528-530] Induction in Continuous Media 463 



The left-hand member is equal, by Stokes' Theorem ( 438) to 



fff./az 8F\ fdX dZ\ /dY dX\) , 



MM + m + ft ^ r &> 



JJ [ VB^ 82 y VS-z OSBJ \ox oy )} 



the integration being over the same area as that on the right hand of equa- 

 tion (463). 



Hence we have 



[[(jidZ dY da\ fdX dZ db\ /dY dX dc\) , 

 J IF" ~ a^ + ~dt) + m Vbz ~~ fa + dt) + U ( ~fa ~ "3" + dt)\ 



This equation is true for every surface, so that not only must each inte- 

 grand vanish, but it must vanish for all possible values of I, m, n. Hence each 

 coefficient of I, m, n must vanish separately. We must accordingly have 



_da_dZ_SY_ 



(it c}ii r) z 



db dX dZ /AG-\ 



-i- = -^ ^- (46o), 



dc 8F dX 



530. The relation between the electric forces and the changes in the 

 magnetic field can be expressed also in a different manner. The surface 

 integral of equation (462) can, by 443, be replaced by a line integral round 

 the boundary, and we* have 



where F, G, H are the components of the vector-potential ( 443) given by 



dff dGr 

 tt = ^ ^r-, etc (467), 



oy dz 



dN 



so that on equating expression (461) to -y- we obtain 



at 



+ ^^-M F +^^+^ + ^;r^ = .-.(468). 

 dt J ds \ dt J ds \ dt J ds) 



This equation is true when the integral is taken round any closed circuit. 



Let us denote the same integral, taken from any fixed point to a point 

 P by - M>\ so that 



( P \( X+ ^\f + f Y+ dGYj [+ ^ + dHY^ ds = 

 J o \\ dt J ds \ dt J. ds \ dt J ds) 



We have not specified the path which is to be chosen in going from to 

 P, but this is unnecessary because it follows from equation (468) that ^ will 



