
¹ (a² + ¥^)® (c² +¥)
=
(a² — c²) ®
√
if (a² + \)³ = (a² — c²)*v; therefore
dv
v² (v² − 1) '
−
A₁ = 4π
−
e
(1 − e²) († log
v+1 1
-
Ax
where e is the excentricity of the generating ellipse. Also
Bλ = С₁ = 2παс²
df
› (a² + 3^)* (c² +¥)²
4π (1 − e³)
e³ 3
.(12),
୮
dv
(v² – 1)²
2π (1-e²)
ע
v + 1\
—
- 1
log
……..(13).
ע
e
3
(ii) If we put a=b> c, so that the surface becomes a
planetary ellipsoid we obtain
Aλ = Вλ = 2πа²с
df
c) λ (a² + y)² (c² +4)*
Απα
dv
L (1 + 292,
V
(1 + v²) ² ›
if (c² +λ)* = (a² — c²)³v;
(a² - c²
therefore
2π (1 − e)
Aλ=
* (
e
3
(cot¹y - 13211)
:).....
ν
.(14),
+
4π (1 − e²)³
Cλ=
e³
со
cot¹»)..
…………….(15).
ע
It will be observed that in the case of an ovary ellipsoid
v=e¹; where e' is the excentricity of the generating ellipse of
the confocal ellipsoid which passes through the point (x, y, z);
and that in the case of a planetary ellipsoid
= √1-e²/e'.
= ע
1
-
A
144
KINEMATICS OF MOVING SOLIDS.
149. If c=0 the planetary ellipsoid becomes a disc, and
₁ = 0; hence a disc which moves parallel to itself cuts through
the liquid without producing any motion.
1
To find the velocity potential when the disc is moving perpen-
dicularly to its plane, we observe that at the surface v = 0; hence
when c and v are small c = av, therefore
therefore
C - 4π
π
a
:)-
2π c
4π
α
$ =
2w
-
π
Z
G
cot¹v).
If μ, v are elliptic coordinates, this equation may be written¹
2wa
(16),
φ
(1 − v cot¹v) μ
π
2wa
q₁ (v) P₁ (µ).
π
By § 99 (14) and § 110 (31), the velocity perpendicular to the
hyperboloid μ = const. is
με
1
1 – µ³ dp
2w
1
2
2
v² + μ² dμ
2
x² + μ
π
2
μ (1 − v cot˜¹v).
At all points in the plane z =0 which do not lie on the disc,
µ = 0, and the velocity perpendicular to this plane
2w
πν
(1 − v cot¯¹v),
which becomes infinite when v = 0. The velocity is therefore
infinite at the edges, as we should expect since the liquid is
supposed to move according to the electrical law of flow.
The solution for a stream flowing past a fixed disc behind
which there is a region of dead water, has not yet been dis-
covered.
1 The function qn (v) is a spheroidal harmonic of the second kind, and is equal to
In
( − 1)†(n+1) Q₂n (‹v) where Qn (v)
Qn (w) where Qn (v) is a zonal harmonic of the second kind. The
potential at an external point of any distribution of electricity upon an oblate -
spheroid which is symmetrical with respect to the axis of the spheroid, can be
expanded in a series of terms of the type In (v) Pn (µ).
ROTATING ELLIPSOIDAL CAVITY.
145
150. To find the velocity potential when liquid is contained in
an ellipsoidal cavity which is rotating about its centre.
1 1
dx₁ = py=(-b).
dn руг
b2
Here
Assume
X₁ = Ayz.
Then
dx₁_py dx₁
pz dx1
dn
+
b2 dy c² dz
=
Apyz (2 + 1).
Equating these two values of dx/dn, we obtain
Hence
b² - c²
Α
b² + c²·
b² - c²
Χι
b² + c²
yz.
This value of X satisfies Laplace's equation, and is such that
X1
the velocities are finite and continuous at all points of the
liquid. Hence
φ=ωι
2
c³ — a²
a y z + w z c² + a²
ba- c²
b² + c²
yz
a² - b²
2
z x + w 3 a² + b² x y ......(17).
151. Let us now suppose that the space between two
concentric coaxial and confocal ellipsoids is filled with liquid, and
that the inner and outer ellipsoids are suddenly moved with
velocities U and V respectively parallel to the axis of z¹.
Let the accented and unaccented letters refer to the outer and
inner ellipsoids respectively; and let
$=Mz+ NC₁z.
The surface conditions are
аф
= Up z
dn
аф
2,
dn
c²
From the first equation we obtain
M + N (C − 4π) = U,
-
M + N (C − 4π) = V,
U (C' — 4π) – V (С — 4π)
(C″
and from the second
(C
M =
C - C
U-V
N=
C-C'
whence
1 Greenhill, "Fluid motion between confocal elliptic cylinders and confocal
ellipsoids," Quart. Journ. vol. xvi. p. 227.
i
B.
10
146
KINEMATICS OF MOVING SOLIDS.
and therefore
φ
U (C′ – 4π) – V (C — 4π) — (U — V) C› ₂ ………….. (18).
— ≈
-
C' - C
If the outer ellipsoid were rotating about the axis of z with
angular velocity 2, and the inner with angular velocity w, the
surface conditions would be
1 1
Q
dx-o pay (3-3), dx = pay (2-1).
dn
= w
α dn
We must therefore assume
$=Mxy+N (Bλ — Aλ) xy.
From the first equation we obtain
1
1
b α
ய
a'2
α
{M + N(B− A)} ( ~ + } }) — 4wN ( ¦ − 1) = ~(1 − 1)
and from the second
1 1
b
2
{M + N (B− A')} ( { ' + '}',) − 4 πN (11-1) -n (1-1)
12
2
=
a
which determine the constants M and N.
12
α
(19),
152. We shall next investigate the motion of a liquid about
an indefinitely thin spherical bowl¹.
Let a be the radius of the sphere of which the bowl forms
a part, O its centre, c the radius of the small
circle which forms the rim of the bowl, A the
pole of this circle which will be called the
vertex of the bowl, Q any point on the bowl;
also let V be the potential at P of a distribu-
tion of matter of density σ on the bowl. Then

n
R
Now
Therefore
v = []
fods
PQ °
PQ² = r² + a² — 2ar cos n.
dv
dr
[John P QⓇ
(r — a cos n) dS
PQ³
hence
cos edS
1 d (Vr) = [[ PQ®
α
dr
where e=π- OQP. The right-hand side of this equation is the
magnetic potential at P of a complex magnetic shell of strength σ.
Proc. Lond. Math. Soc. vol. xvI. p. 286.
SPHERICAL BOWL.
147
153. Let us now suppose that the motion of an infinite liquid
is caused by any system of sources, sinks, or vortex filaments; let
⚫ be velocity potential due to this system (which we shall call the
external system) when the bowl is absent; and let & be the velocity
potential after the bowl has been introduced. Then we may put
$ = N + P,
Φ,
where is to be determined.
If the bowl is fixed, which for the present we shall suppose to
be the case, the surface condition is
when r = a.
bowl.
dQ
dr
аф
dr
This condition is to be satisfied on both sides of the
Now, if we remove the bowl, and substitute over its surface a
sheet composed of doublets, whose axes are in the directions of the
radii passing through them, and whose strength σ, per unit of area,
is such that the normal velocity at every point of the sheet is
equal and opposite to the normal velocity due to P, all the con-
ditions of the problem will be satisfied. But the velocity potential
of such a sheet of doublets is analytically equivalent to the
magnetic potential of a complex magnetic shell of the same
strength, which occupies the position of the bowl, and whose
positive side coincides with the sink side of the sheet of doublets;
hence the problem is reduced to finding the potential and strength
of such a magnetic shell when the normal component of the
magnetic force at the surface of the shell is given.
Now we have shown that, if V be the potential of a surface
distribution of matter upon the bowl of density σ, then
Ω
1 d (Vr)
a dr
;
also, if N, and N, be the values of N at two contiguous points just
outside and just inside the shell respectively, then
Ω-Ω = 4πσ.
The magnetic force at the surface of the bowl is
dQ
1 d² (Vr)
dr
α dr²
1 (d
dV
2
(1
μ²)
α αμ
by Laplace's equation.
1 ď²
d² V
+
2
αμ 1 - µ³ dy³j
μ
10—2
148
KINEMATICS OF MOVING SOLIDS.
3
Now the value of the magnetic force at the surface of the bowl
can always be expanded in a series of spherical surface harmonics
Y; hence, if
n
a² Y
n
dQ
dr
n
V=
and therefore if
=
1
Σ, Γ
n
1);
n (n + 1)
dQ
dr
at the surface, the corresponding value of Vat the surface is
V = a²Σ
Y₂
n (n + 1)
(20)
.(21).
The formula (21) fails when n = 0; the only case, however,
which is necessary for our purpose to consider, is when the mag-
netic force is symmetrical with respect to the axis of the bowl, and
has a constant value F at its surface. In this case,
do
F =
dr
therefore
1 d
2
a² dμ (1 − µ³)
-
d V
dμ'
1 + μ
+B.
I-μ
V=Fa² log (1 − µ³) + ½ A log
Now V must not be infinite when µ = 1, therefore
A = Fa²,
and the value of V may be written.
V = Fa² log a (1 + µ).
But, if an infinite straight line extending from the centre of the
bowl to be electrified with line density Fa², its potential is
· Fa² log r (1 +µ).
Hence V is the potential of the induced charge when the bowl
is under the action of a positively electrified line extending from
the centre to∞o. If, therefore, x be the potential of the bowl,
under the action of a positive charge of unit intensity, situated at
a point on the axis distant u from the centre, and on the negative
side of it.
V = Fa² xdu.
SPHERICAL BOWL.
149
154. The preceding result enables us to find the velocity
potential due to a source situated at the centre of the bowl. In
this case
therefore
therefore
m
Φ
dQ m
dr = a³,
2
α
φ
α
m [® d (rx) du -
m
γ
•
155. To find the velocity potential due to the motion of the bowl
in an infinite liquid.
(i) Consider the case of motion parallel to the axis.
If the liquid were flowing from right to left past the bowl, the
velocity at infinity being equal to w, then
and
whence
at the surface.
Φ
Wz
$ =L₂-wz,
ΦΩ
= w cos e
dr
Hence, if the bowl is moving parallel to its axis with velocity u,
Now, by (21),
$3 = Qz
√x
- ½ wa² cos 0
at the surface. V is therefore the potential of the induced charge,
when the bowl is placed in a uniform field of force parallel to its
axis whose potential is waz + const., whence
1 d (Vr)
$3
α dr
(ii) Let the bowl be moving perpendicular to its axis with
velocity v, and let the plane from which the angle is measured
contain the direction of motion; then if o' be the velocity potential,
dø
dr
= v cos y sin 0,
therefore
V
=
va³ cos y sin
at the surface. V' is therefore the potential of the induced charge,
when the bowl is placed in a uniform field of force perpendicular to
a plane containing its axis.
150
KINEMATICS OF MOVING SOLIDS.
(iii) Let the bowl be rotating about an axis.
It is clear that, if the bowl were rotating about an axis through
the centre of the sphere of which it forms a part, the bowl would
simply cut its way through the liquid without producing any
motion. Now, a rotation about any other axis is equivalent to a
rotation about a parallel axis through the centre, together with a
velocity of translation perpendicular to the plane containing the
centre of the bowl, and the original axis of rotation; hence the
motion of the liquid due to the rotation of the bowl is equivalent
to that due to a properly chosen motion of translation.
156. It thus appears from the preceding articles that the
velocity potential due to the motion of the bowl in a liquid,
depends upon the electro-static potential of an electrified bowl,
which is placed in a field of force whose potential is known. We
shall now show how to find this potential, when the field of force
is symmetrical with respect to the axis¹.

S
α
C
Let ACB be a section of the bowl through
its axis, I the centre of the sphere of which
the bowl forms a part, also let AIC=a,
PIC = 0, IA = a, AB= 2c.
B
If in the equation
1
Ө
(1 − 2h cos 0 + h²)
ια
=1+P₂h+Ph² +......
we put h=e" and equate the real and imaginary parts of the
resulting expressions, we obtain
cos a+ P₁ cos ¾a+P₂ cos ža+
1
2
1
√2 (cos a
cos e)
sin ½ a+P₁ sin §x + P₂ sin §2 + ...... =0
12
(22),
when >α. But if <a, the first series is zero, and the second
t
series = {2 (cos 0 — cos
- os a)}˜¯
1 Ferrers, "On the distribution of electricity on a bowl," Quart. Journ. vol. xvIII.
p. 97.
:
SPHERICAL BOWL.
151
Differentiating the second series with respect to a, we obtain
cos a +3P, cos è̟x + 5 cos §a +
according as 0 < or >a.
sin a
or 0 ......(23),
√2 (cos 0 — cos a)
Multiplying (23) by 2 cos † (2n+1)a, we obtain
cos na + cos (n + 1) x + 3P, {cos (n − 1) a + cos (n + 2) a} +
+ (2n + 1) P₂ {1 + cos (2n + 1) a} + etc.
√/2 sin a cos
(2n + 1) a
or 0,
(cos 0 —
-
cos a
cos α)
a)#
according as < or > a.
If we suppose <a, and integrate both sides with respect
to a, between the limits π and a, we shall find that the series
1 8-x
Σ (2s + 1)
4π2α 8=0
1
Απα
sin (n − s)a, sin (n+s+1)
π
n
S
+
n+s+1
da
tra [ (2n + 1) P. + sina cos + (2n + 1) a dz]
n
π Ja
(cos — cos a)³
P.
P,
.(24).
But if we suppose >a and integrate with respect to a
between the limits a and 0, we shall find that the series in
question vanishes. It therefore represents the density of a certain
distribution of electricity in the bowl. The potential of this
distribution is
S=x
V
1
Σ
sin (n − s) a
+
π
8=0
n-s
sin (n+s+1) a] (a\s+1
n+s+1
α
P¸…………..(25),
if r>a; but if r<a we must interchange a and r and multiply
the result by a/r.
To find the value of V at the surface of the bowl, we must put
r = a, and differentiate with respect to a; we thus obtain
dV
dz
= 2cos (2n + 1) a {cos a + P, cos 3a + P, cos § 52 + ......}
√2 cos (2n + 1) a
= 0
π (cos a
COS
0) $
0 <α,
0> α,
2
by (22).
152
KINEMATICS OF MOVING SOLIDS.
√2
ra
Hence
V=
cos (2n+1) a
dx
0 > a,
π
cos 0)
= F(0)
=
0 。 (cos a
C
o <α.
To determine F(0), let a = in the series (25) for V and
we obtain V P₂
=
The series on the right-hand side of (25) is the potential of
the bowl when placed in a field of force whose potential at the
surface of the bowl is equal to – P₂, and the density is given by
(24); and since the potential of every field of force which is
symmetrical with respect to the axis of the bowl can be ex-
panded in a series of zonal harmonics, we can determine the
potential and density of the bowl when placed in any such field.
157. In order to obtain the potential when the bowl is
placed in a field of force whose potential is waz, we must put
n = 1 in the series (25) and multiply the result by - wa², hence
V
wa² S=∞
Σ
2π
S=0
sin (s-1) a sin (s + 2) a
s-1
+
s+ 2
In order to sum the first series, we have
1
h² (1 − 2hµ + h²)¹ ˜¯h²
therefore
1
ՊՐ
8+1
P¸…………..(26).
1
Р.
+
1
h
2
+P₂+ ......P„h”−² + &c.,
n-2
Ph"-1
n-1
+ &c.
const. + P₁log h + P₂h+
h
dh
h² (1 − 2hµ+ h²)
− (1 − 2µh¯¹ + h˜²)³ — µ sinh˜¹
h-¹
cos
-
sin
Putting h successively equal to ae/r and ae¯"/r, subtracting,
and putting S, for the first series in (26), we obtain
ια
2ır² a¯¹ S₁ = − (a² — 2arµ€¯¹ª +r²€˜21a)§ +(a² — arµ€™ +r² (21α)
2αγμε
- la
re
μα
sinh
-1
a cos e
La
a sin e
sinh-
re"
-
a cos 07
Let
a sin e
=λ
...(27).
a²+r² cos 2a — 2ar cos a cos 0 = λ cos 2x,
r² sin 2a-2ar sin a cos = λ sin 2x,
r² + a² − 2ar cos (a — 0) = p²,
r² + a² – 2ar cos (a + 0) = q².
Then
SPHERICAL BOWL.
153
λ = pq
and the first two terms of (27)
But
= 2√λ sin x.
47* sin³ a — (p −q)² = 2 (λ
(p − q)² = 2 (λ — a² — r² cos 2a + 2ar cos a cos 0)
= 4λ sin² X.
Hence the first two terms
= ± ɩ {4y² sin³ a − (p − q)²} ³.
In order to find the value of the last two terms, let us denote
the quantity in square brackets by - 2.
Since
cosh (sinh¨¹ m — sinh¯¹n) = √√/(1+ m³) (1 + n²) + mn,
we easily obtain
Δια
ια
a² sin² 0 cos 24 = (22 - 2are cos + a²)
0
ια
× (r²-21α-2are- cos 0+a²)
− (p² + a² cos² ( − 2ar cos a cos 0)
=λ = √ (p² + q²)+a² sin² 0,
$
therefore
y = sin¹
-1
9-p
2a sin
= sin-¹
2r sin a
p + q
therefore
α
2r²
S₁ = ± {4r² sin² a − (p − q)²}* +
a² cos 0
2r sin a
−
sin-¹
p²
p+q
The second series can be summed in a similar manner, and we
shall finally obtain,
wa
2c
V₁
r cos e sin-¹
2π
± } {4c² − (p − q)²} §
p+q
3
2r sin a
+
cos e sin-¹
p+q
¿ {4r² sin² a — (p − q)"}³] …..(28).
±
158. If the positive signs be taken, this is the potential at all
points within the space bounded by the plane passing through the
rim of the bowl, and that portion of the sphere passing through
the centre and rim of the bowl, which lies outside the bowl.
The potential for the space enclosed by the bowl and the
plane through its rim is obtained by changing the inverse sine in
154
KINEMATICS OF MOVING SOLIDS.
the first term to π- sin, and taking the negative sign before
the second term, and the positive sign before the fourth term.
The potential for the remaining portion of space is obtained
by changing the inverse sine in the third term to π-sin¯,
and taking the positive sign before the second term, and the
negative sign before the fourth term.
159. We cannot employ an analogous method for determining
the potential when the bowl is placed in a field of force perpen-
dicular to a plane containing the axis, since no analytical theorem
has been discovered for obtaining the potential of a bowl which is
placed in a field of force whose potential is a tesseral harmonic
sin (mp + €m) P™ (cos 0).¹
1
The solution can however be obtained by the following in-
direct method. If we put n = 0 in (25), and sum the resulting
series, we shall obtain the potential of an uninfluenced electrified
bowl. Invert the result with respect to a point P in the plane
containing the rim of the bowl, whose distance from the centre is
equal to ƒ, and multiply the result by – m. We shall thus obtain
the potential when the bowl is under the influence of a positive
charge m at P. Now if we place a negative charge m at
a point P' in PO produced such that OP'=f, and make the two
charges move off to infinity, whilst the product 2m/f² remains
constant and equal to ¿va, the field of force will ultimately become
a uniform field of force perpendicular to a plane containing
the axis whose potential is va sin ◊ cos y, where ✈ is the angle
which the plane through the axis and the point (r, 0, ↓) makes
with some fixed plane through the axis. The resulting expression
for V' will be the potential of the bowl when placed in this field
of force.
The result of this process is,
cos y sin r sin¯¹
F
2cr
p + q * (p + q)³ {(p + q)² — 4c² } ?
2
2c
-1
{(p+q)º 4cºj³
sin" a}$]
va
V'
=
2π
2r sin a
2a2c
+
sin-¹
-1
干
​p
p+q
r (p + q)²
{(p + q)² — 4r² sin² a}
1 If an electrified circular disc is placed in a field of force whose potential is
F (r, 0) sin (p+e), the potential of the induced charge can be obtained by Bessel's
Functions, see Proc. Camb. Phil. Soc. vol. v. p. 425; and thence by inversion, we
can obtain the potential of an electrified spherical bowl when placed in a field of
force of the above form.
SOLID OF REVOLUTION.
155.
The inverse sines and the double signs before the second and
fourth terms must be interpreted in the manner explained in the
preceding article. (See Proc. Lond. Math. Soc. XVI. p. 296.)
The preceding expressions for the velocity potential make the
velocity infinite at the edge of the bowl, and therefore the
motion represented by the formulae could only be approximately
realised in practice.
160. In order to obtain the motion of a liquid in which a
solid is moving by means of the velocity potential, it is necessary
to find a potential function which satisfies an equation at the
surface of the solid which involves the first derivatives of p, and
this circumstance creates a difficulty which has proved insuperable,
excepting in the case of an ellipsoid, an anchor ring¹, and a
spherical bowl. But if the solid is one of revolution which is
moving parallel to its axis, the motion can be determined by
means of Stokes' current function, which Rankine has shown has
a definite value at the surface of the solid.
2
Taking the axis of z as the axis of revolution, let w, u be the
velocities of the liquid parallel and perpendicular to the axis of z;
the surface condition is
lw + mu = IV,
where V is the velocity of the solid, or
1 dy do
+
a da ds
1 dy dz
w dz ds
do
V
ds
Integrating along a meridian curve, we obtain
два
¥ = {√∞ ².
2
Now satisfies the equation
d² d³↓ 1 dy
dz² da @ da
+
In this put = x, and we obtain
d²x d'x 1 dx_
@ do
+
dz + da*
+
0.
Χ
= 0,
(29).
which shows that x sin & is a solution of Laplace's equation; hence
(29) may be written
x sin & = Vy.
Hicks, "On Toroidal Functions." Phil. Trans. 1881, p. 609.
2 Phil. Trans. 1871.
156
KINEMATICS OF MOVING SOLIDS.
+
Hence if U be the electric potential of the induced charge,
when the solid is placed in a uniform field of force perpendicular
to a plane containing the axis and whose potential is
Vy, then
Uw coseco will be the current function when the solid is moving
with velocity V parallel to its axis.
In the case of a sphere
U
Va³y
Va❜ sin o
23
2p³
Va³ 2
Va³ sin² 0
Therefore
зва
273
2r
EXAMPLES.
1. An ellipsoidal shell is filled with liquid and rotates uni-
formly about a given diameter; prove that the path of every
particle of liquid relatively to the ellipsoid will be an ellipse whose
plane is conjugate to the given diameter; and that every particle
will sweep out, about the centre of its elliptic path, equal areas in
equal times.
2. Liquid flows past the solid ellipsoid (x/a)²+(y/b)² + (z/c)²=1,
the velocity at infinity being uniform and parallel to x. Prove
that the lines of equal pressure on the surface of the ellipsoid are
its curves of intersection with the cone y³/b* + z²/c* = x²/A*, where
A is a variable parameter.
3. Liquid is bounded by the ellipsoid (x/a)²+(y/b)² + (z/c)² = 1.
If the surface undergo a uniform torsion about a principal axis,
prove that the instantaneous velocity potential is proportional to
xyz for the liquid in the interior of the ellipsoid, and to
-
аф
da²
c²) do 3 + (c² — a³)
аф
αφ
+(a²-b²)
db²
dc²
2 xyz,
• = √
for the external space, where
འ
λ √{(a² +λ) (b* +λ) (c² +λ)} *
αλ
EXAMPLES.
157
4. Prove that the velocity potential due to a source of
strength m, placed at a point on the axis of a circular disc and
distant ƒ from it, at points on the side of the disc on which the
source is situated, is
dP
&
[o de af-
m
f dz (№² +ƒª — 2ƒ2)*'
where P is the potential of the induced charge when the disc is
under the action of a charge m, situated at a point on the axis on
the other side of the disc, and whose distance from it is f
5. The ellipsoid (x/a)² + (y/b)² + (z/c)² = 1 is surrounded by an
infinite mass of water and rotates about the axis of x.
Prove that
the component velocities of any particle of water parallel to the
axes will be respectively proportional to
dM dN
dN dN
dL
dL dL
dM
dz
dy' dx
dz' dy
dx
b2
C²
x²
y²
where L=
1
λ
+ √
c² + √
2
-
by
2
2
b² + √
z²
ลง
+2
Cz
+4
:)7 dy
P,
dy
M = 2b³ay [, (a² +¥) (b* + ¥) P'
αψ
λ (c²+¥) (a²+¥) P
N = − 2c² zx
·√x
where
P = √(a² + y) (b² + ¥) (c² + ¥),
and λ is the positive root of the equation
202
y
مج
+
+
1.
a² + λ
b² + λ
c² + λ
1
Prove also that if the ellipsoid be filled with water, the values
of L, M, N with 0 instead of λ for the inferior limit, will similarly
determine the velocity of any internal particle of water.
6. A sphere of radius a which is surrounded by an infinite mass
of liquid, is strained uniformly so that e, f, g are the principal
components of strain after unit time. Prove that the velocity
potential of the initially resulting motion is
d²
d
d² 1 1_a³
g
− b a² (• — ' + ƒ & + 9 € ) ! - a' (e + f +9)
-
ale
dx²
dy³
r
3r
158
KINEMATICS OF MOVING SOLIDS.
