Is Hilbert Stress-energy tensor always same as Belinfante stress-energy tensor?

The canonical stress-energy (SE) tensor arises from Noether’s Theorem by employing the conserved currents associated with translational symmetries.

It's defined as $$T^{ab}=\frac{\partial \mathcal{L}}{\partial(\partial_a\phi)}\partial^b\phi+\mathcal{L}\eta^{ab}$$

However in general canonical SE tensor is not symmetric. In fact for a SE tensor $T_{ab}$, $T_{ab}+\partial^c\chi_{cab}$ is also a SE tensor for any $\chi_{cab}=-\chi_{acb}$.

So given a canonical SE tensor, we can always construct a symmetric SE tensor, called Belinfante SE tensor.

There is another way to define SE tensor in QFT in curved spacetime. That is Hilbert SE tensor which is defined as

$$T_{ab}=\frac{-2}{\sqrt{-g}}\frac{\delta(\mathcal{L}\sqrt{-g})}{\delta g^{ab}}$$

So Hilbert SE tensor is also a symmetric tensor.

My questions are

1. Is Hilbert SE tensor always same as Belinfante SE tensor? If yes, how to prove.

2. If the question (1)'s answer is yes, is Hilbert SE tensor or Belinfante SE tensor the unique symmetric SE tensor that you can construct?

1. Well, firstly, we have to assume Lorentz covariance and general covariance of the theory. For non-relativistic theories all bets are off. Secondly, in case of fermions, one needs to generalize the Hilbert SEM tensor from variation wrt. a metric to a variation wrt. a vielbein, see e.g. my Phys.SE answer here. Then the generalized Hilbert SEM tensor is the canonical SEM plus the Belinfante-Rosenfeld improvement term. A proof is sketched in my Phys.SE answer here and links therein.

2. No, a symmetric SEM tensor is not unique. It is in principle possible to add improvement terms that respects the symmetry and the conservations laws.

Yes, provided we work with Torsion-free connections. This is explained in the original papers of Belinfante and Rosenfeld that are cited on the Wikipedia page.

You can see the torsion-free necessity by using a vierbein formulation. Just varying the vierbein alone usually gives the asymmetric Noether tensor. If you define the spin current by the variation of the spin connection, and link the spin connection to the metric by the torsion-free condition, then the spin-connection variation generates precisely the extra Belinfante-Rosenfeld terms.

Belifante tensor is only symmetric for solutions of equations of motion. I.e. It is symmetric only on-shell. But the Hilbert stress energy tensor is symmetric off-shell, by definition. Therefore the two tensors whenever they are equal they are equal only for solutions of equation of motion. When considering the transformation of say electromagnetic field under translation, if you use $F^\prime_{\mu\nu}(x^\prime)=F_{\mu\nu}(x)$ and then make the translation parameter spacetime dependent $x^\prime=x+\epsilon(x)$, you can obtain the canonical stress energy tensor which can be improved to Belifante tensor which is symmetric on shell. But to begin with if you take $x^\prime=x+\epsilon(x)$, and take the transformation of the field strength as that of a second rank tensor under a general coordinate transformation(I.e. given by its Lie derivative) then you can obtain the Hilbert stress energy tensor. Using equation of motion, you can show that one is equal to the other I.e. They are equal only on shell.

• I don't think the equation of motion is required to be satisfied for symmetry. Symmetry follows from Lorentz invariance. So, as long as the action functional is Lorentz invariant, we are OK. The eq of motion (or something equivalent like integrating over the fields) is required for conservation of course -- that is what Noether's theorem is all about. – mike stone Jan 1 '17 at 15:45

The canonical stress-energy tensor defined (with metric notation $\eta=diag(-1,1,1,1)$) $$T^{ab}_{\texttt{(cano)}} = -\frac{\partial \mathcal L}{\partial (\partial_a \phi_I)}\partial^b\phi_I +\mathcal L g^{ab} = -\frac{\partial \mathcal L}{\partial (\partial_a \phi_I)}g^{bc}\partial_c\phi_I +\mathcal L g^{ab}$$

Suppose locally we can find a scalar field such that $$-\partial_a \phi^I = e^I_a\;,$$ where $I,J,...$ are the internal frame indices. We then have the covariant stress-energy tensor \begin{eqnarray} T_{ab} &=& - \frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{ab}} = -\frac 1 e \frac{\delta S_m}{\delta e^{Ib}} e^I_a \;, \\ &=& - \frac{1}{e}\bigg\{ \frac{\partial \mathcal L_m}{ \partial e^{Ib}} e^I_a e + \mathcal L \frac{\partial e}{\partial e^{Ib} } e^I_a \bigg\} \;, \\ &=& -\frac{\partial \mathcal L}{\partial e^{Ib}}e^I_a + \mathcal L e^I_a \frac{\partial e^{Jc}}{\partial e^{Ib}} e_{Jc}\;,\\ &=& -\frac{\partial \mathcal L}{\partial e^{Ib}}e^I_a + g_{ab} \mathcal L\;,\\ &=& -\frac{\partial \mathcal L}{\partial (\partial^b\phi^{I})}\partial_a(\phi^I) + g_{ab} \mathcal L\;,\\ &\equiv& T_{ab}^{\texttt{(cano)}} \end{eqnarray} So I think that we can always link the Hilbert stress-energy tensor to Belinfante–Rosenfeld stress–energy tensor if locally we can find the scalar function $\phi$ such that $-\partial_a \phi^I = e^I_a$. At least, in the case of GR, but I'm not sure for the case of modified gravity such as in the case of metric-affine gravity.

Note also that: In GR the symmetry group is diffeomorphisms which not fits in to the first Noether's theorem but the above derivation, we shift to the internal space with symmetry group is $SO(3,1)$ which is a finite-dimensional group, where the first Noether's theorem is applicable.

Note that: $S_m=\int d^4x \sqrt{-g} \mathcal L_m$

• Quibble to the answer (v5): The second condition does not seem to be generically valid. – Qmechanic Feb 7 '17 at 18:00
• That 's the point I also concern about. The second condition. – Saksith Jaksri Feb 7 '17 at 19:52