Page:ComstockInertia.djvu/10

 the longitudinal energy of the system, and

$W-W_{L}=W_{T}\,$

the transverse energy of the system, we may rewrite equation (14) as

and hence

This gives the total momentum of any isolated, 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

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 $$v_{1}=0$$, which from symmetry of the system must be two-thirds the total energy $$W$$. Therefore

if second order terms he 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 $$\left(\tfrac{v_{1}}{V}\right)^{2}$$ appreciable.

It should be noticed that in equation (17) second order terms may enter in either ($$W_{T}$$) or its derivative with respect to $$v_{1}^2$$. 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