7. A sphere of radius a is surrounded by an infinite mass of
liquid. If the surface of the sphere be suddenly moved with
normal velocity eyz+fzx + gxy, prove that the velocity potential
of the resulting initial motion is
where
8. Given that
— aº (eyz +fzx + gxy)/3r³,
p² = x² + y² + z².
x= a(cosh a + cos ẞ — cosh y),
y=
4a cosh ja cos }3 sinh h
z = 4a sinh ta sin 8 cosh y,
transform the equation of continuity into the form
Φ
+(cosh y + cosh a) aga+ (cosh a – cos 8)
do
dy²
=
2
= 0,
do
(cos B+ cos y)
da²
and show that the surfaces for which a, B, y are constant are
confocal paraboloids.
with
Hence show that the velocity potential for infinite liquid
streaming past the fixed hyperbolic paraboloid B = ß₁,
velocity V parallel to the axis of x at infinity, is given by
ß
p = V (x − a ẞ sin ß₁),
-
and write down the corresponding values of when the fixed
surface is the elliptic paraboloid aa₁, or y = 71°
9. The axes of an ellipsoid which is filled with liquid vary
with the time in such a manner that the volume of the ellipsoid
remains constant; prove that the velocity potential of the
liquid is
$ = ½ (àx²/a+by²/b + ċz²/c).
10. The axes of an ellipsoid which is surrounded by an un-
limited liquid vary with the time in such a manner that the
ellipsoid always remains similar to itself; prove that
αψ
$ = − ↓ abc (à/a+ b/b+ è/c) [~ √ (a² + ¥) (bª + y) (c° + ¥) ·
11. Determine the initial motion of liquid outside an ellip-
soid, when component velocities (i) px, py, pz; (ii) pyz, pzx, pxy
are imparted to every point of its surface; where p is the perpen-
dicular from the centre on to the tangent plane at x, y, z.
7
CHAPTER VIII.
ON THE GENERAL EQUATIONS OF MOTION OF A SYSTEM
OF SOLID BODIES MOVING IN A LIQUID.
161. WHEN a number of solid bodies are moving in an in-
finite liquid, the motion of the solids is most easily determined by
regarding the solids and liquid as constituting a single dynamical
system, and then employing Lagrange's equations. But as the
methods and formulae employed are different according as the
motion of the liquid is cyclic or acyclic, it will be convenient
to consider these two cases separately.
Acyclic Motion.
162. The following notation will be employed; let
Um Um Wm ; Pm, Im rm be the linear and angular velocities
respectively of any solid Sm, along and about axes fixed in the
solid.
,
m
Pm'' Pm" › Pm" ; Xm Xm", Xm" the velocity potentials of the
liquid, when the solid S is moving with unit linear and angular
velocities respectively along and about axes fixed in Sm, and all
the other solids are at rest.
m
the velocity potential due to the motion of S when all the
other solids are at rest.
Y the velocity potential of the whole motion.
M.. the mass of S.
m
m'
m
160
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
:
From § 142 (1) it follows that
Φ um&m' + vm&m" + WmPm" + PmXm' + ImXm'"' +rmXm"'...(1),
m
for at the surface of Sm, dom/dn is equal to the normal velocity of
Sm, and is zero at the surfaces of each of the other solids; whence
m
also,
Ψ=ΣΦ
By § 85 (20) if T be the kinetic energy of the liquid
dy
T = − Z p
tp ffx an
dS,
//
dn
where the integration extends over all the solids; whence
T
dÞ
1
ipffv
[/v (da, ds₁ +
dn 1
d❤
2 dS₂
dn
¿§½ + ...).
·(2).
Substituting the values of ¥, Þ₁, ò̟,... in this equation, it
appears that T is a homogeneous quadratic function of the
velocities. If (umum), (Um, vm) denote the coefficients of um², umv,
&c. we obtain
Φ,
1
do₁
dn
1
(u,u,) = − p [f¢,
2 (u,u,) = − p [[p, do',
2p
j
dø
2 dn
1
ds,
doi
2 dn
1
-
dn
as,
2 dS,
as, − p [fo, do,
pfft,
So, do
=
dS₁ =
&c. &c.
2p
2
do 2 dS,
1 dn
...(3).
These equations at once follow from Green's Theorem, and
from the fact that do,'/dn is zero at the surfaces of all the solids
except S₁.
163. If all the solids are free, each solid will possess six degrees
of freedom, and its position will therefore be determined by six
independent coordinates. The velocities of each solid can be
expressed in terms of these generalised coordinates and their time
fluxes by means of the ordinary methods of Rigid Dynamics, and
the kinetic energy of the liquid will therefore be expressible as a
homogeneous quadratic function of the generalised velocities of
the solids. The coefficients of the velocities will be functions of
the generalised coordinates, and of quantities which determine the
form and dimensions of the solids. Their values cannot be found
without a knowledge of the velocity potential of the liquid, and
they have been determined only in a few cases.
LAGRANGE'S EQUATIONS.
161
The kinetic energy T' of the solids can be found by the usual
methods, hence if T be the kinetic energy of the solids and liquid,
T=T+T'..
·(4),
from which it is evident that T is a homogeneous quadratic func-
tion of the velocities of the solids.
164. Since the coordinates of individual particles of liquid do
not enter into the expression for the kinetic energy, it will be ne-
cessary to establish the legitimacy of the employment of Lagrange's
equations in the present case. The application of these equations
is a particular case of the theory of Ignoration of Coordinates.
Let the position of a dynamical system be determined by
means of a system of coordinates 01, 02..., X1, X2 ...; and let us
suppose that the coordinates x do not enter into the expression
for the potential and kinetic energies. Since
dT
dx
0,
dV
= 0.
dx
Lagrange's equation corresponding to x will be
whence
d dT
0,
dt dx
dT
dx
= const. = к……..
:(5).
The constant is the generalised component of momentum
corresponding to x; and there will be as many equations of the
type (5) as there are coordinates x. Now whatever the motion of
the system at any particular period may be, it can evidently be
produced instantaneously from rest by the application of a system
of impulsive forces, which must be equivalent to the momentum of
the system at the particular period. If however the motion of the
system is such that it could always be produced from rest or
destroyed, without the application of the impulse components
corresponding to x-in other words if the velocities could be
produced or destroyed solely by means of impulsive forces arising
from the connections of the system,-all the constants
zero, and (5) becomes
will be
dT
0.....
dx
...(6).
By means of (6) all the velocities & can be eliminated from T;
B.
11
162
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
if T denote the result of this elimination, then since Ø and
enter into T through x, we have
dT
dT dT dx, dT dx² + ·
+
+
đó đó đx để đxe để
dT
dė
by (6). Similarly
dT' dT
de
de •
d dT"
dT
d dT
dT
dV
Therefore
dt dė
do
dt de de
ᏧᎾ
which shows that when
tion T from which the
Lagrange's equations.
= 0, we may employ the modified func-
y's have been eliminated in forming
Now if the dynamical system consists of a number of moving
solids together with the liquid in which they are immersed, and
which either extends to infinity or is bounded by fixed solids; and
if the motion of the liquid is solely due to that of the solids
moving about in it, we have shown in §§ 85 and 89 that its
motion will be acyclic and irrotational, and that it could be
instantaneously produced or destroyed by means of a proper
system of impulsive forces applied to the solids and boundaries
alone: also since neither the kinetic nor potential energy contains
the coordinates of individual particles of liquid, the preceding
investigation shows that the equations of motion may be obtained
by forming Lagrange's equations by means of the expression for
T given by (4), which contains the coordinates and velocities of the
solids alone.
If the momenta are not zero, Lagrange's equations in their
ordinary form cannot be employed. The modified function which
must be used in this case will be determined in § 173.
165. The system of impulsive forces which must be applied to
the solids to produce the actual motion at any period, when com-
pounded into a single force and a couple about the line of action
of the force, is called by Sir W. Thomson the "Impulse of the
Motion."
If all the solids are free and the liquid extends to infinity and
is at rest there, the Impulse of the Motion is equal to the momentum
of the system; and if no impressed forces are in action, it must be
HAMILTONIAN TRANSFORMATION.
163
constant in magnitude and direction throughout the motion. But
if the liquid has fixed boundaries, the impulse of the motion is
equal to the difference between the momentum of the system, and
the impulsive forces arising from the pressures exerted by the fixed
boundaries.
When there is circulation and the liquid extends to infinity
and is at rest there, the impulse of the motion is equal to the
impulse of the forces which must be applied to the solids, together
with the impulses which must be applied to the barriers in order
to produce the cyclic motion.
15
1
166. Let p be the pressure of the liquid, l, m, n, the direction
cosines of the normal to S; E₁, N₁, Š₁ ; λ₁, M₁, v₁ the force and
couple constituents of the impulse which must be applied to S,
in order to produce the actual motion from rest, then,
§₁ = M¸µ¸+ ſſpl¸dS
Mu
ૐ
= M₁u₁ - pffy do, as₁.
dT
But
du₁
Therefore
Similarly
1
dn
(µ‚µ‚) u¸+ (u,v,) v₁ + ...
-pffy do as
P
દ
мо
λ
dT
du
1
dT
dp₁
1
dn
>
>
&c.
&c.
v₂+
………….(7).
Since T is a homogeneous quadratic function of the velocities
of the solids,
dT
dT
2T=u₁
du
+ V₁ dv
+...
1
= µ‚§₁ + v₁₂+
1
Differentiating with respect to §, on the hypothesis that ₁,
....... are the independent variables, we
dT
2
= u₁₂+ §₁₂
αξι
1
1
obtain
dv₁
du ₁ + n ₁ dn
Writing out (7) in full, we obtain
1
+…..
§₁ = {M, + (u,u,)} u, + (u,v,) v, +
n₁ = (u,v,) u¸ + {M, + (v,v,)} v¸ +
1
1
1
11-2
164 EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
Differentiating these equations with respect to ₁, on the
supposition that ₁, ....... are the independent variables, we
obtain
dv,
du₁
1={M₁+(u,u,)}
de.
+ (u,v₁) dž'
1
du,
dv₁
1
0 = (u₁v₁) + {M₁ + (~‚³‚)} JE,'
dE.
&c. &c.
V
αξι
Multiplying these equations by u₁, v₁……. respectively and adding,
...
we obtain
Whence
u =
E dE.
dry + ni dn₁
dv₁
1
+....
dT
dT
de
U₁
P₁
=P, &c.
αλ
Equations (7) and (8) are well-known dynamical relations.
(8).
Kirchhoff's Equations.
167. When a single solid moves in an infinite liquid, the
equations of motion may be obtained, as Kirchhoff has shown', by
expressing in an analytical form the fact that the rates of change
of the component linear and angular momenta of the system along
and about three rectangular axes fixed in the solid are respectively
equal to the components of the impressed forces and couples along
and about these axes.
Since we are dealing with a single solid we may drop the
suffixes and put w₁, w₂, w, for the angular velocities of the
solid.
3
3
If E, n, be the component linear momenta along, and X, μ, v
be the component angular momenta about three rectangular axes
which are moving with angular velocities 0,, 0, 0, about them-
selves, of any dynamical system whatever; and if X, Y, Z and
L, M, N be the components parallel to and about the axes of the
forces and couples respectively which act upon the system, it is
known that the equations of motion of the system are
2
1 Vorles. über Math. Phys. p. 60.
2 Hayward, Trans. Camb. Phil. Soc. vol. x.; see also Besant's Dynamics, § 232.
KIRCHHOFF'S EQUATIONS.
=X
¿— no₂+50,= X
3
ǹ− 50₁+§0¸= Y
1
8
¿ — §0₂+n0, = Z
i-wη+ζ-μθς + νθο
8
165
.(9),
=
L
μαζωξ-νθι + λθς = Μ
1
v − v§ + un − x0₂+µ0¸= N
1
where u, v, w are the component velocities parallel to the axes, of
the origin of coordinates.
Since these equations are true for any dynamical system
whatever, they will hold when the motion of the liquid in which
the solid is immersed is cyclic or rotational or both; but the
analytical expressions for the momenta §, n, &c. will depend upon
the particular kind of motion of the liquid.
When the motion of the liquid is irrotational and acyclic,
the momenta are determined by (7); also if the motion is
referred to the principal axes of the solid 0₁ = w₁, O₂ = W₂, Oz = Wg,
and the equations of motion become
1
2
29
d dT
dT
+ w
dt du
8 dv
dT
= X
2 dw
d dT
dT
dT
dt dv
w₁
1 dw
+ wg du
Y
d dT
dT
dT
W
dt dw
2 du
+w₁ dv
Ꮓ
.....(10).
d dT
dT
dT
dT
dT
dt dø,
พ +v
dv dw
W z dwz
+w₂ dws
dg
Ꮮ
d dT
dT
dT
dT
dT
dt dw
И +w
dw
du
1
d dT
dt dos
dT dT
w i dw s
dT
+@3
M
dw₁
dT
ข + u
du
dv
2
w z dw、
+w₁
N
1 dw 2
These are Kirchhoff's equations of motion for a single solid
moving in an infinite liquid.
Geometrical Equations.
168. We must now express the velocities in terms of the six
coordinates, which determine the position of the solid.
Let x, y, z be the coordinates of the centre of inertia O of
the solid referred to three fixed rectangular axes. Through O
166
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
draw OX, OY, OZ parallel to the fixed axes, and let OA, OB, OC
be the principal axes of the solid at 0.

B
Ꮎ
X
D
E
The angular velocities are given by the equations (Routh's
Rigid Dynamics, vol. 1. § 256)
@. = sin - sin cos &
ω1
W₂
= cos + sin 0 sin &
@₂ = &+ & cos ✪
ய
3
Ө
!
.(11).
Also the component velocity of O in the direction of OD is
u cos - v sin = (x cos y + y sin ) cos 0 — ż sin 0,
and in the direction of OE is
-
u sin + v cos &=
sin + cos Y.
Solving these equations, and observing that w is the component
velocity of O in the direction of OC, we obtain
u = x (cos 0 cos & cos-sin & sin ↓)
φ
+ý (cos e cos & sin
+ sin & cos ) – 2 sin 0 cos
v = ✰ (cos e sin & cos
+ cos & sin y)
(12).
-ý (cos e sin o sin
cos cos y) +ż sin 0 sin &
20 = ở sin 0 cos n +ỷ sin 0 sin + 2 cos Đ
169. The preceding equations may be considerably simplified
in the case of a solid of revolution.
Let OC be the axis of revolution, OX, OY, OZ three straight
lines parallel to axes fixed in space, let w be the velocity of O
I
GEOMETRICAL EQUATIONS.
167
along OC, u, v the velocities at right angles to OC in and perpen-
dicular to the plane ZOC. Then
u = x cos ↓ cos 0 + ý sin ↓ cos ◊ — ż sin ◊
v = − ¿ sin ♣ + ý cos ¥
w = x cos y sin ✪ + ý sin y sin 0 + ż cos 0
..(13).

B
Z
NC
Ø
A
E
Xx
Also if w₁, w,, w, be the angular velocities about OA, OB, OC
w₁ =- sin 0, w₁ = 0,
2
2
0
w₁ = $+ cos ...... (14),
3
where the plane COE is fixed in the body.
The velocities of each of the solids can be expressed in a
similar manner by means of equations (11) and (12), or (13) and
(14); hence if we can obtain the values of the coefficients in terms
of the coordinates, the motion can be completely determined.
170.
Cyclic Motion.
We must now consider the more general problem of the
motion of any number of solids, each one of which has several
apertures through which circulation takes place¹.
The following additional notation will be employed. Let
= velocity potential of the whole motion.
Y = do. due to motion of solids alone.
Ω=
= do.
due to cyclic motion.
1 Proc. Camb. Phil. Soc. vol. vI. p. 117.
168
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
m
>
,
Pm', Pm"› Pm"; Xm Xm" Xm", the velocity potentials of the liquid,
Φί',
when the solid S is moving with linear and angular velocities
respectively along and about axes fixed in S, and all the other
solids are at rest and there is no circulation.
σm² σm'σm"... the areas of the apertures of S
Km, Km', K,
m
wm, wm', wm
"
//
the circulations through them.
m'
the velocity potentials due to unit circulations
through the apertures of Sm, when all the solids are at rest.
m'
m
Ý Ý, "... the fluxes through the apertures of Sm
relative to S
Pm the velocity potential due to the motion of Sm and the
circulations through its apertures, when all the other solids are
at rest.
By Thomson's extension of Green's Theorem, it is known that
the motion at any period could be instantaneously produced from
rest, by the application of suitable impulses to each of the solids,
together with uniform impulsive pressures mp, mp
K applied to
every point of the barriers σm, m... respectively. Let Xm, Ym, Zm;
Im, Mm, Ñm be the force and couple components of the impulse
along and about axes fixed in Sm, which must be applied to Sm
=
X,
m'
...
Let §m, Nm, 5m; λm› µm› ''m'; §m' › Nm'... be the components of the
impulses which must be applied to each of the barriers of S; also
let mΣm &c.; Xm=Xm+m &c., and let X, Pm, Zm; Lm:
Mmm be the generalised components corresponding to um,
of the momentum of the cyclic motion, when all the solids are
M
at rest.
m
ย
m
Let Mm be the mass of S, T the kinetic energy of the liquid,
T that of the whole motion. It will be shown that T is the
sum of two homogeneous quadratic functions of the velocities and
circulations respectively. Let these be denoted by T and f
respectively, and let (umum), (mm) denote the coefficients of
U Um
um², umum, &c.
2
Since the w's are the velocity potentials due to unit circulations
round circuits which cut the apertures to which they correspond
once only, when all the solids are at rest, they must satisfy the
following conditions.
(i) At all points of the liquid Vo= 0, and w and its first
derivatives must be finite and continuous at all points of the liquid,
and must vanish at infinity.
KINETIC ENERGY WHEN THERE IS CIRCULATION. 169
(ii) At the surface of each solid dw/dn= 0.
(iii) w must be a monocyclic function whose increment is
unity for all circuits which cut the barrier to which it corresponds
once only, and zero for all circuits which do not cut this barrier.
It therefore follows that
Þm= Um$m' + vmPm" +WmPm"”' +PmXm' + 9mXm” +™mXm"" +KmWm
+ km² wm² +
(15),
and that
$ = ΣP₁ = Y + N.
171. The kinetic energy of the liquid is
tp // $ do ds + + Σ kp [[dh
T = - ≤ p || $ dn
do,
dn
where the first integral is taken over the surfaces of all the solids,
and the second over all the barriers. Since dP/dn at the surface
of S is equal to the normal velocity of Sm, and is zero at the
surfaces of each of the other solids,
m
T = − t p
# P //
dÞ
Φ 1
dn
do ds₂+
2
dn 2
do
ds₂+.......)
κασ,
K
+pffd («, do,¸ +x, do' + ... кdo + ...).
We can now show that
Φι
(u,u,) = − pff¢,
1
dn
dò̟i dS₁›
dn
1
αφί
1
S
2 (u,v) = − p [[Þ, do, as, - pff; dias,
ffp,
2 dn
=-2pff, dd,
dn
'
ds,
1 dn
2 (u,v) = - pff, dd, ds, + pffddi do, =0,
2 (µ‚k,) =
P
dn
dn
=-pff∞, do ds + pffdd do, = 0,
dp
- p√]
2 dn
dw₁ do
(1,1,) = p ff dw,
dn
1
1
do₁
1
dn 2
1
dw.
do₁
dn
1,
2 (k,*,) = p[] dwi do₁ + p[[dan, do;' = 2pff doo
dn
pff
2
... (16).
The above equations can be at once established by Thomson's
extension of Green's Theorem. For if in equations (25) and (26)
170
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
1
1
of § 88, we put = w,,,', then since w, is a monocyclic
function whose increment is unity for all circuits which cut the
barrier σ, once, and zero for all other circuits, and 4,' is a single
valued function, we obtain
[[∞, do, as - [] do, do, = [[ +, das, as.
1 dn
dn
Now do/dn is zero at the surfaces of all the solids except S₁,
and do/dn is zero at the surfaces of all the solids, whence the
third of equations (16) follows at once. The others can be proved
in a similar manner; hence the products of velocities and circu-
lations do not enter into the expressions for the kinetic energy of
the system, and we may therefore put
T=Z+K
where I is a homogeneous quadratic function of the velocities of
the solids alone, and f is a similar function of the circulations.
172. If p be the pressure and l, m, n, the direction cosines
of the normal to S₁,
dT
But
=
:
du₁
JJ plas
X₁ = M₁u₁+ || pl₁dS,
11
1
= M₁u, - pff & do as₁.
(µ‚µ‚)µ‚ + (µ‚v¸) v₂+
1
dn
αφι
pff (up; + v,p,'" + ....) dpi as,
P
ρ
11
αφι
Ω
dn
& dobi as₁ + pff ndóias,
φ
dn
do,
do
1
φ dS₁ + p ffdd, Σ (xdo),
- Pff & deb
ρ
dn
dn
where the summation refers to corresponding products, and ex-
tends to all the barriers; hence
Also
dT
X₁ = du
P
αφί
ffdbi & (sdo)..
dn
§₁ = k¸pffl¸do, &c. &c.
..(17).
where l₁, m, n, are the direction cosines of the normal to the
barrier σ; whence
1
ૐ
§₁ = Σ5,=p [[ Σ, (xldo),
MODIFIED LAGRANGIAN FUNCTION.
171
where the summation Σ, extends to the barriers of S, only; also
1
dT
1
X₁ =  + §‚
du,
ρ
SS
do'
dn
(xdo) + §¸ ………………..(18).
From (17) we see that the component impulse corresponding to
which must be applied to S, in order to keep it at rest, when
the cyclic motion is generated by the application of proper impulses
to the barriers of all the solids is
ρ
ff dø/dn.Σ (xdo); and
therefore by (18) the generalised component of momentum X, cor-
responding to u, of the cyclic motion when all the solids are reduced
to rest, is
*₁ = } − p ffddi Σ (xdo) = pff.Σ, (xldo) – p ffdd Σ (xdo).. (19),
whence
αφί
dn
dT
X₁ =
du,
+X,
Similarly it can be shown that
dn
(20).
where
and
dT
dp₁
+ L₁
dxí
dn
λ
·
K
1
L₁ = x, − p //dx₁ Σ (xdo),
x₁ = Σλ₁ = pƒƒ Σ₁ [x (ny—mz) do]
.(21),
(22).
173. We must now obtain an expression for the modified
Lagrangian function.
Let the coordinates of a dynamical system be divided into two
groups and x, the latter of which does not enter into the
Ø
expression for the energy of the system. Since the kinetic energy
is a homogeneous quadratic function of the velocities ... x...,
we may put
2T=(00)0+2(00¸)ÔÒ¸+….. 2(0x)ėx+….. (xx) x²+2(xx₁)XX₁+….. (23).
1
In this expression none of the coefficients involve x, and
Lagrange's equation corresponding to x, gives
K
dT
dx
= const. = x, &c.
where x is the generalised component of momentum corresponding
to x; writing these equations out in full, we obtain
x = (0x) 0 + (0₁x) 0₁ +......(xx) x + (XX₁) X₁ +…………..)
.
*, = (x) Ỏ + (0,x) 0, +....(XX) x + x) +…....
ė
.(24),
1
X₁
172
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
the number of equations being equal to the number of the co-
ordinates X.
Let P, P₁ ... be the portions of к, ₁,
involve the y's, then P, P.
1'
alone, and (24) may be written
which do not
are linear functions of the O's
(XX) X + (XX₁) X₁ +......= x - P
(XX₁) X + (X₁X₁) X₁ +
K
K₁
-
P
.(25).
If ▲ denote the determinant
(xx), (xx₁), (XX2),
A
(XX₁), (X₁X₁), (X1X2),
(XX2), (X1X2), (X2X2),
the solution of (25) may be written
d▲
d▲
; (x − P) + d (xx₁) (*₁ − P₁) + ·
Ax = d(xx) (x
ds
dA
(26).
Αχι
(KP) +
d(xx)
d(X₁X₁)
(k₁ − P₁) + ...
If therefore we put
1 dA
(KK) =
=
A d(xx)'
1 d▲
(KK₁) =
&c.
▲ d (xx₁) '
+...
(26) may be written
x:
dk dB
dP
dP
dk dp x₁ =
dk dP
Χι
dк₁
dР
dP 1
&c......
2P = (kk) P² + 2 (kk₁) PP,+
2K = (kk) k² + 2 (kk¸) kk¸ +
(27),
(28).
Let T be the portion of T which is
independent of x; then,
of the velocities,
since Tis a homogeneous quadratic function
dT dT
2T=0 +0,
1
do
dė,
dT
dT
dT
+.. x
+ X1
+.
dx1
dx
1
dT
Ө
dė
1
Ө
oda
+
dė₁
+ 0 {(0x) x + (0x₁) X₁ + …………..}
+ 0, {(0,x) x + (0,x)
· 2T + (x + P) x + (k, +
= 2T +2K − 2P+ E (
1
+…}+…
P₁) X, +
dk
☆ (1
P
dk
K
dP
dP
•
MODIFIED LAGRANGIAN FUNCTION.
173
Writing out the last term in full, it is easily seen from (27)
that it vanishes; and therefore since P is a homogeneous quadratic
function of the velocities è alone, it follows that T is equal to the
sum of a homogeneous quadratic function of the velocities 0,
together with a similar function of the momenta κ.
We may
therefore put
where
T = Z + K......
T=T-P.....
(29),
.(30).
Let ✪ be the generalised component of momentum correspond-
ing to 0, and let be the value of after the velocities have
been destroyed by means of proper impulses applied to the system.
