Page:PoyntingTransfer.djvu/7

Rh angle between its direction and that of $$\mathfrak{H}$$, the magnetic intensity, the direction cosines L, M, N of the line perpendicular to the plane containing $$\mathfrak{E'}$$ and $$\mathfrak{H}$$ are given by

$L=\frac{R'\beta-Q'\gamma}{\mathfrak{E'}\mathfrak{H}\sin\theta};\ M=\frac{P'\gamma-R'\alpha}{\mathfrak{E'}\mathfrak{H}\sin\theta};\ N=\frac{Q'\alpha-P'\beta}{\mathfrak{E'}\mathfrak{H}\sin\theta};$

so that the surface integral becomes

$\frac{1}{4\pi}\iint\mathfrak{E'}\mathfrak{H}\sin\theta(Ll+Mm+Nn)dS$

If at a given point $$dS$$ be drawn to coincide with the plane containing $$\mathfrak{E'}$$ and $$\mathfrak{H}$$, it then contributes the greatest amount of energy to the space; or in other words the energy flows perpendicularly to the plane containing $$\mathfrak{E'}$$ and $$\mathfrak{H}$$, the amount crossing unit area per second being $$\mathfrak{E'}\mathfrak{H}\sin\theta/4\pi$$. To determine in which way it crosses the plane take $$\mathfrak{E'}$$ along $$Oz$$, $$\mathfrak{H}$$ along $$Oy$$. Then

$\begin{array}{ccccccc} & & P'=0, & & Q'=0, &  & \frac{R'}{\mathfrak{E}'}=1,\\ \\ & & \alpha=0, &  & \frac{\beta}{\mathfrak{H}}=1, &  & \gamma=0,\\ \\\mathrm{and\ if}\ \sin\theta=1\\ & & L=1, &  & M=0, &  & N=0.\end{array}$|undefined

If now the axis $$Ox$$ be the normal to the surface outwards, $$l = 1$$, $$m = 0$$, $$n = 0$$, so that this element of the integral contributes a positive term to the energy within the surface on the negative side of the $$yz$$ plane; that is, the energy moves along $$xO$$, or in the direction in which a screw would move if its head were turned round from the positive direction of the electromotive to the positive direction of the magnetic intensity. If the surface be taken where the matter has no velocity, $$\mathfrak{E'}$$ becomes equal to $$\mathfrak{E}$$, and the amount of energy crossing unit area perpendicular to the flow per second is

$\frac{\mathrm{electromotive\ intensity\times magnetic\ intensity\times sine\ included\ angle}}{4\pi}$|undefined

Since the surface may be drawn anywhere we please, then wherever there is both magnetic and electromotive intensity there is flow of energy.

Since the energy flows perpendicularly to the plane containing the two intensities, it must flow along the electric and magnetic level surfaces, when these exist, so that the lines of flow are the intersections of the two surfaces.

We shall now consider the applications of this law in several cases.