Page:ComstockInertia.djvu/7

 except that we here have to deal with, besides the forces in the constraint, only the electromagnetic force on electricity embedded in the constraint, and we have nothing to do with hypothetical stresses in the free aether.

If ($$\rho$$) represents the electric density and ($$\Im_{x}$$) the $$x$$-component of the total electromagnetic force on unit charge embedded in the constraint, we have

{{MathForm2|(6)|$$\left.\begin{array}{l} \frac{\partial X_{x}}{\partial x}+\frac{\partial X_{y}}{\partial y}+\frac{\partial X_{z}}{\partial z}=-\rho\Im_{x}\\ \\\quad=-\rho X-\frac{\rho}{V}(v_{y}\gamma-v_{z}\beta)\\ \\\quad=-\rho X-(k_{y}\gamma-k_{z}\beta)\end{array}\right\} $$,}}

where ($$k$$) is the density of convection current caused by the movement with velocity ($$v$$) of the electricity of density ($$\rho$$).

Making use temporarily of the vector terminology for the sake of brevity and calling the electric force ($$E$$), the magnetic ($$H$$), and the sign [ ] denoting the vector product, we have

$-\rho\Im_{x}=-\rho E_{x}-[kH]$.

Since div $$E = 4\pi\rho$$ and $$curl\ H-\tfrac{1}{V}\tfrac{\partial E}{\partial t}=4\pi k$$,

Now it is an easily verifiable identity that

and hence, remembering that div $$H = 0$$, equation (7) becomes