The momenta κ will evidently be unaffected by these impulses,
but the velocities & will be affected, since the impulse required to
destroy will produce reactions arising from the connections of
the system which will change the values of the y's. Now
dT dT
+(0x)x+(0x₁)x₁+
dė
dė
da
dkdP
+(0x)
+.
dė
dk dP
whence
(= (0x) JK
dR
+ (OX₁) JK₁
ds
+
(31),
dT dT
dB
and therefore
+®−Σ (0x)
do
dė
dP
dT
dP dP
+0-Σ
do
dė dP
dT
+℗.....
(32),
dė
d dT d dZ do
whence
+
đt đồ đt dò
(33).
dt
It appears from (31) that the momentum is a function of
the momenta and the coordinates only.
K
Again
dT dZ dk
+
de do do
Now since enters into & through κ, we have
dk dk dk dk d₁ +
1
dk
DK
+.
de
do'
+
dᎾ - de de " d-, de
174
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
..
where the symbol /de operates on the coefficients and not on
the momenta. Differentiating (26) with respect to 0, we obtain
dP
0 = (xx) (dx - db) + (xx) (dxy + d)
+...
+ (x − P) √o (kk) + (k¸ − P₁) √ (kk¸) +………….. (34).
do
Multiplying the system of equations of which (34) is the type
by к, к.... respectively and adding, we obtain
dk
dâ dк dⓇ dк,
+
dЯ dk
dê
do
de de " de, ᏧᎾ
1
+ .....
df dPdf dP. 1
dê de dê, do
DK a dk d dk
+2
P
•
=0 ...... (35).
-P
de de de
Multiplying the equations of which (31) is the type by 0, 0,...
respectively and adding, we obtain
ds
Σ(®ė) = P
+P
dk
d&
1 JK₁
+..
1
dᏢ
whence
Σ(®0) = Σ
do
(dp dx)
dk
+ Σ
+ΣΡ
de de
>
dk dk
Q
therefore (35) becomes
+
ᏧᎾ
do ¯ do Σ(℗0) = 0,
dT dT dDK
Q
whence
+
do de de
de
Σ (60).
We may now drop the symbol /de on the understanding that
the momenta è are to be treated as constants, and Lagrange's equa-
tions become
d dZ
dt dė
+
do
dē dz d
+
d
d V
Σ (®Ø) +
= : 0.
dt do do
do
do
Since and
do not contain , the modified function is
L=L+E (ŌÖ) − k + V ............…………….(36).
If the velocities ... be expressed in terms of new velocities u...,
and be the new momentum corresponding to u after the u's have
been destroyed, it can easily be shown that,
Σ (℗0) = Σ (Xu).
MODIFIED LAGRANGIAN FUNCTION.
175
For let
=
ġ – Au+ A‚µ‚ + ½µë
then
И
1
22
+
dė/du=A, dė/du, = A₁, &c.
dė
dTi dT
also by (32),
Σ Θ
= x.
du
du du
therefore
dė
dė
Σ(Xu)=(u
du
1
+ Us du s
®,
+) + @, (u do,
+.
du
.....) +0.
&c.
Σ (Θό),
whence (36) may be written
L=T+Σ(Xu) - K+ V
(37).
174. We have therefore obtained a form of Lagrange's equa-
tions, which can be employed when the kinetic energy is expressed in
terms of the velocities corresponding to the coordinates by which
the position of the system is determined, and the constant momenta
corresponding to the time fluxes of the ignored coordinates.
Now by § 89, when a liquid of density p occupies a multiply-
connected region, circulation can be generated by means of a
uniform impulsive pressure xp applied to every point of one of
the barriers which must be drawn to make the region simply
connected, and the circulation thus generated cannot be destroyed
excepting by the same process as that by which it has been
produced. It therefore appears that the product of the circulation
and the density is a quantity in the nature of a generalised com-
ponent of momentum.
Hence in order to determine the motion of a number of
perforated solids in an infinite liquid, we must first calculate by
means of (16) the quantities T and ; the former of which is the
kinetic energy due to the motion of the solids alone, and is
therefore a homogeneous quadratic function of their velocities, and
must be expressed in terms of the generalised coordinates and
velocities of each solid; and the latter of which is a similar
function of the circulations. The quantity X in (37) is evidently
the generalised component corresponding to u, of the momentum
of the cyclic motion which remains after all the solids have been
reduced to rest, and its value is given by (19) or (22), according
as it is in the nature of a force or a couple.
1 In this term T is supposed to be expressed in terms of the velocities
u... and x....
176
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
175. We can now ascertain the physical meaning of the
generalised velocity x which corresponds to the momentum xp.
1
Let be the flux through the aperture σ, of S₁ relative to
S. Then if l, m, n, be the direction cosines of the normal to σ,
аф
↓ = [] {db −
dn
1, (u¸ + q¸z−r,y) —m, (v,+r,x− p¸²)
-n (w, +p,y-q,o)} do,
1
But
dn
ρ
dn
ffdy do₁ + [fan do, - (u,5, + v,n,+w,&,+P;^,+9,#,+~,;";)/~,p.
1
191
• [[dan do¸ = (x,x,) x¸ + (x,1,') xj' +….....
K
k,'
1
d&
If therefore we put
dk,
a₁ = §,/«‚p=[[¿,do¸, B₁=n,/«,p,_r₁ = 5,/«,p,
ૐ
1,
a₁ =λ/x,p=[[(n¸y—m¸²) dơ¸ b₁=µ‚ׂ‚¢‚=v«/«‚p,
we obtain
dy
dn
do
-
-
1
1
1
1
1
1
1 dk
+
p dk,
.(38).
Now if T be expressed as a quadratic function of all the
momenta
1 dT
P dx = x.
But
27 = Συ
dT
du
+2K = Σu (X −X) + 2K
*)
(39),
by (20). Hence in order to obtain x, we must differentiate (39)
with respect to ₁, on the hypothesis that the momenta X are
constant, and that u is a function of x; whence by (19) and (22),
2
dT
1
dZ du
Σ
dk₁
du dr
+ p
JJane do₁ +
ds
do₁+2
dn
dk
(40).
APPLICATION TO A SINGLE SOLID.
From (20) we obtain
dv
0 = Σ (u, v) Tx₂ + ², p− P¸
1
do,
dn
do
1'
dv
0 = Σ (u,v)
dk₁
ρ
1&2 do 21
dn
177
where the summation extends to all the unsuffixed letters in-
u₁.
འང
cluding v = u₁.
Multiplying these equations by u, v,... respec-
tively and adding we obtain
[[dy do₁ = 0,
+(9,9, +8,%, +8,2%, + &,p,+8b, q₁ + cr₁) p − p [fd
dZ du
Σ
du dê¸
1
dT
whence by (38) and (40)
AK₁
1
1
whence
1
Hence the flux through the aperture σ, relative to the solid
S is the generalised velocity corresponding to the momentum p.
This theorem was discovered by Sir W. Thomson¹.
176. We shall now apply the preceding results to determine
the motion of a single solid having only one aperture.
If u, v, w; w¸, w, w, be the linear and angular velocities of
the solid, along and about axes fixed in the solid, and
velocity potential due to the circulation
the
where
T=Z++KK²,
K = pffd
do.
dn
*
= кр
Also by (19) and (22)
Kpff(1_dds) do, &c.
dn
L = kp
κρ
ny
mz
dxì
dn
dx₁) do,
dT
X=
+X &c. .
du
(41),
L= T + Xu + Yv+Zw+ Lw, + Mw, + Nw¸− k + V…..(42).
3
1 Proc. Roy. Soc. Edin., vol. VII. p. 668.
B.
12
178 EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
Z
In this case the quantities X... are evidently constants, and
we can either obtain the motion by expressing u, v... in terms
of i….. by (11) and (12), or by (13) and (14), and then employing
Lagrange's equations; or since X, Y... are the components of the
momentum of the system along and about the axes of the solid,
we may substitute their values in (9) from (41), and thus determine
the motion by Kirchhoff's equations.
Motion of a System of Cylinders.
177. If we endeavour to calculate the right-hand side of (37),
in the case of the two-dimensional motion of a number of cylinders
in an infinite liquid, when there is circulation round some or
all the cylinders, it will be found that some of the terms become
infinite. In order to avoid this difficulty, we must describe an
imaginary fixed circular cylinder in the liquid, the radius of
whose cross section is a very large quantity c, and then calculate
the value of L for the space contained between the moving
cylinders and the outer one, omitting all the terms which vanish
when c becomes infinite. It will then be found on substituting
the value of L thus obtained in Lagrange's equations and per-
forming the differentiations, that all the terms which become
infinite with c disappear, and we thus obtain the equations of
motion of the cylinders¹.
178. The calculation of L can most easily be effected by
employing the current function instead of the velocity potential,
for the former function is always single valued unless any sources
or sinks exist in the liquid.
1
Let u, v, be the component velocities of any cylinder S, along
rectangular axes fixed in the cylinder, w, its angular velocity,
K₁ the circulation round any closed circuit which embraces this
cylinder once only.
Let the centre 0 of the cross section of the outer cylinder be
the origin, and let x, y, be the co-ordinates of the centre of inertia
of the cross section of S, referred to rectangular axes fixed in space;
x, y, the co-ordinates of the same point referred to moving axes
through O which are parallel to the directions of u,, v. Also let
19 1
1
1 Proc. Camb. Phil. Soc., vol. vi. p. 135.
X
MODIFIED LAGRANGIAN FUNCTION FOR CYLINDERS.
179
be the current function, and be the velocity potential of the
cyclic motion when all the cylinders are at rest.

y
y'
0
C
Χ
In the figure let CA, CB be the axes of any one of the cylinders
along which u, v, are measured, then
x = p ff dx, da'dy'
dy
=-p√x
dx'
de de + p [√x de de].
ds
ds
ds
where the first integral is to be taken once round the circum-
ference of the cross section of the outer cylinder, and the square
brackets denote that the second integral is to be taken once round
the circumferences of the cross sections of each of the moving
cylinders.
At the surface of each of the moving cylinders x is constant,
hence the second integral vanishes, therefore
x₁ = - p√x da ds.
1
ds
Let (r', ') be polar co-ordinates of a point referred to Ox' as
initial line, then at a sufficient distance from 0, x can be expanded
in a series of the form
Therefore
1
x = − m log r' — —- (A¸ cos &' + 1¸ sin 0') — ...
рс
["{
2π
[** {m log c + 1 (A, cos 0 + B, sin 0) + ... sin o do
Ꮎ
Similarly
πρ.
.(43).
ρ
[fdx, da'dy' = -
--Jx dv ds
√x ds
dy'
= πρ
1
.(44).
12-2
180
EQUATIONS OF MOTION OF A SYSTEM OF SOLIDS.
Again, if, be the angular momentum about C of the cyclic
motion
{(x'
x
-
dy
=
£2, − − p ] [
— a¸ ) d x + (y' — v₁) */ }
- 3.) dx} dedy
= - pff { x d x + y²
dx\
dx'dу' — πρ (¤‚' + B₁y‚').
dy')
1
By Stokes' theorem the double integral
ds
bp [radx de + bp [fradx da]
dn
dn
The first integral = pc'm, the second integral may be written
7
–
3p [frºdít/ds.de]
hence
Also
dQ
ds
[fredo de] -
12₁ = πрc²m - tp | fr²
The integral
2T
28
=
=
ds ·πр (A₁x,'+ B₁у,')…………….(45).
p fx dx ds - p [√x
2π
dn
pc [" x dx do +
= pc f®
0
pcf":
1
dr
ds
dx de ]
dQ
ds
p[[x dn da]
ds
dx do + pΣ (xx).
X dr
αθ ρΣ
cos @ + sin
= p[*" {m log c + ( cos
Whence
X
=
C
m
1
C
2
(A cos 0 + sin 0)} de == 2πpm² log c.
§ = πpm² log c + } pΣ (xx)
Hence we finally obtain
L = T +πpΣ (Av – Bu) + Σ (N∞)
(46).
πpm² log c — {pΣ (xx) + V_......(47).
If we substitute the preceding expression for L in Lagrange's
equations and perform the differentiations, it will be found that
the terms πpc'm in 2, and 7pm² log c disappear; we may therefore
write
2
L = T +πpΣ (Av – Bu) + Σ (Nw) − ¿pΣ (xx) + V
12 = - tp // m² dn
......(48).
πp (Ax' + By') …………..(49).
dsds-πp (Ax' + By)
MODIFIED LAGRANGIAN FUNCTION FOR CYLINDERS.
181
179. The quantity which does not depend on the cyclic
motion, can be obtained by the ordinary methods. With respect
to the other terms we must first obtain the values of X and Ω;
we must then draw from O a series of lines parallel to the
directions of u,, u,..., and take each of these lines successively
as the initial line, and expand x in a series of the form
X
m log r
1
(A cos 0 + B sin 0) —- ...
which will determine the values of the A's and B's.
The velocities u, v and the co-ordinates x', y' expressed in terms
of x, y, the co-ordinates of C referred to fixed axes, and the angle
O which CA makes with Ox, are given by the equations
u = * cos 0 + y sin 0,
x' = x cos 0 + y sin 0,
ย
y'
x sin 0 + y cos Đ
x sin + y cos es
•
(50).
When there are several cylinders, the value of x at the surfaces
of the different cylinders is a function of their forms and positions,
and is therefore a function of the co-ordinates; when there is
only one cylinder the value of x at its surface is an absolute
constant.
CHAPTER IX.
ON THE MOTION OF A SINGLE SOLID IN AN INFINITE
LIQUID.
180. WHEN a single solid is moving in an infinite liquid
whose motion is irrotational and acyclic, the kinetic energy of the
solid and liquid is a homogeneous quadratic function of the com-
ponent velocities of the solid alone, and is therefore of the form ;
2T = Pu² + Qv² + Rw² + 2P'vw + 2Q'wu + 2R'uv
2
2
2
+Aw₁² + Bw₁₂+Cw¸² + 2A'∞¸¸ + 2B'w₂w₁ + 2C'w‚w₂
+ 2w₁ (Lu+ Mv + Nw)
1
+2w₂ (L'u + M'v + N'w)
3
+2w¸ (L'u + M″v+ N″w)
.(1),
t
3
where u, v, w; w₁, w, w, are the component linear and angular
velocities of the solid.
If the motion is referred to the principal axes of the solid, the
quantities P, Q, R are called the effective inertias of the solid
parallel to the axes; and the quantities A, B, C are called the
effective moments of inertia about the axes. Their values are
determined by the equations
P=M- pffolds &c. &c.
A = I₁— p√√x₁(ny — mz) dS &c. &c.,
1, - -
1
....
where M is the mass of the solid, I, its moment of inertia about
the axis of x, and ..., X₁... the constituents of the velocity
potential.
The other coefficients depend solely upon the form of the
solid and the density of the liquid; their values are given by
§ 162. (3).
>
MOTION OF A SPHERE.
183
181. When the form of the solid resembles that of an ellipsoid,
which is symmetrical with respect to three perpendicular planes
through its centre of inertia, and the motion is referred to the
principal axes of the solid at that point, the kinetic energy must
remain unchanged when the direction of any one of the component
velocities is reversed; hence the kinetic energy cannot contain
of the products of the velocities, and must therefore be of the
any
form;
•
2
2
2
2T = Pu² +Qv² + Rw² + Aw¸³ + Bw,² + Cw¸² ......(3).
3
If in addition, the solid is one of revolution about the axis of z,
the kinetic energy will not be altered if u is changed into v, and
W2 whence PQ, A = B, and
@1
into w
29
3
2T = P (u² + v²) + Rw² + A (w¸² + w₁₂²) + Cw¸³..……..(4).
Although every solid of revolution must be symmetrical with
respect to all planes through its axis, it is not necessarily sym-
metrical with respect to a plane perpendicular to its axis. The
solid formed by the revolution of a cardioid about its axis is an
example of such a solid. In this case the kinetic energy will be
unaltered when the signs of u, v or w, are changed, and also when
u is changed into v and w, into ; hence in this case
2
w
3
2
2T = P (u² + v²) + Rw³ + A (w¸² + w,³) + Cw¸² + 2Nw (w₂+ w₂)…..(5).
3
If the solid moves with its axis in one plane, (say zx), v and w,
must be zero, and the last term may be got rid of by moving the
origin to a point on the axis of z whose distance from the origin
is - N/R. This point is called the Centre of Reaction.
We shall now consider some special cases.
Motion of a Sphere.
182. Let a sphere of radius a, density σ, and mass M be pro-
jected with velocity V in an infinite liquid of density p; and let
the sphere be acted upon by a constant force Z perpendicular to
the initial direction of projection.
Let the axis of x be in the direction of projection, and that of z
in the direction of the force, then
a³
$ = − 1 =~/23 (ux + wz)
2T = P (u² + w²),
•
184
MOTION OF A SINGLE SOLID.
where
P = M − pfføldS
=
=
M+πpa" ["cose sin ede
πρα
M + ¿M',
where M' is the mass of the liquid displaced. Therefore
2T = (M + {M') (x² + ¿²)
and Lagrange's equations give
d dT
d dT
0,
Z.
dt dx
dt dż
Integrating we obtain
whence
and
hence
(M + 1M')x = const = (M + M') V
x = V ...
(M + 1M') ¿ = Zt
(M + {M') z = {Zť²
.(6),
·(7).
Since remains constant and equal to its initial value, it
follows that if a sphere which is acted upon by no forces, is pro-
jected in any direction with given velocity, it will continue to
move along that direction with the velocity of projection. The
same result can also be shown to be true in the case of any solid
which is symmetrical about an axis, and which is projected
parallel to that axis. This however is altogether contrary to ex-
perience, and the reason of this discrepancy between theory and
observation is, that we have assumed the liquid to be frictionless,
whereas all liquids with which we are acquainted are more or less.
viscous. The viscosity gives rise to a retarding force by which the
solid and liquid are gradually reduced to rest, and the kinetic
energy is converted into heat.
The motion of a sphere in a viscous liquid will be considered in
the second volume.
Equation (7) shows that the only effect of the liquid is to pro-
duce an apparent increase in the inertia of the sphere, whose
amount is equal to half the mass of the liquid displaced.
When the sphere is moving under the action of gravity
Z = (M – M') g; therefore
P
gt. .
σ + 1 p
MOTION OF A SPHERE.
185
Hence the sphere will describe a parabola in the liquid with
vertical acceleration g (σ − p)/(σ + ½p) ¹.
183. In the preceding investigation we have assumed that the
liquid always remains in contact with the sphere; but it may
happen that the pressure becomes negative at some point of the
sphere, in which case a hollow would be formed in the liquid.
If the sphere is moving with constant velocity V in a straight
line,
Ф=
Va³x
27.3
also since the origin to which is referred is in motion with
velocity V,
P II
p
p
dx
· 19²
II
+ V² (2 cos²0 - §)
ρ
where II is the pressure at infinity. Hence if
II < § V²p,
p will become negative when lies between a and π-a, where
a <π, and a belt of liquid will be thrown off and violently dis-
turbed motion will ensue. For a discussion of the subsequent
motion, see a paper by Sir W. Thomson, Phil. Mag., March, 1887.
184. A sphere of radius a and mass M is contained within a
fixed concentric sphere of radius c, and the intervening space is filled
with liquid of density p which is initially at rest. If an impulse I
be applied to the inner sphere, prove that its initial velocity w is
equal to
Σπρα (c + 203)
3 (c³- a³)
I {M.
IM +
Let
ф
$ = (=
A
2
Br) c
+ Br cos 0.
Then
аф
dr
= w cos 0 when r= a,
do
= 0
when r = c.
dr
Greenhill, Mathematical Tripos, 1877.
186
MOTION OF A SINGLE SOLID.
was cs
was
3
Therefore
A
B
2 (c³ — a³) '
3
C
was
C³
and
$
c³ - a³
3
2r2
+r)c
+ cos 0.
ρφ,
Now if p be the impulsive pressure on inner sphere p pp,
therefore
Mw=I+pffp cos OdS
= I –
= I
-
πρωα (c + 2α)
3
C' a³
["
π
cos20 sin ede
2πρα (c + 2a) w
3 (c³ — a³)
Motion of a Cylinder.
185. When a right circular cylinder is projected in an infinite
liquid which is at rest, and no forces are in action, it will move
(as will presently be shown) in a straight line with uniform velocity,
and the only effect of the liquid will be to produce an apparent
increase in the inertia of the cylinder, which is equal to the mass of
the liquid displaced. There is however an important difference
between the motion of a cylinder and of a sphere, since the space
outside a cylinder is a doubly connected space, and hence circula-
tion round the cylinder is possible.
We shall therefore consider the problem in its most general
aspect¹.
Let a be the radius of the cylinder, (r, 0) the polar coordinates
of any point referred to its centre; (x, y) the coordinates of the
same point referred to fixed axes, (a, B) the coordinates of the
centre of the cylinder, (u, v) its component velocities referred to
the fixed axes; the circulation. Then
a²
κθ
$
(u cos 0 + v sin 0) +
2π
a²
2
u (x − a) + v (y – B)
K
+
tan-19-B
2π
X α
:
(x − a)² + (y − ß)²
1 Lord Rayleigh, "On the irregular flight of a tennis ball," Mess. Math., vol. vii.
p. 14; Greenhill, "Note on previous paper," Mess. Math., vol. 1x. p. 113.
MOTION OF A CIRCULAR CYLINDER.
187
Now
whence we easily find
$
α
r
¿ = u,
B=v,
2a2
p2
a²
(ù cos 0 + v sin 0) + 2ª (u² + vº) — 20² (u cos 0 + v sin 0)²
K
22
+ (u sin 0 – v cos 0)
2πη
and therefore when r = a
-
$ = a (ù cos 0 + v sin 0) + u² + v² − 2 (u cos 0 + v sin 0)²
K
Also
+ (u sin + v cos 0).
Σπα
аф
q²:
1 do
+
dr
r do
Therefore when r = a
2
K²
K
q² = + (u sin 0 – v cos 0) + u² + v² .
Απα
2
πα
If gravity be the only force in action, and the axis of y be
drawn vertically upwards, the pressure is determined by the
equation
P
ρ
3
2
K²
δπα
2
K
a (ù cos 0 + v sin 0) + — (u² + v²) + + (u sin ◊ — v cos 0),
2 (u cos 0 + v sin 0)² + g (3+ a sin 0) = const.
Let X, Y be the forces parallel to the axes due to the pressure,
then
X
- √
2π
ap cos ede,
Y=
-S
** up sin 0d0,
whence omitting the terms which are independent of e, and which
therefore vanish when integrated round the circle, we obtain
2π
X = ap
K
πα
(u sin 0 — v cos 0) — a (ù cos 0 + i sin ()
Similarly
— 2 (u cos 0 + v sin 0)² + ga sin e}
- κρυ - παρά
Y = κρυ-παρύ + gpa
2
sin e cos ede,
.(8).
..(9).
Hence if σ be the density of the cylinder, the equations of
motion are
πσαύ = X,
==
πσαοϋ = Υ - πσgα
2
188
MOTION OF A SINGLE SOLID.
?
which by (8) and (9) become
кр
2
πα
(p + o) i + KD
v = 0
(o
кр
(p + o) v − KP u + (σ − p) g = 0
-
πα
2
-
We draw the following conclusions from (10),
=
.(10).
(i) Let x=0, g 0. In this case the acceleration vanishes
and the velocity is constant. Hence if the cylinder is projected
with any velocity, it will continue to move along the direction of
projection with this velocity, and the only effect of the liquid will
be to produce an apparent increase in the inertia of the cylinder
which is equal to the mass of the liquid displaced.
(ii) Let x = 0. In this case the horizontal velocity is constant,
and the cylinder will describe a parabola with vertical acceleration
g (σ − p)/(o+p).
(iii) Let g0: and let the initial velocity be parallel to y
and equal to V. Putting xp/Ta² (p+o) =λ, and integrating (10)
we obtain,
πα
И
- V sin λt,
-1
a=
Vλ¹ cos λt,
v = V cos λt,
-1
B = Vλ¹ sin λt.
If therefore the cylinder is projected with velocity V in any
direction, and no external forces are in action, it will describe a
circle in the same direction as that of the cyclic motion.
(iv) When neither g nor к are zero, the integrals of (10) are
U
(o - p) g
(σ + p)X
V sin λt,
v = V cos xt,
V
V
α
+ cos λt,
B
sin Xt,
+
λ
(o - p) gt
(στρ)λλ
and therefore the cylinder describes a trochoid moving from right
to left with mean velocity (σ - p) g/(o + p) λ.
186. The preceding results may also be obtained by Lagrange's
equations; for with the notation of § 178,
T = {πα² (p + σ) (x² +ÿ²),
also if (r', ') be current coordinates
whence
x=
|
K
2π
K
2π
log {(r′ cos 0′ − x)² + (r′ sin 0′ − y)²} *
K
log r'+ (x cos 0' + y sin '),
2πr'
A: Kx/2π,
Ky/2π.
MOTION OF AN ELLIPTIC CYLINDER.
189
Taking for a moment the origin at the centre of the cylinder,
the value of 2 is
dx
12--[[ d x r'drdo
=
K
2π
dr
rdrde
ffrare
= {K (c² — a²),
whence is constant, and the value of L is
L= }ma (i + y) + xp ( y − y ),
whence equations (10) at once follow.
187.
Let us now suppose that the cross section of the cylinder
is any curve, which does not possess cusps projecting into the
liquid', and that there is no circulation. The kinetic energy will
be a homogeneous quadratic function of the velocities u, v, w, and
by changing the directions of the axes we can make the term uv
disappear. We shall however for simplicity confine ourselves to
the case in which the cross section is a curve (such as an ellipse),
which is symmetrical with respect to two perpendicular straight
lines through its centre of inertia. In this case all the products
will disappear, and
2T=Pu² + Qv² + Aw²
(11).
