448 
MR, 0. HEAVISIDE ON THE FORCES, STRESSES, AND 
expresses a translational activity. Using them all in (108) it becomes 
e 0 J + h 0 G = (Q + U + T) + q(E/> + H* — VU C - VT’ + VJB + VDG) 
+ div VEjHj + g (U c + T m ).(114) 
It is clear that we should make the factor of q be the complete translational force. 
But that has to be found; and it is equally clear that, although we appear to have 
exhausted all the terms at disposal, the factor of q in (114) is not the complete force, 
because there is no term by which the force on intrinsically magnetised or electrized 
matter could be exhibited. These involve e 0 and h 0 . But as we have 
q (Vj„B + VDg 0 ) = - (ej 0 + hg 0 ),.. . (115) 
a possible way of bringing them in is to add the left member and subtract the right 
member of (115) from the right member of (114) ; bringing the translational force to 
f, say, where 
f=E /> + H<r-VtJ e -VT + V(J+j 0 )B +V(G + glJ )D.(116) 
But there is still the right number of (115) to be accounted for. We have 
- div (Veh 0 + Ve 0 h) = ej 0 f hg 0 4- e 0 j 4 h 0 g,.(117) 
and, by using this in (114), through (115), (116), (117), we bring it to 
A 
e 0 J + h 0 G = (Q + U+ T) + fq - (e 0 j + h 0 g) + div (VE,H, - Veh 0 - Ve 0 h) + ^ (U c + U) ; (118) 
or, transferring the e 0 , h 0 terms from the right to the left side, 
e 0 J 0 + h ( |G 0 = Q + U + T + fq + div (YEjHj — Veh 0 — Ve 0 h) 4- ^ (U<- +T M ) . . (119) 
Here we see that we have a correct form of activity equation, though it may not be 
the correct lorm. Another form, equally probable, is to be obtained by bringing in 
Veil; thus 
di7 Veil = h curl e — e curl li = — (ej + hg) = q (VjB 4 VDg), .... (120) 
which converts (119) to 
e 0 J 0 + hoG 0 = Q + U + T 4 Fq + div (VE^, - Veh - Veh, - Ve 0 h) + | (U c 4- T*) . (121) 
where F is the translational force 
F = E/i + H<t - VU f - VT U + V curl H. B + V curl E.D, . . . . (122) 
