Background to my question: the flat case

I am interested in the following (part of a) Lagrangian involving spinors in a general curved spacetime: $$L=\frac{1}{2}\overline{\psi}\left(i\gamma^{\alpha}\nabla_{\alpha}-\mu_{0}\right)\psi$$ where the covariant derivative acting on fermions is defined using the vielbeins (as described in S. Weinberg, Gravitation and Cosmology, John Wiley&Sons, New York, 1972):

$$\nabla_{\alpha}=V_{\alpha}^{\,\,\,\,\mu}\partial_{\mu}+\frac{1}{2}\sigma^{\beta\gamma}V_{\beta}^{\,\,\,\,\nu}V_{\alpha}^{\,\,\,\,\mu}\left(\partial_{\mu}V_{\gamma\nu}-\Gamma_{\nu\mu}^{\lambda}V_{\gamma\lambda}\right)$$ where $$\sigma^{\alpha\beta}=\frac{1}{4}\left[\gamma^{\alpha},\gamma^{\beta}\right]$$

To find the energy-momentum tensor (EMT) associated with this Lagrangian, I used the following relation (derived from the standard definition of the EMT, along with the relation $g_{\mu\nu}\left(x\right)=V_{\,\,\,\,\mu}^{\alpha}\left(x\right)V_{\,\,\,\,\nu}^{\beta}\left(x\right)\eta_{\alpha\beta}$):

$$T_{\mu\nu}=-g_{\mu\nu}L-g_{\mu\rho}V_{\,\,\,\,\nu}^{\beta}\left(x\right)\frac{\delta L}{\delta V_{\,\,\,\,\rho}^{\beta}\left(x\right)}$$

I began by finding the EMT in flat spacetime, by varying around Minkowski (that is, taking $V_{\,\,\,\,\alpha}^{\beta}=\delta_{\,\,\,\,\alpha}^{\beta}+\epsilon_{\,\,\,\,\alpha}^{\beta}$ and expanding to order $\epsilon$). Using various manipulations, identities on $\gamma$ and $\sigma$, and the equations of motion, I obtained:

$$T_{\mu\nu}^\flat= \frac{i}{8}\overline{\psi}\left(\gamma_{\mu}\overrightarrow{\partial}_{\nu}+\gamma_{\nu}\overrightarrow{\partial}_{\mu}\right)\psi-\frac{i}{8}\overline{\psi}\left(\gamma_{\mu}\overleftarrow{\partial}_{\nu}+\gamma_{\nu}\overleftarrow{\partial}_{\mu}\right)\psi -\eta_{\mu\nu}\left(\frac{1}{2}\overline{\psi}i\gamma^{\alpha}\partial_{\alpha}\psi-\frac{\mu_{0}}{2}\overline{\psi}\psi\right) $$

where $\flat$ denotes flat spacetime.

My question: generalizing this result to curved spacetime

My original aim was to find the EMT in curved spacetime. The long way would be to return to the definition of $T_{\mu\nu}$ as I wrote above, and perform the full derivation in curved spacetime. But having already computed $T_{\mu\nu}^\flat$, I turned to the simple solution: Take $T_{\mu\nu}^\flat$ (as given above), write it in a fully covariant manner (which is easy: just replace $\partial_\mu$ with $\nabla_\mu$ and $\eta_{\mu\nu}$ with $g_{\mu\nu}$). Then, since it is a covariant equation written in one coordinate system (local Cartesian coordinates, in which the flat and the covariant expressions coincide), it may be transformed to hold in any coordinate system, yielding the general expression in curved spacetime.

This results in:

$$T_{\mu\nu}= \frac{i}{8}\overline{\psi}\left(\gamma_{\mu}\overrightarrow{\nabla}_{\nu}+\gamma_{\nu}\overrightarrow{\nabla}_{\mu}\right)\psi-\frac{i}{8}\overline{\psi}\left(\gamma_{\mu}\overleftarrow{\nabla}_{\nu}+\gamma_{\nu}\overleftarrow{\nabla}_{\mu}\right)\psi -g_{\mu\nu}\left(\frac{1}{2}\overline{\psi}i\gamma^{\alpha}\nabla_{\alpha}\psi-\frac{\mu_{0}}{2}\overline{\psi}\psi\right) $$

This is indeed the correct expression for the covariant EMT of the given Lagrangian.

