494: Lord Rayleigh on the Influence of Obstacles 



Then, if Hx be the potential due to the sources at infinity 

 other than the spheres, V^Ha, and 



so that the specific conductivity of the compound medium 

 parallel to a is 



i-st <«» 



We will now show how the ratio 13,/H is to be calculated 

 approximately, limiting ourselves, however, for the sake of 

 simplicity to the case of cubic order, where ct = fi = y. The 

 potential round P, viz. 



A + A 1 r.Y 1 + A 3 r 3 Y 3 + ..., 



may be regarded as due to H.r and to the other spheres Q 

 acting as sources of potential. Thus, if we revert to rectan- 

 gular coordinates and denote the coordinates of a point rela- 

 tively to P by x, y, z, and relatively to one of the Q's by 

 x\ y' ', z' } we have 



Ao+(A l -H)*+A s (^-J«- ? )+... 



H .. ™ v. A'' 3 - § a'S* 



^B^^+BsS^--^ +..., . (61) 



T" 7 



in 



which 



#' = #-£ y'^y-y. z' = Z-%, 



if £> Vi f De the coordinates of Q referred to P. The left side 

 of (61) is thus the expansion of the right in ascending powers 

 of x, y, z. Accordingly, A x — H is found by taking d/clx of 

 the right-hand member and then making x, y, z vanish. In 

 like manner 6A 3 will be found from the third differential 

 coefficient. Now, at the origin, 



A A = -A A = - A ~£- p 9 -z? 



dxr' 3 d% r' 3 df p 3 p b ~' 



in which 



?=?+*?+.?. 



It will be observed that we start with a harmonic of order 

 1 and that the differentiation raises the order to 2. The law 

 that each differentiation raises the order by unity is general ; 

 and, so far as we shall proceed, the harmonics are all zonal, 