Let the liquid be initially at rest, and let the solid be set in
motion by means of an impulse F. This impulse is equivalent to
a linear impulse F applied at the centre of inertia of the cylinder,
together with a couple about its axis. Let be the initial
angular velocity due to the couple, ẞ the angle which the direction
of the impulse F makes with the initial direction of u.
If ◊ be the value of this angle at any subsequent time, the
Principle of Conservation of Linear Momentum gives,
or
PuF cos 0, Qu-Fsin 0.
Substituting in (11) we obtain
P
cos
2
F¹2
+
ᎪᎾ
sin² 0
Q
2
+4j F¹2
=
cos² B sin²
+
³)+
+ AN²,
P
sin³
AỪª = AN² + F² ( 1 − 1 ) (sin² 0 — sin² §)
(sin² 0 – sin² ß) ………….. (12).
1 Greenhill, "On the motion of a cylinder through a frictionless liquid under no
forces," Mess. Math., vol. xx. p. 117.
190
MOTION OF A SINGLE SOLID:
Let Q>P; then if
Q-P
<F sin ẞ √ APQ
ß
I will vanish, and the cylinder will oscillate; but if
>Fsinß2
Q-P
APQ
Ø will never vanish, and the cylinder will make a complete
revolution.
Case I. When the cylinder oscillates, (12) may be written
0 = I √ sin² 0 — sin² a
where I = F √ (Q − P)/APQ, I² sin² ß — N² – I² sin² a.
(13),
Equation (13) shows that can never be < a nor>π- a through-
out the motion, hence the axis of least effective inertia (i.e. the
longest diameter of the cross section) will oscillate between the
angles a and π-α.
a. The cylinder will therefore move so that
its flattest side tends to turn itself towards the direction of
motion.
Let
then (13) becomes
cose
= cos a sin o,
αφ
It
‡π √/(1 − cos² a sin² )
and therefore
cos ✪ = cos a sn (K + It)
(14),
and the period of oscillation is 4K/I.
Let (x, y) be the coordinates of the centre of inertia of the
cross section referred to fixed axes, then
whence
¿ = u cos 0 — v sin 0, y = u sin 0 + v cos §,
F
Q
2
x
4 = { + F ( − 1 ) cos² 8
e Ө
y= F(-) sin cos 0
6
(15).
These equations show that the centre of inertia of the cross
section of the cylinder moves along a straight line parallel to the
direction of F with uniform velocity F/Q, superimposed upon
which is a variable periodic velocity, and at the same time vibrates
..
*
MOTION OF AN ELLIPTIC CYLINDER.
191
perpendicularly to this line. This kind of motion frequently
occurs in hydrodynamics, and a body moving in such a manner is
called by Thomson and Tait a Quadrantal Pendulum¹.
Substituting the value of 0 from (14) in terms of t in (15), and
integrating, we shall obtain the values of x and y in terms of t,
and the equation of the path will be obtained by eliminating t
from the resulting equations.
Case II. When the cylinder makes a complete revolution, let
ΑΩ
F²
F2/1
AN² + (-1) 008' - (-1)
cos² B=
k2 P
then it is easily seen that k < 1, and (12) becomes,
whence
I
• = — (1 − k² cos³ 0)*
-
cos ✪ = sn (K – It/k),
choosing the constant so that vanishes with t. Hence the
solution can be continued as before.
Case III. This is the limiting case between I. and II.
Here
and therefore
ΑΩ
An' = F(-1) sin' 9,
G
P
0 = I sin 0
n² ß,
It = log tan 10.
Therefore
dy IA
de
F
cos 0,
IA
y
sin 0,
F
dx
F
IA
cosec e
do ~ PI
sin 0,
F
F
x =
Pi log tan 10+
IA
cos 0.
F
Putting IA/F= c, and eliminating we obtain the equation of
the path, viz.
F
log
y
√c²
x = 17 108 6 + √o — y + √o - y²,
PI c
c² - y²
¹ Natural Philosophy, vol. 1. part 1. § 322.
192
MOTION OF A SINGLE SOLID.
გ
FIG. 1.
α
n
FIG. 2.
0
IX
Χ
yjp
FIG. 3.
d
X

MOTION OF AN ELLIPSOID.
193
The curves described by the centre of inertia of the cylinder
in the three cases, have been traced by Greenhill and are shown in
the figures 1, 2, 3 of the accompanying diagram.
If the cylinder is projected parallel to the longest diameter of
its cross section and be slightly displaced, it appears from (12) that
its motion will be the same as that considered in Case III.
The values of P, Q, A for an elliptic cylinder are,
whence QP.
pb
=
ρα
P= M (1+0), Q = M (1+0).
σα
+ b² + P (a² − b³)²)
A = = z M {a² + v²
2oub
ob
188. If the cross section is a curve such as a cardioid, which
is symmetrical with respect to only one straight line through its
centre of inertia, which we shall take as the direction of u, the
kinetic energy will be
2T = Pu²+Qv² + Aw² + 2Lwu,
and if we transfer the origin to a point on the axis of y whose dis-
tance from the origin is -L/P, the kinetic energy will be
2T= Pu² + Qv² + ( A
and the previous results apply.
(4-13).
P
w²,
Motion of an Ellipsoid¹.
189. If a solid which is symmetrical with respect to three
planes through its centre of inertia, which are mutually at right
angles, is set in motion along one of its principal axes, and there
are no forces in action, it will continue to move along that direction
with uniform velocity. Similarly if it be set in rotation about a
principal axis, it will continue to rotate about that axis with
uniform angular velocity, provided the solid does not contain any
apertures through which circulation takes place.
¹ Greenhill, "Fluid motion between confocal elliptic cylinders and confocal
ellipsoids," Quart. Journ., vol. xvi. p. 227.
B.
13
194
MOTION OF A SINGLE SOLID.
Let us now suppose that the solid is set in motion by means of
an impulse F, whose direction is inclined at an angle a to the axis
OC of the solid.

B
Ꮎ
Ө
X
E
LLI
A
If the axis of z coincides with the direction of the impulse,
and no forces are in action, the component momentum parallel to
z must be equal to F, and the components parallel to x and y
must be zero throughout the motion, whence
Pu=- Fsin e cos &,
Qv=F sin 0 cos &,
Rw= F cos 0.
Substituting these values of u, v, w in (3) we obtain,
2
2T = F" {sin²0 (cos + sin³4)
P
cos20)
2
2
+ +Aw₁²+Bw¸² + Cw¸²... (16).
R
The motion is therefore the same as that of a rigid body
rotating about its centre of inertia, under the action of a system of
forces whose potential is
coste.
·cos20
+ F² {sin² 0 (Cos² + sin'd)
P
+
R
190. Let the solid be moving without rotation along one of its
principal axes which coincides with the direction of the axis of x,
and be slightly disturbed from its state of steady motion.
Let u=u+u' be the new velocity parallel to a after disturb-
In the beginning of the disturbed motion, u', v, &c. are all
small quantities, and Kirchhoff's equations give
ance.
Pi=0,
A¿₁ = 0,
Q v
Pujw3
Rw= Pu₂w₂,
29
Bó₂ = (R − P) uw,
Cŵ=(P-Q) u¸v.
MOTION OF AN ELLIPSOID.
195
P(P-Q)
Hence
Qü +
u₁₂ v = 0,
Ꮯ
P(P-R)
Ri +
u³w = 0.
C
The motion will therefore be unstable unless P is greater than
either Q or R.
191. The only solid for which the quantities P, Q, R, A, B, C
have been determined is the ellipsoid.
From § 180 (2),
P = M -p ffp₁lds,
A = 1, − p SSX₁ (ny — mz) dS.
Hence if we write
A' = 2παbс
we obtain from § 147
1°
αλ
。 (a² +λ)ª (b² + λ)* (c² + λ)¹ '
eff A'st_ds,
P=M-PZ
ρ
4π
=M-A dadydz
A =
= M +
- 4π
M'A'
4π – A'
√
by § 7 (9), where M' is the mass of the liquid displaced. Similarly
A − I¸₁ = } M (b² + c²) — pa' Sſyz (ny — mz) dS,
-
1
· M (b² + c²) — pa' SSS (y² — 2²) dx dydz,
—
= ƒ {M (b³ + c³) +
M′ (b³ — c²) (C′ – B′)
4π (b³ — c²) + (B′ − C′) (b² + c²
Since C'>B'>A', it follows that R> Q> P, whence in the
case of the ellipsoid the least axis is the only direction of stable
motion.
192. When the motion is such that two of the axes always
remain in a plane, the equations of motion can be integrated; for
if these axes be the directions of u and v, we have w=0, w₁ = 0,
w₁ = 0, and
W
2T = Pu² + Qv² + Cwz3,
the kinetic energy is therefore of the same form as in the case
of the cylinder considered in § 187.
13-2
196
MOTION OF A SINGLE SOLID.
Under the same circumstances, when the solid is symmetrical
with respect to two perpendicular planes through its centre of
inertia, the kinetic energy is of the form
2T = Pu² + Qv² + Aw₂² + 2Luw₂,
3
which is reducible to the previous form.
On the Motion of a Solid of Revolution¹.
193. In considering the motion of a solid of revolution, it will
be convenient to discuss the case of a ring through whose aperture
there is circulation. If in our results we put = 0, we shall
obtain the motion of any solid of revolution ring shaped or not
when there is no circulation.
Let G be the centre of inertia of a plane curve S, OZ any
fixed straight line lying in the plane of S, and let OG be per-
pendicular to OZ. We shall assume S to be symmetrical with
respect to OG, but otherwise it may be of any form, provided
there are no singular points capable of giving rise to sharp edges;
and the ring will be supposed to be generated by the revolution of
S about OZ. Then O will be the centre of inertia of the ring, OZ
its axis of unequal moment, which will be called the axis of the
ring; and the circle described by G will be called the circular axis
of the ring.
Let the ring be introduced into an infinite liquid which
is at rest, and held fixed; let the circular aperture be closed up
by means of a plane diaphragm, whose area is σ; and let cyclic
irrotational motion be generated by applying to every point of this
diaphragm a uniform impulsive pressure «p, where p is the density
of the liquid, and then let the diaphragm be removed.
The velocity potential of this cyclic motion will be
where
= κΩ,
is a monocyclic function whose cyclic constant is unity,
and is the circulation, round any closed circuit, which embraces
the ring once only.
The resultant momentum of the cyclic motion will be parallel
1 Proc. Camb. Phil. Soc., vol. vi. p. 47.
:
1
MOTION OF A RING.
197
to the direction of the impulsive pressure on the diaphragm, and
equal to Z; and the energy to K³, where
2
= Kρσ — кpffQndS,
K=p√]
ΦΩ
ffdo do,
dv
and ʼn is the z-direction cosine of the normal to the ring drawn
outwards, and dS an element of its surface.
If the ring be set in motion, the kinetic energy and momentum
of the ring and liquid will be determined by the equations
2
2
2T = P (u² + v³) + Rw² + A (w¸² + w¸³) + Cw¸² + K«³…………..(17),
2
= Pu,
n = Bv,
C=Rw + Z
λ=Aw₁₂, µ = Aw₂,
λ=4ων
v=Cw,
.(18).
}
Since the liquid is incapable of producing a couple about the
axis of the ring, w, const. = ∞ throughout the motion.
=
Hence, if the ring be let go after the cyclic motion has been
generated, it will remain at rest; for the only possible motion will
be in the direction of its axis, and consequently
therefore
2T=Rw² + Co² + Kx²= its initial value,
W = 0.
194. Let the ring be set in motion by means of an impulsive
couple G about any diameter OB of its circular axis.
B
Z

Ө
A
The axis OC of the ring will evidently move in a fixed plane,
which is perpendicular to the axis of the couple. Let ◊ be the
inclination of OC to OZ at time t; u, w the velocities of O along
OA and OC.
198
MOTION OF A SINGLE SOLID.
The principle of Conservation of Linear Momentum gives,
whence
- sin + cos 0=Z,
E cos 0+ sin 0 = 0,
Pusin 0
Rw=-Z (1-cos e)
}
(19).
If ż, ✰ be the velocities of O along and perpendicular to OZ,
then
x = u cos 0 + w sin 0,
ż = w cos u sin 0.
-
Therefore
Z
R
1
✰ = (-) sin cos 0 –
0. sin
R
7.
+Z
P
G
1 1
Z
..(20).
cos20
cos e
R P
Ꭱ
2
Also
2T=Pu² + Rw² + AÒ² + Kʲ = const.
Substituting the values of u and w from (19) we obtain,
1
272
AÒª = A œ² – س ( 1/2 + 1/1
– Z²
R
)
+
R
cos @ + Z³ (1)
2
P R
1)
cos20...(21)
=ƒ(0) say,
where is the initial value of Ø.
W
The character of the motion depends upon the roots of the
equation ƒ(0) = 0, which we shall now consider.
The roots are

7
Z2
2
+
Aw2
Ꭱ
P2
(
1
PR
cos
Z
G
1
P R
)
Case I. Let R> P.
In order that the roots may be real, we must have
R
w< Z√√ AP (R −P) ·
*
If this condition be satisfied, one root will be positive and <1,
and the other will be negative and less than — 1. Hence Ở will
vanish when has some value ẞ lying between 0 and 7, and the
ring will oscillate between the angles ß and — ß.
But if
MOTION OF A RING.
Ꭱ
w>% √√ AP (R−P)
W.
199
both roots will be imaginary, and will never vanish. Hence the
ring will make a complete revolution.
Case II. Let P> R.
o
In this case both roots are real, and one of them is positive and
<1 provided w is sufficiently small; but if ∞ is sufficiently large
both roots will be negative and <-1. In order that one root
should not be <-1, it is necessary that
@<
2Z
JAR
If this condition be satisfied, the ring will oscillate between the
angles ẞ and -ß, where ẞ lies between 0 and ; but if
B
ß
ω>
27
JAR'
the ring will make a complete revolution.
195. In order to find the period of oscillation or revolution,
as the case may be, we must express @ in terms of t.
Case I. R> P.
(i) Let the roots be real and equal to p and -q, where
q>1> p>0.
Equation (21) may be written
where
Let
where
Then
where
Ò² = M² (cos 0 − p)(cos 0+q),
Z²
M²
2
cos Ꮎ
do
(cos 0 − p) (cos + q)
D
-
APR (R−P).
1-D cos²
1 + D cos²
1-p
1 + p
2 √D sin dø
1 + D cos²
(1 − p)(1 + q)
(1 − D cos² ¿)½ (1 — k² sin² 4),
k² _ (q − 1) (1 − p)
2 (p + q)
Ф)
4
t
200
Therefore
therefore
MOTION OF A SINGLE SOLID.
Mdt
=
2
dp
√(1 +p) (1 + q) √(1 − k² sin² µ)
$=am It,
I=¿M √(1+p) (1 + q).
where
Therefore
cos 0:
1+ p − (1 − p) cn² It
1+p+ (1 − p) cn² It'
and the period of a complete oscillation is 4K/I.
(ii) Let the roots be imaginary and equal to p±ıq.
Then
2
Ở = M * {(cos 0 — p) +g}.
Let
cos
=
1-D+(1+D) cos
1+D+ (1 − D) cos
Then
do =
2√Ddp
and
1 + D + (1 − D) cos &'
{(cos0− p)²+q²} {1+D+(1−D) cos&}={1−D−p(1+D)}²+q²(1+D)²
+2 cos [(1 − p)² + q³ − D² {(1+p)² + g'}] + [{1+D−p (1 — D)}²
2
·
+q² (1-D)³] cos² p.
Hence, if
D2
(1 − p)² + q²
2
(1 +p)² + q² ›
the coefficient of cos will vanish; substituting this value of D,
we obtain,
do
√{(cos 0 − p)²+q²}
1
αφ
{(1+p² + q²)² - 4p²}³ √(1 — k² sin² 4
1
where
k2
2
- [ 1 +
Hence
where
1 — p² — q²
~ ]⋅
{(1+p² + q³)² - 4p²)
$ = am I't,
I' = M {(1 + p² + q²)² — 4p²}* ;
and we finally obtain
q²)
tanº 30 = {(1+p)² + 7°
((1 − p)² + q³ ³ 1 — cn I't
((1+p)² + q²) * 1+ cn I't'
and the time of a complete revolution is 4K/I'.
201
-
Case II. P> R.
MOTION OF A RING.
In this case both roots are real, and one root is always negative
and numerically greater than unity.
(i) Let the roots be p and -q, where q>1>p>0. The
transformation is the same as in Case I. sub-case (i).
(ii) Let the roots be -p and -q, where q>1>p>0.
Then
where
j2
2
Ở = M * (cos 0 + p) (cos 0 +q),
M²
2
2
(P – R).
APR
In this case we employ the same transformation, but must put
D
1+ p
1-p'
k²
_
1+q+D(q − 1)
-
D(g-1) (q − 1)(1 − p)
2 (g-p)
(iii) Let the roots be -p and -q, where q>p> 1.
We must put
cos A
where
Ꭰ
1-D sin²
1+ D sin² o'
D=_P-1
p+1'
k² = (p − 1) (q − 1)
(p + 1) (q + 1)
In order to obtain the path described by the centre of inertia
O of the ring, we must substitute the value of 0 in terms of t in
(20), and integrate the result.
We can however ascertain the character of the motion of O
without integrating (20). For differentiating (21) we obtain
R
e
Z2
Аё
AÏ =
2
R
sin - Z³ (1-1) sin cos 0.
Аё
Therefore
x
Z'
and
A
x =
Also the value of ż may be written
(Ô – ).
1
7
P
PR
1) cos²
cos² 0 +
2
Ꭱ
8+ // cos 0].
202
MOTION OF A SINGLE SOLID.
The term in square brackets has its greatest value when 0 = 0,
in which case =0; hence can never become negative. The
motion of O is such that O moves along the initial direction of the
axis of the ring with a uniform velocity, superimposed upon which
is a variable periodic velocity; and at the same time vibrates
perpendicularly to this line.
196. Since the momentum due to the circulation alone is
always perpendicular to the plane of the ring, it follows that if a
ring initially at rest be set in motion by means of a couple about a
diameter, the direction of this momentum will be changed; and
the opposition which the liquid exerts against this action on the
part of the ring, will produce a couple tending to oppose the rotation
of the ring. Also, since the impressed couple can produce no effect
on the linear momentum of the system, it follows that the effect of
changing the direction of the momentum due to the circulation,
will be to cause the ring to move with a velocity of translation,
which gives rise to a linear component of momentum of the whole
system, such that the resultant of the latter and Z (whose direction
has been changed) must be a momentum equal to Z, and whose
direction coincides with the original direction of Z.
197. We shall now investigate the stability of the motion of a
ring, which is moving parallel to its axis in the direction of the
cyclic motion.
When the motion is steady
C=Rw+= const. =y,
v=Cw₂ = const.
ξ===μ= 0.
=
ΟΩ,
In order to obtain the disturbed motion, we must have recourse
to Kirchhoff's equations of motion; we shall also suppose that the
co-ordinate axes are fixed in the ring.
Putting for shortness
P(Z-y)
Z=y+
Ꭱ
the equations of disturbed motion are,
Pú - PQv + yw, = 0,
Pv − yw₁ + PQu = 0,
A œ₁ + Zv + (C − a) Qw, = 0,
1
Αώ, - Zu-(C- α) Ωω, = 0.
2
1
t
STABILITY OF STEADY MOTION OF A RING. 203
Putting u=u'e, &c. the equation for determining p is,
Pp
- PQ
0
Y
ΡΩ
Pp
Υ
0
0
Ꮓ
Ар
Z 0
— (C — A) N
(C – A) N
Ap
=
= 0,
or
A³P²p*+P[2ZAy+{(C−A)²+A*} PQ²]p² + {P (C−A)Q²+Zy}²=0.
If Zy is positive both values of p² are real and negative, and
the motion will be stable; but if Zy be negative, the motion will
be unstable unless
Q² >
2AZY
P {(C − A)² + A²}
If = 0 the roots are
Zy
p = ± i
AP'
.(22).
and the criterion depends altogether on the sign of Zy. Now
Zy = y² - Pyw.
(i) Hence if κ and w are both positive, y will be positive and
K
Zy>0 if R>P,
but if R<P, Zy will not be positive unless
Z>(P− R) w.
(ii) If x is positive and w negative - w', y=Z-Rw';
hence if > Rw', then Zy> 0; but if Z< Rw', Zy will not be
positive unless
(R − P) w' > Z,
which requires that R>P.
(iii) If = 0 and N is not zero,
K
Zy = R (RP) w².
Hence if R> P the motion will be stable; but if R <P the
motion will be unstable unless
2AR (PR)
Ω>ω P {(C − A)² + A*}*
198. Another kind of steady motion may be obtained by setting
the ring in motion by means of a couple G about a diameter of its
circular axis, and at the same time applying an impulse Z in the
opposite direction to that of the cyclic motion.
204
MOTION OF A SINGLE SOLID.
The effect of the latter impulse is to destroy the linear
momentum of the system, hence
Therefore
=0, C=0.
Z
u =0, w
R.
Ꭱ
·
G = AÒ.
Kirchhoff's 5th equation gives
μ = const. = G
The motion of the ring is such that its centre of inertia O,
describes a circle about a fixed axis parallel to the axis of the
couple, through which the plane of the ring always passes. If r
be the distance of O from this axis,
therefore
Z
Ꭱ
Gr
v=r8= ;
Α
AZ
r =
RG
In order to determine the stability, we must put
= Pu,
n = Pv,
C=Rw,
λ = Aw₁,
μ=G+Aw₂,
v = 0,
Z
G
W
พ
Ꭱ
+w,
W2
A
+ @2
in Kirchhoff's equations of motion, where the quantities u, v, &c.,
on the right-hand sides of these equations, are small quantities in
the beginning of the disturbed motion. Also, if the axes are fixed
in the ring,
G
= 0
0,
1
19
A
+@g
3
0₁ = 0,
and the equations of disturbed motion are
RG
Pu+ W =
A
= 0,
Pi=0,
PG
Rw
Aw, +
1
Αώ,
U =
A
PZ
R
= 0,
v = 0,
u = 0.
PZ
R
HELICOIDAL STEADY MOTION.
From the first and third equations we obtain
Gt
w = w' sin
The fifth equation gives
Rw'
P
Zw
ω
sin
G
The second and fourth give
COS
A
+a),
(Gt+a).
(Gt + a)
+ a)+const.
v = const.,
PZ
@1
vt + const.
AR
205
These equations show that the motion is stable for all displace-
ments which do not tend to remove the centre of inertia from the
plane of its motion; but the motion is unstable for all displacements
which tend to produce this effect. If the disturbance is such that
v=0, the disturbed motion will still be stable, but the axis of
rotation will be shifted through a certain angle.
199. A third kind of steady motion, which is helicoidal, is
obtained by first communicating to the ring an arbitrary angular
velocity about its axis; secondly by applying an impulsive
couple G about an axis inclined at an arbitrary angle a to the axis
of the ring; and thirdly by applying a determinate impulse in the
plane of the axes of the ring and couple.
In order that steady motion may be possible, it is necessary
that v and therefore ʼn should be zero throughout the motion. This
η
Z

0
X
B
A
condition may be secured by means of an impulsive force whose
components in the direction of X and Z are -Zsin a, and F.
206
MOTION OF A SINGLE SOLID.
The equations of momentum are
whence
(cos + sin ) cos yn sin = 0,
(cos + sin ) sin + cos y = 0,
- Esin 0+(cos 0=F+Z cosa;
§ = − (F+Z cos x) sin ✪)
n=0
=(F+Z cos a) cos 0
(23).
Since the components of momentum parallel to the axes of X
and Y(which are fixed in direction, but not in position because (
is in motion) are zero throughout the motion, the angular momen-
tum about OZ is constant, whence
– A∞, sin 0 + CN cos 0 = G + CD cos a.
The equation of energy gives
2
Pu² + Rw² + A (w¸² + Ö²) = const.,
putting Z=F+Z cos a, this becomes
Z2 sin² 0
+
P
2
{Z cos - Z}, {G+ CD (cos a cos 0)}*
0 Z}²
+
Ꭱ
A sin² 0
C
+ A0² = const. =
= const. = its initial value.....
This equation determines the inclination ✪ of the axis.
(24).
•
(25).
200. So far our equations have been perfectly general, we
shall now introduce the conditions of steady motion. These are
0
= α,
is = µ,
Ô = 0 = 0...........
Aμ sin² α = G.....
whence (24) becomes
.(26),
(27).
Differentiating (25) with respect to t, and using (26) and (27),
we obtain
A
2
Αμ' cos a - Ωμ +
1 1
RP
Z3Z
Z2 cos a
Ꭱ
=0...(28).
In order that steady motion may be possible, we must have
C²² > 4ZA cos a
R
1 1
P
Z cos a
Z
Ꭱ
.(29).
Hence, if R>P steady motion will always be possible, but if
P> R, steady motion will be impossible unless the condition (29)
is satisfied.
✯ = (u cos @ + w sin @) cos y = {7 (1/2 − 1)
STABILITY OF HELICOIDAL MOTION.
