192 Mr. F. L. Hitchcock on 



it follows that 



\% 2 \ 4 p^fx — X [m. 2 x — m 1 )\ + /x (m 2 x — m { )fx 



= %n A 4- m 2 S(\x\ + fxx/x + vx v ) 

 — 2;/?x — m 2 q ; 





11a) 



the last term of (11) may be written cv%\ ; therefor 

 SVx v ma y {l ^ s0 De expanded thus 



S\7 X v=SV(c\) 



which by comparison give? 



SVx=-(2 mi -W-^f + J J, • • (13) 



and so from (9) 



>*' VlTJ . 111 _ *■ — - 



2_^ ^_„2 



(14) 



b\ Because d(c/x) = dc . /x + cd/x and /a is a unit-vector, it is 

 clear we may write the value of d/x by inspection of (8), 

 dropping the component along \x and dividing the rest by c. 

 This gives 



dX}Y\7v = X&dp(- —X-SvV/x./x-c^+vSdpx/*. (15) 



The differentials of yy and U^, that is of cX and \, are 

 easily expressed in terms of yjr and x- -^ or 



dyy = d[^\]v . v) =cY/xxdp — Yvyjrdp ) 



the first term on the right is the same as cv&\xdp and the 

 last term is the same as — \&dp\lr'fi + fi$dpilr f \ : therefore 



d x v= -\Sdp\7c+ fiSdpyy\ + cvS\xJp- ■ ( 16 ) 



For dTJx v we have only to drop the component of dxv along 

 \, and divide the rest by c. This gives 



dUxv^-pSdpf'X + vSdpx'X (17) 



