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 + h, and then making & = oc, we get the 

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

 distance h, viz. 



1 (xvii), 



\ '"' / 



a result due to Stokes. 



This also follows from (vi) and (xv), by putting 6 = a, u' = U, c= Zh, 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 r, $, o> and ?', #', a>' 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 



dd> A dd>' 



^=-cos0, -. = 0, forr=a, 



and ^ = 0, ^,'= - cos ff, for r' = b. 



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

 A would be 



i -a COS B. 

 2 r 2 



Since r cos 6r f cos ff, 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 



, 3 6 3 cos & 

 t -r, 