If x, y, z be the co-ordinates of 0, we have
0
Ꮓ
207-
Z
COS a
P
R
sin a cos pt,
1
1
Z
y = (u cos 0 + w sin 0) sin y
cos a
Р
R
R
sin a sin ut,
ż = w cos 0 — u sin 0 = Z
sin² a cos² a
+
2
Z cos a
P
R
R
whence the centre of inertia describes the helix
1
1
Z
XC
Z
cos a
sin a sin ut,
R
P
R
μ
1
1
1
y
Ꮓ
cos a
R
sin a cos pet,
μ
2
a
Z cos
2
P
R
Ꭱ
R P
2
= {z (sin² + cos a) - 008 a) t.
This last result may be at once obtained from the fact that the
impulse of the motion must consist of wrench about a fixed axis¹.
201. To examine the stability differentiate (25) with respect
to t, we thus obtain
AÖ + ƒ (0) = 0.
Hence the motion will be stable or unstable according as ƒ' (a)
is positive or negative.
Now
1 1
f(0)=3Z
sin 20+
PR
CR
+
A sin @ {G+CQ (cos a
therefore
ZZ sin 0
R
{G+CQ(cos a−cos)}
p² =ƒ' (a) = Aµ³ (1+2 cos² a)
cos
A sin³ @ {G+CQ (cos a — cos ()}²;
- A sin³ 0
C2Q2
1
−3CDµ cos a +
Α΄
R
Z2 1) cos
cos 2a +
+22
cos α.
R
Eliminating
2 4
by means of (28) we obtain
A³µ³µ³ = A²µ¹ + Aµ³Z Z
{z (1/1) - -
1)(1 − 3 cos' a) +
R P
27
R
cos a}
1
1
Z
2
+ZZ
Cos a
R P
R
¹ An elementary demonstration of the results of this article when there is
no circulation, has been given by Greenhill; Quart. Jour., vol. xvii. p. 86.
208
MOTION OF A SINGLE SOLID.
The condition that p² should be positive is easily found to be
that
(25Z
R
R P
+ Z (1/2 − 1) (1 − 3 cos a)
(2Z
Ꭱ
Z
1 11
R P
(1 + 3 cos
3 cos a)}
should be positive.
If there is no circulation = 0, Z = F, whence the condition
becomes
F2
α
F² (1-1) (9 cosa - 1) > 0,
R P
1
3
which requires that a should lie between cos¹ and 0, or between
π- cos¹ and π.
The azimuthal motion of a solid of revolution when there is no
circulation, has been worked out by Prof. Greenhill in the Quart.
Jour., vol. XVI. pp. 247-254; and another investigation by him
by means of Weierstrass's Functions will be found in the
Appendix.
General Motion of a Solid.
202. Having discussed the preceding special cases of motion
we shall pass on to discuss certain general theorems relating to
the motion of a single solid.
If the form of the solid is similar to that of a two bladed screw
propeller of a ship, which is symmetrical when turned through two
right angles about the axis of z, the kinetic energy must be
unaltered when the signs of u, v, w₁, w, are all changed, whence
2T = Pu² + Qv²+ Rw² + 2R'uv+Aw,+ Bw+Cw+2Cw,w₂
2
3
+2w¸ (Lu+Mv)+2w, (L'u + M’v)+2N'w¸w…………..(30).
If the solid resembles a four bladed screw propeller which is
symmetrical when turned through any multiple of a right angle,
the kinetic energy must be unaltered when —v, u, @2, w, are
written for u, v, w,, w, respectively, whence
2
2T =P (u² + v²) + Rw² + A (w¸² + w₂²)+Cw¸²
1
2
2
3
1
+2L (w¸u+w¸v)+2M (w¸v−wu)+2N″∞¸w………….(31).
In this expression the term wv - wu can be got rid of by
moving the origin along the axis of z.
GENERAL MOTION OF A SOLID.
209
If the solid is symmetrical with respect to itself when the axes
of x and y are turned through any given angle a in either direction,
it can be shown that if (1) be transformed by putting
Ꮳ
u = u' cos 0 — v′ sin 0,
w₁ = w₁' cos 0 — w½' sin 0,
v=u' sin 0 + v' cos 0,
w₂=w,' sin 0+w, cos 0,
the condition that the transformed expression for T should be
unaltered when is put equal to a or -a, is that T must be of the
form (31).
-
This kind of symmetry is called helicoidal symmetry.
X
Let us now suppose that there is another axis situated anywhere,
with respect to which the solid possesses helicoidal symmetry.
Since the form of (31) is not affected by turning the axes of x and y
through any angle, we may suppose them placed so that the other
axis of helicoidal symmetry lies in the plane cz. Turning the
axes of x and z round that of y through a certain angle o, the
new axis of a will be the axis of helicoidal symmetry, and the
expression for the energy will be of the form (31) but with the axes
of x and z interchanged; whence
2
2T =P (u² + v²+w²) + A (w¸² + w₂² + w¸³)
3
+2L (uw¸ + vw₂+ww¸)…………….(32).
A solid of this kind is called by Sir W. Thomson an isotropic
helicoid¹.
203. When a solid is set in motion along a given direction, it
will not in general continue to move along that direction: similarly,
if the solid be set in rotation about a given axis, it will not in general
continue to rotate about that axis. We shall however show that
there are always three directions mutually at right angles, such
that if the solid is set in motion along any one of them without
rotation and then left to itself, it will continue to move along this
direction with uniform velocity.
When there are no impressed forces, Kirchhoff's equations of
motion, § 167, are satisfied by putting ∞, ∞, ∞,= 0, and u, v, w
w。=w¸=
all constant, and
w
1
1 dT 1 dT 1 dT
2
3
u du
v dv
w dw'
1 Proc. Roy. Soc. Edinburgh, vol. vII. p. 384.
Hydrokinetic Symmetry," Quart. Journ. vol. xx. p. 261.
See also, Larmor, “On
B.
14
210
MOTION OF A SINGLE SOLID.
whence 2T= Pu² + Qv² + Rw² + 2P'vw + 2Q'wu + 2R'uv,
and
Pu+R'v+Q'w _ R'u + Qv+P'w _ Q'u + P'v + Rw
И
V
W
These equations show that the resultant velocity must be
in the direction of one of the principal axes of the ellipsoid
Px²+Qy²+ Rw² + 2P'yz +2Q'zx+2R'xy = const.,
which proves the proposition.
204. It is shown in treatises on Statics that every system of
forces is reducible to a wrench; that is to say a single force, and
a couple whose axis coincides with the direction of the force. The
ratio of the couple to the force is called the pitch of the wrench.
Similarly the motion of every rigid body is reducible to a twist
about a certain screw; that is to say a velocity of translation
along a certain line which is called the axis of the screw, together
with a rotation about that axis. The ratio of the linear to the
angular velocity is called the pitch of the screw.
If in § 203 the axes of coordinates coincide with the three
directions of permanent translation, the impulse is determined by
the equations
М
dT
dT
Pu; λ
Lu,
du
dw,
and therefore consists of a wrench of pitch L/P.
205. The above motion is not the only permanent steady
motion of which the solid is capable: for if the velocities and there-
fore the momenta are constant, Kirchhoff's first three equations
of motion give
ૐ
ๆ *
h......
W1
W2
W3
and the last three combined with these give
λ - hu
μ - hv
W2
(33),
V hw = k...............(34).
W3
Equation (33) expresses the condition that the axes of the screw
and wrench should be parallel, the condition that they should be
coincident is
λωι - ξω _ μω, - ηυ
2
ω,
2
which by (33) is equivalent to (34).
vw, — {w
ω
STEADY MOTION OF A SOLID.
211
Hence there exists a simply infinite system of possible steady
motions, each of which consists of a twist about a certain screw.
The pitches of the screw and the wrench are in general
different; if be that of the former and ' that of the latter
K
whence
λξ + μη + νζ_ wu + wu +w w
k
K =
3 +
{² + n² + 5²
2
w₁² + w₁₂² + w.
2
h'
3
k = h (x' — x).
2
2
And the expression for the kinetic energy becomes
21 = ξu+ην + ζω + λω, + μω, + νω;
= (x + K') hw³,
1
where w is the resultant angular velocity.
The values of h and k are not independent, for if the three
directions of permanent translation be chosen for the axes of
coordinates, and we substitute in (33) and (34) the values of §, n, §
&c. obtained by putting P, Q, R' equal to zero, we shall have the
following system of equations
(A − k) w₁ + C'w₂+ B'w₂+ (L− h) u + Mv+ Nw= 0
&c.
&c.
(L − h) w₁ + L'œ₂+ L''w₂+ Pu= 0
&c.
&c.
.(35).
Substituting the values of u, v, w from the last three equations
in the first three, it will be found that (35) are of the form
aw₁ + y w₂ + B'∞¸= kw,,
1
2
3
yw₁ + ßw₂ + a'∞¸ = kw¿
1
8
B'w₁ + a w₁₂+ yw₂ =
1
2
whence k is determined by the equation
kw,
a-k,
Y'',
B'
Y'
B-k,
á
=
= 0.
B',
ά,
Y-k
The roots of this equation are all real; hence to every value of
h there are three values of k, which are all real; and the axes
of the three screws are mutually at right angles but do not in
general intersect.
14-2
212
MOTION OF A SINGLE SOLID.
206. We shall now show that when the impulse of the motion
consists of a couple only, the motion of the solid consists of a
motion of translation combined with a motion of rotation, which is
the same as that of a certain ellipsoid which rolls upon a certain
moveable plane.
Taking the axes of permanent translation as the axes of
coordinates, we have === 0 throughout the motion; hence
Pu + Lw₁ + L'w₂+L"∞,= 0 &c. &c.
wz
3
Aw₁ + C'∞₂+ B'w + Lu+Mv+ Nw=λ &c. &c.
1
3
If we eliminate u, v, w from the last three equations by means
of the first three, it will be found that
where
λ=
do
dw'
do
μ
dw₂'
ע
do
dw3
2
20 = Pw¸² + Qw¸² + Rwz² + 2}'w₂w; + 2Q´w;w¸ + 2R'w¸w„…….(36),
1
L2 M2 N2
1
P = A
-
&c., &c.,
P Q
R
LL"
M'M"
N'N"
P'= A' -
&c., &c.
P
Q R
The equations of motion are
λ = w‚µ — w‚v &c. &c.
...
·(37).
In equations (37) let us change the directions of the axes
which are fixed in the body, so that they coincide with the principal
axes of the quadric
Px² + Qy²+Rz² + 2P'yz + 2Q'zx + 2R'xy = const.
If this be done, and the equation of the quadric referred to
these axes is
we shall have
and (37) becomes
ax² + By²+yz² = 1,
x' = aw₁, µ' = Bw₂, v' = yw;',
aw' – (B − y) w,' w;' = 0, &c.
-
whence the motion of rotation is obtained by making the above
mentioned quadric roll on the plane
r tuy tuz = const.,
whose direction is fixed in space (since λ, µ, v are constant), with
an angular velocity proportional to the length OI of the radius
vector drawn from the origin to the point of contact I.
EXAMPLES.
213
The motion of translation is obtained by making the plane and
quadric move through space with a velocity whose components are
given by =0, n = 0, $= 0.
The theorems of the last two articles are taken from a paper
by Prof. Lamb, Proc. Lond. Math. Soc. vol. VIII. p. 273.
EXAMPLES.
1. Apply Lagrange's equations to determine the equations of
motion of an anchor ring; and thence obtain the theorem that the
flux through the aperture relative to the ring, is the generalized
velocity corresponding to the product of the circulation and density
of the liquid.
2. If A and B be the forces required to act per unit of time, in
order to generate unit velocity perpendicular and parallel respec-
tively to the axis of an ellipsoid of revolution in an infinite liquid,
and if G be the couple required to act per unit of time in order to
generate unit angular velocity about an equatoreal axis, prove that
the kinetic energy of the ellipsoid and the liquid is
2
† (Au²+Av² +Bw² + Gw¸² + Gw² +Cw¸²)
1
with Euler's notation, C being the polar moment of inertia of the
solid.
Express T in terms of Lagrange's coordinates x, y, z, 0, 4, 4;
and prove that if the axis of z be parallel to the impressed impulse
F, then
.
1 11
∞ = − F
A
B
sin e cos e cosy,
1
ÿ = − F
sin e cose sin,
A
B
ż
* = F(sin" e
sin Ꮎ
2
cos Ꮎ
+
Α
B
1820), & + cos 0 = w
Ꮎ
Gi sin² 0 + Cw, cos 0 = E
&
3
(a constant),
´sin² 0
2
cos² 0\
GO® + Gi*sin 0 +C+F
2
+
A
B
9) = 2T.
214
MOTION OF A SINGLE SOLID.
3. In the midst of an infinite mass of liquid at rest, is a sphere
of radius a, which is suddenly strained into a spheroid of small
ellipticity. Find the kinetic energy due to the motion of the
liquid contained between the given surface, and an imaginary
concentric spherical surface of radius c; and show that if this
imaginary surface were a real bounding surface which could not
be deformed, the kinetic energy in this case would be to that in
the former case in the ratio
c³ (3a5 +2c³) : 2 (c³ — a³)².
4. A pendulum with an elliptic cylindrical cavity filled with
liquid, the generating lines of the cylinder being parallel to the
axis of suspension, performs finite oscillations under the action of
gravity. If be the length of the equivalent pendulum, and l' the
length when the liquid is solidified, prove that
l' — l =
ma2b2
h (M+m) (a² +b²) ·
where M is the mass of the pendulum, m that of the liquid, h the
distance of the centre of gravity of the whole mass from the axis of
suspension, and a, b the semi-axes of the elliptic cavity.
5. Find the ratio of the kinetic energy of the infinite liquid
surrounding an oblate spheroid, moving with given velocity in its
equatoreal plane, to the kinetic energy of the spheroid; and denot-
ing this ratio by P, prove that if the spheroid swing as the bob of
a pendulum under gravity, the distance between the axis of the
suspension and the axis of the spheroid being c, the length of the
simple equivalent pendulum is
(1+P) c + 2a²/5c
1-p/o
where a is the equatoreal radius, σ and p the densities of the
spheroid and liquid respectively.
6. A pendulum has a cavity excavated within it, and this
cavity is filled with liquid. Prove that if any part of the liquid
be solidified, the time of oscillation will be increased.
7. Prove that if a number of solids be moving freely under
their mutual attractions in an unbounded liquid, the impulse of
the motion remains constant.
י
EXAMPLES.
215
8. The space between two infinitely long coaxial cylinders of
radii a and b respectively, is filled with liquid of density p, and the
inner cylinder is suddenly moved with velocity U perpendicular to
the axis, the outer one being kept at rest. Show that the resultant
impulsive pressure on a length l of the inner cylinder is
πραlU
b² + a² 2
b² -— a²·
2
9. An elliptic cylindrical shell, the mass of which may be
neglected, is filled with water, and placed on a horizontal plane
very nearly in the position of unstable equilibrium with its axis
horizontal, and then let go. When it passes through the position
of stable equilibrium, find the angular velocity of the cylinder, (i)
when the horizontal plane is perfectly smooth, (ii) when it is
perfectly rough; and prove that in these two cases, the squares of
the angular velocities of the cylinder are in the ratio
(a³ — b²)² + 4b² (a² + b²) : (a² − b²)²,
2a and 26 being the axes of the cross section of the cylinder.
10. A solid ellipsoid of density σ is placed inside a fixed con-
centric, confocal, and similarly situated ellipsoidal shell, and the
space between them is filled with liquid of density p. Supposing
that the whole matter attracts according to the Newtonian law,
and that σ >p, show that when the solid ellipsoid is slightly
displaced parallel to its greatest axis, the time T of a small
oscillation is given by
p
2pabc
T²p (σ − p) A/2π =σ +P - abc (2 - A') — a'b'c' (2 — A)'
where a, b, c and a', b', c' are the semi-axes of the outer and inner
ellipsoids, and
A
1 = √。
୮
0
abcdλ
{(a² +λ)³ (b² +λ) (c² + x)}* °
11. The space between two coaxial cylinders is filled with
liquid, and the outer is surrounded by liquid, extending to infinity,
the whole being bounded by planes perpendicular to the axis. If
the inner cylinder be suddenly moved with given velocity, prove
that the velocity of the outer cylinder to that of the inner will be
in the ratio
2b³c²p: p(a²b² — a²c² + b² + b²c²) + σ (a² − b²) (b² — c²),
where a and b are the external and internal radii of the outer
cylinder, σ its density, c the radius of the inner cylinder and p the
density of the liquid.
216
MOTION OF A SINGLE SOLID.
12. The ellipsoid (x/a)² + (y/b)² + (z/c)² = 1, is filled with liquid
originally at rest, and rotates uniformly about an axis through its
centre of inertia: prove that the surfaces of equal pressure are
given by the equation
(A, B, C, A', B', C′)(x, y, z)² =λ,
where
A
2
2
(c² — a²) (3c² + a²) w₂² _ (a² — b²) (3b² + a²) w²
(c² + a²)²
(a² + b²)²
2
3
b²c² + c²a² + a²b² — 3aª
A'
(c² + a²) (a* +b²)
2
WqW gr
and @, @„, w, are the component angular velocities of the ellipsoid.
19
3
13. In the last example prove that if the ellipsoid be set in
rotation and then left to itself, the components of the velocity of
the liquid relatively to the ellipsoid are
x
Ў
2a³w.y 2a³w,z
a²+c²'
2b³∞,x
2
2
a²+b²
2
2b³∞,z
=
b² + c²
b² + a²,
2c²∞,x
c²+a²
d•
2c²w₁y
c² + b²'
and that if the ellipsoid revolves about a fixed axis after
2bcw,
1 +
b² + c²
2
2caw₂
c² + a²
2
+
2
(Zabo
2abwⓇ -}
2
a² + b²
revolutions of the ellipsoid, every particle of liquid will be in the
same position relatively to the ellipsoid.
14. A closed vessel filled with liquid of density p, is moved in
any manner about a fixed point 0. If at any time the liquid
were removed, and a pressure proportional to the velocity potential
were applied at every point of the surface, the resultant couple
due to the pressure would be of magnitude G, and its direction in
a line OQ. Show that the kinetic energy of the liquid was pro-
portional to pw G cos 0, where w is the angular velocity of the
surface, and the angle between the direction of w and OQ.
15. A solid cylinder of radius a immersed in an infinite liquid,
is attached to an axis about which it can turn, whose distance
from the axis of the cylinder is c, and oscillates under the action
EXAMPLES.
217
of gravity. Prove that the length of the simple equivalent
pendulum is
1 a² + c² (1 + p/o)
c (1 − p/o)
>
o and p being the densities of the cylinder and liquid.
16. A light cylindrical shell whose cross section is an ellipse
filled with water is placed at rest on a smooth horizontal plane in
its position of unstable equilibrium. If it is slightly disturbed,
prove that it will pass through its position of stable equilibrium
with angular velocity w, given by the equation
2
w²
=
8g
a² + b²
(a + b)² (a - b) *
17. A quantity of heavy heterogeneous liquid is placed inside
an ellipsoid, which is then moved so that the density of the liquid
is always the same function of the depth. Prove that a certain
cone coaxial and concyclic with the reciprocal ellipsoid, moves so
as always to have one of its generators vertical.
18. Liquid of density p is contained between two confocal
elliptic cylinders and two planes perpendicular to their axes. The
lengths of the semi-axes of the inner and outer cylinders are
c cosha, c sinh a, ccosh ß, c sinh 6 respectively. Prove that if the
outer cylinder be made to rotate about its axis with angular
velocity, the inner cylinder will begin to rotate with angular
velocity
P
Op cosech 2 (B - a)
coth 2 (ẞ − a) + σ sinh 42'
where σ is the density of the cylinder.
19. A circular cylinder of mass M, whose centre of inertia is
at a distance c from its axis, is projected in an infinite liquid under
the action of gravity. Prove that the centre of inertia of the
cylinder and the displaced liquid will describe a parabola, while
the cylinder oscillates like a pendulum of length
{(M+M') k² + M'c/2M'c,
where M' is the mass of the liquid displaced, and k is the radius
of gyration of the cylinder about its axis.
20. The space between two coaxial similar and similarly
situated elliptic cylinders is filled with liquid, and the cylinders
are rotating with uniform angular velocity w. Find what would
be the new angular velocity if the liquid were suddenly solidified.
218
MOTION OF A SINGLE SOLID.
21. A hollow vessel of the form of an equilateral prism filled
with liquid, is struck excentrically by a given blow in a plane
perpendicular to the axis and bisecting three edges; find the
initial motion of the vessel.
22. A cylinder whose cross section is an ellipse is moving in
an infinite liquid. Prove that when there is circulation round the.
cylinder, its equations of motion are
d
dt
d
(Pu cos 0 - Qu sin 0 + кру) = X,
dt (Pu sin ✪ + Qv cos ✪ — êpx) = Y,
d² e
C (P − Q) uv = N,
dt2
where (x, y) are the coordinates of the centre of the cross section,
X, Y the components of the impressed forces parallel to fixed
axes, N is the impressed couple about the axis of the cylinder, u, v
are the component velocities of the cylinder parallel to the major
and minor axes of its cross section, and ✪ is the angle which the
major axis makes with the axis of x.
23. Prove that helicoidal steady motion is always possible
when a planetary ellipsoid is moving in an infinite liquid; but it
is not possible in the case of an ovary ellipsoid, unless the ratio of
the angular momentum of the ellipsoid about its polar axis, to its
component velocity along this axis is greater than 2√RA (1 − R/P);
where R and P are the effective inertias of the ellipsoid about its
polar axis, and an equatoreal axis and A is its effective moment of
inertia about the latter axis.
24. A solid of revolution of mass M, is rotating in any
manner about its centre of inertia, in an infinite liquid. Prove
that if it is allowed to descend under the action of gravity, its
vertical velocity at time t will be equal to
/sin20 cos A
(M — M') +
P R
gt,
where M' is the mass of the liquid displaced; and is the
inclination of the axis of the solid to the vertical at time t.
Obtain the differential equation for determining de/dt.
CHAPTER X.
ON THE MOTION OF TWO CYLINDERS.
207. We have shown in Chapter V. that, when two cylinders
are moving in a liquid of density p, the kinetic energy of the
whole motion is
2T
=
(M+P) (u² + v³) + (M′ +Q) (u'² + v´²) + 2L (uu' — vv′),
where M, M' are the masses of the cylinders; u, v, u', v' their com-
ponent velocities perpendicular to and along the line joining their
centres. The values of the coefficients are given' by equations (73)
of § 123 or (74), (75) and (76) of § 124; and are functions of the
distance between the cylinders alone.
208. We shall now apply these formulae to the consideration
of the motion of a cylinder in a liquid bounded by a fixed plane,
when there is no circulation2.
When two equal cylinders are projected with equal velocities
perpendicularly to the line joining their centres, it is clear that
during the subsequent motion, the velocities of each cylinder
perpendicular to this line will remain equal, and that their veloci-
ties parallel to this line will be equal and opposite. Hence the
plane which is perpendicular to this line and bisects it will be fixed
in space, and there will be no flux across it. One of the cylinders
may therefore be removed, and the above mentioned plane sub-
stituted in its place; we shall thus obtain the motion of a cylinder
in a liquid which is bounded by a rigid plane.
1 See Errata.
2 Hicks, "On the motion of two cylinders in a fluid," Quart. Journ., vol. xvi.
p. 193.
220
MOTION OF TWO CYLINDERS.
Let the axis of a lie in the plane, and be perpendicular to the
axis of the cylinder; the kinetic energy of the liquid will be ob-
tained by putting a = B, 0, 0,= √q; u = u', v = − v' in equations
=
1
2
(74), (75) and (76) of § 124 and halving the result.
☛ be the density of the cylinder, and a its radius
where¹
2T = {(P + L) + πа³o} (u² + v³)
=
R (u² + v²)………...
{1
R = παρ 1 + 2Σ,
1+2
m
Hence if
……..(1),
∞ (1 − q)³ q™)
+ πασ
(1 — qm+1)³)
2
If no external forces act upon the system, the energy, and also
the momentum parallel to x, are constant; the latter condition gives
or
dT
= const. = G,
du
Ru= G ..
(2).
Since T and G are both constant, the equations of motion may
now be written
Ru= G
R (u² + v²) = 2T)
..(3).
Differentiating with respect to t and remembering that R is a
function of y alone, we obtain
1 dR
i +
2R dy
(v³ — u²) = 0
..(4).
Now R is necessarily positive; also y = a cosh a = {a (1 + q)/q*,
therefore R decreases as y increases; hence dR/dy is negative, and
therefore has always the same sign as v² — u².
u². Let U be the
resultant velocity, & the angle which its direction makes with the
axis of y, then
U² dR
cos 24.
2R dy
If therefore the direction of motion makes with the axis of y an
angle lying between 1 and 2, the acceleration from the plane
will be negative and the cylinder will be attracted towards the
plane, but if this angle lies between 0 and 1 or 2 and π, the
acceleration will be positive, and the cylinder will be repelled from
the plane.
Also since u = G/R, and R decreases as y increases, u increases
as the cylinder moves from the plane, and vice versâ.
1 The value of P+L in terms of elliptic functions will be given in the Appendix.
CYLINDER MOVING PARALLEL TO A PLANE.
U2 dR
If we put
2R dy
=ƒ
the component accelerations are
ú= ƒ sin 24,
i = ƒ cos 24.
221
209. If the cylinder be initially in contact with the plane,
and be projected perpendicularly from it, u = 0, and
v² = 2T/R = v³R/R,
where the suffixes denote the initial values of the quantities.
