# Dirac field and stress-energy tensor density

I read somewhere that stress-energy tensor density is a symmetric tensor. But if I take the Dirac Field tensor:

$$T^{\mu \nu}=i \psi^\dagger \gamma^0 \gamma^\mu \partial^\nu \psi$$

How could I demonstrate this property?

• Comment to the question (v2): Note that there are different definitions of the stress-energy tensor, of which not all are symmetric. – Qmechanic Nov 10 '13 at 10:59
• – Prahar Feb 4 '15 at 16:32

There is a lot of ambiguity in the definition of the stress-energy tensor. The stress-energy tensor is a conserved current, and like all conserved currents it is only defined up to a total divergence. I assume this $T_{\mu \nu}$ was calculated using the canonical prescription $T^\mu_\nu=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^i)}\partial_\nu \phi^i-\mathcal{L}\delta^\mu_\nu$ (you seem to be missing the second piece, or you are dealing with a massless field). The canonical tensor is not symmetric for fields with spin. Essentially, the intrinsic angular momentum is also contributing to T. So you find a term $S^\lambda_{\mu \nu}$ satisfying $\partial_\lambda S^\lambda_{\mu \nu}\approx T_{[\mu \nu]}$ (S is antisymmetric in its first two indices, and thus has vanishing divergence) and add it to the canonical tensor. See this worked out in detail here http://en.wikipedia.org/wiki/Belinfante%E2%80%93Rosenfeld_stress%E2%80%93energy_tensor
This procedure might seem a little random, but of course what you really should be doing is obtaining T from $T^{\mu \nu}=\frac{\delta S}{\delta g_{\mu \nu}}$ as in general relativity. This $T$ will always be symmetric, and is in fact the same as the Belinfante tensor. However, there is still ambiguity in this procedure. In order to obtain T this way, you have to "covariantize" the theory, promoting the metric to a dynamical field. This covariantization is ambiguous: you may couple the metric to the curvature non-minimally. These couplings vanish in the flat space limit, but can still affect the expression for T. But at least this expression will always be symmetric.
• The enrgy momentum tensor is usually defined "on shell", then the piece with $\delta^\mu_\nu {\mathcal L}$ vanishes. The Belinfante tensor, by an algebraic miracle related to the $\gamma$-matrix algebra, is simply the symmetrized version of the Noether canonical tensor. – user72382 Feb 4 '15 at 16:19