What is the correspondence between the conserved canonical energy-momentum tensor, which is $$ T^{\mu\nu}_{can} := \sum_{i=1}^N\frac{\delta\mathcal{L}_{Matter}}{\delta(\partial_\mu f_i)}\partial^\nu f_i - \eta^{\mu\nu}\mathcal{L}$$ (the four conserved Noether currents corresponding to four possible spacetime translations)

where $\{f_i\}_{i=1}^N$ are the $N$ matter fields in the theory, and we assume $f_i\mapsto\alpha^\nu\partial_\nu f_i$ for translations,

and stress-energy tensor from the Einstein-Hilbert action, which is: $$ T_{\mu\nu}\equiv-\frac{2}{\sqrt{-g}}\frac{\delta\mathcal{L}_{Matter}}{\delta g^{\mu\nu}} $$

In particular, how do you get that the two are equal (are they?) for Minkowski space, for which there is no variation in the metric?

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    $\begingroup$ Even though you are interested in a case where the metric is fixed, say Minkowski, you should still perform the variation with $g$ free as the metric is a dynamical field (in GR). Then in the final answer you may specify to a particular metric. $\endgroup$
    – Winther
    Jun 18, 2014 at 21:07
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    $\begingroup$ See also discussion in Peskin & Schroeder page 683. $\endgroup$
    – PPR
    Jul 16, 2014 at 22:51
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    $\begingroup$ Related: physics.stackexchange.com/q/27048/2451 and links therein. $\endgroup$
    – Qmechanic
    Jan 3, 2015 at 16:24
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    $\begingroup$ The question (v2) states: [...] stress-energy tensor (SET) from the Einstein-Hilbert (EH) action. Note that the SET in the Einstein field equations (EFE) is typically from a matter action, not from the EH action. $\endgroup$
    – Qmechanic
    Jan 3, 2015 at 16:55
  • $\begingroup$ You may find interesting: arxiv.org/abs/math-ph/0208003 and arxiv.org/abs/math-ph/0306020. $\endgroup$
    – Fede LA
    Sep 29, 2016 at 19:15

2 Answers 2


You should think at the way the Noether current is obtained. When an infinitesimal symmetry transformation is made spacetime dependent, that is the parameters $\omega^a$ that control the symmetry are taken as functions of the spacetime point $\omega^a=\omega^a(x)$, the action is not longer left invariant $$ \delta S=-\int d^D x\, J^{a\,\mu}(x)\partial_\mu \omega^a(x) $$ but rather it provides the definition of the current $J^{a\,\mu}$ that is conserved on-shell.

Now, let's look at the case of the energy momentum tensor: in this case, the translations $x^\nu\rightarrow x^\nu+\omega^\nu$ are made local $x^\nu\rightarrow x^\nu+\omega^\nu(x)$ so that $$ \delta S=-\int d^D x\, T_\nu^\mu(x)\,\partial_\mu \omega^\nu(x)\,. $$ Actually, one looks for a symmetric tensor $T^{\mu\nu}=T^{\nu\mu}$ so that one can rewrite the expression above in the following form $$ \delta S=-\frac{1}{2}\int d^D x\, T^{\nu\mu}(x)\,(\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\,. $$ Now, here is the catch: if we were to transform the spacetime metric $g_{\mu\nu}$ (equal to $\eta_{\mu\nu}$ in the case at hand) as if $x^\nu\rightarrow x^\nu+\omega^\nu(x)$ was just an infinitesimal change of coordinates, that is $$g_{\mu\nu}\rightarrow g_{\mu\nu}-(\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\,,$$ then the action (rendered coordinates independent by the inclusion of the metric in the usual way such as $d^D x\rightarrow d^D x \sqrt{|g|}\,,\ldots$) would be left invariant $$ \delta S=-\frac{1}{2}\int d^D x\, (\partial_\mu \omega_\nu(x)+\partial_\nu\omega_\mu)\left. \left(\sqrt{|g|}\,T^{\mu\nu}(x)+2\frac{\delta S(x)}{\delta g_{\mu\nu}}\right)\right|_{g_{\mu\nu}=\eta_{\mu\nu}}=0\,. $$ From this equation it follows that the current associated with spacetime translations can be written as $$ T^{\mu\nu}=\left. -\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g_{\mu\nu}}\right|_{g_{\mu\nu}=\eta_{\mu\nu}} \qquad \mbox{evaluated on the bkg }g_{\mu\nu}=\eta_{\mu\nu}\,. $$ It should be apparent that this definition gives a symmetric energy-momentum tensor that matches the one appearing the Einstein's equations. From the derivation above it should also be clear that the alternate versions of $T_{\mu\nu}$ arise because the definition of $T^{\mu\nu}$ via the variation of the action when the translation is made spacetime dependent does not uniquely fix it. For example, given a valid $T_{\mu\nu}$, one can always define another one $T^{\mu\nu}\rightarrow T^{\mu\nu}_B=T^{\mu\nu}+\partial_\rho B^{\rho\mu\nu} $ with an arbitrary $B^{\rho\mu\nu}=-B^{\mu\rho\sigma}$ which also gives $$ \delta S=-\int d^D x\, T^{\mu\nu}(x)\partial_\mu\omega_\nu(x)=-\int d^D x\, T_B^{\mu\nu}(x)\partial_\mu\omega_\nu(x) $$ up to an integration by parts. Einstein's equations break this degeneracy and nicely identify ``the'' energy-momentum tensor.


