# Sign mistake in the energy momentum tensor of the Klein-Gordon Equation

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)

• The quantity $\partial^\mu \phi$ depends on the metric, so you need to account for that too. (After all, it is defined in the first place by raising $\partial_\mu \phi$.) After you account for the two resulting extra terms you'll get the right sign, essentially because $-1 + 1 + 1 = 1$. – knzhou Dec 27 '18 at 16:47
• (sorry misunderstood your answer) – Francesco Bernardini Dec 27 '18 at 23:07

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.