FLUXES OF ENERGY IN THE ELECTROMAGNETIC FIELD. 
477 
(e 0 j 0 .+ h„Gr (1 ) + cony [V (E - e„) (H - hj + q (U + T)] 
= (Q + U + T) + Fq + conv %q 
= (Q + u + T) + 2a Vq, .(215) 
where Q is the conjugate of the stress vector, F the translational force, and Q, U, 
and T the rate of waste and the stored energies, whatever they may be. 
Comparing with the preceding equation (214), we see that we require 
2av^ = (Q-EC - HK) + 
+ 
+ [E.DV.q - (ED - U) div q] + [H.BV.q - (HB - T) divq] 
(216) 
Now assume that there is no waste of energy except by conduction ; then 
Also assume that 
Q = EC + HK. 
(217a) 
0U ?D 0T 0B 
-= E — - - — H — 
dt dt * dt df 
(21 7b) 
These imply that the relation between E and D is, for the same particle of matter, 
an invariable one, and that the stored electric energy is 
U = | D E,/D.(218) 
J 0 
where E is a function of D. Similarly, 
T = j B H dB .(219) 
expresses the stored magnetic energy, and H must be a definite function of B. 
On these assumptions, (216) reduces to 
2aV 3 = [E.DV.q - (ED — U) dir qj + [H.BV.q — (HB — T) div q j, 
from which the stress-vector follows, namely, 
Or, 
P N = [E.DN - N (ED - U)] + [H.BN - N (HB - T)]. 
P N = (VDVEN + NU) + (VBVHN + NT). . . . 
( 220 ) 
( 221 ) 
( 222 ) 
Thus, in case of isotropy, the stress is a tension U parallel to E combined with 
a lateral pressure (ED — U); and a tension T parallel to H combined with a lateral 
pressure (HB — T). 
The corresponding translational force is 
F = E div D + DV.E - V (ED - U) 
+ H div B + BV.H - V (HB - T), 
which it is unnecessary to put in terms of the currents. 
(223) 
