# Why is the stress-energy tensor symmetric?

The relativistic stress-energy tensor $T$ is important in both special and general relativity. Why is it symmetric, with $T_{\mu\nu}=T_{\nu\mu}$?

As a secondary question, how does this relate to the symmetry of the nonrelativistic Cauchy stress tensor of a material? This is apparently interpreted as being due to conservation of angular momentum, which doesn't seem connected to the reasons for the relativistic quantity's symmetry.

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In the most general context of special relativity, one may define the tensor so that it's not symmetric. There are various special situations in which the symmetry may be proven. In GR, $T^{\mu\nu} =\partial {\mathcal L}_{\rm matter} / \partial g_{\mu\nu}$ which is manifestly symmetric. –  Luboš Motl Jun 19 '13 at 14:28
@LubošMotl: Why not make that into an answer so I can upvote it? –  Ben Crowell Jun 19 '13 at 14:29
We had redundant tags for stress-energy-tensor and energy-momentum-tensor. Many questions had both tags. I went through and changed all the energy-momentum-tensor tags to stress-energy-tensor. –  Ben Crowell Jun 19 '13 at 14:44
@Ben Crowell: Please read my message in hbar chat. –  Qmechanic Jun 19 '13 at 14:53

## 1 Answer

Here is my own answer to the first part of the question. I don't know the answer to the second part.

Let's pick a local set of Minkowski coordinates $(t,x,y,z)$. Then $T_{\mu\nu}$ represents a flux of the $\mu$ component of energy-momentum through a hypersurface perpendicular to the $\nu$ axis. For example, say we have a bunch of particles at rest in a certain frame, and consider $T_{tt}$. The time component $p_t$ of the energy-momentum vector is mass-energy. Since these particles are at rest, their mass-energy is all in the form of mass. If we make a hypersurface perpendicular to the $t$ axis, i.e., a hypersurface of simultaneity, then all these particles' world-lines are passing through that hypersurface, and that's the flux that $T_{tt}$ measures: essentially, the mass density.

This makes it plausible that $T$ has to be symmetric. For example, let's say we have some nonrelativistic particles. If we have a nonzero $T_{tx}$, it represents a flux of mass through a hypersurface perpendicular to $x$. This means that mass is moving in the $x$ direction. But if mass is moving in the $x$ direction, then we have some $x$ momentum $p_x$. Therefore we must also have a $T_{xt}$, since this momentum is carried by the particles, whose world-lines pass through a hypersurface of simultaneity.

More rigorously, the Einstein field equations say that the Einstein curvature tensor $G$ is proportional to the stress-energy tensor. Since $G$ is symmetric, $T$ must be symmetric as well.

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