Page:LorentzGravitation1916.djvu/58

Each value of $e$ occurring twice, i.e. combined with the two values different from $e$ which $a$ can take, we have in addition to (118)

$-2\int {\frac {\partial \nu }{\partial n}}d\sigma$

so that (117) becomes

$E_{2}={\frac {c}{2\varkappa }}\int {\frac {\partial \nu }{\partial n}}d\sigma$

As now outside the sphere

$\nu =-{\frac {\varkappa }{r}}\int \limits _{0}^{a}\varrho \ dr$

we have for every closed surface that does not surround the sphere $E_{2}=0$, but for every surface that does

$E_{2}=2\pi c\int \limits _{0}^{a}\varrho \ dr$

(119)

As to $E_{1}$ we remark that substituting (65) in (41) and taking into consideration (64) we find,

$G=\varkappa T,\ Q=\varkappa {\sqrt {-g}}T$

(120)

From this we conclude that $E_{1}$ is zero if there is no matter inside the surface $\sigma$. In order to determine $E_{1}$ in the opposite case, we remember that $G$ is independent of the choice of coordinates. To calculate this quantity we may therefore use the value of $T$ indicated in § 56, which is sufficient to calculate $E_{1}$ as far as the terms of the first order. We have therefore

$G={\frac {\varkappa }{r^{2}}}\varrho$

and if, using further on rectangular coordinates, we take for ${\sqrt {-g}}$ the normal value $c$,

$Q={\frac {c\varkappa }{r^{2}}}\varrho$

From this we find by substitution in (114) for the case of the closed surface a surrounding the sphere

$E_{1}=-2\pi c\int \limits _{0}^{a}\varrho \ dr$

This equation together with (119) shows that in (113) when integrated over the whole space the terms of the first order really cancel each other. In order to calculate those of the second order