444 Journal of the Asiatic Society of Befigdl, [July, 1907- 



Putting V=u^ P^ (cos 6)^ we have, since 



1 d / . .dp 



sine d9 



si^e !f^^ + «(«-|-l)P„ = 0. 



«„, given by — (/'-l) — -« (n + 1) n^ = 



V =2J„ P„ (cos ^) P„ (r) + B„ P„ (cos 9) Q„ (r), 



where P's and Q's are Legendre's functions of the first and second 



kind. 



2. It would be also useful at the stage to expand (distance) —^ 



in these harmonics : 



For this we have, writing 



pS = £e* 4- T^s -^ 2% 



= h^ (?^-siu2^) 



A^^p^ + p'^ — 2pp' [cos a cos a' + sin a sin a' cos ^^^ ] 



where Asdistance between two points (/j, p'), 



also, p cos a = jj 



p sin a=\/ar* + ^* = ^\/r*— I sin ^, &c. 



/. A»=A«[r»"-sin« ^+r'«-sin3 d'~2 rr' cos ^ cos ^' 



"2 v/r»-l %/r'» ^ 1 sin B sin ^' cos (^-6')]. 



Now putting cos ^ = ^ &c., and cos (<?"<^') = l5 for terras in- 

 dependent of ^, D, the corresponding value of A 



^_r{r««l) + (r'«-.l]+/x« + /x'3-2rr'/x/ 



^' L "2v/r»-l v^r'2-1 v/l-/ii8 v/l-./2 



In order to expand D-^ put y' = l, / = 1 ; and if D^ is the cor- 

 responding value of D, 



r 



/. obviously, ~J^ 2(2 « + l) P^ (/.) P^ (^0 Q^ (r) Q>'). 



3. To find the potential due to a circular current at a point 

 P, m any s^yslem oE harmonics. 



