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]}
 ↑ 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