Calculating Energy-Momentum Tensor in GR

[Markus Kahn]

Homework Statement

During the derivation of the Einstein field equations via the variation principle we defined the energy momentum tensor to be
$$T_{\mu\nu} \equiv -2\frac{1}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu\nu}}.$$
Assume now that
$$S_M = \int d^4x \sqrt{-g}(\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2),$$
where ##\phi## is the scalar field. $Calculate ##T_{\mu\nu}##.

Relevant Equations

All given above

My attempt was to first rewrite ##S_M## slightly to make it more clear where ##g_{\mu\nu}## appears
$$S_M = \int d^4x \sqrt{-g} (g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2).$$
Now we can apply the variation:
$$\begin{align*}
\delta S_M
&= \int d^4x (\delta\sqrt{-g}) (g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\
&= \int d^4x (-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu} )(g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\
&= \int d^4x \sqrt{-g}\left(- \frac{1}{2}g_{\mu\nu} [g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2] + \nabla_\mu\phi\nabla_\nu\phi \right)\delta g^{\mu\nu}\\
&= \int d^4x \frac{\sqrt{-g}}{2}\left(\nabla_\mu\phi\nabla_\nu\phi+\frac{1}{2}g_{\mu\nu}m^2\phi^2 \right)\delta g^{\mu\nu}.
\end{align*}$$
We then have
$$\frac{\delta S_M}{\delta g^{\mu\nu}} = \frac{\sqrt{-g}}{2}\left(\nabla_\mu\phi\nabla_\nu\phi+\frac{1}{2}g_{\mu\nu}m^2\phi^2 \right) \quad \Rightarrow \quad T_{\mu\nu} = - \left(\nabla_\mu\phi\nabla_\nu\phi+\frac{1}{2}g_{\mu\nu}m^2\phi^2 \right)$$
Is this sensible? I'm not really sure if I performed the variation correctly, so I would appreciate if someone could give it a look.

 

[TSny]

Assume now that
$$S_M = \int d^4x \sqrt{-g}(\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2),$$
where ##\phi## is the scalar field. $$

Usually there is a factor of 1/2 multiplying ##\nabla_\mu\phi\nabla^\mu\phi##. I just wanted to verify that for your problem there is no such factor of 1/2.

$$\begin{align*}
\delta S_M
&= \int d^4x (\delta\sqrt{-g}) (g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\
&= \int d^4x (-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu} )(g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi
\end{align*}$$

In the second line you have too many ##\mu##, ##\nu## indices which is causing problems in your derivation.
In the second line, I suggest writing ##(g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi-\frac{1}{2}m^2\phi^2)## with different summation indices, e.g., as ##(g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2)##

 

[Markus Kahn]

You are absolutely right, there should be a ##1/2## in front of the first term... I completely overlooked this. With this in mind we have
$$
\begin{align*}
\delta S_M
&= \int d^4x (\delta\sqrt{-g}) (\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi\\
&= \int d^4x (-\frac{1}{2}\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu} )(\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi \\
&= \int d^4x \sqrt{-g}\left( -\frac{1}{2}g_{\mu\nu}\left[\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2\right] + \nabla_\mu\phi\nabla_\nu\phi \right)\delta g^{\mu\nu}
\end{align*}
$$
Which then would give
$$T_{\mu\nu} = g_{\mu\nu}\left[\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2\right] -2 \nabla_\mu\phi\nabla_\nu\phi.$$
I don't see how this could be simplified further (if it can), maybe contracting one of the covariant derivatives with the metric, but other than that...

 

[TSny]

$$
\delta S_M = \int d^4x (\delta\sqrt{-g}) (\frac{1}{2}g^{\alpha\beta} \nabla_\alpha\phi\nabla_\beta\phi-\frac{1}{2}m^2\phi^2) + \int d^4x \sqrt{-g} \delta g^{\mu\nu} \nabla_\mu\phi\nabla_\nu\phi$$

That missing factor of 1/2 should also appear in the second integral above.

Getting all the signs right is a headache. I haven't checked them carefully. Make sure you have all your signs correct in the original expression for the action.

I don't see any way to simplify the answer further.