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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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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. –  Winther Jun 18 at 21:07
    
See also discussion in Peskin & Schroeder page 683. –  PPR Jul 16 at 22:51

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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.

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