So, suppose I have an action of the type: $$ S =\int \text{d}^4 x\sqrt{-g}( \frac{i}{2} (\bar{\psi} \gamma_\mu \nabla^\mu\psi - \nabla^\mu\bar{\psi} \gamma_\mu \psi) +\alpha \bar{\psi} \gamma_\mu \psi\bar{\psi}\gamma_\nu \psi g^{\mu\nu})$$ Where $\psi$ is a fermionic field and the rest has the usual meaning ($\alpha$ is a coupling constant). Now, if I write down the Canonical energy momentum tensor, i find $$ \tilde{T}_{\mu\nu}= \frac{\delta L}{\delta \nabla^\mu\psi} \nabla_\nu\psi+ \nabla_\nu\bar{\psi} \frac{\delta L}{\delta \nabla^\mu\bar{\psi}}- g_{\mu\nu} L = 2i\bar{\psi} \gamma_{(\mu}\nabla_{\nu)}\psi -g_{\mu\nu} L $$

But, if I write the Einstein's tensor in general relativity i get $$T_{\mu\nu}=\frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu\nu}}=2 i\bar{\psi} \gamma_{(\mu}\nabla_{\nu)}\psi + 2 g \bar{\psi} \gamma_\mu \psi\bar{\psi}\gamma_\nu \psi- g_{\mu\nu} L$$

The two are obviously different. So, which one should i use in the Einstein's equations? The problem comes when you write an interaction term of the type $A_\mu A^\mu$, where $A$ is some current. Because otherwise the two tensor coincide. The first energy momentum is the one invariant under translations, so it is the one satisfying $$\nabla_\mu \tilde{T}^{\mu\nu} = 0$$ While the second satisfy the same identity only if $$\nabla_\mu A^\mu = 0$$ Basically my question is, which one of the two should be used in the Einstein's equations? $G_{\mu\nu} = \kappa \overset{?}{T}_{\mu\nu}$ Or am i doing something wrong and the two tensor do actually coincide?

  • 1
    $\begingroup$ I haven't checked your calculations, so you might well have done this already, but have you included the variation of the $\gamma$s in your calculation of $T_{\mu\nu}$? You'll need a vielbein formulation of the action. $\endgroup$ Jan 27, 2015 at 16:38
  • 2
    $\begingroup$ In any case, the right thing to use in Einstein's equations is certainly what you get from $\frac{\delta S_\text{matter}}{\delta g^{\mu\nu}}$: the variation of the whole action (with Einstein-Hilbert included) with respect to the metric should vanish. $\endgroup$ Jan 27, 2015 at 16:40
  • $\begingroup$ @Holographer Actually you're right, i didn't, and that's a good idea, thanks :) I'll let you know what i find.. $\endgroup$
    – MrFermiMr
    Jan 27, 2015 at 20:01

2 Answers 2


$\require{cancel}$I) OP is considering Dirac fermions in a curved spacetime. OP's action has various shortcomings. The correct action reads$^1$

$$ S~=~\int\!d^nx~ {\cal L}, \qquad {\cal L} ~=~e L, \qquad L~=~T-V,\qquad e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|}, $$ $$ T~=~\frac{i}{2} \bar{\psi} \stackrel{\leftrightarrow}{\cancel{\nabla}} \psi, \qquad V~=~ \alpha j^a \eta_{ab} j^b, \qquad j^a~:=~ \bar{\psi} \gamma^a\psi,\qquad \bar{\psi}~:=~\psi^{\dagger}\gamma^0,$$ $$ \bar{\psi}\stackrel{\leftrightarrow}{\cancel{\nabla}}\psi ~:=~ \bar{\psi}\stackrel{\leftrightarrow}{\cancel{\partial}}\psi +\frac{1}{2} \omega_{c, ab}~\gamma^{cab}\psi ~=~\bar{\psi}\left[\gamma^c\stackrel{\rightarrow}{\nabla_c} -\stackrel{\leftarrow}{\nabla_c}\gamma^c\right]\psi, $$ $$\stackrel{\rightarrow}{\nabla_c}\psi ~:=~ \stackrel{\rightarrow}{\partial_c}\psi +\frac{1}{4} \omega_{c, ab}~\gamma^{ab}\psi, \qquad \bar{\psi}\stackrel{\leftarrow}{\nabla_c} ~:=~ \bar{\psi}\stackrel{\leftarrow}{\partial_c} -\frac{1}{4} \bar{\psi}~\gamma^{ab}\omega_{c, ab},$$ $$\stackrel{\leftrightarrow}{\cancel{\partial}} ~:=~ \gamma^c\stackrel{\rightarrow}{\partial_c} - \stackrel{\leftarrow}{\partial_c}\gamma^c, \qquad \stackrel{\rightarrow}{\partial_c}~:=~E^{\mu}{}_c \stackrel{\rightarrow}{\partial_{\mu}}, \qquad \stackrel{\leftarrow}{\partial_c}~:=~ \stackrel{\leftarrow}{\partial_{\mu}}E^{\mu}{}_c,$$ $$ \partial_{\mu}~:=~\frac{\partial}{\partial x^{\mu}}, \qquad \gamma^{ab}~:=~\frac{1}{2}[\gamma^a,\gamma^b], \qquad \gamma^{abc}~:=~\frac{1}{2}\{\gamma^a,\gamma^{bc}\}_+. \tag{1} $$

