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Consider the stress-energy-momentum tensor $$T_{\alpha \beta}=(\nabla_\alpha \phi )\nabla_\beta \phi -\frac{1}{2}g_{\alpha \beta}((\nabla^\nu \phi ) \nabla_{\nu} \phi +m^2 \phi^2$$ where the smooth, real-valued function $\phi(x)$ on the spacetime satisfies the Klein-Gordon equation $$\nabla^\alpha\nabla_\alpha\phi-m^2\phi=0.$$ The task is to show, that $T_{\alpha \beta}$ is conserved.

I've come as long as to show, that $$\nabla^\alpha T_{\alpha \beta}=(\nabla_\alpha\phi)\nabla^\alpha\nabla_\beta\phi-\frac{1}{2}((\nabla_\alpha\phi)(\nabla_\beta\nabla^\alpha\phi)+(\nabla^\alpha\phi)\nabla_\beta\nabla_\alpha\phi).$$ I have a little bit trouble to see the last step. I believe, that $(\nabla_\alpha\phi)(\nabla_\beta\nabla^\alpha\phi)=(\nabla_\alpha\phi)\nabla^\alpha\nabla_\beta\phi$ may be ok, because $\phi$ is a scalar function and I could assume a Levi-Civita connection? That $(\nabla_\alpha\phi)\nabla^\alpha\nabla_\beta\phi=(\nabla^\alpha\phi)\nabla_\beta\nabla_\alpha\phi$ would be true by interchanging $\nabla^\alpha$ and $\nabla_\alpha$. Am I allowed to do that, and if, why?

The second part of the question asks, whether $T_{\alpha \beta}$ satisfies the weak energy condition.

I again am assuambly quite far in my calculation, but I can't figure out the last step. I have come as far as: $$T_{\alpha \beta}T^\alpha T^\beta=(\underbrace{\nabla_\alpha \phi T^\alpha)^2}_{\geq 0} \underbrace{-\frac{1}{2} g_{\alpha \beta}T^\alpha T^\beta}_{\geq 0}((\nabla^\nu\phi)\nabla_\nu\phi+m^2\phi^2)\geq 0.$$ So it basically remains to show, that $(\nabla^\nu\phi)\nabla_\nu\phi+m^2\phi^2\geq 0$. Any ideas?

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  • $\begingroup$ If it's a real scalar field, $m^2 \varphi^2$ is obviously >= 0. Same with every component of the sum of $(\partial_x \varphi)^2$ (It's applied to a scalar field so no Christoffel symbols) $\endgroup$ – Slereah Jul 2 '15 at 12:43
  • $\begingroup$ Oh wait, I guess the timelike component is negative. Not sure for that part. $\endgroup$ – Slereah Jul 2 '15 at 12:57
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For the first question, you do not need to interchange $\nabla_{\alpha}$ and $\nabla^{\alpha}$. We simply have $$ \nabla_{\alpha}\varphi\nabla^{\alpha}\nabla_{\beta}\varphi \\ = \eta^{\alpha\gamma}\nabla_{\alpha}\varphi\nabla_{\gamma}\nabla_{\beta}\varphi \\ = \nabla^{\gamma}\varphi\nabla_{\gamma}\nabla_{\beta}\varphi \\ = \nabla^{\alpha}\varphi\nabla_{\beta}\nabla_{\alpha}\varphi $$

where in the last step it was used that for torsion-free $\nabla_{\alpha}$ one has $[\nabla_{\alpha},\nabla_{\beta}]\varphi = 0$.

For the second question, it is easiest to prove the weak-energy condition by first picking a Lorentz frame $(x^0,\vec{x})$ adapted to $T^{\alpha}$. By this I mean choose a Lorentz frame such that $T^{\alpha} = (1,0,0,0)$. Then $\nabla_{\alpha}\varphi T^{\alpha} = \partial_0 \varphi$ so that $$ T_{\alpha\beta}T^{\alpha}T^{\beta} = (\partial_0\varphi)^2 + \frac{1}{2}[-(\partial_0\varphi)^2 + (\vec{\nabla}\varphi)^2 + m^2 \varphi^2] \\ = \frac{1}{2}[(\partial_0 \varphi)^2 + (\vec{\nabla}\varphi)^2 + m^2\varphi^2]\geq 0 $$

as desired. Since this holds in one Lorentz frame it must hold in all Lorentz frames.

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  • $\begingroup$ Thank you very much! For the first question, I didn't really mean to actually interchange $\nabla_\alpha$ and $\nabla^\alpha$.But you provided a nice way to see it, thank you! So for the connection, I actually have to assume $\nabla$ is a Levi-Civita connection? Using the Lorentz-frame it gets quite clear, thank you again! $\endgroup$ – Bultar Jul 3 '15 at 9:01
  • $\begingroup$ Yes you must assume that. $\endgroup$ – FenderLesPaul Jul 3 '15 at 16:04

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