Since q = 0 when y = ∞, the limiting value of R is πа² (p+o).
When y = a, q=1; in order to find the value of R, let q = 1-λ,
where X is a small quantity which ultimately vanishes: then
Ro/πα
R。/πа² = p
{1
1 1 1
32 42
1+2 +22 +
= p (}π² − 1) +σ.
+
·)}+
to
Whence the ratio of the initial to the terminal velocity is
2
ρ
√3π²p + (σ − p).
(σ + p)
210. When the direction of projection is not perpendicular to
the plane, the direction of the velocity at any subsequent time is
given by the equation
cot &=v/u= ±√ Rp − 1,
where p=2T|G, and the upper or lower sign must be taken
according as the initial value of is < or > 7. Let cot be
&
initially positive, so that the cylinder is projected from the plane,
then since R diminishes to the limit wa² (p +σ) it follows that if
πа³р (p+σ) < 1, there will be some point which is determined by
the equation Rp = 1, at which cot = 0, and where the cylinder will
consequently be moving parallel to the plane. During the subse-
quent motion cot & will be negative, and the cylinder will approach
the plane and R will increase. The quantity √Rp 1 continually
increases as R increases, and hence will increase from 7 and the
-
Hence the cylinder will
cylinder will ultimately strike the plane.
or will not strike the plane according as wa²p (p + σ) < or > 1.
If πa³р (p +σ)=1, and a be the initial value of 4,
cot a = √√/ (R。/πα² − p − o)/(p + o);
222
MOTION OF TWO CYLINDERS.
whence a cylinder projected at an angle > a will meet the plane at
an angle
-1
tan¯¹ √{πα²р (π³p + σ − p) −1},
ρ
and a cylinder projected at an angle < a will move, when at an
infinite distance from the plane in the direction
cot¹ √{πα³р(p+o) − 1}.
If the direction of projection is equal to a, the cylinder when
at an infinite distance will move parallel to the plane.
211. Let one of the cylinders be fixed whilst the other moves
independently.
Let (r, 0) be polar coordinates of the centre of the moving
cylinder referred to the centre of the fixed cylinder as origin; if
R=P+M; then
2T = R (r² + r²Ø²).
Since R is independent of 0, we must have
dT
const. = h,
dė
or
Rr²0 = h.
Also since
d dT dT
0,
dt dr
dr
we obtain
1 dR
j- rj²
(j‚² — p² j²).
2R dr
the angle which its direction
Let U be the resultant velocity,
makes with the radius vector; the radial acceleration
f=
U² dR
2R dr cos 2p.
Since R decreases as r increases dR/dr is negative; hence the
cylinder will be repelled when & lies between 0 and 1 or between
π and π; and will be attracted if & lies between 1 and π.
212. If the cylinders be initially in contact, and one of them be
projected with velocity V along the line joining their centres, then
j² = 2T/R, V² = 2T/R。.
Therefore
js2
R
M+P
y²= R
M+P
223
CYCLIC MOTION.
If the cylinders are equal it can be shown in a similar manner
as before, that
or
Po
2
2
= πа² (1π² — 1) p,
P_= παρ,
∞
jo
whence
V
πρτσ- ρ
στρ
Cyclic Motion.
213. Let us now consider the motion of two equal cylinders
round which there is circulation in opposite directions, and which
are initially projected with equal velocities parallel to Ox.
—
Let A and B be the common inverse points of the two cylinders,
a the radius of either of them, u, v and u, v their velocities
parallel and perpendicular to Ox, y the ordinate of the centre of
the cylinder A; also let the circulation round A be in the contrary
directions of the hands of a watch.
It is known from the theory of rectilinear vortices, which will
be explained in Vol. II., that the cyclic motion is the same as
A
n
20
B
would be produced by two rectilinear vortices of circulations
and situated at A and B, hence with the notation of § 178,
the value of
by § 121.
Χ will be
K
x=-2π
BP
AP κη
log
2π'
224
MOTION OF TWO CYLINDERS.
Also, if a be the value of ʼn at the surface of the cylinder A,
and AB=2c,
and
a = c cosecha, y=c coth a
2
Σ (kx) = k³a/π.
(5),
Since this kind of cyclic motion could be produced by applying
a uniform impulsive pressure xp to every point of that portion of
AB which lies between the cylinders, we must have =0. Let
(r, 0) be the polar coordinates of P referred to O, then
r² + c² + 2rc sin
K
r² + c² - 2rc sin
X
log
4π
whence
Therefore
A=0, B=- Kс/π.
KC
sin 0 + &c.,
Πλ
L= T + 2ксρи — к²р²/2π + V
K
Also if M, M' be the masses per unit of length of either of the
cylinders, and of the liquid displaced,
T = R(u² + v²).
m
R = M' { 1 + 2 (1 − q)² Σ
9″
+ M
1
(1 − qm+1) ²;
where q = €-2α.
2a
If we suppose the cylinder B to be replaced by the fixed plane
Ox which forms the boundary of the liquid, the value of L must
be halved, and the equations of motion of the cylinder A will be
(
d
dt
(+
dT
du
+kpc) = X .........
..(6),
d dT
14
dT
dc K² da
2
dt dv
dy
+
кри
dy 4π dy
Y
(7).
From (5) we obtain
C=N
y² — a² and y = a cosh a,
dc
da
1
therefore
coth a,
dy
dy
whence (7) becomes
d dZ
dT
K
k2
½
흐
​dt dv
dy
κρu coth a+
=
Y
·(8).
4пс
Let us now suppose that gravity is the only force in action, and
that the plane boundary Ox is horizontal, forming, so to speak, the
bed of the ocean; (6) and (8) respectively become
Ru+кρc=const. = h
·(9).
dR
κρ
Rv + ½ (v² - u²)
кри coth a+
(M-M') g
dy
4πC
STEADY MOTION OF A CYLINDER.
225
These equations are satisfied by v=0, u and y constant, pro-
vided u satisfies the quadratic
κερ
гриз -
ру³ — κри coth a+
+(M −M')g=0………….………..(10),
4πC
where pdR/dy. The roots of this quadratic will be real
provided x²p² coth² a >p
K²p
K
+ 4 (M — M') } ... (11).
and
πC
CASE (i). Since p is positive the roots will always be real if
M'> M
2
к³ρ <πс (M' - M) g.
K
In this case the liquid is denser than the cylinder, and one of
the roots of (10) will be positive and the other negative, and the
positive root will be numerically greater than the negative root.
Hence there will be two cases of steady motion, in one of which
velocity of the cylinder will be in the same direction as that of
the liquid, due to the circulation at points between the cylinder
and plane; and in the other the velocity will be in the opposite
direction; also the velocity in the former case will be greater than
in the latter.
CASE (ii).
M'>M, x²p> 4πс (M' — M) g.
In this case the roots of (10) will be both real and positive
provided (11) is satisfied; hence the velocity in the two cases of
steady motion will be in the same direction as that due to the cir-
culation.
CASE (iii).
M> M'.
In this case the cylinder is denser than the liquid, and the
roots of (10), if real, must be both positive, hence the two
velocities must be in the same direction as that due to the cir-
culation.
CASE (iv). If either g=0 or M = M', (11) becomes
прс
coth² a > p.
Here both roots of (10) are positive, and the two velocities
must be in the same direction as that due to the circulation.
This case has been discussed by Mr W. M. Hicks'.
1 Quart. Journ. vol. xvII. p. 194.
B.
15
+
1..
226
MOTION OF TWO CYLINDERS.
CASE (v). Suppose that the cylinder is reduced to rest, and then
let go. Since u and v are initially zero, the initial acceleration is
1
4Rπс
{4πc (M — M') g + k²p}…………………….(12).
Hence if the liquid is denser than the cylinder it is possible
for the right-hand side to vanish; in which case the cylinder will
remain in equilibrium under the combined action of gravity
and the pressure due to the cyclic motion.
If the plane formed the upper boundary of the liquid the sign
of g in these five cases would have to be reversed.
215. The results of the last two cases may be inferred from
general reasoning.
We have shown in § 14, that the product of the velocity of a
liquid and the cross section of a tube of flow, is constant through-
out the length of the latter. Now in Case v. where the cylinder is
at rest, the tubes of flow are circles, and those portions of them
which lie between the cylinder and the plane will be more com-
pressed than the portions which lie on the remote side of the
cylinder; hence the velocity of the liquid at points between the
cylinder and the plane will be on the whole greater than at points
which lie on the opposite side of the cylinder, and consequently
the pressure on the side of the cylinder nearest the plane will be
less than that on the remote side, and therefore the cylinder will
be attracted towards the plane. If the cylinder is less dense than
the liquid, and the plane forms the lower boundary of the liquid,
the effect of gravity will be to repel it from the plane, and hence
there must be a certain position in which the two forces balance
one another, and in which the cylinder will be in equilibrium.
If on the other hand the plane forms the upper boundary of the
liquid, there will be a position of equilibrium, provided the
cylinder is denser than the liquid.
216. In Case IV. let the cylinder be moving with a small
velocity u parallel to the plane, and in the same direction as that
of the circulation between the cylinder and the plane. Let the
cylinder be reduced to rest by impressing on the whole liquid a
velocity u equal and opposite to that of the cylinder. At points
between the cylinder and the plane, the reversed velocity u of the
liquid and the velocity due to the circulation will be in opposite
EXAMPLES.
227
directions, whilst at points on the other side of the cylinder they
will be in the same direction. Also by § 14 each velocity will
be on the whole greater at points between the cylinder and plane,
than on the opposite side of the cylinder. Hence if u be small
enough, the cylinder will be attracted towards the plane, and
therefore if u increase from zero, a certain critical value u, will be
reached, at which the cylinder is neither attracted nor repelled,
but will be in equilibrium. In this case the resultant velocity at
points between the cylinder and plane, will be in the opposite direc-
tion to that on the other side of the cylinder.
2
If u continue to increase, the cylinder will at first be repelled
from the plane, but ultimately a second critical value u₂ will be
reached, at which the resultant of u, and the velocity due to the
circulation at points between the cylinder and the plane will on
the average be equal to the same quantity on the opposite side of
the cylinder, and there will be another position of equilibrium. In
this case the resultant velocity of the liquid at points between the
cylinder and the plane will be the same direction as that on the
other side of the cylinder.
If u exceeds this second critical value the cylinder will thence-
forth be attracted. The two critical values of u are evidently the
roots of the quadratic obtained by putting g = 0 in (10).
EXAMPLES.
1. A cylinder of radius a is surrounded by a concentric
cylinder of radius b, and the intervening space is filled with
liquid. The inner cylinder is moved with velocity u and the
outer with velocity v along the same straight line; prove that the
velocity potential is
φ
b³v - a²u
b² - a²
r' cos 0 +
(v — u) a²b² cos
(b² — a²) r
2. A long cylinder of given radius is immersed in a mass of
liquid bounded by a very large cylindrical envelope. If the
envelope be suddenly moved in a direction perpendicular to the
cylinder with velocity V, the cylinder will begin to move with
velocity V, provided the density of the cylinder be three times
that of the liquid.
15-2
228
MOTION OF TWO CYLINDERS.
3. Two infinite parallel cylinders in an infinite liquid are
projected with given velocity; (i) in opposite directions along a
line at right angles to their axes, (ii) in the same direction per-
pendicular to this line. Prove that they experience in the first
instance a repulsion from one another, and in the second instance
an attraction towards one another.
If their radii are indefinitely small in comparison with one
another, prove that their motion is initially the same as that of
two rectilinear vortices of equal and opposite strengths.
4. A solid cylinder with flat ends is fixed between two parallel
planes, and a cylindrical shell of the same length can slide freely
between the planes. If the space between the cylinder and shell
is filled with liquid, and the shell is placed so as to be coaxial
with the cylinder and then jerked in any direction with velocity
V, prove that the resultant impulse on the cylinder is
2MVb² (a² — b²),
where a and b are the radii of the cylinder and shell, and M is the
mass of the liquid which the cylinder displaces.
5. The space between a moveable cylinder and a fixed excentric
cylinder is filled with liquid. If the moveable cylinder be initially
projected with given velocity, perpendicular to the line joining its
centre with that of the fixed cylindrical boundary, determine its
motion, (i) when there is no circulation, (ii) when there is circu-
lation.
6. Examine the stability of the steady motion of a cylinder
parallel to a fixed plane, discussed in § 214.
!
CHAPTER XI.
19
ON THE MOTION OF TWO SPHERES¹.
2
1
217. WHEN two spheres are in motion in an infinite liquid,
the velocity of each sphere may be resolved into three components
U₁ V₁ W₁; U₂ V₂ w₂, where u, u, are the component velocities of
V2,
the spheres along the line joining their centres; and v₁, w₁; v₂, wa
are the component velocities parallel to two straight lines at right
angles to one another, which are perpendicular to the line joining
the centres of the two spheres. It would therefore at first sight
appear, that the kinetic energy of the liquid must contain twenty-
one terms, but it can easily be shown that twelve of these terms
must vanish. For let us suppose that v₁, w₁, u, v₂ are each zero,
U V 2
and consider the term involving uw,. The kinetic energy on
2
1 The present chapter has been taken from the following papers by Mr Hicks:
"On the Motion of Two Spheres in a Fluid," Phil. Trans. 1880, p. 455.
"On the Problem of Two Pulsating Spheres in a Fluid,” Proc. Camb. Phil. Soc.
vol. III. p. 277, and vol. IV. p. 29;
and a paper by the author,
"On the Motion of Two Spheres in a Liquid and allied Problems," Proc. Lond.
Math. Soc. vol. xvII. p. 369.
References may also be made to the following papers:
Stokes. "On some cases of Fluid Motion," Trans. Camb. Phil. Soc. vol. vII.
p. 105.
Bjerknes. Forhand. Skand. Naturfors, Christiania 1868, and Forhand. Vidensk.,
Christiania 1871 and 1875.
G. Forbes. "Hydrodynamic analogies to Electricity and Magnetism,” Nature,
vol. xxiv. p. 360.
Bertin. "Phénomènes Hydrodynamiques inversement analogues à ceux de
l'Électricité et du Magnétisme,” Ann. de Chimie et de Phys. (5) xxv. p. 257, 1882.
Pearson. "On the Motion of Spherical and Ellipsoidal bodies in Fluid Media,”
Quart. Journ. vol. xx. p. 60.
Herman. "On the Motion of Two Spheres in a Fluid and allied Problems,”
Quart. Journ. vol. xxii. p. 204.
230
MOTION OF TWO SPHERES.
account of the symmetry of the motion, must clearly be unaltered
if the direction of w, be reversed, and this requires that the
coefficient of u,w, should be zero. By similar reasoning it can be
shown that all the other coefficients must vanish, except those
2
2
2
2
1
2
U¡
2)
of u², u², v², v², w², w², u₁₂, vv, ww,; and also that the co-
efficients of v², v2, vv, must be respectively equal to those of
2
W.
2
w₁², w‚², w¸w,
19 2
2
2
1
2
Hence the kinetic energy of the system may be written
T={(Au¸² — 2Bu̟¸µ₂+ Cu¸²)+¿A′ (v¸² +w‚²)
2
1
2
1
2
+B' (v¸v₂+w¸w₂)+{C′ (v²+w¸²),
where the six coefficients are functions of the distance between
the centres of the two spheres and their radii.
The values of A, B and C must be determined by supposing
that the motion of the spheres is along the line joining their
centres, and those of A', B', C' by supposing that the motion is
perpendicular to this line.
Motion along the Line of Centres.
218. Let A and B be the centres of the spheres, a and b their
radii, c the distance between their centres.

B F
D
A
2
Let ₁ be the velocity potential when A is moving with
velocity u₁ along BA and B is at rest; ₂ the velocity potential
when B is moving with velocity u, along the same direction and
A is at rest. By § 162 the velocity potential of the whole motion
is ₁+₂, and the kinetic energy of the liquid is
29
do₁
1
چھ
d&2 dS2
T = − t p / fp, db, as, - pff 4, dh; as, - tp ft. Un
аф
1 dn
T22
= T₁₁+2T 12
2T₁₂+ T
11
1
2 dn
dn
.(1).
MOTION ALONG THE LINE OF CENTRES.
231
In order to find the value of T we shall employ the method of
images.
-
If B were absent, the velocity potential due to the motion of
A, would be the same as that of a positive doublet¹ at A of
strength qua³, whose axis coincides with BA. By § 53 the
image of this in B, is a negative doublet situated at the inverse
point F, where BF.BA = b², and whose strength is — u₁a³b³/2c³.
This latter doublet will have an image in A, and so on ad infi-
nitum. Hence the kinetic energy of the liquid due to the motion
of the sphere A, will be the same as that due to two infinite
systems of doublets, both of which lie respectively within each
sphere.
219. Let Pr
be the distance of the nth image in A from A, µ₂
its strength; and let σ be the distance of the nth image in B
from A, v, its strength. The part of T, due to μ will be
n
σ„ n
Ꮎ
11
"πpa³u̸, ["μ, (a cos ✪ +p₁) sin é cos e de
πρακ
0
n
2
(a² + p₂² + 2αp₂ cos ()
n
- πραυ μη
୮"
(În + ax) xdx
-1 (a² + p„² + 2αp„x)
p₂²+2αp¸x)**
But
(r+ax) x dx
d
1
x dx
-1
(a³ + r² + 2arx)
dr
-1 (a²+r²+2arx)³
d
1
dr 3a²r² {(a+r) (ar — a² — µ‚²)
G
±(a− r) (a² + r²+ar)}.
n
When r = p₁<a, the integral is equal to
d 2r 2
dr 3a²
3a2
……….(2).
But when rσ > a, it equals
n
d 2a
4a
dr 3r²
3p³
Therefore Ty= πρωτομ - που ενσ
11
Now
=
N n
- αν σ.
ανασ
.(3).
1 A doublet is considered positive when its source end is at the positive
extremity of its axis. If m be its strength, its velocity potential is -mr-2 cos 0.
232
MOTION OF TWO SPHERES.
1
Hence if M₁ be the mass of the liquid displaced by the
sphere A,
2
11
T₁₁ = M,u,² (1+3Σ, Ha
+
Mo
..(4).
This is the kinetic energy due to the surface integral of A's
motion over itself.
Again,
vn = — b³µn_1/(C — P₂-1); Pn=a²/o, co₁ = b²/(c — P₁₂)...(5),
whence
'n
-
37.3
a³
fl
n-1
3
n'
b³ pn μn-1
Pn
n
3
σn³ (C — P₂-1)³ a³ (c — P₂-1)³
3n
((c -
3
3
3
Mo......(6).
PmPn-1... P1
Pn-1) (C-Pn-2)... (c — |
(c - p₁) c) Mo
Eliminating σ from (5) we obtain
n
a²c=0.....
CPnPn-1 — (c² — b³) Pn - a²ρn-1+a²c = 0.………………………….(7).
11
220. The formulae of the preceding section enable us to
obtain an approximate value of T₁, as far as c-12 without much
difficulty, but in order to obtain the complete solution we must
solve (7). To do this, put p₁ = u₁+x, and choose a so as to make
the constant term vanish, and we obtain
2
cx² − (a² + c² — b³) x + a²c = 0
………….(8).
Let F, F, be the common inverse points of the two spheres, O
the middle point of FF,; also let FF₁ = 2λ, 0A =r₁, OB=r„, then
therefore
also
2
r² - λ² = AF,. AF = a²
2
r² — λ² = b²,
2
2
1
r₁² — r² = a² — b²;
-
r₁ + r₂ = c,
.(9).
therefore
1
r₁ = (a² + c² − b²)/2c
Let P be any point on the sphere A, and let the constant
ratio F₂P/FP be denoted by q₁, and let q, be the similar constant
for the sphere B. Then since the triangles PFA and FPA are
similar,
and (8) becomes
2
q₁ = F,A/a = (r₂+λ)/a = a/(r, − λ),
1
q₂ = b/(r₂+x) = (r, − λ)/b,
x² - 2r₁x + a² = 0,
MOTION ALONG THE LINE OF CENTRES.
1
2
n
233
19
the roots of which are î₁ =r₁+λ, x=r¸- λ. Putting p₂=u₂+x₁,
equation (7) may now be written
U₁₂-1 — (x — a³/c) u„ + (x, − a²/c) u = 0.
2
-¹
Now a² = x,,, whence, writing v, for u,, we obtain
X₂ (c − x¸)
V₁₂
n
1
x₁ (c-x₂
ย
n-1
C
In this equation
c-x
1
r2
- λ
2
C-x
ro + λ
2
x ₁ _ r₁ + λ
હ
1
C
r₁
- λ
C
=
x₁ (c − x₂) ¯¯ (r₁ + λ) (r½ + x) *
1
1
Whence putting q=q/q₁, we obtain
2
C
Vn+1 — q² vn
(r₁ + λ) (r₂ + λ)'
2
V
v₁ = Eq²" — —λ~¹,
the solution of which is
hence
n
P₂ = aq₁+ (Eq² — —λ˜¹)~¹.
But p=0 when n= 0, therefore
=
1
1
η-λ
E =
1
2λ
1
r₁ + λ¯2λ (r₁ + x)
therefore
Pn
1
-2
Also
1
2λ (r₁ +λ) ¯¯ 2λq,²
2λ
2n
P₁ = aq₁ — 1 — q™” q₁
1 - q²
= (r₁ — λ) 1 — q² q,
2n
21
C − Pn = r₁ + r₂ − r₁ − λ + 2x/(1 − q²n q˜²)
2
= (³₂+λ) 1
2
1 − q²n+2
2n
29
__ q ( 1 − q²¯¹³ q, ¹³) __ ¶Pμ-1 (say);
q²n−² qi
therefore
bpn
a (c-P₂)
therefore
n
μn
2n
8
Pn-1 Pn-2 · · · Po) 3
PnPn-1······P₁
3
((1 − 9, *) q″) ³
1
2n
Ho
21-1
Pa
n
Мо
A
234
MOTION OF TWO SPHERES.
i
If therefore we put
we obtain
Q (4, 3, 2) = (1 − q ) ³ Σ (1
2
qr
2n -2
..(10),
……….(11).
T₁ = {M‚µ‚² {1+3Q (q,², q)}
11
Similarly if the sphere B were moving with velocity u, along
BA whilst A is fixed, it can be shown that
2
2
T₂ = {M¸u¸² {1+3Q (92, 9)}
22
.(12).
12
221. We must now calculate the quantity T₁₂ which is the
surface integral of B's motion taken over A, and which by Green's
theorem is equal to the surface integral of A's motion taken over
B. We thus obtain,
T₁₂
μn
12
18
1
афг
2 dn
0
= − bp √ √ 4, dh¹ ds₁ = — πpa³u, "Þ, sin & cos de.
σ
- πρακ
Let p denote the distance from A of the nth image of B
in A, its strength; also let o, denote the distance of the nth
image in B from A, v, its strength; then remembering that the
original doublet is in B, we obtain
µn' = − (a/o'n-1)³ v'n-v µ‚' = — (a/c)³ v.' — — §a³b³c¯˜³u₂...(13).
Hence
·
OC V
n
=
T= - πραμ Στο ( - ) + πρι, Σαμα
12
1
σ
n
n
Also
2πρυ, Σιμε
a³bs
при иг C³
n
(14).
P₁' = a²/c, σ = b²/(c - P₁), P₂ = a³ /(c — σ,')...... (15),
2
whence, proceeding as before, it will be found that
Mn
bpn
; (c — p',
-
3n-3
3
n-1
n-
Pr' P'n-1... P₂
((C — P'n-1)………….. (C — f
and it can be shown as before that
}
8
µ½''…………………….(16),
P₁ = aq₁+ (Eq²" — —λ˜¹)~¹.
MOTION ALONG THE LINE OF CENTRES.
Whence, determining E by the condition that
shall find
P₂' = aq, — 2λ (1 − q²)˜¹ = (r, − λ)
Therefore
1 − q² q₁
1 – q²n
2n
2
c — p'n-1 = (r₂ + λ)
1 - qq,
2n
2
bon
2
2n−2
1 - q²n-2 ›
(1-92-2)
a (c = p.) = 2 (1 - 2²)
P´n-i
n
— q²) q″-¹'
2n
q
8
2n
μέ.
Pi
=
235
a³/c, we
n
1 - q²n
Now from (9) we obtain
q²
2
(r, — λ) (r, — λ)
-
1 − q² =
therefore
If therefore we put
we obtain
(~½ + λ) (r₁ + λ)'
2
2λc
1
(r₂ + λ) (r₁ + λ)
Q₁ (9) - " ( 229 )
=
βι Στ
T12
2λcq
ab
2λα"
3
…………….(17),
.(18).
Hence, if m₁, m, be the masses of the two spheres, the kinetic
energy of the whole motion when the spheres are moving along
the line joining their centres is
2
T= † (Au¸² — 2Buu₂+ Cu¸³)
(19),
where
A = m₁ + 3M, {1 + 3Q (q,˜¯¹², q)})
-1
.(20).
1
1
C = m₂ + ¿M¸ {1 + 3Q (₂, q)}
2
B = 2πρu¸µ₂Q₁ (q)
1
The three coefficients A, B, C can be shown to diminish as the
distance between the spheres increases; for when c and therefore
λ is large,
q₁ = (r₁ +λ)/a = 2λ/a,
q2=b/(r₂+x) = b/2λ,
q = ab/4x²,
ultimately, and therefore A, B, and C diminish as c increases.
Also, since T is essentially a positive quantity, AC > B².
236
MOTION OF TWO SPHERES.
