194 Mr. F. L. Hitchcock on 



tangential component ; the term v^^v equals CiV%\, by (1) 

 or by (15). Thus the first term of (19c) becomes 



S (2%/i — m 2 fjL)(y]rp + CjV^X) ; 



multiply] ng, and noticing that f-iy = \ while Sv^X,^ju= — ra l5 



S(2p£ — m^n^ii— 2 ^ra j — c,m 2 S A \A 



is the product. We next obtain from (15) 



V i u=XS,i X ^ + ) u(e 1 -S\xj*)-''- 5 • • (20) 



and here again we are not concerned with the normal com- 

 ponent. Thus the second term of (19c) equals 



s [>*2 2 — 2»<!)//,— »» 2 %/*] [>$ /*x/* +/*( c i — SA >X/*)1 



= —c 1 m 2 ' 2 + ?«o 2 S/V%yLt -f 2c 1 m 1 — 2m 1 S\x/jL — c i m 2 Sj J L'Xf J u. 



The third term of (19c) is the same as — — * SA^/a. Collecting 



results and noticing that S (\%\ + /x%/x) = — m 2 , Ave find that 

 all the terms containing Cj cancel out, and the result is 



S(2x-™ 2 ) /i f /1 +SA. x/ *(™/-2m 1 + ^) = 0, (19d) 

 which by (lice) may be written 



S(2%— w2 2 )/i^/A— SA%/*SV%i'=0, . . (19<?) 

 Again, because of the identity 



(t-tv= y (vvv^ 



we shall have the following expansions : — 



S(2^-m 2 )yu,^ A 6=S(2^-??z,^)(i|r' A 6-^VVV I/ ) 



= S (2%— m^^'fi + 2SfixfjffWv 

 = S(2 % -m^fi^fi + 2SA %A 6SvV Wv 

 = S(2 X — m 2 >fV + 2SA %/ ,(S V % v + V 2 Vv), 

 and by using this result in (19c), 



S(2 % -m 2 )^ / x + S\ %/ ,(SV%v-f 2Y 2 Vv) =0. . (19/) 



Finally, by adding (19c) and (19/) 



S^(^ + f , )(2%-^ 2 )/i + 2S\ X /^V 2 Vv = 0, . (19c/) 

 where the only operation involving the second differential 