My question is, is my reasoning valid or does it require further justification? E.g., can't there be extra terms in the covariant expression that vanish in flat spacetime, hence not allowing the full construction of $T_{\mu\nu}$ from $T_{\mu\nu}^\flat$? What guarantees there are no such terms in this case?

  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/161821/2451 $\endgroup$
    – Qmechanic
    Commented Jun 1, 2023 at 1:45
  • $\begingroup$ @Qmechanic I don't think it is a duplicate. I'm asking about the validity of a specific reasoning that leads from the flat case to the curved case in a very simple way, saving the need to perform further calculations. $\endgroup$
    – Whyka
    Commented Jun 1, 2023 at 9:20
  • $\begingroup$ @Qmechanic In the meanwhile I also realized the answer to my question, and I will be happy to share it, if possible. $\endgroup$
    – Whyka
    Commented Jun 1, 2023 at 9:22

2 Answers 2


I'm not used to working with fermions in curved spacetime, so I can't say I know how to do the complete calculation. Nevertheless, there does seem to be some subtleties in the argument. When working with a non-minimally coupled scalar field, there is a term in the stress tensor proportional to the Einstein tensor (I wrote this pdf a while ago showing the detailed computations, but it is a straightforward exercise in GR). As such, this term would not appear in the flat spacetime expression, but would be necessary in curved spacetime. Similarly, other non-minimal curvature couplings could interfere with the argument.

In the particular case of minimally coupled fermions, I wouldn't expect these sorts of difficulties to occur, since all occurrences of the metric in the Lagrangian would be in the covariant derivative, and hence your argument seems to make sense. Every place in the curved spacetime Lagrangian in which the metric would appear is already there in the flat spacetime Lagrangian, for no curvatures are appearing to make terms vanish. If there were some non-minimal couplings, the story would be completely different.

There could be, of course, other nuances concerning fermions I'm not taking into account due to not working so often with them.

  • $\begingroup$ Hi @Níckolas Alves, thank you for the answer. I agree with the subtleties, and I pointed to the key issue in my answer. Indeed, in the case discussed here, there is no problem in the reasoning presented in the question. $\endgroup$
    – Whyka
    Commented Jun 1, 2023 at 21:34

I realized the answer is yes: it is justified to make the transition from $T_{\mu\nu}^\flat$ to $T_{\mu\nu}$ via local Cartesian coordinates in this case. The key is to note whether there are second derivatives of the metric involved in $\delta L$.

At a general point in curved spacetime, one can always construct a system of local Cartesian coordinates, in which the metric and its first derivatives vanish. Second derivatives do not generally vanish, and these may introduce curvature-related terms (such as $R$ or $R_{\alpha\beta}$) as occurs in the scalar, non-minimally-coupled case. Hence, in the latter case (of a scalar field with non-minimal coupling), one will have to perform the full derivation directly in curved spacetime $-$ it will not be possible to reconstruct the curved case from the flat case through local Cartesian coordinates.

In the current case, however, there aren't any second derivatives of the metric: evidently, $\delta L$ only involves the metric and its first derivatives. Then, the Minkowski expression $T_{\mu\nu}^\flat$, computed by perturbing around flat spacetime, should be identical to $T_{\mu\nu}$, computed in local Cartesian coordinates in curved spacetime. The argument continues as spelled out in the question: one may rewrite the expression for $T_{\mu\nu}^\flat$ in a covariant manner that still holds in local Cartesian coordinates, and then transform this tensorial relation to any other coordinate system, yielding the general result for $T_{\mu\nu}$ in curved spacetime.

Edit: I now see @Níckolas Alves has already given an answer, pointing to the same subtleties as me.

I would like, then, to restate my point: the key is in the second derivatives of the metric, because local Cartesian coordinates do not nullify those. That is the difference between the flat case and the local Cartesian curved case. If that difference doesn't exist (i.e., no second derivatives of the metric are involved), then the reasoning given in the question holds. And that is indeed the case here.


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