30 Derivation of Symmetrical Gravitational Fields. 



Then 



l 2 = c 2 hxi + a ; h = czkxi + ft; h = cjcx x + 7, 



where a, ft, 7 are constants.) The constants C 2 , C 3 , C 4 or 

 *, /3, 7 respectively are not independent. We have 



C2O3G4 = c 2 + 6*3 + c 4 or a + yS + 7 + log & = 0. 



There remains the equation 



Hr TT X TT v, 



2 XI3 -0.4 



h" + h' 2 +h"+h'*+h"+h' 2 = o. 



Again eliminating l ?J and l 4 , 



I C 2 - 6' 2 J 2 I C 2 C 2 2 J 



This shows (a) that if c 2 + c 3 H-^ 4 = then Z 2 ' = (giving 

 the trivial Euclidean case) unless c 2 2 + c 3 2 + c 4 2 = 0, in which 

 case all the equations are satisfied. 

 Hence we have the solution 



Ids) 2 = (d^y + e^^d^) 2 + e**°i(dx 8 )* + e»n (dw±) 2 

 = \ {dr) 2 + r*>* (dx 2 ) 2 + r 2c * (da*)* + r** (^ 4 ) 2 , 



where r=e Xl , and c 2 , c 3 , c 4 are three constants satisfying the 

 two equations 



c 2 + c 3 + c 4 = 0; c 2 2 + c 3 2 + c 4 2 = 0. 



(We have suitably modified the units in which #,, #2, ^3* #4 

 1 lire measured.) 



(b) If c 2 + c 3 + c 4 gfc we have 



_^ = ^ 2 + ^3 2 -fC 4 2 = ^ 2 -f6' 3 4-C 4 

 / 2 ' 2 C 2 (6' 2 -f6' 3 H-C 4 ) 6' 2 



from a previous equation ; yielding 



C2C3 + C3C4+C4C2 = 0. 



TT tt rr 



(This relation is equivalent to tt^/ + tt 1 / + ttj = if none 



±1 2 11 3 ri 4 



of the c's =0.) If, then, we choose three constants c 2 , c 3 , c 4 



