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

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

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

So consider the action

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

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):

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

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

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

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

## 1 Answer

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.

• (sorry misunderstood your answer) – Francesco Bernardini Dec 27 '18 at 20:11