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.
 A: 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.
A: Here we shall only discuss general relativistic diffeomorphism-invariant matter theories in a curved spacetime in the classical limit $\hbar\to 0$ for simplicity. In particular, we will not discuss the SEM pseudotensor for the gravitational field, but only the stress-energy-momentum (SEM) tensor for matter (m) fields $\Phi^A$.  We emphasize that our conclusions will be independent of whether the EFEs are satisfied or not. 
I) On one hand, the basic Hilbert/metric SEM tensor-density$^1$ is manifestly symmetric$^2$
$$\tag{1} \sqrt{|g|}T^{\mu\nu}~\equiv~{\cal T}^{\mu\nu}~:=~-2\frac{\delta S_m}{\delta g_{\mu\nu}},$$
cf. a comment by Lubos Motl. However, note that the basic definition (1) is not applicable to e.g. fermionic matter in a curved spacetime, cf. Section II.
Diffeomorphism invariance leads (via Noether's 2nd theorem) to an off-shell identity. Using the matter eqs. of motion (eom)
$$\tag{2} \frac{\delta S_m}{\delta \Phi^A}~\stackrel{m}{\approx}~0, $$
the corresponding Noether's 2nd identity reads$^3$
$$\tag{3} \nabla_{\mu} T^{\mu\nu}~\stackrel{m}{\approx}~0, $$
cf. e.g. Ref. 1. [Here the $\stackrel{m}{\approx}$ symbol means equality modulo matter eoms. The connection $\nabla$ is the Levi-Civita connection.] Eq. (3) serves as an important consistency check. A matter source $T^{\mu\nu}$ to the EFEs should satisfy eq. (3), cf. the (differential) Bianchi identity.  
II) In the Cartan formalism, the fundamental gravitational field is not the metric tensor $g_{\mu\nu}$ but instead a vielbein $e^a{}_{\mu}$. The generalized Hilbert SEM tensor-density is defined as
$$\tag{4}{\cal T}^{\mu}{}_{\nu}~:=~{\cal T}^{\mu}{}_a~e^a{}_{\nu}, \qquad 
{\cal T}^{\mu}{}_a ~:=~- \frac{\delta S_m}{\delta e^a{}_{\mu}}, $$
which is no longer manifestly symmetric, cf. e.g. Ref. 2.
Next we have two symmetries: local Lorentz symmetry and diffeomorphism invariance. 
Firstly, local Lorentz symmetry leads (via Noether's 2nd theorem) to an off-shell identity. Using the matter eoms (2), the corresponding Noether's 2nd identity reads
$$\tag{5}  {\cal T}^{\mu\nu}~\stackrel{m}{\approx}~{\cal T}^{\nu\mu},$$
i.e the generalized Hilbert SEM tensor-density (4) is still symmetric when the matter eoms are satisfied. 
Secondly, diffeomorphism invariance leads (via Noether's 2nd theorem) to an off-shell identity$^4$
$$\tag{6} d_{\mu}{\cal T}^{\mu}{}_{\nu}
~=~{\cal T}^{\mu}{}_a ~d_{\nu}e^a{}_{\mu}
-\frac{\delta S_m}{\delta \Phi^A}~d_{\nu}\Phi^A. $$
Not surprisingly, eqs. (5), (6), and $(\nabla_{\nu} e)^a{}_{\mu}=0$ imply eq. (3).
III) On the other hand, the canonical SEM tensor-density 
$$\tag{7} \Theta^{\mu}{}_{\nu}
~:=~\frac{\partial {\cal L}_m}{\partial\Phi^A_{,\mu}}\Phi^A_{,\nu} -\delta^{\mu}_{\nu}{\cal L}_m$$ 
is not always symmetric, cf. e.g. this Phys.SE post. The fact that the Lagrangian density ${\cal L}_m$ has no explicit spacetime dependence leads (via Noether's 1st theorem) to an off-shell identity
$$\tag{8} d_{\mu}\Theta^{\mu}{}_{\nu}
~=~-\frac{\delta S_m}{\delta e^a{}_{\mu}}~d_{\nu}e^a{}_{\mu}
-\frac{\delta S_m}{\delta \Phi^A}~d_{\nu}\Phi^A. $$
Recall that the gravitational field, the vielbein $e^a{}_{\mu}$, is not necessarily on-shell. Remember we are doing FT in curved spacetime rather than GR. As a consequence, the first term on the right-hand side of Noether's 1st identity (8) breaks the usual interpretation of Noether's 1st theorem as leading to an on-shell conservation law. [It is comforting to see that it gets restored for a constant vielbein with $d_{\nu}e^a{}_{\mu}=0$.]
IV) Eqs. (4), (6), and (8) imply that
$$\tag{9} d_{\mu}({\cal T}^{\mu}{}_{\nu}-\Theta^{\mu}{}_{\nu})~=~0.$$
One may show that there in general exists a Belinfante improvement tensor-density 
$$\tag{10} {\cal B}^{\lambda\mu}{}_{\nu}~=~-{\cal B}^{\mu\lambda}{}_{\nu},$$ 
such that
$$\tag{11} {\cal T}^{\mu}{}_{\nu}-\Theta^{\mu}{}_{\nu}
~=~d_{\lambda}{\cal B}^{\lambda\mu}{}_{\nu},$$
cf. e.g. my Phys.SE answer here. Note that eqs. (10) and (11) are consistent with eq. (9).
V) Eq. (11) serves as an important consistency check of the Hilbert SEM tensor-density (4) vs. the canonical SEM tensor-density (7). Eq. (11) implies that the two corresponding Noether charges, the energy-momentum $4$-covectors
$$\tag{12} P_{\nu} ~:=~ \int\! d^3x~{\cal T}^0{}_{\nu} \quad\text{and}\quad  \Pi_{\nu} ~:=~ \int\! d^3x~\Theta^0{}_{\nu}   $$
are equal up to spatial boundary terms
$$\tag{13} P_{\nu}-\Pi_{\nu}~=~\int\! d^3x~d_i{\cal B}^{i0}{}_{\nu}~\sim~0,$$
cf. the divergence theorem.
References:


*

*R.M. Wald, GR; Appendix E.1.

*D.Z. Freedman & A. Van Proeyen, SUGRA, 2012; p. 181.
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$^1$ A tensor-density ${\cal T}^{\mu\nu}= e T^{\mu\nu}$ is in this context just a tensor $T^{\mu\nu}$ multiplied with the density $e=\sqrt{|g|}$.
$^2$ Conventions: In this answer, we will use $(+,-,-,-)$ Minkowski sign convention. Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while Roman indices $a,b,c, \ldots,$ are so-called flat indices.
$^3$ Note that eq. (3) is not a conservation law by itself. To get a conservation law, we need a Killing vector field, cf. e.g. my Phys.SE answer here.
$^4$ Here we have assumed that the matter fields $\Phi^A$ only carry flat or spinorial indices, cf. the setting of my Phys.SE answer here. If $\Phi^A$ also have curved indices, there will be further terms in eq. (6) proportional to the matter eoms.