Let there be given a general covariant matter action $$S~=~ \int \! d^4x~ {\cal L}, \qquad {\cal L}~=~e L, \qquad L~=~L(\Phi,\nabla_a\Phi). \tag{1}$$

The main strategy will be to demand that the matter fields $\Phi^A$ carry flat rather than curved indices$^1$. This is achieved with the help of a vielbein $e^a{}_{\mu}$, where

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

$$ e~:=~\det(e^a{}_{\mu})~=~\sqrt{|g|}, \tag{3}$$

and a spin connection $\omega_{\mu}{}^a{}_b$ compatible with the Levi-Civita Christoffel symbols $\Gamma_{\mu\nu}^{\lambda}$,

$$0~=~(\nabla_{\mu}e)^{a}{}_{\nu}~=~\partial_{\mu}e^{a}{}_{\nu} +\omega_{\mu}{}^a{}_b ~e^b{}_{\nu}- e^a{}_{\lambda}~\Gamma_{\mu\nu}^{\lambda}.\tag{4}$$

In other words, the spin connection $\omega_{\mu}{}^a{}_b$ is uniquely given by

$$2\omega_{\mu, ab}~=~2\left(-\partial_{\mu}e_{a\nu} +e_{a\lambda}~\Gamma_{\mu\nu}^{\lambda}\right)E^{\nu}{}_b ~=~-\left(\partial_{\mu}e_{a\nu} +\partial_a g_{\mu\nu}\right)E^{\nu}{}_b -(a\leftrightarrow b)$$ $$~=~-\partial_{\mu}e_{a\nu}~E^{\nu}{}_b-\partial_a e_{b\mu} + g_{\mu\nu}~\partial_a E^{\nu}{}_b -(a\leftrightarrow b),\tag{5}$$

$$2\omega_{c, ab}~:=~2E^{\mu}{}_c~\omega_{\mu, ab} ~=~-f_{cab}-f_{abc}-f_{acb}-(a\leftrightarrow b), \tag{6}$$

$$f_{abc}~:=~\partial_a e_{b\nu}~E^{\nu}{}_c . \tag{7}$$

The covariant derivative of the matter fields is of the form

$$ (\nabla_{\mu}-\partial_{\mu})\Phi^A ~=~ \omega_{\mu}{}^{a}{}_{b} ~(\Delta_a{}^b)^A{}_B~\Phi^B.\tag{8}$$

Due to the antisymmetry of the spin connection $\omega_{c, ab}=-\omega_{c, ba}$, it is always possible to write the covariant derivative of the matter fields as $$ (\nabla_c-\partial_c)\Phi^A ~:=~ E^{\mu}{}_c ~(\nabla_{\mu}-\partial_{\mu})\Phi^A ~=~ \frac{1}{2}\omega_{c,ab} ~(\Sigma^{ab}\Phi)^A,\tag{9}$$ $$(\Sigma^{ab}\Phi)^A~:=~(\Sigma^{ab})^A{}_B~\Phi^B \tag{10} $$

where $(\Sigma^{ab})^A{}_B$ is a representation of the $so(3,1)$ Lorentz Lie algebra

$$[\Sigma^{ab},\Sigma^{cd}] ~=~ \left(\eta^{bc} \Sigma^{ad} - (a \leftrightarrow b)\right) -(c\leftrightarrow d), \qquad \Sigma^{ab}~=~-\Sigma^{ba}.\tag{11}$$

