# 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.

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. – Ollie113 Apr 18 '19 at 11:58

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. – Ollie113 Apr 18 '19 at 11:57