II) The main point is that in order to write down a covariant kinetic term for a Dirac fermion in curved spacetime, we should use a covariant derivative $\nabla_{\mu}\psi$ of a spinor $\psi$, and hence we need a spin connection $\omega_{\mu}{}^a{}_b$. In turn, we need a vielbein

$$g_{\mu\nu}~=~e^a{}_{\mu} ~\eta_{ab}~e^b{}_{\nu}, \qquad e^a{}_{\mu}~ E^{\mu}{}_b~=~\delta^a_b, \qquad E^{\mu}{}_a~e^a{}_{\nu}~=~\delta^{\mu}_{\nu}, \tag{2} $$

which (we for simplicity will assume) is covariantly conserved

$$0~=~(\nabla_{\mu}e)^{a}{}_{\nu}~=~\partial_{\mu}e^{a}{}_{\nu} +\omega_{\mu}{}^a{}_b ~e^b{}_{\nu}- e^a{}_{\lambda}~\Gamma_{\mu\nu}^{\lambda}.\tag{3} $$

Hence the spin connection is complete determined

$$ 2\omega_{\mu, ab}~=~2\left(-\partial_{\mu}e_{a\nu} +e_{a\lambda}~\Gamma_{\mu\nu}^{\lambda}\right)E^{\nu}{}_b ~=~-\left(\partial_{\mu}e_{a\nu} +\partial_a g_{\mu\nu}\right)E^{\nu}{}_b -(a\leftrightarrow b)$$ $$ ~=~-\partial_{\mu}e_{a\nu}~E^{\nu}{}_b-\partial_a e_{b\mu} + g_{\mu\nu}~\partial_a E^{\nu}{}_b -(a\leftrightarrow b),\tag{4} $$


$$ 2\omega_{c, ab}~:=~2E^{\mu}{}_c~\omega_{\mu, ab} ~=~-f_{cab}-f_{abc}-f_{acb}-(a\leftrightarrow b), \tag{5}$$

where we defined

$$f_{abc}~:=~\partial_a e_{b\nu}~E^{\nu}{}_c . \tag{6}$$

III) The kinetic term becomes

$$ T~=~\frac{i}{2}\bar{\psi}\stackrel{\leftrightarrow}{\cancel{\nabla}}\psi ~=~\frac{i}{2}\bar{\psi}\stackrel{\leftrightarrow}{\cancel{\partial}}\psi -\frac{i}{4}\bar{\psi}~f_{abc}~\gamma^{cab}~\psi $$ $$ ~=~\frac{i}{2}\bar{\psi}\left[\gamma^c~E^{\mu}{}_c\stackrel{\rightarrow}{\partial_{\mu}} -\stackrel{\leftarrow}{\partial_{\mu}}E^{\mu}{}_c~\gamma^c \right]\psi -\frac{i}{4}\bar{\psi}~E^{\mu}{}_a~\partial_{\mu} e_{b\nu}~E^{\nu}{}_c ~\gamma^{cab}~\psi. \tag{7}$$

IV) The natural generalization of the Hilbert SEM tensor

$$T^{\mu\nu}~=~- \frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}, \qquad T_{\mu\nu}~=~\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}},\qquad(\leftarrow\text{Not applicable!})\qquad \tag{8} $$

to fermions is given by the formula

$$T^{\mu\nu}~=~-\frac{E^{\mu c}}{2e}\frac{\delta S}{\delta e^c{}_{\nu}}+(\mu\leftrightarrow \nu), \qquad T_{\mu\nu}~=~\frac{e_{c\mu}}{2e}\frac{\delta S}{\delta E^{\nu}{}_c}+(\mu\leftrightarrow \nu).\tag{9}$$

Formula (9) reduces to the standard Hilbert SEM tensor (8) if the action only depends on the vielbein through the metric (2). However formula (9) is more general and is necessary in the case of fermions in curved spacetime.

