dr 



( 263 ) 



1/2 TT 



= -r— — — I sin n— V dr. 



sm ^~^rj 



VI. In the usual way we deduce the iJT, which is planivector 

 potential of the field E^. 



dh = c£ sm n— S/* sin "~2 ^^ 



Force in /--direction : 



^ = I (n — 1) cos ^ . ii)i (r) . CE sin '^—-r sin '^—^(p . sin r d(p =: 

 



■= fi„ (r) . c£ sm «— ly sm "~'y. 



/if = — = f*n (^') SÏW r sin (p = x„ (r) sin <p. 

 dh 



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

 unity -intensity in the origin : 



Hji (r) sin y, 



directed parallel to the acting vortex element and projecting itself 

 on its plane according to the tangent to a concentric circle; <p is 

 here the angle of the radiusvector w^ith the 82)71—2 perpendicular to 

 the vortex element. 



VII. Here too a planivector potential of a vortex element can be 

 deduced, but we cannot speak of a direction propagated parallel 

 to itself, that direction not being uniformly determined in elHptic 

 space; after a circuit along a straight line it is transferred into 

 the symmetrical position with respect to the normal plane on the 

 straight line. 



But we can obtain a vector potential determined uniformly, by 

 taking that of two antipodic vortex elements in the spherical SjJn (in 

 their ^sphere the two indicatrices are then oppositely directed). 



The vector potential in a point of the elliptic Spn then lies in the 

 space through that point and the vortex element; if we regard the 

 plane of the element as equator plane in that space then the plani- 

 vector potential V is normal to the meridian plane: it consists of: 



