( 360 ) 



,, = -- (24) 



a 



From (11) we have 



for T z=T^ . . . u = — I, for T = T, .../<= + 1 ; 

 Further according to (19) and (24) 



cdx du 



aR R—^{.v-{-vt) 



or, if we pnt in tlie vahie of R from (24) 



cdr da ^_., 



— = (2o) 



aR ct{1 — If-) — an—ifr 



On the right iiand we must here express r by ii. According to 

 (24) we have between t and u the quadratic : 



(a n — CTy = R' = (.f + r t)^ + //^ + ~^ 

 If we solve it with respect to c r, we get 



(1 — 1^-) {(■ ry — 2 (a n -\- ,i .c ) c t = r'- — a'- n', 



a u + i3 X rfc 1/ (a u + ,i .r)- + (1 — i^') (?'' — «' u') 



0T=: ^ . (2b) 



1 ~ p^ 



What was proved before for the hmits t^ t^ of the interval of 

 integration viz. that only one of the two possible values is positive, 

 holds as well for the values of r in the interior of the interval. 



The one positive value is obviously given by the positi\e sign of the 

 root in (26). Omitting therefore the negative sign we conclude from (26) : 



c T (1 — i3^) — a n — (i X = -I- y/{a2i-\- ^ x)' + (1 — /?') (/•'— a'u') = 



+ )/'{'^ -^ ct u i3y + (1 - ^') iy' + z^) (27) 

 This happens to be the denominator occurring on the right of 

 (25). Instead of (25) and (21) we therefore write : 

 cdr d n 



~^ ~ l/"(7+ a u ^r + (l-i3^) iy' + z') 



+ 1 



3« f (1— w') du 



4.jt<f = -^j ...... „ , „— r ' • (29) 



1 



This is the rigorous expression of the potential. The integration 

 leads to elementary functions, but becomes rather troublesome. So we 

 content ourselves with an approximation putting a = 0. Then the root 

 in (29) becomes independent of u and we get: 



