Page:PoyntingTransfer.djvu/17

Rh from some fixed plane by $$z$$, the line integral of the M.I. is $$-\tfrac{d\mathfrak{H}}{dz}$$, while the current, being an alteration of displacement, is $$\tfrac{K}{4\pi}\tfrac{d\mathfrak{E}}{dt}$$

Therefore

But since the displacement is propagated unchanged with velocity $$v$$, the displacement now at a given point will alter in time $$dt$$ to the displacement now a distance $$dz$$ behind, where $$dz=vdt$$.

Therefore

Substituting in (2)

$\frac{d\mathfrak{H}}{dz}=Kv\frac{d\mathfrak{E}}{dz}$|undefined

whence

the function of the time being zero, since $$\mathfrak{H}$$ and $$\mathfrak{E}$$ are zero together in the parts which the wave has not yet reached.

If we take the line-integral of the E.M.I. round a face perpendicular to the M.I. and equate this to the decrease of magnetic induction through the face, we obtain similarly

It may be noticed that the product of (4) and (5) at once gives the value of $$v$$, for dividing out $$\mathfrak{E}\mathfrak{H}$$ we obtain

$1=\mu Kv^{2}$

or

$v=\frac{1}{\sqrt{\mu K}}$|undefined

But using one of these equations alone, say (4), and substituting in (1) K for $$\mathfrak{H}$$ and dividing by $$\mathfrak{E}^{2}$$, we have

$\frac{K}{4\pi}=\frac{K}{8\pi}+\frac{\mu K^{2}v^{2}}{8\pi}$|undefined

or

$1=\mu Kv^{2}\,$

whence

$v=\frac{1}{\sqrt{\mu K}}$|undefined