# A problem related to metric variation

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? What am I missing?

• Please see Lie derivative - any standard treatment will show the partial derivatives can be everywhere be replaced by covariant derivatives because the (Levi-Civita) connections cancel – lux Jan 31 at 3:58
• On my phone so I can’t write out equations, but (derivative of inverse metric is equal to minus the raised index version of the derivative of metric. – Prahar Mar 22 at 2:30

• I started with law indices due to canonical defenition of metric:$$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$$ Of course, you can repeat all my calculations in case with upper indices. All calculations will be the same. If you wanna, I can do this calculation for you. – Nikita Feb 14 at 19:25