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?
Answer
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.
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