and Electromagnetic Theory. 3-47 



An energy equation on the mechanical side is obtained by 

 multiplying these by (uvw), viz., 



- S— (uX x + vY x + wZ x ) = P $uX' -f ~ Su(vy-t^8) + Sm ^ . (00) 



This equation is the complement of (56), and the division 

 of -^-2 Vu into two parts is noticeable. Equation (60) may 

 also be got by subtracting (56) from (58), and quoting (37); 

 but -p.- must be used in (56) to transfer to the standpoint 



of (58). 



§ 18. We now pass to the special problem of discontinuities 

 moving in free aether under circumstances which give a 

 steady condition from the moving standpoint. 



The solution of (54) has the form 



_<m_dG X =-^t 1?Z 

 "~ dy dz> dx~~V Dt 



1 / dyjr dir\ 1 / D'H D'G\ 



where fFGH are defined by 

 subject to 



In these 



i we 



write 







d 

 dt' 



= 0, or 



D' 



Dt 



Thus 





" dy ~ 



dG 

 ' dz' 



X'=-^(t-2/)F), . (61) 



where (p, q, r) = (u, ?;, ic)/V, and the other variables are 

 given by a'=ra + rY — qZ, and X' = X + 27— rj3. 



For the integration of J Sdr, i.e. J (i2XX' + IjXoca'jdr, the 

 forms of X'... and a... as gradient and curl are essential, 

 For a volume-distribution, 



\ S«/t = \\ p (Vr - SpF )</t. + j^y SiFrfTft . 



(62) 



