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

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