# Question on energy conservation from the stress tensor of a classical scalar field

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.

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 $$\phi$$.

• Thanks, I forgot that the Levi-Civita connection is torsion free. 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.

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