QipshaqUli
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- TL;DR
- Under the coordinate transformation $\bar x=x+\varepsilon$, the variation of the metric $g^{\mu\nu}$ is:
$$
\delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ g^{\mu\beta}\frac{\partial \varepsilon^{\nu}}{\partial x^{\beta}}+g^{\alpha\nu}\frac{\partial \varepsilon^{\mu}}{\partial x^{\alpha}}
$$
the right hand side is equal to $$- {g^{\mu\nu}}_{,\alpha}\varepsilon^{\alpha}+ {\varepsilon^{\mu,\nu}}+{\varepsilon^{\nu,\mu}}
Under the coordinate transformation $\bar x=x+\varepsilon$, the variation of the metric $g^{\mu\nu}$ is:
$$
\delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ g^{\mu\beta}\frac{\partial \varepsilon^{\nu}}{\partial x^{\beta}}+g^{\alpha\nu}\frac{\partial \varepsilon^{\mu}}{\partial x^{\alpha}}
$$
the right hand side is equal to $$- {g^{\mu\nu}}_{,\alpha}\varepsilon^{\alpha}+ {\varepsilon^{\mu,\nu}}+{\varepsilon^{\nu,\mu}}=\varepsilon^{\mu;\nu}+\varepsilon^{\nu;\mu}$$
I have problem with the proof of the last equality.
$$
\varepsilon^{\mu;\nu}+\varepsilon^{\nu;\mu}=g^{\alpha\nu}{\varepsilon^{\mu}}_{;\alpha}+g^{\alpha\mu}{\varepsilon^{\nu}}_{;\alpha}=
$$
$$
g^{\alpha\nu}({\varepsilon^{\mu}}_{,\alpha}+\Gamma_{\beta\alpha}^{\mu}\varepsilon^{\beta})+g^{\alpha\mu}({\varepsilon^{\nu}}_{,\alpha}+\Gamma_{\beta\alpha}^{\nu}\varepsilon^{\beta})=
$$
$$
\varepsilon^{\mu,\nu}+g^{\alpha\nu}\frac{1}{2}g^{\mu\gamma}(g_{\gamma\beta,\alpha}+g_{\gamma\alpha,\beta}-g_{\beta\alpha,\gamma})\varepsilon^{\beta}+
\varepsilon^{\nu,\mu}+g^{\alpha\mu}\frac{1}{2}g^{\nu\gamma}(g_{\gamma\beta,\alpha}+g_{\gamma\alpha,\beta}-g_{\beta\alpha,\gamma})\varepsilon^{\beta}=
$$
Considering the summation over the repeated indeces each of the three items in both brackets gives the same quantity coupling with the respective indeces as: A(B+C-D)E, ABE=ACE=ADE, then A(B+C-D)E=ACE. I chose ACE
$$
\varepsilon^{\mu,\nu}+\varepsilon^{\nu,\mu}+g^{\alpha\mu}g^{\nu\gamma}g_{\gamma\alpha,\beta}\varepsilon^{\beta}={g^{\mu\nu}}_{,\beta}\varepsilon^{\beta}+{\varepsilon^{\mu}}^{,\nu}+{\varepsilon^{\nu}}^{,\mu}
$$
I have the first term with plus sign, opposite to the original one. What I did wrong?
$$
\delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ g^{\mu\beta}\frac{\partial \varepsilon^{\nu}}{\partial x^{\beta}}+g^{\alpha\nu}\frac{\partial \varepsilon^{\mu}}{\partial x^{\alpha}}
$$
the right hand side is equal to $$- {g^{\mu\nu}}_{,\alpha}\varepsilon^{\alpha}+ {\varepsilon^{\mu,\nu}}+{\varepsilon^{\nu,\mu}}=\varepsilon^{\mu;\nu}+\varepsilon^{\nu;\mu}$$
I have problem with the proof of the last equality.
$$
\varepsilon^{\mu;\nu}+\varepsilon^{\nu;\mu}=g^{\alpha\nu}{\varepsilon^{\mu}}_{;\alpha}+g^{\alpha\mu}{\varepsilon^{\nu}}_{;\alpha}=
$$
$$
g^{\alpha\nu}({\varepsilon^{\mu}}_{,\alpha}+\Gamma_{\beta\alpha}^{\mu}\varepsilon^{\beta})+g^{\alpha\mu}({\varepsilon^{\nu}}_{,\alpha}+\Gamma_{\beta\alpha}^{\nu}\varepsilon^{\beta})=
$$
$$
\varepsilon^{\mu,\nu}+g^{\alpha\nu}\frac{1}{2}g^{\mu\gamma}(g_{\gamma\beta,\alpha}+g_{\gamma\alpha,\beta}-g_{\beta\alpha,\gamma})\varepsilon^{\beta}+
\varepsilon^{\nu,\mu}+g^{\alpha\mu}\frac{1}{2}g^{\nu\gamma}(g_{\gamma\beta,\alpha}+g_{\gamma\alpha,\beta}-g_{\beta\alpha,\gamma})\varepsilon^{\beta}=
$$
Considering the summation over the repeated indeces each of the three items in both brackets gives the same quantity coupling with the respective indeces as: A(B+C-D)E, ABE=ACE=ADE, then A(B+C-D)E=ACE. I chose ACE
$$
\varepsilon^{\mu,\nu}+\varepsilon^{\nu,\mu}+g^{\alpha\mu}g^{\nu\gamma}g_{\gamma\alpha,\beta}\varepsilon^{\beta}={g^{\mu\nu}}_{,\beta}\varepsilon^{\beta}+{\varepsilon^{\mu}}^{,\nu}+{\varepsilon^{\nu}}^{,\mu}
$$
I have the first term with plus sign, opposite to the original one. What I did wrong?