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To do functional derivative of some actions, we need to know a functional differential of metrics $g_{\mu \nu}(x)$. One of the formulae about that is: $$g_{\mu\nu}\delta g^{\mu\nu} = - g^{\mu\nu} \delta g_{\mu \nu},$$but this confused me, because for arbitrary tensors, it is true that$$A_{\mu\nu}B^{\mu\nu}=A^{\mu\nu}B_{\mu\nu}.$$

What is the difference?

I suppose $\delta g^{\mu\nu}$ is a tensor because actions must be scalars, but i don't have any proof.

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3 Answers 3

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The inverse metric tensor $(g^{-1})^{\mu\nu}$ is often written as $g^{\mu\nu}$ when no confusion can arise. Let us not use this shorthand notation here. OP's formula then reads

$$ \delta(g^{-1})^{\mu\nu}~=~-(g^{-1})^{\mu\lambda}\delta g_{\lambda\kappa}(g^{-1})^{\kappa\nu} .$$ We can now raise and lower indices with the metric/inverse metric, e.g. $$\delta(g^{-1})^{\mu\nu}~=~-(\delta g)^{\mu\nu},$$ without running into inconsistencies.

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  • $\begingroup$ Thank you for answering. Now I can see for the metric tensor, we can get $$\delta(g^{-1})^{\mu\nu}~=~-(\delta g)^{\mu\nu}$$ so if i substitute this for my formula $$ g_{\mu\nu}(\delta g)^{\mu\nu} = g^{\mu\nu} (\delta g)_{\mu \nu}$$ so there is no inconsistency. Thank you very much. $\endgroup$
    – Keyflux
    May 31, 2019 at 13:54
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Use $g^{\mu\nu}g_{\nu\lambda}=\delta^\mu_\nu$. Now for any matrix $G$ and its inverse $G^{-1}$ we have $$ 0= \delta(\mathbb I)= \delta (G^{-1} G)= (\delta G^{-1}) G + G^{-1}\delta G. $$ Multiply by $G^{-1}$ from the right to get $$ \delta G^{-1} = - G^{-1}\delta G G^{-1}. $$ Thus $$ \delta g^{\mu\nu}= - g^{\mu \rho} \delta g_{\rho\sigma} g^{\sigma \nu} $$

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  • $\begingroup$ Thank you for answering. The formula which I wrote has no living indices so this derivation is educational for me. Thank you, but how can I understand the difference between this and the formula for arbitrary tensors in my question? In other words, why $$g_{\mu\nu}\delta g^{\mu\nu} = +g^{\mu\nu} \delta g_{\mu \nu}$$ is not true while $$A_{\mu\nu}B^{\mu\nu}=A^{\mu\nu}B_{\mu\nu}$$ is true. Sorry to bother you again. $\endgroup$
    – Keyflux
    May 31, 2019 at 13:36
  • $\begingroup$ Sorry, now I understood. I didn't understand the difference between $\delta(g^{\mu\nu})$ and $(\delta g)^{\mu \nu}$. Thanks. $\endgroup$
    – Keyflux
    May 31, 2019 at 13:56
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$\delta g^{\mu \nu}$ is definitely a tensor, but you have to be careful when you have variations of a tensor. To be clear, it is $\delta (g^{\mu \nu})$, not $(\delta g)^{\mu \nu}$. Then you must express $\delta (g^{\mu \nu})$ in terms of $\delta (g_{\mu \nu})$ as given in your first equation. You do this by taking contractions with the metric. This is where the difference comes in. Precisely, we have

\begin{aligned} \delta (g^{\mu \nu}) &= \delta (g^{\alpha \mu} g^{\beta \nu} g_{\alpha \beta})\\ &= \delta (g^{\alpha \mu}) \bar{g}^{\beta \nu} \bar{g}_{\alpha \beta} + \bar{g}^{\alpha \mu} \delta(g^{\beta \nu}) \bar{g}_{\alpha \beta} + \bar{g}^{\alpha \mu} \bar{g}^{\beta \nu} \delta(g_{\alpha \beta}) \end{aligned}

where an overbar on the metric $\bar{g}_{\mu \nu}$ indicates that the metric takes its background value.

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  • $\begingroup$ Thank you for answering. Sorry for a basic question, what is the definition of metric's "background" value? I suppose $\bar{g}^{\mu\nu} = g^{\mu\nu}(x)$ where $x$ is a point in area which $g^{\mu\nu}$ is defined, is it right? $\endgroup$
    – Keyflux
    May 31, 2019 at 13:20
  • $\begingroup$ @Keyspire Think of perturbation theory. In going from first equality to the second, I used the Leibniz product rule and applied $\delta$ to only one of the three metrics inside the brackets each time. The other two are the values of $g_{\mu \nu}$ around which we are perturbing the metric. I chose this value of $g_{\mu \nu}$ to be $\bar{g}_{\mu \nu}$, that is, which satisfies background equations of motion. But anyway, this is not a rule and, in general, you can perturb around some given $g_{\mu \nu}$. $\endgroup$
    – Avantgarde
    May 31, 2019 at 13:57
  • $\begingroup$ Thank you, I understood now. Can we raise and lower indices with $\bar{g}$ still now? If so, we obtain $\delta (g^{\mu\nu})~=~-(\delta g)^{\mu\nu}$. I suppose it's true because the form of function $g$ varied but the properties under general coordinate transformations is still unchanged. $\endgroup$
    – Keyflux
    May 31, 2019 at 14:08
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    $\begingroup$ @Keyspire As written in my answer, you raise and lower indices with respect to the full metric, and then take the variation. If you define $\delta g_{\mu \nu} \equiv h_{\mu \nu}$, then $\delta g^{\mu \nu} = - h^{\mu \nu}$. $\endgroup$
    – Avantgarde
    May 31, 2019 at 14:35

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