Derivation of Symmetrical Gravitational Fields. 27 

 Now on putting H 4 =^H 2 ' in (D') we get 



2H 2 H 2 " + H 2 /2 =:0, 

 so that 1 = 27*, and our solution of the 



set of differentia] equations is 



H 1 = l; H 3 = H 3 sinff 2 ; H 4 = 5H/ ; 



where H 2 satisfies the simple differential equation 



2u(3 



H 2 ' 2 = l 



H* 



It is now convenient to introduce a new variable r which 

 is a function of ac x alone, and which is defined in such a way 

 that H 2 = r ; then 



, A 2a/3 _ TT , dr A 2&.B 



* = V l - r ** H2 = di\ = V 1 ~ ~T' 

 and 



(d s y = (d^y + R^dx.y+E^dxsy+R^d^) 2 



With an obvious modification of the notation this is the 

 classical result. 



In the paper referred to at the beginning of this paper, 

 what is done is to determine at-i such a function of r that 

 B 2 = rH 1 . 



We have d.v l = H 1 dr=B, 2 ( ^; whilst ^= \/ 'l- 2 *^. 

 So ti at 



dld 2 _ dr 



kA-S 



whence 



H 2 -«/3 + vH2 2 ~2«/3H 2 = Or, 



* If we put <j)" = 0, (p'= constant, and y=0. 



t Another solution (corresponding to <£'=:constant) is 



