Suppose that there exists some arbitrary scalar field $\phi$ s.t. $\phi$ has an associated energy-momentum tensor given by $$T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi-\frac12g_{\mu\nu}(-\frac12\partial_\alpha\phi\partial^\alpha\phi-V(\phi))$$ where the matter lagrangian $$\mathcal{L}_m\equiv-\frac12\partial_\alpha\phi\partial^\alpha\phi-V(\phi).$$ Given this definition for $T_{\mu\nu}$, can this definition be extended to a matter field $\psi$, just replacing $\phi$ with $\psi^{(x)}$? I was told that this definition can be used for any field that has a lagrangian description, however this is the definition for strictly a scalar field and Im just curious as to whether this can be extended to the matter fields.
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No, this expression is valid only for the particular case of the scalar field with the Lagrangian: $$ \mathcal{L} = -\frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - V (\phi) $$ Quite general way to derive the expression for stress-energy tensor is via computing variation with respect to the change of metric. Given a Lagrangian: $$ S = \int d^{D} x \sqrt{|g|} \mathcal{L} $$ The Hilbert stress energy tensor is computed as follows: $$ T_{\mu \nu} = \frac{1}{\sqrt{|g|} } \frac{\delta S}{\delta g^{\mu \nu}} $$ Another way is to compute the stress-energy Tensor from the Noether's theorem explicitly, as in the derviation in Wikipedia article, but this tensor won't be manifestly symmetric.
Since you are speaking about matter fields, for fermions situation is more tricky. The covariant expression for action involves the vierbein, and in order to compute the stress energy-tensor one has to compute variation with respect to this vierbein (see Belinfante–Rosenfeld stress–energy tensor): $$ T_{cb} ^{(0)} \eta^{ca} e^{*b}_\mu= \frac{1}{\sqrt{g}}\left.\left(\frac{\delta S_{\rm eff}}{\delta e^\mu_a}\right)\right|_{\omega^{ab}_\mu} $$
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