# Variation of the metric tensor

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## Summary:

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 $$- {g^{\mu\nu}}_{,\alpha}\varepsilon^{\alpha}+ {\varepsilon^{\mu,\nu}}+{\varepsilon^{\nu,\mu}} ## Main Question or Discussion Point Under the coordinate transformation \bar x=x+\varepsilon, the variation of the metric g^{\mu\nu} is:$$
\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}
$$I have the first term with plus sign, opposite to the original one. What I did wrong? ## Answers and Replies Related Special and General Relativity News on Phys.org Orodruin Staff Emeritus Science Advisor Homework Helper Gold Member You do not need to use the expression for the Christoffel symbols. All that is needed is the metric compatibility of the connection$$
\nabla_\mu g^{\nu\rho} = \partial_\mu g^{\nu\rho} + \Gamma^\nu_{\mu\sigma} g^{\sigma\rho} + \Gamma^\rho_{\mu\sigma} g^{\nu\sigma} = 0.