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