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A problem related to metric variation

Physics Asked on December 3, 2021

Under the coordinate transformation $bar x=x+varepsilon$, the variation of the metric $g^{munu}$ is:
$$
delta g^{munu}(x)=bar g^{munu}(x)-g^{munu}(x)=-frac{partial{ g^{munu}}}{partial x^{alpha}}varepsilon^{alpha}+ g^{mubeta}frac{partial varepsilon^{nu}}{partial x^{beta}}+g^{alphanu}frac{partial varepsilon^{mu}}{partial x^{alpha}}
$$

the right hand side is equal to $$- {g^{munu}}_{,alpha}varepsilon^{alpha}+ {varepsilon^{mu,nu}}+{varepsilon^{nu,mu}}=varepsilon^{mu;nu}+varepsilon^{nu;mu}$$
I have problem with the proof of the last equality.
$$
varepsilon^{mu;nu}+varepsilon^{nu;mu}=g^{alphanu}{varepsilon^{mu}}_{;alpha}+g^{alphamu}{varepsilon^{nu}}_{;alpha}=
$$

$$
g^{alphanu}({varepsilon^{mu}}_{,alpha}+Gamma_{betaalpha}^{mu}varepsilon^{beta})+g^{alphamu}({varepsilon^{nu}}_{,alpha}+Gamma_{betaalpha}^{nu}varepsilon^{beta})=
$$

$$
varepsilon^{mu,nu}+g^{alphanu}frac{1}{2}g^{mugamma}(g_{gammabeta,alpha}+g_{gammaalpha,beta}-g_{betaalpha,gamma})varepsilon^{beta}+
varepsilon^{nu,mu}+g^{alphamu}frac{1}{2}g^{nugamma}(g_{gammabeta,alpha}+g_{gammaalpha,beta}-g_{betaalpha,gamma})varepsilon^{beta}=
$$

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
$$
varepsilon^{mu,nu}+varepsilon^{nu,mu}+g^{alphamu}g^{nugamma}g_{gammaalpha,beta}varepsilon^{beta}={g^{munu}}_{,beta}varepsilon^{beta}+{varepsilon^{mu}}^{,nu}+{varepsilon^{nu}}^{,mu}
$$

I have the first term with plus sign, opposite to the original one. What I did wrong? What am I missing?

One Answer

I have done such calculation, I think it will be helpful:

Answered by Nikita on December 3, 2021

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