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