# Page:LorentzGravitation1916.djvu/41

$\left(\frac{\partial\mathrm{L}}{\partial g^{ab}}\right)_{x}=\frac{1}{2}\sqrt{-g}T_{ab}$

in the differentiation on the left hand side the coordinates of the material points are kept constant. To show that $T_{ab}$ and $\mathfrak{T}_{c}^{b}$ satisfy equation (66) we must now show that

$-\mathrm{L}-\sqrt{-g}V_{c}^{c}=2\sum(a)g^{ac}\left(\frac{\partial L}{\partial g^{ac}}\right)_{x}$

and for $b\ne c$

$-\sqrt{-g}V_{c}^{b}=2\sum(a)g^{ab}\left(\frac{\partial\mathrm{L}}{\partial g^{ac}}\right)_{x}$

If here the value (72) is substituted for $\mathrm{L}$ and if (70) is taken into account, these equations say that for all values of $b$ and $c$ we must have

 $2\sum(a)g^{ab}\left(\frac{\partial H}{\partial g^{ac}}\right)_{x}+V_{c}^{b}=0$ (76)

Now this relation immediately follows from a condition, to which $\mathrm{L}$ must be subjected at any rate, viz. that $\mathrm{L}dS$ is a scalar quantity. This involves that in a definite case we must find for $H$ always the same value whatever be the choice of coordinates.

§ 45. Let us suppose that instead of only one coordinate $x_{c}$ a new one $x'_{c}$ has been introduced, which differs infinitely little from $x_{c}$, with the restriction that if

$x'_{c}=x_{c}+\xi_{c}$

the term $\xi_{c}$ depends on the coordinate $x_b$ only and is zero at the point in question of the field-figure. The quantities $g^{ab}$ then take other values and in the new system of coordinates the world-lines of the material points will have a slightly changed course.

By each of these circumstances separately $H$ would change, but all together must leave it unaltered. As to the first change we remark that, according to the transformation formula for $g^{ab}$, the variation $\delta g^{ab}$ vanishes when the two indices are different from $c$, while

$\delta g^{cc}=2g^{cb}\frac{\partial\xi_{c}}{\partial x_{b}}$

and for $a\ne c$

$\delta g^{ac}=2g^{ca}=g^{ab}\frac{\partial\xi_{c}}{\partial x_{b}}$

The change of $H$ due to these variations is

$2\frac{\partial\xi_{c}}{\partial x_{b}}\sum(a)g^{ab}\left(\frac{\partial H}{\partial g^{ac}}\right)_{x}$