FLUXES OF ENERGY IN THE ELECTROMAGNETIC FIELD. 
449 
which is perfectly symmetrical as regards E and H, and in the vector products 
utilises the fluxes and their complete forces, whereas former forms did this only 
partially. Observe, too, that we have only been able to bring the activity equation 
to a correct form (either (119) or (122)) by making e 0 J 0 be the activity of intrinsic 
force e 0 , which requires that J 0 should be the true electric current, according to the 
energy criterion, not J. 
^ 18 . Now, to test (119) and (121), we must interpret the flux in (121), or say 
Y = VE 1 H 1 - Veil - Veh 0 - Ve (l h, 
(123) 
which has replaced the Poynting flux when q = 0, along with the other 
changes. Since Y reduces to VEjH, when q = 0, there must still be a Poynting 
flux when q is finite, though we do not know its precise form of expression. There 
is also the stress flux of energy and the flux of energy by convection, making a total 
flux 
X = W + q (U + T) - 20? + q (U 0 + T 0 ), .(124) 
where W is the Poynting flux, and — 2Q q that of the stress, whilst q (U 0 + T 0 ) 
means convection of energy connected with the translational force. We should 
therefore have 
e 0 J 0 + h<A> = (Q + U + T) + (Q 0 + U 0 + T 0 ) + div X .(125) 
to express the continuity of energy. More explicitly 
e 0 J 0 + h 0 G- 0 — Q + U + T + div [W + q (U + T)J 
+ Q 0 + U 0 + T 0 + div [- 202 + q (U 0 + T 0 )].(126) 
But here we may simplify by using the result (69) (with, however, f put = 0), making 
(126) become 
e 0 J 0 + h„G 0 = (Q + U + T) + Fq + Sa + div [w + q (U + T) - 2 ftj], . . . (127) 
where S is the torque, and a the spin. 
Comparing this with (121), we see that we require 
W + q (U + T) — 2 Ctg = VE 1 H 1 - Veh - Ve 0 h - Veh 0 , .(128) 
with a similar equation when (119) is used instead ; and we have now to separate the 
right member into two parts, one for the Poynting flux, the other for the stress flux, 
in such a way that the force due to the stress is the force F in (121), (122), or the 
force f in (119), (116); or similarly in other cases. It is unnecessary to give the 
failures; the only one that stands the test is (121), which satisfies it completely. 
MDCCCXCII.—A. 3 M 
