# Page:PoyntingTransfer.djvu/17

359
OF ENERGY IN THE ELECTROMAGNETIC FIELD.

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

Therefore

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

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

Therefore

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

Substituting in (2)

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

whence

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

the function of the time being zero, since ${\displaystyle {\mathfrak {H}}}$ and ${\displaystyle {\mathfrak {E}}}$ 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

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

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

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

or

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

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

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

or

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

whence

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