Page:PoyntingTransfer.djvu/4

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But since f=\tfrac{K}{4\pi}P, with corresponding values for g and h, and a=\mu\alpha, b=\mu\beta, c=\mu\gamma, substituting, the energy becomes

\frac{K}{8\pi}\iiint\left(P^{2}+Q^{2}+R^{2}\right)dx\ dy\ dz+\frac{\mu}{8\pi}\iiint\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)dx\ dy\ dz (1)

Let us consider the space within any fixed closed surface. The energy within this surface will be found by taking the triple integrals throughout the space.

If any changes are taking place the rate of increase of energy of the electric and magnetic kinds per second is

\frac{K}{4\pi}\iiint\left(P\frac{dP}{dt}+Q\frac{dQ}{dt}+R\frac{dR}{dt}\right)dx\ dy\ dz+\frac{\mu}{4\pi}\iiint\left(\alpha\frac{d\alpha}{dt}+\beta\frac{d\beta}{dt}+\gamma\frac{d\gamma}{dt}\right)dx\ dy\ dz (2)

Now Maxwell's equations for the components of the true current are

u=p+\frac{df}{dt},\ v=q+\frac{dg}{dt},\ w=r+\frac{dh}{dt},

where p, q, r are components of the conduction-current.

But we may substitute for \tfrac{df}{dt} its value \tfrac{K}{4\pi}\tfrac{dP}{dt}, and so for the other two, and we obtain

\left.\begin{array}{c}
\frac{K}{4\pi}\frac{dP}{dt}=u-p\\
\\\frac{K}{4\pi}\frac{dQ}{dt}=v-q\\
\\\frac{K}{4\pi}\frac{dR}{dt}=w-r\end{array}\right\} (3)

Taking the first term in (2) and substituting from (3) we obtain

\begin{array}{c}
\frac{K}{4\pi}\iiint\left(P\frac{dP}{dt}+Q\frac{dQ}{dt}+R\frac{dR}{dt}\right)dx\ dy\ dz=\iiint\left\{ P(u-p)+Q(v-q)+R(w-r)\right\} dx\ dy\ dz\\
\\=\iiint(Pu+Qv+Rw)dx\ dy\ dz-\iiint(Pp+Qq+Rr)dx\ dy\ dz\end{array} (4)

Now the equations for the components of electromotive force are (Maxwell vol. ii, p. 222)