( 119 ) 



From this ensues : a vector field is determined uniformly by its 

 rotation and its divergency. 



VII. So, if we can indicate elementary distributions of divergency 

 and of rotation, the corresponding vector fields are elementary fields, 

 i. e. the arbitrary vector field is an arbitrary space-integral of such 

 fields. 



For such elementary fields we find thus analogously as in a Euclidean 

 space (1. c. p. 74 seq.) : 



1. a field Ey^, of which the second derivative consists of two 

 equal and opposite scalar values, close to each other. 



2. a field E^, of which the first derivative consists of equal 

 planivectors in the points of a small circular current and perpendicular 

 to that same current. 



At finite distance from their origin tlie fields E^ and E^ are here 

 again of the same identical structure. 



VIII. To indicate the field E^ we take a system of spherical 

 coordinates and the double point in the origin along the axis of the 

 system. Then the field E.^ is the derivative of a potential: 



cos (p 

 sinh^r 

 It can be regarded as the sum of two fictitious "fields of a 

 single agenspoint", formed as a derivative of a potential — 1 -|- coth7\ 

 which have however in reality still complementary agens at infinity. 



IX. The field E^ of a small circular current lying in the equator 

 plane in the origin is outside the origin identical to the above 

 field E^. Every line of force however, is now a closed vector 

 circuit with a line integral of 4.t along itself. We shall find of this 

 field E^ a planivector potential, hing in the meridian plane and 

 independent of the azimuth. 



In order to find this in a point P with a radius vector r and 

 spherical polar distance </? we have but to divide the total current 

 between the meridian plane of P and a following meridian plane 

 with difference of azimuth dO-, passing between P and the positive 

 axis of revolution, by the element of the parallel circle through P 

 over dd-. For, if ds is an arbitrary line element through P in the 

 meridian plane making with the direction of force an angle f, \ï dli 

 is the element of the parallel circle, 2 the above mentioned current 

 and H the vector potential under consideration, we find : 



d^ ■=z dh . Xds sin r, 



