Page:CunninghamExtension.djvu/20

96 or

$$H'_{s}\delta S=\frac{\lambda}{\beta\left(1-\frac{vw_{r}}{c^{2}}\right)}h'_{s}\delta s.$$

Thus, since $$\delta S/\delta s$$ does not depend on the physical properties of the medium, and is consequently continuous in crossing the surface of discontinuity, H' will be continuous in those directions in which h' is continuous, and discontinuous in those directions in which h' is discontinuous.

Exactly the same is to be proved of $$E'=E-\frac{1}{c}[BW]$$ by means of (10'). Taking the component of D perpendicular to the same element, we have

$$\begin{array}{ll} D_{R}\delta N-D_{N}\delta R= & \lambda d_{r}\left\{ \delta_{n}+\frac{vw_{n}\delta_{r}}{c^{2}\left(1-\frac{vw_{r}}{c^{2}}\right)}\right\} \\ \\ & -\frac{\lambda\delta r}{1-\frac{vw_{r}}{c^{2}}}\left\{ d_{n}\left(1-\frac{vw_{r}}{c^{2}}\right)+\frac{v}{c^{2}}d_{r}w_{R}-\frac{v}{c}h'_{n_{1}}\right\} \end{array}$$

where $$h'_{n_{1}}$$ denotes the component of $$h'$$ in the direction perpendicular to r and n.

Thus, on simplifying,

$$D_{R}\delta N-D_{N}\delta R=\lambda\left(d_{r}\delta_{n}-d_{n}\delta_{r}\right)-\frac{\lambda v\delta r\ h'_{n_{1}}}{c\left(1-\frac{vw_{r}}{c^{2}}\right)}.$$

Thus D will be continuous in any direction in which d is continuous, provided the component of $$h'$$ perpendicular to that direction is continuous.

Hence, with what has just been proved as to the continuity of h', it follows that the condition of the continuity of the normal component of d is conserved ; and in the same manner it may be shewn of the normal component of b.

9. Let the physical quantities considered in § 3 be now treated in the light of the inversion in four-dimensional space.

If the elements ds, dS are corresponding elements of area of which dS is at rest in the (R, T) system, and (l, m, n), (L, M, N) are the direction cosines of the respective normals, by considering the projections of the