10 Dr. D. F. Comstock on the 
the longitudinal energy of the system, and 
w-w L =w T 
the transverse energy of the system, we may rewrite equation 
(14) as 
Wi W=V 2 M-v 1 W + 2?; 1 Wl+Vi 2 M, . . (15) 
and hence 
M = 2 ^>i (16) 
V 
■(-{?}) 
Mass = 
T/iZ5 £/u*£S £7*e fotoZ momentum of any isolated, nwmng, 
purely electrical system, which has on the average the same 
internal structure, in terms of its transverse energy, i. e., the 
energy represented by the components of the electric and 
magnetic forces which are perpendicular to the velocity of 
the system. The mass of the system is then 
<™-A W T ( /rA 2 1 -) , 
dW 
T 
HW 
If, as we have assumed in deriving this expression, the 
system possesses the same momentum for uniform translation 
in any direction, this formula for the mass can contain terms 
of even powers only in the ratio of the velocity of the system 
to the velocity of light. If we neglect terms of the second 
and higher orders W T has the same value as for ^ = 0, which 
from symmetry of the system must be two-thirds the total 
energy W. Therefore 
Mass = |^, (18) 
if second order terms be neglected. This formula would 
apply with extreme accuracy for the electromagnetic mass 
of ponderable bodies, for no such bodies have in nature a 
velocity large enough to make l^\ appreciable. 
It should be noticed that in equation (17) second order 
terms may enter in either (W T ) or its derivative with respect 
to i?i 2 . In fact such terms do enter for two reasons. In the 
first place, the setting of the body in motion requires work 
and hence adds new energy, through a second order term; 
and secondly there is an effect due to the change which 
