July 25, 1902.] 



SCIENCE. 



125 



When any radius, vector OP revolves 

 about the axis of s, it remains in the 

 first (or physical) space until 61=2 -:. It 

 then crosses the branch membrane and 

 enters the second fold of the Riemann space. 

 In the problem before us, a second revolu- 

 tion brings the radius vector into the first 

 fold again. It is to be understood that each 

 fold fills all space. Two underlying points 

 have the same r and the same z, but their 

 ff coordinates diflier by 2^ or some odd 

 multiple of 2tt. Two points whose vec- 

 torial angles differ by an even multiple of 

 27r are in the same fold. 



The problem now is to find a Laplace's 

 function which shall vanish on each plane 

 and at infinity, and shall have only one pole 

 in the original physical space between the 

 planes. 



Let (rj, ^1, 2j) be the coordinates of the 

 assigned pole, and (r, &, z) those of the 

 current point. Then 1/B is a solution of 

 Laplace's equation, where 



iJ2= (« — 2i)2-f-r2 + V — 2rr, C08(fl — «i). 



Dr. Sommerfeld first replaces 6^ by an 

 arbitrary parameter a, and denotes the 

 result by B'. He then multiplies 1/E' by 

 an arbitrary function /(«), and integrates 

 with regard to a. The result is still a solu- 

 tion of Laplace's equation. By a proper 

 choice of the function /(«), and of the 

 range of integration, he obtains a function 

 of (r, d, z) satisfying all the conditions. 

 He takes the two-valued function 



/(«) = 



and puts 





-da, 



then regards a as a complex number, and 

 performs the integration in the a-plane 

 around a eontoiir enclosing the point a = 0^ 

 and excluding the other points where the 

 integrand becomes infinite. The function 



itj thus obtained becomes infinite at the 

 pole (fj,, Oj^, Zj^) but does not fulfill the con- 

 dition of vanishing on the two planes. 

 Next he forms a similar function w, for 

 the pole P, situated at {r^, 0^, z^), and so 

 on. The required Green's function is 



U=Vi Vn + W3 «4 + I's Wo- 



The poles of u^ and u^ would, under 

 ordinary circumstances, both lie in the 

 given region, but the pole of u^ is given 

 such a vectorial angle as to bring it into 

 the second fold of the Riemann space. The 

 function n has then only one pole for the 

 physical region defined by < (9 < 2 w. 



Moreover, ti vanishes for points on the 

 two planes, and fulfills all the other con- 

 ditions for Green's function. 



Thus we see how a function, which would 

 be two-valued and bi-polar if restricted to 

 the given physical region, becomes single- 

 valued and uni-polar in the Riemann 

 space. We may say that the second fold 

 of this space is a refuge for the second 

 value and the second pole. Care has to be 

 taken to use the proper values for 6 when 

 the indicated operations are being per- 

 formed. The difficulties of the problem are 

 thus reduced to those of the integral cal- 

 culus. 



In the more general case in which the 

 angle of the planes is v.rj'm, there are 2m 

 poles in the circuit (one in each angle 

 t/«i), of which n are in the given region. 

 The Riemann space is then n-fold. 



Sommerfeld has worked out at length 

 the very interesting case in which the angle 

 between the planes is 2-. The region is; 

 then bounded by the surfaces « = ± 00 ,. 

 r = CO , ^ = 0, 0=1-; the last two being: 

 the two faces of an infinite half plane with 

 a straight edge. The assigned pole and its 

 image are both in the given region; hence 

 the corresponding Riemann space is two- 

 fold; and the required solution is 



,7=« («,)-«(-«,) 



