Page:PoyntingTransfer.djvu/17

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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}}