Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/417

Rh Let $$C$$ be the specific conductivity of the medium, $$K$$ its specific capacity for electrostatic induction, and $$\mu$$ its magnetic permeability.

To obtain the general equations of electromagnetic disturbance, we shall express the true current $$\mathfrak{C}$$ in terms of the vector potential $$\mathfrak{A}$$ and the electric potential $$\Psi$$.

The true current $$\mathfrak{C}$$ is made up of the conduction current $$\mathfrak{i}$$ and the variation of the electric displacement $$\dot{\mathfrak{D}}$$, and since both of these depend on the electromotive force $$\mathfrak{E}$$, we find, as in Art. 611,

But since there is no motion of the medium, we may express the electromotive force, as in Art. 599,

But we may determine a relation between $$\mathfrak{C}$$ and $$\mathfrak{A}$$ in a different way, as is shewn in Art. 616, the equations (4) of which may be written

Combining equations (3) and (4), we obtain

which we may express in the form of three equations as follows―

{{numb form | $$ \left. \begin{align} \mu \left(4 \pi C + K \frac{d}{dt} \right) \left( \frac{dF}{dt} +  \frac{d\Psi}{dx} \right) + \nabla^2 F + \frac{dJ}{dx} &= 0, \\ \mu \left(4 \pi C + K \frac{d}{dt} \right) \left( \frac{dG}{dt} +  \frac{d\Psi}{dy} \right) + \nabla^2 G + \frac{dJ}{dy} &= 0, \\ \mu \left(4 \pi C + K \frac{d}{dt} \right) \left( \frac{dH}{dt} +  \frac{d\Psi}{dz} \right) + \nabla^2 H + \frac{dJ}{dz} &= 0. \end{align} \right\} $$|(7)}}

These are the general equations of electromagnetic disturbances.

If we differentiate these equations with respect to $$x$$, $$y$$ and $$z$$ respectively, and add, we obtain

If the medium is a non-conductor, $$C=0$$, and $$\nabla^2 \Psi$$, which is proportional to the volume-density of free electricity, is independent of $$t$$. Hence $$J$$ must be a linear function of $$t$$, or a constant, or zero, and we may therefore leave $$J$$ and $$\Psi$$ out of account in considering periodic disturbances.