222. The general formulae (20) are too complicated to be of
much use, we shall therefore obtain approximate values of A, B
and C as far as c-12.
From (5) and (6) we obtain
whence
P₁ = a²c/(c² — b²),
3
M₁ = (bpt) = (abbr)
ac
b² P2P 1
2
(21);
b²
•
Мо
also from (6)
Мо
μ₂ (a²c (c-p₁))
From (7)
P2
therefore
C-P₁
P₂ (cp₁ − c² + b²) = — a² (c — p₁),
2
a²
a² (c² — b²)
2.29
@_b³ - cp, — (ob¹) — a'd'
1
a²b²
3
whence
Мог
Мо
(c² — b²)³ — a³
-
(22).
The last expression varies as c12, whence expanding the values
of μ₁/μ。, μ₂/μ。 in powers of c¹, and neglecting higher powers than
c-12, we obtain
3a³b³
A = m₁ + }M, {1 +
св
(1
362
6b4
1+
+
C
c²
+10
1166
Cε
and the value of C can be obtained by interchanging a and b.
To determine B to the same order, we obtain from (16)
με
bp₂
a (c - P₁))
3
a³b³
(c² — a² — b²)³
3
whence
B=
2πρα
C³
{1
+ +
св
၉၆
a³b³¸ 3a³b³ (a² +b²)
(23),
Collecting our results, the values of A, B and C as far as
6
(
3b26b4
66 1166,
c-12 are
3a³b³
A
= m₁ + {M₁ {1 +
Cε
1 + +
c²
+
C¹
11609)}
3a³b³
2
C = m₁₂+ &M, 1 +
2
2πρα
C8
B = 2πpa²b³ {1 + a²b² + 3a²b³ (a* + ¿9)}
6
C
св
(1
3a² 6a 11a
1 +
c²
+ +
C
св
(24).
:
MOTION PERPENDICULAR TO THE LINE OF CENTRES. 237
Motion perpendicular to the Line of Centres.
223. When the spheres are moving perpendicularly to the line
joining their centres, the kinetic energy may be determined by
the method of images without much difficulty, provided we
neglect powers of c¹ higher than the eighth; but if it is desired to
carry the approximation to a higher degree, the successive images
become exceedingly complicated, and it is better to employ a
different method, which will be explained later on.
19 2
α
3
224. Let v₁, v, be the velocities of A and B perpendicular
to AB. If B were absent the velocity potential due to A's motion
would be the same as that due to a positive doublet at A, of strength
va³, whose axis is perpendicular to AB. By § 54 the image
of this in B, is a positive doublet of strength va³b³½³ situated
at the inverse point F, together with a negative line doublet
extending from F to B, whose strength at any point P is
— va³BP/bc per unit of length. Hence the successive images
consist of a series of single doublets and line doublets, and
evidently become exceedingly complicated.
1
Let Χ be the angle which any plane through AB makes with
the direction of motion of the spheres, r the distance of any
doublet element from A, μ its strength. The kinetic energy will
be given by an expression of the same form as (1), whence
the part of T₁ depending on μ will be
π
11
3
2π υγρα μ sin® θ cos" χαθαχ
2
(r² + a² + 2ar cos 0)
3
Π
sin Ꮎ ᏧᎾ
=
- πραυμ
0
(µª² + a² + 2ar cos 0)
The value of this integral is
p² + a²
3
a³µ³
1
-
{r + a + (r− a)} − 3 a³µ³ {(r + a)³ ‡ (r − a)³},
3
in which the upper or lower sign is to be taken according as
r> or <a. Hence the value of the integral is
a³, a >r; and r³, r>a
11
and therefore the part of T, depending on μ is πρμν, oг
πρμν¸³/r³, according as r < or > a.
>
1
238
MOTION OF TWO SPHERES.
Letv and σ be the strengths and distances from A, of the
doublets within B due to A's motion, and μ the strengths of the
doublets within A. Then
3 8
Τα = πρυ Σ (μ + να σ3)
11
μ
= } M,¸v, Σ (~+~-~²).
}M₁ ', 3
Now every v produces in A an image consisting of a doublet
of strength va³/σ³ at a distance a² from the centre of A, together
with a negative line doublet extending from the doublet image to
the centre of A, and whose line strength at a point whose distance
from A is x, is vx/ao. Hence the whole amount of the image
is
Also every μ
particular v, hence
(~)* .
ע
·()' - Z (@)' - + (@)"
ע
2ασ σ
except μ forms part of an image of some
М
Σ
=
2Σ 3
2μ
α
as
Therefore
Ꭲ,
11
(μ +3μ)
11 2a³
2
= +M,,* {1+ 3, ()}
Mv
11
.(25).
225. In order to find the term involving v,v,, we must find
the portion of the kinetic energy due to B's motion over A and
double the result.
Since the original doublet is in B, every v except v forms part
of an image of some μ, whence if the accented letters refer to
the images of B's motion
hence
Σ
ν
3
2x
μ'
ν' α
།
3
T₁ = 3πpv¸ Σ (v'a² + µ')
T.. πρυ,
12
=
=
- 2πρυ, Στα μ
προ,υ, Στ
μ'
.(26),
CALCULATION OF THE IMAGES.
239
and we therefore obtain
where
T= ¿ (A'v¸² + 2B′v¸v₂+('v¸³),
n
A'=m, + ¿M, {(1 + 32" (~~)})
1
1
C' = m₁₂ + + M₁₂ { 1 + 32, (+)
M¸ +3Σ
2
Β' = 2πρι Σ. (μ.)
n
.(27).
226. We shall now calculate the values of the coefficients
when all the images are omitted except μ,,,, μ.
The image of A in B consists of a doublet of strength μb/c at
F, together with a negative line doublet from F to B of strength
- μox/bc per unit of length; also BF = b²/c, AF = (c² — b³)/c.
The image in A of the doublet at F is a doublet at a point F'
whose strength is
Mas μα ε
c³АƑ¹³ — (c² — b²)³
3
.(28),
where AF" = a²c/(c² — b²), together with a negative line doublet
from F" to A whose whole amount is
μ b³ [AF ydy
Ο
с
0
аAF
2 (c² — b²js
(29).
AP.AQ
In order to find the whole amount of the image of the line
doublet between B and F, let P be any point in BF, Q a point on
AF" such that AP. AQ= a; also let BP = x, AQ= y; then
y (cx) = a². The doublet element - μxdx/bc at P, produces a
doublet element - μxa³dx/bc (c-x) at Q, together with a line
doublet from Q to A whose whole amount is
3
a2
μ xdx
bc
c-x
ydy
3
μа³xdx
0
a (c − x)¯¯¯ 2bc (c — x)³ *
Therefore the whole amount of the line doublet is
2bc
adding (28), (29) and (30) we obtain
Hoar 190 (C-20)³
xdx
(c
μα
3
4c² (c² — b²)²
·(30),
Again
μί
Но
0
ab
3
=
— b²
νο
ac 2c³
0
3
2c² (c² — b²)²)
3
[а² xdx_vα³
xdx _ va³
a³b³
ठ
(31).
.(32),
240
MOTION OF TWO SPHERES.
whence substituting from (31) and (32) in (27), we obtain
ab 3
3a³b³
4c² (c² — b²)²)
A' = m, + ¿M, {1 + 4 (¿?—— b³)” — 40° (0 = 6')}
b2
2
C' =
m₂ + } M₂ { 1 + $ ( 6, 0 a³)®
c² —
2
3a³b³
4c² (c² — a²)²
१
...(33).
B'
πρα
C³
-12
The second ratio μ/μ is of the order c¹², and the next term in
Hence (33) gives the correct values of A'
and the expression for the kinetic energy
B' is of the order c.
and C as far as c-10,
derived from (33) is correct as far as c.
227. We shall now explain a different method for obtaining
approximate values of the coefficients¹. The approximation is
carried as far as c12, but it could without much additional labour
be carried to a higher order if desired.
It will first be necessary to establish the following proposition.
え
​P

B
Θ'
7
Ө
A
M
m
In the figure, let PM, AM=z, BM=2', AB = c, cos 0=µ,
cos' =µ'; also let Pm (u) be an associated function of degree n
and order m, whose origin is A, and axis is AM; and let P' (u')
denote a similar function having the same axis and whose origin is
B. Then we shall prove that, when r<c,
(n+m)! (n + m + 1)!
Pm-
2m!
m
(2m+1)! c
P'm
pm
n
p'n+1 − (n − m)! cn+m+1
−)³ (n + m + s)! (r`
+
pm.+
and when r' < c,
P",
Pm+...
m+1
(2m +8)! (6) P... + .....
m+8
n
(34),
(−)n-mpm
p'm
¯(n+m)!
pn+1
(n − m)! cn+m+1
Pim +
(n+m+1)! r'
m
+
2m!
(n+'m + s)! (r' \ ³
P'm
(2m+1)! c m+1
(2m + 2) 1 (5) Pa...
........
(2m+s)! C
m+8
1 Proc. Lond. Math. Soc. vol. xvIII. p. 371.
.(35).
•
TRANSFERENCE THEOREM.
.241
It is known that P can be expressed in either of the forms¹
2
n--m
M (1 − µ³)¹m (" {µ + √ µ³ − 1 cos ¿}*¯* sin³™ µdė,
-
or
M (1 − µ²)àm
[]
where
M
Therefore
p'm (u')
n
p'ntl
=
=
Ma"
Μπ
m
-
sin2m do
2
фаф
。 {μ + √µ³ − 1 cos p}n+m+1
(n+m)!
(nm)! 1.3... (2m - 1) π
S
Π
π
=)
sin2m do
(2′ + iw cos $)n+m+1
sin2m do
- M=" [". =
2
2
[c + r {µ + √μ² - 1 cos p}]n+m+1
g
whence, if λ = µ+ √μ² − 1 cos ò̟, and r < c,
["{1 - (n+m+1)
+
(n+m+1)(n+m+2)
n
Pim (μ')
p'n+1
Ma" ["
π
ηλ
c²+m+1
с
(−)³ (n + m + 1)...(n + m + s)
+
whence, by the first form of Pm, we obtain
n
2!
(~~) -
(? +
C
sin2modo;
(n + m + 1)! ~ pm + ...
(2m + 1)! c
m+1
P'm
доть
n
(n+m)!
r'ntl
+1
Pm
(n − m)! cn+m+1
2m!
m
(−)' (n+m+s)!
..+
(2m + s)!
(
mts
Pm+.
In order to obtain the second equation, let us change 0 and
into their supplements; then, since
P {cos (π-0)} = (-)"-" Pm (cos 0),
n
we obtain
(−)n-m pm (µ)
pr+1
-1
p'm
(n − m)! cn+m+1
(n+m)!
2m!
P'm +
m
(n + m + 1)! r
(2m + 1)!
P'm
c
m+1
(n + m + s)! (r'
+
P'm.. +
(2m+s)!
m+8
+...]
The corresponding formulae when r>c or r' > c could be easily
obtained, but they are not required for the present investigation.
1 These formulae will be proved in the second volume. See also Heine, Kuge-
functionen, ch. Iv.: Mess. Math., vol. xш., p. 147.
B.
16
242
MOTION OF TWO SPHERES.
1
2
228. Let ø, be the velocity potential of the liquid when A is
moving with velocity v,, whilst B is kept at rest, and let o, be the
velocity potential when B is in motion and A is fixed. Then if p
be the velocity potential of the whole motion,
$ = $₁ + $₂......
1
(36).
The problem is therefore reduced to the determination of $₁,
for when this is known, o, can be written down by symmetry.
2
1
Let Χ be the angle which a plane through AB and any point P
makes with the plane through AB which contains the directions
of motion of A and B; also let Q, Q' be written for P1 and P'¹.
Then, in the neighbourhood of A, 4, must be expressible in the
form of the series
1
n
22
v₁ a³ Q₁ + A₁r +
+₁₂(~+
a³
2p²
2a5
Q₁ + A₂ (r² +
2
33
+
cos x (37),
for this value of 4, satisfies the surface condition
1
dr
V₁ sin e cos X.
α
In the neighbourhood of B, 4, must be expressible in the
form
В
12
COS
4₁ = { B₁ (~' + 2 ~ ) Q;' + B₂ (~'² + 2,5%) 2 + ... cos. x....(38),
Φι
B₁(r'
12
for the value of 4, satisfies the surface condition
1
do₁
1 = 0.
dr' b
/-1
The series consisting of powers of r¹ and r'→¹ are convergent at
all points outside the two spheres, but the series consisting of
powers of r and r' will be divergent if r and r' be sufficiently
great; but we shall only require these latter series in the neigh-
bourhood of the two spheres where they are convergent.
The kinetic energy consists of a series of terms of the form
2π
0də
1
[* dx ["Q,v,a" cos² x sin² Ode = πa³v¸ ['_¸ (1 − µ³)
1
-1
1
= 2πα²º₁) -1°
αμ
n
d.P
άμ
αμ
Απαν.
(n = 1)………………………..(39),
3
=0 (n any other value).
APPROXIMATE VALUE OF THE VELOCITY POTENTIAL.
243
Hence the terms involving Q₂, Q₂, &c. contribute nothing to the
energy, and we may therefore, in writing down the final value of
Þ₁, reject all terms except those involving Q, or Q,.
229. Dropping the factor cosx for the present, we should
have, if B were absent,
Φι
v₁a³Q₁
2p2
Putting m = 1, n = 1 in (35), the value of this near B is
v₁a³
200 (2
r'Q₁ +
+
C
r¹³Qs', r'¹Q₂
C¹
+
+...).
From (38) it follows that, in order to make the velocity at B
vanish, we must add the series
v‚a³ (b³Q‚'‚ 2b³Q₂, 3b’Q½'
1
1 +
+
+
2c³ 2r'2 3cr'3 4c2'4 50% + ...).
Transforming this latter series back again to A by (34), and
retaining the important terms only, the value of o, near A becomes
8
1
b2 96 46
Φι
v₁a³Q₁
2r.2
_ _ va² Q₁ v₁a³b³ (1
C6 4
+ √² + 4+c ²
+ Qr + Q₂n²
C6
4c7
In order to satisfy the surface condition at A, add the terms
v₁a³b³ (1
C6
b² 9b4 46° Q₁a³, v,α³b³Q₂
+ + +
c² 4c
св
+
C6 2r² 6c'
Neglecting powers of c¹ higher than the twelfth, the value of
these added terms near B is
Adding the terms
v₁ab³ /1
2c⁹
145
+
b2
Q'r' - Y₁₂ af³
20¹ Q'r'.
1
11
v₁ab³ (1
409
(
b2
+ +
c²
(40),
omitting Q,, &c., and restoring cos x, the value of the velocity
potential near B becomes
v₁a³
Φι
1
2c³
{1+
a³b³
+
4c6
a³b³ (a² + b²)`
C8
b³
+
2r'²
1
Q'cos x...(41).
The first term of (40) on transformation becomes
— v₁ab®Q₁r/16c¹²,
16-2
244
MOTION OF TWO SPHERES.
whence the value of o, near A is
1
c²
9b
+
4c¹
Φι
v₁a³Q₁
2r2
1
cos X
v₁a³b³ (1 b2
C6 4 + 02
3
X
272) 2,
2
+
(a³ + 64b³) b³)
16c6
Q₁ cos X.....
…………..(42).
The values of ò̟, at A and B can be written down by symmetry;
whence, if T be the kinetic energy of the system
2
2
2T= A´v¸² + 2B′v₁v½ +C'v¸²,
where
Α'
Φι
dd, as,
-
A² = m₁ - pff p₁ dan
m.
1
= m₁ + ½ M₁ 1 +
1
3a³b³ (1 b2 9bª, b³ (a³ + 64b³)
C6 4 C² 4c⭑
+ + +
16c6
C' = m₂+ M₂ 1+
3a³b³ (1
a² 9aª
3
+
c²
+ +
4c
a³ (b³+64a³)
16c6
...(43),
½
2
B' = - p√ √ p₁ dhe ds,
ρ
Φι dn
C6
πρα
a³b³
C³
1+ +
4c6
a³b³ (a² + b²)
C8
2
where m₁, m₂ are the masses of the spheres A and B; M,, M,
those of the liquid displaced by them, and p is the density of the
liquid.
The values of A', B', C′ have been calculated by Mr Herman
as far as c-15
230. We shall now apply the preceding results to obtain the
solution of some problems.
If a sphere is projected in a liquid which is bounded by a fixed
plane, we must put a = b, u₁ = U₂ = u, v₁ = v₂ =v; then
Из
V 1
2T = (A + B) u² + (A′ + B′) v²,
and, if higher powers than c be neglected, we obtain from (24)
A+B=m+‡M(1 +
and (43)
3a³ Зав
+
C³
၉၆
(44),
A' + B′ = m + ¿M ( 1 +
3a
3a
+
3
2c³
4c6
י!
equation
MOTION OF A SPHERE PARALLEL TO A PLANE. 245
where fc is the distance of the sphere from the plane. Lagrange's
gives
d dT dT
2
0
dt du
dc
d
d
(A + B) ù = v²
dc
(A' + B′) − w² / (A + B).
u²·
dc
(A + B) u = V2 sin' 0
{sin
d
Also, since the momentum parallel to the plane is constant
(A' + B′) v = const. =
G.
Let V be the resultant velocity of the sphere,
its direction makes with the normal to the plane, then
d
the angle which
2
dc
(A' + B') - cos² 0 (A + B)
dc
+ B)}
9MV2a3
2a³
2c¹
Ma ((1+24) cos' 0 -1 (1 + sine.
3
2)
If, therefore,
2 (c³ + 2a³)
tan a =
;
c³ + a³.
3
it follows that, whenever the direction of motion makes with the
<«
normal to the plane an angle which is <x or > π-a, the sphere
will be repelled from the plane; but, whenever this angle lies be-
tween a and π-a, the sphere will be attracted. Also, since A' + B
increases as c diminishes, the velocity parallel to the plane will be
accelerated when the direction of motion lies between a and π
and retarded when this direction makes with the normal an angle
<aor>π-a. If, therefore, the sphere be projected parallel to
the plane, it will ultimately strike it.
M
a;
We have shown in § 208 that in the case of a cylinder a = 1π,
hence in the case of a sphere a>. The discussion of the sub-
sequent motion of a sphere projected in any given direction in a
liquid bounded by a fixed plane, can be carried on in the same
manner as in the corresponding case of a cylinder, but it must be
recollected that the preceding values of the coefficients may not
give correct results if the sphere gets too close to the plane.
231. Let X, Y be the forces upon the sphere, arising from
the pressure of the liquid, then
X = m = m
mù
Y = mi
d
u² dc (A + B) } . /(A + B),
{v² de
(A′ + B') — u²
dc
d
dc
2muv (A' + B') . /(A' +B').
246
MOTION OF TWO SPHERES.
From (44) we obtain
d
dc
(A + B) 9 Ma² (1+24²),
2c
de (4' + B) = - 9 Ma² (1+2)
dc
4c¹
whence neglecting higher powers than cwe obtain
X=
9 Mma³
(2m + M) c²
u² - 1 v² +
a³ [u² (4m — M) + v² (M — m)]
(2m + M) c³
-m)]},
9 Mmuva³
Y=
(4m — M) a³
1+
(2m + M) c*
2 (2m + M) c³
}}
232. Let us now suppose that the sphere A is a pendulum
performing small oscillations along AB about its mean position,
whilst the sphere B is free to move.
Let A be the mean position of A, B the initial position of B:
A', B' their displaced positions, and let AA'=x, BB' = y, AB = c;
A'B'p. Then p=c+x-y and if u is the force required to
maintain the oscillation of A, the equations of motion are
dA'
A'ï – B'ÿ – (œÿ — † ¿³)
dB'
dC
+ y²
dp
dp
Zy² + мох = 0,
dp
dA'
dB'
dC
C'ÿ – B'ï + ¹ x²
x2
dp
dp
xy)
0,
dp
where the accents denote the values of A, B, C at time t.
To obtain a first approximation, neglect squares and products
of small quantities, and we find
(AC − В²) ï + µCx = 0,
-
Cij - Bx = 0.
If therefore the sphere A is initially displaced to a distance
x and then let go, the integrals are
x=x。 cos kt,
where k² =μC/(AC – B²).
У
=
Bx
(cos kt - 1),
C
Since y is negative and increases numerically so long as a lies
between a and -x, it follows that to a first approximation B
is repelled from A so long as A is moving away from its initial
position A', and attracted when A is returning to A'.
SMALL OSCILLATIONS.
247
233. In order to obtain a second approximation, we must
take into account the squares of small quantities. Let
Y
=
Bx
(cos kt -1)+z,
where z is at least of the order 2. Then
A′ = A + (x − y)
Therefore B's equation of motion becomes
{0 + (x − y)
B
p = c -
(3 − 1).
Bx 0
X
cos kt +
Bar +=
%,
0
C
dA
&c.
dc
y) da
3)
Bx k²
cos kt
dc
C
{B
dB
+ { B + (x − y)
xk² cos kt — x²k²
dc
272² (dc
dB
dA
sin² kt
dc
B2
x²²
ne
B\ dC
sin² kt = 0.
202 C dc
Neglecting cubes of small quantities, this equation may be
written
Cz=f+ L cos kt + M cos 2kt,
2
where
f=
0
k²x² (dA 2B dB B² dC
+
4 dc c dc c² dc
2
A
-x, d (4-5)
k²x
4 dc
If we only take into account the first terms in A and B, which
is equivalent to neglecting the twelfth and higher powers of c¹,
we obtain from (4) and (21)
3M
1
A = m₁ + {M₁ +
M₁+
IM1
1 2μ。
= m₁ + ½ M₁ +
2πρα
(c* — b²)³ ›
dA
therefore
dc
|
B2
12πра b³c
(c² - b²) ›
πρα
C (2o+p) c
4
3
61
d B2
therefore
56πρ ab
dc C
(20+ p) c²
248
MOTION OF TWO SPHERES.
where σ is the density of the sphere B; whence
ƒ= 3πра b³k²x²
2
C
3p
(c² – b²)³ (2o+p)
The term f indicates that the sphere B, in addition to its
vibratory motion, will be attracted towards or repelled from the
sphere A, according as ƒ is positive or negative. Hence there
will be repulsion when
i.e. when
3p
>
20+ P
CS
(c² — b²)4 ›
b
c>
{1 − 4/} (2o/p + 1)}
which can only happen when σ <p or the density of the sphere B
is less than that of the liquid.
If therefore the sphere B is denser than the liquid it will in
general be attracted, but if the density of the sphere is less than
that of the liquid there will be a critical point, beyond which
there will be repulsion, and within which there will be attraction,
this critical distance is given by
C =
b
{1 − √/ } (2o/p + 1)}* '
-1
Since this result has been obtained on this assumption that c
is so large compared with a and b, that powers of c above the
twelfth may be neglected, it fails to give a correct result if with a
given density, c comes out nearly equal to b. If σ/p=9 then
c = 7·648b.
This theorem is due to Sir W. Thomson; the preceding
demonstration is due to Mr Hicks.
On the Pulsations of Two Spheres.
234. The term pulsation is applied to denote a periodic
change of volume; and the problem which we shall now investi-
gate is the following:-Let there be two spheres in a liquid, whose
centres are fixed, and which are composed of some elastic material
such as india rubber; let each sphere be compressed or expanded
into a concentric sphere and then let go; it is required to deter-
mine the motion.
PULSATIONS OF TWO SPHERES.
249
If the spheres were composed of some highly elastic material,
the inequality of the pressure of the liquid upon their surfaces
would produce a deformation which would cause their surfaces to
cease to be spherical; we shall therefore suppose the rigidity of
the spheres to be sufficiently great to render such deformations
inappreciable.
1
235. If ₁ be the velocity potential of the liquid when the
sphere A pulsates, and B does not; and ₂ be the similar quantity
when A and B are interchanged,
$ = &₁ + $₂•
2
1
Let a and b be the radii of the spheres A and B, c the distance
between their centres. If B were absent the value of ☀, would be
-a'a/r, for this value of satisfies the boundary condition
do/dr=a. This is the velocity potential due to a source of
strength a³å situated at the centre of A, and by § 52 the image of
this in B will be a source of strength a'ba/c at the inverse point P,
together with a line sink extending from the inverse point to
the centre of B, of strength a³å/b per unit of length. Putting
m= a²bȧ/c, ƒ = b²/c, the strength of the source at P is m, and that
of the line sink from B to P is — m/f per unit of length; and by
§ 55 the image of these in A is an arrangement of the same kind.
Hence, and ₂ will be the velocity potentials of two infinite
systems of sources and line sinks, which respectively lie within
each of the spheres.
2
2
2
236. Taking the density of the liquid as unity, let F₂ be the
resultant of the pressure of the liquid on B towards A, then
F₂ =- Sfp cos 0dS
2
4
πb³¨ ($ + V²) sin 20d0,
where V is the velocity of the liquid at the surface of B; let
Then
P = ["
Q
=
0
o sin 20 de,
["p² sin 20d9.
["¿ sin 20 = P.
250
MOTION OF TWO SPHERES.
In order to find the portion of F, which depends upon V², let
v=b³b, then V²= v²/bª + (dp/bd0)²; and since v is constant over
the surface of B, the portion of F, depending upon it is zero,
whence, denoting the portion of the pressure depending upon V²
by I, we have
["
Π
2
Π
2
I = 1πb² [" V² sin 20d0 = ±π [" (de)* sin 20 de
sin 20+
df2
do²
de
cos 20 do
) do
π
S
0
π
φ
By Laplace's equation
d²
do
+ 24) sin 20 d0 — §π [¿¹]".
