( 259 ) 



'sin "— V . 1 :=: o sin «— 'r (fl) 



dr I dr \ 



sm «— 'r . =c t sin "—V dr. 



dr 



= c \ sin "" 



ƒ■■"■■ 





If the field E^ is to be composed out of the function F^ (r) then 

 the opposite point of the centre may not have a finite outgoing 



vector current; we therefore put I sin "— V dr =i 0, so that we get 



- =. ; I sin «— ir dr, 



sm n— V ^ 



dF, 

 dr 



which corresponds to the above result. 



IV. The field E^ of a small vortex «— ^sphere according to Spn~i, 

 perpendicular to the axis of the just considered double point, is iden- 

 tical to that field Ei outside the origin; but now each force line is 

 closed and has a line integral k,i along itself. 



According to the method of C ^ VII we shall find of this field 

 E^ the planivector potential H, lying in the meridian plane and depen- 



dent only on r and cp ;so that it is a iX We find : 



dh = ce sin ^~-r sin ^'~'^<p. 

 Force in r-direction : 



I sin " 



~V dr 



cot V 

 (n — 1) cos (fi \ — [- —— . - — ——, \ = {n—l) cos (p . «>„ (r.) 



9 

 2 z=. I (n — 1) cos <p ojn {r) . cs sin «—2^ ^j^ "—2^ . sin r dtp ■=. 







z=. oi^^r . cs sin " — V sin " — ^(f. 



■ ^ . . _ 



H = —- =2 On (r) sin r sin g) zn'An (^) sin (p. 



dh 



From this ensues for the force of a plane vortex element with 

 unity-intensity in the origin : 



Xn (r) sin (p, 



