# Page:LorentzGravitation1916.djvu/32

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derivatives and we shall determine the variation it undergoes by arbitrarily chosen variations $\delta g_{ab}$, these latter being continuous functions of the coordinates. We have evidently

$\delta Q=\sum(ab)\frac{\partial Q}{\partial g_{ab}}\delta g_{ab}+\sum(abe)\frac{\partial Q}{\partial g_{ab,e}}\delta g_{ab,e}+\sum(abef)\frac{\delta Q}{\partial g_{ab,ef}}\delta g_{ab,ef}$

By means of the equations

$\delta g_{ab,ef}=\frac{\partial}{\partial x_{f}}\delta g_{ab,e}$ and $\delta g_{ab,e}=\frac{\partial}{\partial x_{e}}\delta g_{ab}$

this may be decomposed into two parts

 $dQ=\delta_{1}Q+\delta_{2}Q$ (42)

namely

 $\delta_{1}Q=\sum(ab)\left\{ \frac{\partial Q}{\partial g_{ab}}-\sum(e)\frac{\partial}{\partial x_{e}}\frac{\partial Q}{\partial g_{ab,e}}+\sum(ef)\frac{\partial^{2}}{\partial x_{e}\partial x_{f}}\frac{\partial Q}{\partial g_{ab,ef}}\right\} \delta g_{ab}$ (43)
 $\begin{array}{c} \delta_{2}Q=\sum(abe)\frac{\partial Q}{\partial x_{e}}\left(\frac{\partial Q}{\partial g_{ab,e}}\delta g_{ab}\right)+\sum(abef)\frac{\partial}{\partial x_{f}}\left(\frac{\partial Q}{\partial g_{ab,ef}}\delta g_{ab,e}\right)-\\ \\ -\sum(abef)\frac{\partial}{\partial x_{e}}\left\{ \frac{\partial}{\partial x_{f}}\left(\frac{\partial Q}{\partial g_{ab,ef}}\right)\delta g_{ab}\right\} \end{array}$ (44)

The last equation shows that

 $\int\delta_{2}QdS=0$ (45)

if the variations $\delta g_{ab}$ and their first derivatives vanish at the boundary of the domain of integration.

§ 35. Equations of the same form may also be found if $Q$ is expressed in one of the two other ways mentioned in § 33. If e.g. we work with the quantities $\mathfrak{g}^{ab}$ we shall find

$(\delta Q)=\left(\delta_{1}Q\right)+\left(\delta_{2}Q\right)$

where $\left(\delta_{1}Q\right)$ and $\left(\delta_{2}Q\right)$ are directly found from (43) and (44) by replacing $g_{ab}$, $g_{ab,e}$, $g_{ab,ef}$, $\delta g_{ab}$ and $\delta g_{ab,e}$ etc. by $\mathfrak{g}^{ab}$, $\mathfrak{g}^{ab,e}$ etc. If the variations chosen in the two cases correspond to each other we shall have of course

$(dQ)=\delta Q$

Moreover we can show that the equalities

$\left(\delta_{1}Q\right)=\delta_{1}Q,\ \left(\delta_{2}Q\right)=\delta_{2}Q$

exist separately.[1]

1. Suppose that at the boundary of the domain of integration $\delta g_{ab}=0$ and $\delta g_{ab,e}=0$. Then we have also $\delta\mathfrak{g}^{ab}=0$ and $\delta\mathfrak{g}^{ab,e}=0$, so that

$\int\left(\delta_{2}Q\right)dS=0,\ \int\delta_{2}QdS=0$

and from