# Page:LorentzGravitation1916.djvu/33

The decomposition of $\delta Q$ into two parts is therefore the same, whether we use $g_{ab},g^{ab}$ or $\mathfrak{g}^{ab}$.

It is further of importance that when the system of coordinates is changed, not only $\delta QdS$ is an invariant, but that this is also the case with $\delta_{1}QdS$ and $\delta_{2}QdS$ separately.[1]

We have therefore

 $\frac{\delta_{1}Q'}{\sqrt{-g'}}=\frac{\delta_{1}Q}{\sqrt{-g}}$ (46)

§ 36. For the calculation of $\delta_{1}Q$ we shall suppose $Q$ to be expressed in the quantities $\mathfrak{g}^{ab}$ and their derivatives. Therefore (comp. (43))

 $\delta_{1}Q=\sum(ab)M_{ab}d\mathfrak{g}^{ab}$ (47)

if we put

$M_{ab}=\frac{\partial Q}{\partial\mathfrak{g}^{ab}}-\sum(e)\frac{\partial}{\partial x_{e}}\frac{\partial Q}{\partial\mathfrak{g}^{ab,e}}+\sum(ef)\frac{\partial}{\partial x_{e}\partial x_{f}}\frac{\partial Q}{\partial\mathfrak{g}^{ab,ef}}$

Now we can show that the quantities $M_{ab}$ are exactly the quantities $G_{ab}$ defined by (40). To this effect we may use the following considerations.

We know that $\left(\tfrac{1}{\sqrt{-g}}\mathfrak{g}^{ab}\right)$ is a contravariant tensor of the second

1. $\int(\delta Q)dS=\int\delta Q\ dS$

we infer

$\int(\delta_{1}Q)dS=\int\delta_{1}Q\ dS$

As this must hold for every choice of the variations $\delta g_{ab}$ (by which choice the variations $\delta\mathfrak{g}_{ab}$ are determined too) we must have at each point of the field-figure

$(\delta_{1}Q)=\delta_{1}Q$

2. This may be made clear by a reasoning similar to that used in the preceding note. We again suppose $\delta g_{ab}$ and $\delta g_{ab,e}$ to be zero at the boundary of the domain of integration. Then $\delta g'_{ab}$ and $\delta g'_{ab,e}$ vanish too at the boundary, so that

$\int\delta_{2}Q'\ dS'=0,\ \int\delta_{2}Q\ dS=0$

From

$\int\delta Q'\ dS'=\int\delta Q\ dS$

we may therefore conclude that

$\int\delta_{1}Q'\ dS'=\int\delta_{1}Q\ dS$

As this must hold for arbitrarily chosen variations $\delta g_{ab}$ we have the equation

$\delta_{1}Q'\ dS'=\delta_{1}Q\ dS$