I am struggling to answer an old general relativity exam question, which is as follows:

"Consider a scalar field $\phi(t,x^i)$ with potential $V(\phi)$ on a general spacetime. Its stress tensor is given as $$T_{\mu\nu} = \nabla_\mu \phi \nabla_\nu \phi - \frac{1}{2} g_{\mu\nu} (\nabla^\alpha \phi \nabla_\alpha \phi)-g_{\mu\nu} V(\phi)\tag{1}$$ Using the equation of motion of this scalar field, $$\nabla^\alpha \nabla_\alpha \phi = \frac{dV(\phi)}{d\phi}\tag{2}$$ show that the stress energy is conserved."

The metric convention used is $(-, +, +, +)$ signature.

To answer this I have been trying to show that this stress-energy tensor is divergence free; $$\nabla_\mu T^{\mu\nu}= 0 \Rightarrow \nabla^\mu T_{\mu\nu} = 0.\tag{3}$$ Doing this I get that $$\nabla^\mu T_{\mu\nu} = (\nabla^\mu \nabla_\mu \phi) \nabla_\nu \phi + \nabla_\mu \phi (\nabla^\mu \nabla_\nu \phi) - \frac{1}{2} \nabla_\nu (\nabla^\alpha \phi \nabla_\alpha \phi) - \nabla_\nu V(\phi)\tag{4}$$ I then use the equation of motion on the first term and the chain rule on the final term (justifying that $V(\phi)$ must be a scalar potential) to cancel them both. Then since $(\nabla^\alpha \phi \nabla_\alpha \phi)$ is a scalar I argue that its covariant derivative is simply the partial derivative. This leaves me with

$$\nabla^\mu T_{\mu\nu} = \nabla_\mu \phi (\nabla^\mu \nabla_\nu \phi) - \frac{1}{2} \partial_\nu (\nabla^\alpha \phi \nabla_\alpha \phi).\tag{5}$$

However I cannot seem to get these terms to cancel. If I use the fact that $\phi$ is a scalar field and therefore its covariant derivative is its partial derivative I get that

$$\nabla^\mu T_{\mu\nu} = \partial_\mu \phi (\partial^\mu \partial_\nu \phi) - \frac{1}{2} \partial_\nu (\partial^\mu \phi \partial_\mu \phi) - \partial_\mu \phi g^{\mu\rho}\Gamma^\sigma_{\rho\nu} \partial_\sigma \phi\tag{6}$$

Which I still cannot cancel.

What am I doing wrong? Should I relax the divergence free condition and simply consider $$\partial^\mu T_{\mu\nu}\tag{7}$$ instead? I have been stuck on this problem for hours and it's really bugging me.


2 Answers 2


Hints: The fully covariant calculation $(\nabla_{\mu}T)^{\mu\nu}=\ldots=0$ works with the help of the following rules:

  1. $(\nabla_{\lambda}g)_{\mu\nu}~=~0$ (since the Levi-Civita connection $\nabla$ is compatible with the metric).

  2. $\nabla_{[\mu}\nabla_{\nu]}\phi~=~0$ (since the Levi-Civita connection $\nabla$ is torsionfree).

  3. EL equation for $\phi$.

  • $\begingroup$ Thanks, I forgot that the Levi-Civita connection is torsion free. $\endgroup$
    – Ollie113
    Apr 18, 2019 at 11:57

Your last equation is correct, and it does cancel. In case this question ends up being marked as "homework and exercises," I'll just give a few hints.

An easier way

Maybe the most important hint is that there's an easier way. Start with your fifth equation, namely $$ \nabla^a T_{ab} =\nabla_a\phi(\nabla^a\nabla_b\phi) -\frac{1}{2}\partial_b\big((\nabla^a\phi)(\nabla_a\phi)\big). $$ To make the remaining steps easier, follow this rule: use only covariant derivatives even when acting on scalars, because that way you can raise/lower indices without worrying about what the derivatives do to the implicit factors of the metric. In particular, rewrite your fifth equation as $$ \nabla^a T_{ab} =\nabla_a\phi(\nabla^a\nabla_b\phi) -\frac{1}{2}\nabla_b\big((\nabla^a\phi)(\nabla_a\phi)\big) $$ and then see what happens when you use the product rule in the last term. The result is $\nabla^a T_{ab}=0$, as desired.

The original way works, too

Your last equation $$ \nabla^a T_{ab} = \partial_a\phi(\partial^a\partial_b\phi) -\frac{1}{2}\partial_b(\partial^a\phi\partial_a\phi) -\partial_a\phi g^{ad}\Gamma^c_{db}\partial_c\phi $$ is also correct, but showing that it actually is zero takes a bit more work. Here are a few hints about how to do it:

  • Use explicit factors of the metric to write all partial derivatives with subscripts.

  • Use the product rule to expand all of the partial derivatives. You'll get a term that involves a partial derivative of the inverse metric.

  • After cancelling the $\partial\partial \phi$ terms using the symmetry of $g^{ab}$, use the fact that the last term is symmetric in $a\leftrightarrow c$, together with the identities $$ g^{ad}\Gamma^c_{db}+g^{cd}\Gamma^a_{db} = g^{ad}g^{ec}\partial_b g_{de} $$ and $$ \partial M^{-1}+M^{-1}(\partial M) M^{-1}=0. $$ The last identity holds for any matrix $M$, such as the matrix with components $g_{ab}$. The result is $\nabla^a T_{ab}=0$, as desired.

  • $\begingroup$ Thanks for the tip, it made things a lot easier to keep the covariant derivatives instead of replacing them with partial derivatives. $\endgroup$
    – Ollie113
    Apr 18, 2019 at 11:58

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