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

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

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

$(\delta _{1}Q)=\delta _{1}Q$

1. 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

   $\delta _{1}Q'\ dS'=\delta _{1}Q\ dS$