Page:LorentzGravitation1916.djvu/57

${\mathfrak {t}}_{4}^{'4}={\frac {1}{2\varkappa }}\left\{-Q+\sum (abfe){\frac {\partial }{\partial x_{e}}}\left({\frac {\partial Q}{\partial g_{ab,fe}}}g_{ab,f}\right)\right\}$

(113)

where we must give the values 1, 2, 3 to $e$ and $f$.

The gravitation energy lying within a closed surface consists therefore of two parts, the first of which is

$E_{1}=-{\frac {1}{2\varkappa }}\int Q\ dx_{1}dx_{2}dx_{3}$

(114)

while the second can be represented by surface integrals. If namely $q_{1},q_{2},q_{3}$ are the direction constants of the normal drawn outward

$E_{2}={\frac {1}{2\varkappa }}\sum (abfe){\frac {\partial Q}{\partial g_{ab,fe}}}g_{ab,f}q_{e}d\sigma$

(115)

In the case of the infinitely feeble gravitation field represented by $\lambda ,\mu ,\nu$ (§ 57) both expressions $E_{1}$ and $E_{2}$ contain quantities of the first order, but it can easily be verified that these cancel each other in the sum, so that, as we knew already, the total energy is of the second order.

From $Q={\sqrt {-g}}G$ and the equations of § 32 we find namely

${\frac {\partial Q}{\partial g_{ab,fe}}}={\frac {1}{2}}{\sqrt {-g}}\left(2g^{ab}g^{fe}-g^{bf}g^{ae}-g^{af}g^{be}\right)$

(116)

so that we can write

$E_{2}={\frac {1}{4\varkappa }}\int {\sqrt {-g}}\sum (abfe)\left(2g^{ab}g^{fe}-g^{bf}g^{ae}-g^{af}g^{be}\right)g_{ab,f}q_{e}d\sigma$

The factor $g_{ab,f}$ is of the first order. Thus, if we confine ourselves to that order, we may take for all the other quantities these normal values. Many of these are zero and we find

$E_{2}=-{\frac {c}{2\varkappa }}\sum (ae)\int g^{aa}\left(g_{aa,e}-g_{ae,a}\right)q_{e}d\sigma$

(117)

Here we must take $a=1,2,3,4$; $e=1,2,3$, while we remark that for $a=e$ the expression between brackets vanishes. For $a=4$ the integral becomes $\int {\frac {\partial \nu }{\partial x_{e}}}q_{e}d\sigma$ do, which after summation with respect to $e$ gives

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

(118)

$n$ representing the normal to the surface. If $a$ and $e$ differ from each other, while neither of them is equal to 4, we can deduce from (110) and (109)

$g_{aa,e}-g_{ae,a}={\frac {\partial \nu }{\partial x_{e}}}$