II) The covariant Euler-Lagrange equations for the matter fields $\Phi^A$ read

$$0~\stackrel{m}{\approx}~\frac{\delta S}{\delta \Phi^A} ~=~\frac{\partial {\cal L}}{\partial \Phi^A} -{\cal P}^{\mu}_A \stackrel{\leftarrow}{\nabla_{\mu}} , \qquad {\cal P}^{\mu}_A \stackrel{\leftarrow}{\nabla_{\mu}} ~:=~{\cal P}^{\mu}_A \stackrel{\leftarrow}{\partial_{\mu}} -{\cal P}^{\mu}_B~\omega_{\mu,ab}~(\Sigma^{ab})^B{}_A,\qquad \tag{12}$$

where the Lagrangian momenta are

$${\cal P}^{\mu}_A ~:=~\frac{\partial {\cal L}}{\partial (\partial_{\mu}\Phi)^A} ~=~E^{\mu}{}_a ~{\cal P}^a_A,\qquad {\cal P}^a_A~:=~\frac{\partial {\cal L}}{\partial (\nabla_a\Phi)^A}. \tag{13}$$

[Here the $\stackrel{m}{\approx}$ symbol means equality modulo matter eoms.]

III) The Belinfante improvement tensor-density is defined as

$$ 2{\cal B}^{\lambda\mu,\nu} ~:=~{\cal H}^{\lambda,\mu\nu}-{\cal H}^{\mu,\lambda\nu}-{\cal H}^{\nu,\lambda\mu}~=~-(\lambda \leftrightarrow \mu),\tag{14} $$

or inversely

$$ {\cal H}^{\lambda,\mu\nu}~=~{\cal B}^{\lambda\mu,\nu}-{\cal B}^{\lambda\nu,\mu}~=~-(\mu \leftrightarrow \nu),\tag{15}$$


$${\cal H}^{\lambda,\mu\nu}~:=~{\cal H}^{\lambda,ab}~E^{\mu}{}_a~E^{\nu}{}_b\qquad {\cal H}^{\lambda,ab}~:=~{\cal P}^{\lambda}_A~(\Sigma^{ab}\Phi)^A. \tag{16} $$

IV) The variation of the matter action $S$ wrt. to the vielbein yields

$$\delta S ~=~\int \! d^4x\left[L~\delta e +e\frac{\partial L}{\partial(\nabla_c\Phi)^A}~ \delta (\nabla_c\Phi)^A\right] ~=~\int \! d^4x\left[L~\delta e +{\cal P}^c_A~\delta (\nabla_c\Phi)^A\right] ,\qquad \tag{17} $$


$$\delta S -\int \! d^4x\left[L~\delta e + {\cal P}^c_A~\delta E^{\mu}{}_c~ \partial_{\mu}\Phi^A\right] ~\stackrel{(17)}{=}~ \frac{1}{2}\int \! d^4x~{\cal P}^c_A~\delta \omega_{c,ab}~(\Sigma^{ab}\Phi)^A$$ $$~\stackrel{(16)}{=}~ \frac{1}{2}\int \! d^4x~{\cal H}^{c,ab}~ \delta \omega_{c,ab} ~\stackrel{(6)+(14)}{=}~\int \! d^4x~{\cal B}^{cb,a}~ \delta f_{cab} $$ $$~=~\int \! d^4x~{\cal B}^{cb}{}_a~ \delta f_c{}^a{}_b ~\stackrel{(7)}{=}~\int \! d^4x~{\cal B}^{\lambda b}{}_a\left[ \partial_{\lambda} e^a{}_{\nu}~\delta E^{\nu}{}_b +\partial_{\lambda} \delta e^a{}_{\nu}~ E^{\nu}{}_b \right].\tag{18}$$

