# Page:LorentzGravitation1916.djvu/33

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

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

We have therefore

 ${\displaystyle {\frac {\delta _{1}Q'}{\sqrt {-g'}}}={\frac {\delta _{1}Q}{\sqrt {-g}}}}$ (46)

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

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

if we put

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

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

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

1. ${\displaystyle \int (\delta Q)dS=\int \delta Q\ dS}$

we infer

${\displaystyle \int (\delta _{1}Q)dS=\int \delta _{1}Q\ dS}$

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

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

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

${\displaystyle \int \delta _{2}Q'\ dS'=0,\ \int \delta _{2}Q\ dS=0}$

From

${\displaystyle \int \delta Q'\ dS'=\int \delta Q\ dS}$

we may therefore conclude that

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

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

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