1
$\begingroup$

Recently I understood that the energy momentum tensor can be calculated by:

\begin{equation} T_{\mu \nu}=\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu \nu}}.\tag{1} \end{equation}

So consider the action

\begin{equation} S_m=\int d^4x\sqrt{-g}\times \frac{1}{2}(g^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi-m^2\phi^2).\tag{2} \end{equation}

Here I'm using the $(+,-,-,-)$ Minkowski sign convention. Varying this action with respect to $g^{\mu\nu}$, I obtained (the factor of $\frac{1}{2}$ dropped temporarily):

\begin{equation} \begin{split} &\quad \frac{\delta \sqrt{-g}}{\delta g^{\mu \nu}}(g^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi-m^2\phi^2) + \sqrt{-g} \frac{\delta}{\delta g^{\mu \nu}}(g^{\alpha \beta}\partial_{\alpha}\phi \partial_{\beta}\phi)\\ &=-\frac{\sqrt{-g}}{2}g_{\mu\nu}(g^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi-m^2\phi^2) + \sqrt{-g}\left(\frac{1}{2}\delta_{\mu}^{\alpha}\delta_{\nu}^{\beta}+\frac{1}{2}\delta_{\nu}^{\alpha}\delta_{\mu}^{\beta} \right)\partial_{\alpha}\phi \partial_{\beta}\phi \\ &=\sqrt{-g}\left( \partial_{\mu}\phi \partial_{\nu}\phi - \frac{1}{2} g_{\mu \nu}(g^{\alpha \beta}\partial_{\alpha}\phi \partial_{\beta}\phi-m^2\phi^2)\right) \end{split}\tag{3} \end{equation}

which yields: $$ T_{\mu \nu} = \partial_{\mu}\phi \partial_{\nu}\phi - \frac{1}{2} g_{\mu \nu}(g^{\alpha \beta}\partial_{\alpha}\phi \partial_{\beta}\phi-m^2\phi^2)\tag{4} $$ and is consistent with the one given by Noether theorem.

But if I write the action as

\begin{equation} S_m=\int d^4x\sqrt{-g}\times \frac{1}{2}(g_{\mu \nu}\partial^{\mu}\phi \partial^{\nu}\phi-m^2\phi^2)\tag{5} \end{equation}

then the identity: $$ \delta g_{\mu \nu} = - g_{\mu \alpha} g_{\nu \beta} \delta g^{\alpha \beta}\tag{6} $$ since to be introducing an additional minus sign so that the energy momentum tensor is $$ T_{\mu \nu} = -\partial_{\mu}\phi \partial_{\nu}\phi - \frac{1}{2} g_{\mu \nu}(g^{\alpha \beta}\partial_{\alpha}\phi \partial_{\beta}\phi-m^2\phi^2).\tag{7} $$

Is it that $$g_{\mu \nu}\partial^{\mu}\phi \partial^{\nu}\phi \neq g^{\mu \nu}\partial_{\mu}\phi \partial_{\nu}\phi\tag{8}$$ in GR or have I done something wrong? (I have never studied GR before, so much appreciate if anyone find other stuff I've done wrong)

$\endgroup$
3

1 Answer 1

2
$\begingroup$

The point is that the derivative $$\partial^{\mu}~:=~g^{\mu\nu} \partial_{\nu}, \qquad \partial_{\nu}~:=~\frac{\partial}{\partial x^{\nu}},$$ by definition, cf. above comment by user knzhou.

So the metric dependence in OP's action (5) is not really different from the metric dependence in OP's action (1). When the implicitly written metric dependence in OP's action (5) is taken properly into account, the two approaches agree.

$\endgroup$
1
  • $\begingroup$ (sorry misunderstood your answer) $\endgroup$ Commented Dec 27, 2018 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.