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?

  • $\begingroup$ 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, 2015 at 12:43
  • $\begingroup$ Oh wait, I guess the timelike component is negative. Not sure for that part. $\endgroup$
    – Slereah
    Jul 2, 2015 at 12:57

1 Answer 1


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.

  • $\begingroup$ 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, 2015 at 9:01
  • $\begingroup$ Yes you must assume that. $\endgroup$ Jul 3, 2015 at 16:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.