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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.[1]

We have therefore

\frac{\delta_{1}Q'}{\sqrt{-g'}}=\frac{\delta_{1}Q}{\sqrt{-g}} (46)

§ 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))

\delta_{1}Q=\sum(ab)M_{ab}d\mathfrak{g}^{ab} (47)

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}}

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

  1. \int(\delta Q)dS=\int\delta Q\ dS

    we infer

    \int(\delta_{1}Q)dS=\int\delta_{1}Q\ dS

    As this must hold for every choice of the variations \delta g_{ab} (by which choice the variations \delta\mathfrak{g}_{ab} are determined too) we must have at each point of the field-figure


  2. This may be made clear by a reasoning similar to that used in the preceding note. We again suppose \delta g_{ab} and \delta g_{ab,e} to be zero at the boundary of the domain of integration. Then \delta g'_{ab} and \delta g'_{ab,e} vanish too at the boundary, so that

    \int\delta_{2}Q'\ dS'=0,\ \int\delta_{2}Q\ dS=0


    \int\delta Q'\ dS'=\int\delta Q\ dS

    we may therefore conclude that

    \int\delta_{1}Q'\ dS'=\int\delta_{1}Q\ dS

    As this must hold for arbitrarily chosen variations \delta g_{ab} we have the equation

    \delta_{1}Q'\ dS'=\delta_{1}Q\ dS