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Sometimes the stress-energy tensor $T_1^{\mu\nu}$ is defined as $$T_1^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}(\partial^\nu\phi)-\eta^{\mu\nu}\mathcal{L}$$ which for translational symmetries of spacetime satisfies $\partial_\mu T_1^{\mu\nu}=0.$ It is not always symmetric. In general relativity, stress-energy tensor is defined as $$T_2^{\mu\nu}=\frac{\delta S}{\delta g_{\mu\nu}}$$ where $S=\int d^4x\sqrt{-g}\mathcal{L}(\phi,\partial_\mu\phi)$ is the action of the theory. It is always symmetric because the metric tensor $g_{\mu\nu}$ is always symmetric.

  • Can we say that the second definition is more general? If so, can we derive the $T_1^{\mu\nu}$ starting from the second expression $T_2^{\mu\nu}$ in the flat spacetime limit?

Please keep the discussion elementary.

Solidification
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1 Answers1

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By taking a particular form of the Lagrangian, it's possible to reach from $T_2^{\mu\nu}$ in your question to $T_1^{\mu\nu}$. Take the free scalar field theory in a generic manifold. Then the Lagrangian density looks like: $\mathcal{L} = \frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi \partial_{\nu}\phi + m^2 \phi^2$. and the action is:

$S = \frac{1}{2}\int d^4x \sqrt{g}\, (g^{\mu\nu}\partial_{\mu}\phi \partial_{\nu}\phi + m^2 \phi^2)$

Now we use the fact that: $\delta \sqrt{g} = \frac{1}{2}\sqrt{g} g^{\mu\nu} \delta g_{\mu\nu}$ and the integral of your definition $\delta S = \int d^4x\, T_2^{\mu\nu} \delta g_{\mu\nu}$, we can get

$T_2^{\mu\nu} = -g^{\mu\nu} \mathcal{L} + \partial^{\mu}\phi \partial^{\nu}\phi$ which is gonna give you $T_1^{\mu\nu}$ in flat space limit.

If it's the symmetric property of the Stress-Energy tensor you're looking for then there is a way to start with any stress-energy tensor and make it symmetric. This construction is known as Belinfante construction. The basic idea is as follows:

To any stress-tensor you can divergence of an rank 3 tensor which is anti-symmetric in first two indices. $T^{\mu\nu} = T^{\mu\nu}_{old} + \partial_{\sigma}B^{\sigma\mu\nu}$ with the property $B^{\sigma\mu\nu} = -B^{\mu\sigma\nu}$. It's easy to show that the new tensor is also conserved. Then we find an appropriate $B$ that makes the new $T^{\mu\nu}$ symmetric.

Ari
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