( 126 ) 



give when inteo;rated with respect to r an infinite energy at infinity of 

 Spn- So for ?i ]> 3 are excklded by the field property only vector distri- 

 butions which cannot have physical meaning. 



For 7Ï = 2 however the postulation lacks its right of existence ; 

 more sense has the condition (equivalent for ?z > 2 to the field pro- 

 perty) that for given rotation and divergency the vector distribution 

 must have a minimum energy. Under these conditions we shall once 

 more consider the field and we shall find back there too the duality 

 with regard to both derivatives and both potentials. 



IX. Let us consider first of all distributions with divergency only 

 and let us find the potential function giving a minimum energy for 

 given V'- 



We consider the hyperbolical Sp^ as a conform ^representation of 

 a part of a Euclidean Sp^ bounded by a circle; if we then apply 

 in corresponding points of the representation the same potential, we 

 retain equal energies and equal divergencies in corresponding plane 

 elements. So the problem runs : 



Which potential gives within a given curve (in this case a circle) 

 m the Euclidean Si\ under given divergency distribution a minimum 

 energy ? 



According to the theorem of Green we have for this : 



/fbu\ r du ddu r d.du r 



62 I ^— (Zr = I JS" -- . -:^ . dt =z I w . --— . dO — I u w ^ du , dr, 

 \d.vj J dx dx J dv J 



so that, as \7^öit is everywhere within the boundary curve, the 

 necessary and sufficient condition for the vanishing of the variation 

 of the energy is : 



u^O, along the boundary curve. 



For the general vector distribution with divergency only in the 

 hyperbolical Sp^ we thus find under the condition of minimum 

 energy also, that the potential at infinity must be 0. So we find it, just 

 as under the postulation of the field property, composed of fields ^i, 



cos (fi 



derived from a potential -— ; — . 



sink r 



The lines of force of this field E^ have tlie equation, 



sin (f coth r = c. 



Only a part of the lines of force (in the Euclidean plane all of 

 them) form a loop; the other pass into infinity. None of the equi- 

 potential lines, however, pass into infinity ; they are closed and are 

 all enclosed by the circle at infinity as the line of 0-potential. 



