Page:PoyntingTransfer.djvu/17

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

 $-\frac{d\mathfrak{H}}{dz}=K\frac{d\mathfrak{E}}{dt}$ (2)

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

 $\frac{d\mathfrak{E}}{dt}=-v\frac{d\mathfrak{E}}{dz}$ (3)

Substituting in (2)

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

whence

 $\mathfrak{H}=Kv\mathfrak{E},$ (4)

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

 $\mathfrak{E}=\mu v\mathfrak{H}.$ (5)

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

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

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

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

or

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

whence

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