142 PROBLEMS IN THREE DIMENSIONS. [CHAP. V 



If a, but not necessarily 6, is small compared with the shortest distance 

 between the spherical surfaces, we have 



approximately. By putting c b + k, and then making I = oo , we get the 

 formula for a sphere moving perpendicularly to a fixed plane wall at a 

 distance h, viz. 



tf ........................ (xvii), 



a result due to Stokes. 



This also follows from (vi) and (xv), by putting b = a, u = U, c=2k, in which 

 case the plane which bisects AB at right angles is evidently a plane of 

 symmetry, and may therefore be taken as a fixed boundary to the fluid on 

 either side. 



98. When the spheres are moving at right angles to the line 

 of centres the problem is more difficult; we shall therefore content 

 ourselves with the first steps in the approximation, referring, for a 

 more complete treatment, to the papers cited below. 



Let the spheres be moving with velocities v, v in parallel directions at 

 right angles to A, B, and let ?*, $, o&amp;gt; and /, 6 , be two systems of spherical 

 polar coordinates having their origins at A and B respectively, and their polar 

 axes in the directions of the velocities v, v . As before the velocity-potential 

 will be of the form 



with the surface conditions 



and f, = 0, f=-cos ,for/=i. 



If the sphere B were absent the velocity-potential due to unit velocity of 

 A would be 



Since r cos B=r cos 6 , the value of this in the neighbourhood of B will be 



approximately. The normal velocity due to this will be cancelled by the 

 addition of the term 



