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.

We have therefore

§ 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))

if we put

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

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