V) The Hilbert SEM tensor with flat indices then becomes

$$T_{cd}~:=~ E^{\mu}{}_c~ T_{\mu\nu} ~E^{\nu}{}_d~\stackrel{(9)}{=}~-\frac{e_{c\nu}}{2e}\frac{\delta S}{\delta e^d{}_{\nu}} +(c\leftrightarrow d) ~=~\frac{E^{\nu}{}_c}{2e}\frac{\delta S}{\delta E^{\nu d}} +(c\leftrightarrow d)$$ $$~\stackrel{(7)}{=}~\frac{i}{4}\bar{\psi}\left[\gamma_c\stackrel{\rightarrow}{\partial_d} -\stackrel{\leftarrow}{\partial_c}\gamma_d +\frac{1}{2} (f_{cba}-f_{abc}-f_{acb})~\gamma_d{}^{ab} \right]\psi -\frac{1}{2}\eta_{cd}L+(c\leftrightarrow d)$$ $$~\stackrel{(5)}{=}~\frac{i}{4}\bar{\psi}\left[\gamma_c\stackrel{\rightarrow}{\partial_d} -\stackrel{\leftarrow}{\partial_c}\gamma_d +\frac{1}{2} \omega_{c,ab}~\gamma_d{}^{ab} \right]\psi -\frac{1}{2}\eta_{cd}L+(c\leftrightarrow d)$$ $$~\stackrel{(1)}{=}~\frac{i}{4}\bar{\psi}\left[\gamma_c\stackrel{\rightarrow}{\nabla_d} -\stackrel{\leftarrow}{\nabla_c}\gamma_d\right]\psi -\frac{1}{2}\eta_{cd}L+(c\leftrightarrow d). \tag{10} $$

Eq. (10) is the formula for the (generalized) Hilbert SEM tensor of a Dirac fermion in curved spacetime. This is the appropriate matter source term in the EFE, cf. OP's title question (v3). For further details, see also my Phys.SE answers here and here.


$^1$ One may show that the Lagrangian density (1) is real using

$$ (\gamma^a)^{\dagger}~=~ \gamma^0\gamma^a\gamma^0,\qquad (\gamma^0)^2~=~{\bf 1}.\tag{11} $$

Conventions: In this answer, we will use $(+,-,-,-)$ Minkowski sign convention, and Clifford algebra

$$\{\gamma^a,\gamma^b\}_+~=~2\eta^{ab}{\bf 1}.\tag{12}$$

Greek indices $\mu,\nu,\lambda, \ldots,$ are so-called curved indices, while Roman indices $a,b,c, \ldots,$ are so-called flat indices.

  • $\begingroup$ Do you have a reference for equation (9)? $\endgroup$ Jul 23, 2017 at 22:23
  • 1
    $\begingroup$ I'm not sure who historically first wrote eq. (9) down, but if you just want to see it in print, see e.g. Freedman & van Proeyen, SUGRA, eq. (8.48). $\endgroup$
    – Qmechanic
    Jul 24, 2017 at 7:56
  • $\begingroup$ There is a discussion of the Dirac tensor in the Wikipedia page en.wikipedia.org/wiki/Belinfante-Rosenfeld_stress-energy_tensor $\endgroup$
    – mike stone
    Jan 6, 2018 at 17:10

As @Holographer has mentioned in a comment, the correct formula for the stress-tensor that enters the EFE is $$ T_{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{ \delta g^{\mu\nu} } $$ whereas what you are computing is the canonical stress energy tensor. However, there is a subtle relation between the two, which I will elaborate upon here.

Apart from a theory that contains only scalars, the canonical stress tensor is never the one that enters the EFE. This is because, in general, the canonical stress tensor is not symmetric and therefore cannot possibly be the same stress tensor that enters in the EFE. For instance the canonical stress tensor for electromagnetism is $$ (T^{EM}_{\mu\nu})_{\text{canonical}} = F^\rho{}_\mu \partial_\nu A_\rho + \frac{1}{4} g_{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} $$ which is not only not symmetric, but also not gauge invariant. PS - The non-symmetry is due to the spin of the field involved and is closely related to the angular momentum tensor.

However, there is an ambiguity in the construction of the stress tensor (the ambiguity does not change the conserved charges which are physical quantities). This ambiguity allows construction of an improved stress tensor (often known as the Belinfante tensor) that is symmetric and conserved. It is this improved tensor that enters the EFE. (ref. this book)

To see the equivalence, let us recall the standard construction of the stress-tensor. Consider a coordinate transformation $$ x^\mu \to x^\mu + a^\mu (x) $$ Since the original Lagrangian is invariant under translations (where $a^\mu$ is constant), the change in the action under such a coordinate transformation is $$ \delta S = \int d^d x \sqrt{-g} \nabla_\mu a_\nu T_B^{\mu\nu} $$ Now, if the stress-tensor is symmetric then we can write $$ \delta S = \frac{1}{2} \int d^d x \sqrt{-g} \left( \nabla_\mu a_\nu + \nabla_\nu a_\mu \right) T_B^{\mu\nu} $$ Note that the term in the parenthesis is precisely the change in the metric under the coordinate transformation. Thus, $$ \delta S = \frac{1}{2} \int d^d x \sqrt{-g} \delta g_{\mu\nu} T_B^{\mu\nu} \implies \frac{2}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu\nu}} = T_B^{\mu\nu} $$

Thus, we see that the symmetric Belinfante stress tensor is precisely the gravitational stress tensor. Note of course that what I've said holds specifically in a Minkowskian background, since the construction of $T_B^{\mu\nu}$ assumes Lorentz invariance.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.