# Variation of the metric under the coordinate transformation

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 $$\frac{\partial{\bar g^{\mu\nu}(x)}}{\partial x^{\alpha}}\varepsilon^{\alpha}=\frac{\partial{g^{\mu\nu}(x)}}{\partial x^{\alpha}}\varepsilon^{\alpha}?$$

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

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

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

• $\bar g^{\mu\nu}$ is different from $g^{\mu\nu}$. So are their derivatives wrt. $\varepsilon$. They are inconsistent. Jan 30, 2020 at 8:45
• 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. Jan 30, 2020 at 9:19
• I update answer. Jan 30, 2020 at 9:47
• Even more explicitly is it $$\bar g^{\mu\nu}=g^{\mu\nu}+C\varepsilon$$ $$\frac{\partial \bar g^{\mu\nu}}{\partial x^{\alpha}}|_{\varepsilon=0}=\left(\frac{\partial g^{\mu\nu}}{\partial x^{\alpha}}+\frac{\partial C\varepsilon}{\partial x^{\alpha}}\right)|_{\varepsilon=0}?$$ Jan 30, 2020 at 12:43