-
d2f
do²
26 dp
d²
dr
+b² + cot
dr²
do
ᏧᎾ
in which r is to be put equal to b after the differentiations have
been performed. Hence dø/dr = v/b², so that
I = T
1π
But
S
2
dr²
π
$
{(20
d²
+ b²
వాళ్ళ - 26
24) sin 20 + 2 cos² 0
аф
d0 – 1π [8²]".
do
√
cos' 0
do do
do
ao = [
and
S
$
27 ф
dr²
whence
and
when r =
r = b.
sin 20d0 = ("(+
2v
0
¤² sin 20d0+ [¤²³]″ = Q + [6³]",
0
d2f2
sin 20d0=
dr² bª
d² Q
dr²
(45)
dQ
I = { π ( 2 0 P - Q + b²
V
dr²
>
dQ
(b² P + P − b Q + b² ¹² 2 )
- 1
F₁ = π -
1
dr²
(46)
237. Let P₁ be the part of P due to ₁, then if μn be the
strength of any image whose distance from B is r, the portion of
P, due to this is
1
T
sin Ꮎ cos ᎾdᎾ
(b² + r² — 2br cos 0)*
2「"
0
which is equal to
4μ b
n
3p2,
r>b; and
4μr
忆
​362
>
r < b.
:
CALCULATION OF THE PRESSURE.
251
n
Hence if μ be the strength of the nth source image in A from
A, and p', that of the other extremity of the line sink image; the
part of P, due to μ is
1
4μ b
Pr
X
n
3 (c — P₂)²
十​季
​2
μ₁bdx
Let vn
4μb (P₁-P'₂)
3 (c — Pn)² (C — P´n)
(În − P'n) (c — x)²
nɔ
.(47).
n
denote the strength of the nth image in B, σ, o' the
distances of its extremities from B; then the part of P, due to
4v
X
n n
n
Now
V n
therefore
Όμ
σε vxdx
3b2 Jon (On-o'n) b²
2vn (on — oʻn)
362
c - Pn
>
X. - -
σ
n
C
b2
Pn
b2
>
n
c-P₂'
n
2 µ₂b (În — p'n)
3 (c — Pn)² (c — P´n)
ע
(48).
…………….(49).
is
n
Adding (47) and (49) and summing for all integral values of n
from ∞ to 0, we obtain
P₁ = − 220
1
μnb (Pn - P'u)
(c — Pn)² (C — P'n)
…………….(50).
2'
238. In order to find the portion P, of P due to o, we must
remember that the original source is now in B. Let σn, o'n denote
on σ
the distances of the extremities of the nth image in B from B, due
to ₂, then expressing P₂, Pn p'n in terms of vn, on, o'n we shall
obtain
22
P₁ = - 22" ₂ (0₂ = 0.).
ση
2
n
b2
σ
n'
σ
n
Hence P=-2
n
b
μ₂ P´n)
(P.-P) - 2. (0,0)...(51)
n
n
where μ P P' refer to the images of A's motion, and v, σ, σ',
n
μηλ
to those of B's motion.
252
MOTION OF TWO SPHERES.
By direct calculation we easily find
Po = 0,
P₂ =
a²c
c² = b²
>
a²c (c² — a² — b²)
P₂
(c² — b²)² — a²c² ›
po = ∞
a²
also if m, be the mass of the liquid
C
a² (c² — a²)
P'₂ = c(c² — a² — b²)
2
displaced by A,
..(52),
μ₂ = a²à
Но
mr
1
4π'
abm,
4π (c² — b²) '
μ₂ =
a²b²m,
1
4π {(c² — b³)³ — a³c*} ···(53).
The v's and o's can be obtained by symmetrically interchanging
a and b and putting m₂ for m. If we write M, N, for the two
series in the right-hand side of (51), we shall find that
m.b
M
1+
2 Απο
+
a³b³
(c² — a²) (c² — a³ — b²)²
aεbε
{(c² — a²)² — b³c²} {(c² — b³)² — 2a²c² + a² (a² +b²)}ª
and
ma³b
1
N.
2 Απο
(c³ — a²)²
+
+.
·] …..
..(54),
a³f³
— +
| …….(55),
P = − 2 (M₂ + N₂)…………….
........
(c² — a² — b³) {(c² — a²)² - b²c²}²
.]...
.(56).
-2
From the above formulae it appears that M, is of the order c¯²,
and N₂ of the order c´.
2
239. The value of the portion of F, which depends on the
square of the velocity is more difficult to obtain, and we shall
content ourselves with obtaining an approximate value as far as
the term c-5.
Let us now put u=a³à, v=b³b, and let P denote zonal har-
monics when the origin is at A and axis BA, and P' similar
quantities when the origin is at B.
Near the surface of B
n
and
$2
2
1
•₁ = Σ, A, {R" +
Ꭱ
n
บ
R
+ Σ Β
n
b2n+1
(n + 1) R+1 P' + const.,
b2n+1
R" +
(n+1) R+1
n
P'" + const.
APPROXIMATE VALUE OF THE VELOCITY POTENTIAL. 253
Dropping the accents for the present and writing C for the
coefficient of P, in the value of o, we obtain
2
2
Q = 2 ['_ (20_P₂)° µdp.
(ΣΟ
Since P3 is unchanged when —μ is written for
μ,
Hence
Ľ
P²₂ µdµ = 0.
-1
1
CmC,
n
Q = 4ΣC_C_ [ P¸„P₂ µdµ,
μαμ,
n
the summation extending to all positive integral values of m, n
except m = n.
Let
Þ =
m n
P P. άμ.
μ
Then
Φαμ
[ P_P₁ µdµ = [] P_P, dµ + [ & dµ
-1
n
μαμ
-1
-1
= 1,0
Φαμ.
-1
Now (Ferrers' Spherical Harmonics, § 24),
1
n (n + 1)
Φ
(m −n) (m + n + 1)
2n + 1
Pm (Pnti-Pn-1)
m (m + 1)
P₂ (Pm+1 − P
1
2m + 1
- P-1)}.
Hence [Þdu vanishes unless m=n+1 or n−1 and its values
-1
in the two cases are
2 (m + 1)
(2m+1) (2m+3)
2m
and
-
(2m − 1) (2m + 1) *
240.
n
P'
R₂+1
Putting m = 0, r' = R in (34) and (35) we obtain
C
P。
n+1
(-)" P.
P'
n
pr+1
cati +
(n + 1) P¸r (n + 1) (n + 2) P¸r²
C²+2
+
2! c
n+3
(n + 1) P‚'R, (n + 1) (n + 2) PR²
cn+2
+
n+3
2! "+s
C
(57).
+
Now if B were absent, the value of 4, would be
1
И
Φι
is
254
MOTION OF TWO SPHERES.
The value of this near B is
И
Φι
C
(P:
P'R PR
P. +
+
C
c²
(x + ...).
In order to make the velocity at the surface of B vanish, we
must add the series
ub³ / P 2Pb² 3P,'b¹
+
+
C 2R2 3cR³ 40" R$ + ...).
Transforming each term of the last series by means of (57), the
value of ₁ near A becomes
1
Φι
ub³
Pr
P
2c⭑ 0
+
с
Adding the proper series, the value of ò̟, near A becomes
U
ub³ ub³
Φι
r
2c*
20€ + 208 (
(~ + 2) P₁ +
.(58).
The added term produces at B a constant term of the order
c`, which contributes nothing to the pressure, hence the value of
₁ near B is
Фо
U
И
b3
Φι
R+
C
2R²) P
P-
И
3
(R² +
265
3R3
P₂......(59).
-
Changing P, into P, it follows from (58) that the value of
₂ near B is
V
va³ va
b³
ΦΩ
R+
2
R
2c* 2c5
2R2
:) Pi
(60),
1
whence the value of ☀ near B is
V
vas
И
φ
R
C
2c¹
4
U va
+
2c5
b³
U
265
R+
P
2R2
1
C³
R² +
3R3
P,
2
Фо
Putting in this R = b, we obtain
I
ข
น
va³
3b /u
va³
5ub2
+
P'
P'-
b
C
2c
2
2c5
3c³
d² f
also
Therefore
dR2
2v 3/u vas P-10 P-
b³
25"
b c²
Q=2 φ μαμ
4
-1
uv
(
+
2c5
+
u²b v² as
+
v²a
8u2b2
+
C³
2c5 3c5
- &c. ...
\
FINAL VALUE OF THE PRESSURE.
Also by (45)
255
dQ
4
dR2
8
-1
d'o
φ
dR³ μdµ
2uv u² v²a³ 8u2b)
+
+
b² c² + bc³ + b² c 3c5
Restoring the values of u and v, we obtain
b² ď² Q
4 dR2
´2ȧb ab²¿²,
Q = a²b²
+
c²
aboja).
By (54), (55) and (56)
2a²bȧ
2a³b³ j
P =
-
higher powers of c¯¹.
c²
C5
Therefore by (46) the force depending on the square of the
velocity
πα 642
C5
(61),
which varies as c.
Hence
F₂ = · 2πb² (M₂+ NË)-
(M,+N,)
d
a³b4b2
C5
1
The value of F₁ the force on A towards B, can be obtained by
symmetrically interchanging a and b.
241. If we neglect all powers of c¹ above the second
F
2πb² d
c² dt
(a²bȧ).
Let
a = ā + a sin
2πt
T,
2π
b=b+ß sin
-
T
(t − e),
so that a, b denote the mean values of the radii. The mean value
of F. will be
2
2π
b2
dt
F. - - 25 ["bd (a'ab) dé
Ꭲ . .
Tc².
4π ST
To²Jo
Tc²
dt
a²b² ȧb dt
16π³
(ã¯)² aß
[T
T³ ³
3
Tc²
COS
T 2″t 2π
COS
T
2πε
= F
T
=
1'
COS (te) dt
T
.www
י
256
MOTION OF TWO SPHERES.
Hence if the spheres are pulsating in the same periodic time
they will attract one another when their phases differ by less than
a quarter of a period; but if the phases differ by more than a
quarter and less than three quarters of a period, they will repel one
another.
EXAMPLES.
1. An infinite liquid contains a fixed sphere of radius b, and a
sphere of radius a and mass M fastened to a spiral spring per-
forming small oscillations in the line joining the spring to the
centre of the sphere. Prove that if a and b are so small (or c so
large) that we may neglect powers of a/c and b/c above the sixth,
the time of oscillation is
1
T{1
1
M₁ ab
T1+3/ 음 ​2M+M, c²
1
where M, is the mass of the liquid displaced by the moving
sphere, T the time of oscillation if the fixed sphere were removed
from the liquid, and c the mean distance between the centres of
the spheres.
2. An infinite mass of liquid is divided into two parts by an
infinite rigid plane, and a sphere is moving in the liquid in a line
perpendicular to the plane. Explain by general reasoning what
will be the effect of making a circular opening in the plane with
its centre in the line of motion of the sphere, when the sphere is
moving (i) towards the plane, (ii) from the plane.
3. Two equal small spheres of mass m and radius a, which
attract each other with a force equal to the product of their
masses divided by the square of the distance between them, move
in a straight line towards each other in an infinite liquid. If λ is
the ratio of the density of the liquid to that of the spheres, and x
the distance between their centres; prove that so long as (a/x)*
and higher powers can be neglected, the velocity of either sphere is
x √m (1 + žλ)
{œ³ + ±λ (x³ + 3a³)}* '
the motion beginning when the spheres are at an infinite distance
apart.
*
L
EXAMPLES.
257
4. If a spherical vessel of radius a contain a concentric sphere
of radius b and density σ, the intermediate space being filled with
liquid of density p, prove that if the vessel be moved with velocity
U, the concentric sphere will move forward with relative velocity
(p-o) U
}p (a³ + 2b³)/(a³ — b³) + o '
5. An impulse I is applied to one of two spheres, perpen-
dicular to the line joining their centres. Prove that with the
notation of § 229, both spheres will begin to move parallel to the
direction of the impulse and in opposite directions, and that their
velocities v,, v, are determined by the equations
2
I
B' A'C′ – B'² ·
6. Liquid of unit density fills the space between two con-
centric spheres. The outer one whose radius is b and the inner
one whose radius is a, is suddenly distorted in such a manner that
the velocity at any point of its surface is cF(0, 4), with the
condition that its volume remains unaltered. Find the velocity
potential of the liquid, and prove that when F (0, 4) is a zonal
harmonic of degree n, the kinetic energy of the liquid is
2a³ {nb²n+¹ + (n+1) a²n+1} πc²
n (n + 1) (2n + 1) (b²n+¹ — a²n+1) •
7. Liquid is confined within a sphere of radius b; and a solid
sphere of radius a' is moving with velocity v along a radius
of the fixed sphere. Prove that if the distance x between the
centres of the two spheres is small compared with b, the velocity
potential is approximately equal to
1
2r
1
・ua³
+
cose + x
b³
3r²
+
26*) (3 cos²
eos 0
0
-
• − 1)},
the origin being the centre of the fixed sphere.
8. The space between a spherical envelope and a solid
concentric sphere is filled with liquid which is at rest. If the
outer surface is moved so that at each point its velocity is a
spherical surface harmonic Y, prove that the solid sphere will
remain at rest, unless n = 1.
B.
17
258
MOTION OF TWO SPHERES.
9. Prove that the augmented inertia of a ball pendulum of
radius a oscillating in a spherical envelope of radius b is
½M (2a³ + b³)/(b³ — a³)
where M is the mass of the liquid displaced.
10. A string of length ƒ — a is attached to a sphere of radius
a and mass m, by means of some mechanical arrangement which
prevents the sphere from rotating. The other end of the string is
attached to a fixed point, and the system is surrounded by a
liquid of unlimited extent, which is bounded by a fixed plane.
Prove that if the string is initially at right angles to the plane,
and sphere is projected perpendicularly to the string, with velocity
V, the tension of the latter will be equal to
m
1
f
3
3f
(3ƒ (1 −
— cos €) +
0)
(1
C
− ³ƒ cos 0) sin²
3 cos 0) sin' 0}]
√²
3 Ma
2 (2m + M) c³ \c
3
3
9MV ma³ cos³
4
2 (2m + M) c²
where c is the distance of the fixed point from the plane, 0
the angle which the string makes with its initial position, M
the mass of the liquid displaced by the sphere, and powers higher
than care neglected.
APPENDIX.
I.
To prove the equation
p = kp".
The laws of Boyle and Charles show that the pressure, volume, and
temperature of a gas are connected by the relation
p = ᎡᎾ ...
·(1),
where R is a constant, and is the temperature measured from the
absolute zero of the air thermometer, i.e. from – 270° C.
-
Let a quantity dH of heat be communicated to the gas; the effect of
communicating this amount of heat will be to change the pressure,
volume, and temperature of the gas, and since by (1) the volume is a
function of the pressure and temperature we may put
dH = K₂d0 + λdp ...
..(2),
where K, is the specific heat at constant pressure. From (1) we have
de dp dv
+
Ө p v
whence eliminating dp from (2) we obtain
dH = K₂d0 + Xp (°
(do - do),
whence if K, be the specific heat at constant volume
K₂=K₂+
λφ
Ө
.(3),
(4).
Let us now suppose that the gas experiences a small change of
volume but without loss or gain of heat, then dH=0, and (2) becomes
K₂d0 + λdp = 0.
Eliminating 0 and λ by means of (3) and (4), and putting y = K₂| K„,
p
we obtain
dp
Ρ
+Y
dv
= 0 ..
v
.(5).
Now it is an experimental fact that y is independent of the pressure,
temperature or volume, whence integrating (5) we obtain
pv = const.,
or
p = kp³,
where p is the density.
260
APPENDIX.
II. To express the value of R (see page 220) in terms of elliptic
functions.
The value of R is
∞
πa³ 2Σ,
(1-9)²qm
qm+1)³)
R = wa² (1 + 22" (1-2)47)
..(1)
and we have to express this series in terms of elliptic functions. From
§ 124 it follows that the value of R or (P+L) p¬¹ is
∞
Απο Σ
ngn (1 + q²n) + 2nq²n
1 - q²n
∞n (1+qm) qn
=
Απο Σι
- qn
1+gn
=
4πc²Σ n
1-gn
qn
· 1 –g")}.….. (2).
K2L2
K
∞
ngn
Now
π-2
sn² Kx/π =
(K – E) – 2≥1
cos nx.
1 — q²n
Changing x into x + K'/K we obtain
K2
K
π-2
cosecam² Kx/π =
— 2 (K - E) - Σ1
∞n (1 + q²n)
1 − q²n
cos nx.
Adding we obtain
K2
2K
2
(k² sn² Kæ/π + cosecam² Kx/π) = ²= (K – E) – Σ
∞
-
n(1+qn)
1-gn
1 - qn
cos nx.
Also Σn (1 + q″) cos nx = − 1 cosec² 1x +
{(1 + q²) cos x − 2q}
(1 − 2q cos x + q²)²
2
Therefore
Σ0 {n (1 + q″)
1-9
− n (1 + q″) }
· (1 + q) cos nx
2K
π-2
(K – E)
K2
2
77-2
(k² sn² Kx/π + cosecam² Kx/π) + cosec² 1x
q{(1+q²)cosx-2q}
(1 − 2q cos x + q²)²
...(3).
The required series is equal to the limit of the right-hand side of (3)
when x=0, that is
q
2K
R = 4πc²
(K – E) - (1
– (1 − q)².
−
III. Professor Greenhill has kindly worked out the following
investigation of the Motion under no forces of a Solid of Revolution in
Infinite Liquid, by Weierstrass's functions.
Taking the expression (4) for the kinetic energy T of the solid of
revolution and of the surrounding infinite frictionless liquid given in
§ 181, but writing p, q, r instead of w₁, w, wg respectively, then
T= P(u² + v²) + Rw² + }A (p² + q²) + & Cr²;
APPENDIX.
261
and employing this in the equations of motion of § 167, supposing there
are no impressed forces; then since
=
dT
dT
Pu,
=
Pv,
du
dv
dT
dw
dT
dT
dT
Ap,
Aq,
=
dq
dr
Rw;
Cr;
dp
=
the equations of motion become
P
du
dt
-
- Pvr + Rwq
-0........
..(1),
dv
P
- Rup + Pur
.(2),
dt
dw
R
dt
Puq + Pvp
= 0..........
.(3),
dp
A
| ཅུ|ཅུ
dt
A
dq
dt
− ( A − C) qr – (P– R) vw=0…………….
+ (A − C) pr + (P – R) uw=0…………..
(4),
.(5),
dr
C
dt
= 0........
.(6).
Equation (6) shows that r is constant during the motion; and from
the other equations we can obtain three first integrals of the equations
of motion.
First, Pu +v + Rw
du dv
dt dt
dw
dt
+ A
(p dp +9
dq
dr
dt dt
+ Cr
0,
dt
so that
{P (u² + v²) + ↓ Rw² + §A (p² + q²) + ¿Cr² = T……………………. (7),
a constant, the constant value of the kinetic energy during the motion.
Secondly,
so that
ps (u
du
dv
dw
+ v
dt dt
+ R²w
= 0,
dt
F2....
P² (u² + v²) + R²w² = F².
(8),
a constant; and then F represents the resultant linear momentum of
the system.
Thirdly, AP
dt P + u
(
du
dp
dv
dq
dw
+
dt
q + v
dt
+ CR
r = 0,
dt
dt
so that
=
G....
(9),
AP (up + vq) + CRwr
a constant; and then G may be taken to represent the constant angular
momentum of the system.
From equations (7), (8), (9),
P² (u² + v²) = F¹² – R²w²,
-
A (p² + q) = 2T - Cr²- Rw²- P (u² + v³),
H2
=
= 2T - Cr²
P
R
-(1-1) Rw²,
G – CRwr
P (up + vq) =
;
A
262
APPENDIX.
so that from equation (3)
R2
dt
(dw)² = P² (uq - vp)²
=
P² {(u² + v²) (p² + q³) − (up + vq)²}
R²w²\
A
2T Cr²
F2
P R
R2w2
G- CRwr\2
-(-) (27-cr--(-1)) - ( - Rior);
{
-G
A
a quartic function of Rw, so that Rw is an elliptic function of the time
t, which we shall proceed to express by means of the notation of
Weierstrass.
Rw
Putting, for the moment, = x = cos 0, then
dx\ 2
dt
F2/1
F
1
−
1 ( x − 1 ) (x − x) (x − x) (x − x₂) (x — X3),
AR
where ∞, ∞₁, x2, x denote the roots of the quartic in x, arranged in
descending order of magnitude; also
Xo + X1 + X2 + X3 = 0.
D
s-d'
Now put
X - Xo
D s-e1
then
X — X₁ =
X1
s-dd-e₁
D s - ez
X - X₂ =
8-dd-e'
XC
—
- X3
D s-l3
s-dd-e₂'
where é, é, és are the roots of the discriminating cubic of the quartic
483-938-93=0,
9₂ and g, being the quadrinvariant and the cubinvariant.
Then
ds\2 F2/1
D2
(de)² = 12 ( 1/2 - 1 ) D²
dt A R
and we may choose D, so that
-
483-928 - 93
P 4d³ — g₂d — 93´
and then
ds\2
D² = 4d³ — gåd – 93,
(483
F2/1
-
(d) ² (1-1) (45² – 9.8 - 93),
dt AR
so that now, with Weierstrass's notation (Halphen, Traité des fonctions
elliptiques et de leurs applications, Paris, 1886),
W1
8 = p (tw₁/T + w₂),
w₁ and w, denoting the real and imaginary half periods of the elliptic
functions, and the time of oscillation; the imaginary half period w
being added in order to make s oscillate between e, and e, and therefore
x between x and x3.
APPENDIX.
Then the time of oscillation is given by
2
F2/1
AR
P
2).
263
We may write pc instead of d; and use pu instead of p (tw₁/7+ ws)
for brevity, and then D
х
- p'c, and
X — X1
- =
- p'c
pu-pc
- p'c pu-e1
pu- pc pc - C1
- p'c pu- e₂
X — X2
pu-pc pc-Eg'
- p'c pu- es
X
X3
;
pu- pc pc - Cz
and then, as explained in the Proceedings of the London Mathematical
Society, vol. XVII., p. 279, 1886, introducing the function (u, defined by
d
du
Su
-
-ри,
x。 = 2¿c — (2c:
p"c
2
p'c'
C =
14
p" (c + w₁)
p' (c + w₁) '
"
..., X3 =
and p2c, p'2c are the coefficients of x² and x respectively in the quartic
-
(x − x。) (x − x₁) (x − ∞。) (x — X3) ;
also
x = ¿ (u + c) − } (u – c) – (2c = ½
p' (u- c) - p'2c
p (u - c) - p2c
Taking the axis OZ in the direction
(fig. p. 166), then
and
Pu= - F sin 0 cos o, Pv = F sin 0 sin 4, Rw
P (up + vq) = F sin 0 (− p cos & + q sin 4),
of the resultant impulse F
= F cos 0,
=
= F sin² 0
dy
dt
;
so that equation (9) becomes
dy
AF sin² 0
G- CRwr,
dt
=
G-CFr cos 0,
or, using x to denote cos 0,
dy G-CFrx G+ CFr 1
=
dt AF(1-x²) AF 1+x
the equation to determine the azimuthal motion v.
G-CFr 1
+
AF 1-x'
264
APPENDIX.
As explained in the Proc. London Math. Soc., vol. xvII. p. 280,
writing u for tw₁/T+w, this equation becomes
dy
p'a (pu-pc)
=
Li
+ şi
p'b (pu - pc)
—
du (pa-pc)(pu - pa) (pb - pc) (pu - pb)'
a and b being the values of u which make cos
1 or +1, respectively.
αψ
Then 2
diu
p'a
pa - pc
+
p'a
pu - pa
p'b
p'b
pb - pc
pu- pb’
= ¿ (a + c) + ¿ (a − c ) − 25a — § (u + a) + } (u − a) + 25a
+ ¿ (b + c) + ¿ (b − c) + 2¿b − ¿ (u + b) + ¿ (u − b) — 25b,
o (u − a) o (u – b)
4 = i log
where
and
σ (u + a) σ (u + b)
o
+ LiPu,
P = { (a + c) + ¿ ( a − c) + § (b + c) + Č (b − c),
=e
-Pu
lo (u + a) σ (u + b)
σ (u − a) o (u − b) `
-
- -
Taking a point on the axis OC at unit distance from 0, the pro-
jection of the motion of this point on a plane through O perpendicular
to OZ will be given by
x + iy = sin fe¹¹
o (u + a) o
C
σ (u + b)
σ (u + c) o (u − c)
exp (− †Pu).
In a similar manner, by means of the equation
d
d
uv – úv
log (u + iv) = 1 + log (u² + v²) + i
dt
dt
U² + v² 9
d
dt
= 12 + log (u² + v²) – ir + i
up + vq Rw
u² + v² P
>
d
= { log(F² — R²w²) — ir + i
=
dt
G-CRwr Rw
F2 – R²w² A
-
we can express u + iv by means of Weierstrass's σ functions; and the
same method can be applied to the expression of p+iq and also of
x+iy, x and y now denoting the coordinates of O with respect to fixed
axes in a plane perpendicular to the direction of the resultant impulse F.
It will be noticed that the letter u has been used in two senses,
first as expressing a component velocity of translation, and secondly as
an abbreviation for tw₁/T + w; this was unavoidable in order to reconcile
the different notations, but will not be found to lead to confusion.
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