V) The basic Hilbert SEM tensor-density$^2$ is defined as

$${\cal T}^{\mu\nu}~:=~-2\frac{\delta S_m}{\delta g_{\mu\nu}}, \qquad\qquad(\leftarrow\text{Not applicable!})\tag{19}$$

but this formula (19) is not applicable to e.g. fermionic matter in a curved spacetime. Instead the generalized Hilbert SEM tensor-density is defined as

$${\cal T}^{\mu}{}_{\nu} ~:=~-\frac{\delta S}{\delta e^a{}_{\mu}}e^a{}_{\nu} ~=~E^{\mu}{}_a\frac{\delta S}{\delta E^{\nu}{}_a} ~\stackrel{(18)}{=}~\Theta^{\mu}{}_{\nu}+d_{\lambda}{\cal B}^{\lambda\mu}{}_{\nu}, \tag{20}$$

where $\Theta^{\mu}{}_{\nu}$ is the canonical SEM tensor-density

$$ \Theta^{\mu}{}_{\nu} ~:=~ {\cal P}^{\mu}_A~\partial_{\nu}\Phi^A- \delta^{\mu}_{\nu}{\cal L}.\tag{21} $$

The last expression in eq. (20) is the answer to OP's question about the difference between the Hilbert SEM tensor-density (20) and the canonical SEM tensor-density (21). It is given by the Belinfante improvement tensor-density (14).

IV) The Hilbert SEM tensor-density (20) is symmetric on-shell

$$ {\cal T}^{\mu\nu}~\stackrel{m}{\approx}~{\cal T}^{\nu\mu}, \tag{22} $$

cf. e.g. my Phys.SE answer here, which also explains the connection to Noether's theorems.

Eqs. (15), (20), and (22) imply that the antisymmetric part of the canonical SEM tensor-density (21) is

$$ \Theta^{\mu\nu}-\Theta^{\nu\mu} ~\stackrel{m}{\approx}~d_{\lambda}{\cal H}^{\lambda,\nu\mu}.\tag{23} $$


  1. M.J. Gotay & J.E. Marsden, Stress-Energy-Momentum Tensors and the Belinfante-Rosenfeld Formula, Contemp. Math. 132 (1992) 367.

  2. M. Forger & H. Römer, Currents and the Energy-Momentum Tensor in Classical Field Theory: A fresh look at an Old Problem, Annals Phys. 309 (2004) 306, arXiv:hep-th/0307199.

  3. L.B. Szabados, Quasi-Local Energy-Momentum and Angular Momentum in General Relativity, Liv. Rev. Rel. 12 (2009) 4; Section 2.1.1 p. 11.

  4. A. Bandyopadhyay, Improvement of the Stress-Energy Tensor using Spacetime symmetries, PhD thesis (2001); Chapter 2 & 3.

(Hat tip for Refs. 1 & 2: David Bar Moshe. Hat tip for Refs. 3 & 4: Konstantin Konstantinov.)


$^1$ 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. Capital Roman indices $A,B,C, \ldots,$ denote multiple flat or spinorial indices.

$^2$ 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 $e$.

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    $\begingroup$ More references: 5. L.B. Szabados, On canonical pseudotensors, Sparling's form and Noether currents, Class. Quantum Gravity 9 (1992) 2521. The preprint pdf file is available here. $\endgroup$
    – Qmechanic
    Feb 18, 2017 at 14:02
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    $\begingroup$ Another quite comprehensive review arxiv.org/abs/1605.01121 $\endgroup$ Feb 18, 2017 at 18:41
  • $\begingroup$ "but this formula (19) is not applicable to e.g. fermionic matter in a curved spacetime"... why not? $\endgroup$ Jan 25, 2022 at 22:13
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Jan 26, 2022 at 0:37
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    $\begingroup$ Hi @Evan. Thanks for the feedback. I recorded your suggested edit in v3 so we have it for later. Please note that $d_{\lambda}$ is a total derivative while $\partial_{\lambda}$ is an explicit derivative (unless they act on a fundamental target-space variable where they mean the same). $\endgroup$
    – Qmechanic
    Jun 22, 2023 at 17:36

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