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Under the coordinate transformation $\bar x=x+\varepsilon$, the metric in new coordinates is: $$ \bar g^{\mu\nu}(\bar x)=g^{\alpha\beta}(x)\frac{\partial \bar x^{\mu}}{\partial x^{\alpha}}\frac{\partial \bar x^{\nu}}{\partial x^{\beta}}=g^{\mu\nu}(x)+g^{\mu\beta}\frac{\partial \varepsilon^{\nu}}{\partial x^{\beta}}+g^{\alpha\nu}\frac{\partial \varepsilon^{\mu}}{\partial x^{\alpha}} $$

By expanding $\bar g^{\mu\nu}(\bar x)$ to the first order of $\varepsilon$ $$ \bar g^{\mu\nu}(\bar x)=\bar g^{\mu\nu}(x+\varepsilon)=\bar g^{\mu\nu}(x)+\frac{\partial{\bar g^{\mu\nu}}}{\partial\varepsilon^{\alpha}}\varepsilon^{\alpha} $$ The variation of the metric under such a transformation is: $$ \delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ g^{\mu\beta}\frac{\partial \varepsilon^{\nu}}{\partial x^{\beta}}+g^{\alpha\nu}\frac{\partial \varepsilon^{\mu}}{\partial x^{\alpha}} $$

My question is

Why is $\dfrac{\partial{\bar g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}$ considered to be equal to $\dfrac{\partial{ g^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}$ ?

I rewrite it in detail: $$ \bar g^{\mu\nu}(\bar x)=\bar g^{\mu\nu}(x+\varepsilon)=\bar g^{\mu\nu}(x+\varepsilon)|_{\varepsilon=0}+\frac{\partial{\bar g^{\mu\nu}(x+\varepsilon)}}{\partial x^{\beta}}\frac{\partial x^{\beta}}{\partial \bar x^{\alpha}}|_{\varepsilon=0}\varepsilon^{\alpha}= $$ $$\bar g^{\mu\nu}(x)+\frac{\partial{\bar g^{\mu\nu}(x)}}{\partial x^{\alpha}}\varepsilon^{\alpha} $$ And why is $$\frac{\partial{\bar g^{\mu\nu}(x)}}{\partial x^{\alpha}}\varepsilon^{\alpha}=\frac{\partial{g^{\mu\nu}(x)}}{\partial x^{\alpha}}\varepsilon^{\alpha}?$$

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  • $\begingroup$ why in the first equation are not the partial derivatives of epsilon multiplied by metric? $\endgroup$
    – Umaxo
    Jan 30, 2020 at 8:33
  • $\begingroup$ How you got your first equation? $\endgroup$
    – Eli
    Jan 30, 2020 at 8:35
  • $\begingroup$ I corrected it. $\endgroup$
    – Constantin
    Jan 30, 2020 at 8:36

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This is because you work in first order in $\varepsilon$. So you need to consider only the linear terms in $\varepsilon$:

$$ \delta g^{\mu\nu}(x)=\bar g^{\mu\nu}(x)-g^{\mu\nu}(x)=-\frac{\partial{ \bar{g}^{\mu\nu}}}{\partial x^{\alpha}}\varepsilon^{\alpha}+ \frac{\partial \varepsilon^{\mu}}{\partial x_{\nu}}+\frac{\partial \varepsilon^{\nu}}{\partial x_{\mu}} $$

So $g^{\mu\nu}$ and $\bar{g}^{\mu\nu}$ differ only in first order in $\varepsilon$.

And because we work in linear order, we have

$$ \frac{\partial\bar{g}_{\mu\nu}(x)}{\partial x^\alpha} \varepsilon^\alpha= \frac{\partial g_{\mu\nu}(x)}{\partial x^\alpha}\varepsilon^\alpha + O(\varepsilon) $$

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  • $\begingroup$ $\bar g^{\mu\nu}$ is different from $g^{\mu\nu}$. So are their derivatives wrt. $\varepsilon$. They are inconsistent. $\endgroup$
    – Constantin
    Jan 30, 2020 at 8:45
  • $\begingroup$ It looks like in general $$\frac{\partial \bar F(x)}{\partial x^{\alpha}}=\frac{\partial F(x)}{\partial x^{\alpha}}$$ Why they must be equal? $\bar F(x)$ and $F(x)$ are different. $\endgroup$
    – Constantin
    Jan 30, 2020 at 9:19
  • $\begingroup$ I update answer. $\endgroup$
    – Nikita
    Jan 30, 2020 at 9:47
  • $\begingroup$ Even more explicitly is it $$\bar g^{\mu\nu}=g^{\mu\nu}+C\varepsilon$$ $$\frac{\partial \bar g^{\mu\nu}}{\partial x^{\alpha}}|_{\varepsilon=0}=\left(\frac{\partial g^{\mu\nu}}{\partial x^{\alpha}}+\frac{\partial C\varepsilon}{\partial x^{\alpha}}\right)|_{\varepsilon=0}?$$ $\endgroup$
    – Constantin
    Jan 30, 2020 at 12:43

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