# 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.

• 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. Jun 19, 2013 at 14:28
• – user4552
Jun 19, 2013 at 14:32
• @BenCrowell : For the first part of the question, the most natural definition of the stress-energy tensor is : $$T^{\mu\nu}(x) = \frac{2}{\sqrt{- g}} \frac{\delta S_{matter}}{\delta g_{\mu\nu}(x)}$$ Because the stress-energy tensor is the source of gravitation. Jun 19, 2013 at 19:49

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:

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

2. D.Z. Freedman & A. Van Proeyen, SUGRA, 2012; p. 181.

--

$$^1$$ 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.

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.