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Hilbert energy-momentum tensor is generally defined as

$$T_{\mu\nu} \delta g^{\mu\nu} = -2\delta\mathcal{L} + g_{\mu\nu} \mathcal{L} \delta g^{\mu\nu}$$

where $\mathcal{L}$ is the Lagrangian density of the field. Now if the field is "on-shell", then according to this Physics SE answer [comment below Eq. 3] and this paper [Non-Existence of Black Hole Solutions for a Spherically Symmetric, Static Einstein-Dirac-Maxwell System, Eq. 3.8], in case of a "on-shell" Dirac field, the second term in the above expression, $g_{\mu\nu} \mathcal{L} \delta g^{\mu\nu}$ will be zero.

But in these papers [Cosmological model with non-minimally coupled fermionic field (Eq. 8) and Spinors, Inflation, and Non-Singular Cyclic Cosmologies (Eq. 12)], the aforementioned $g_{\mu\nu} \mathcal{L} \delta g^{\mu\nu}$ term for "on-shell" Dirac field has been kept as it is, without making it zero!

Therefore,

  1. Which one is correct? Will the aforementioned term be zero?
  2. If so, then why?
SCh
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  • For the proper approach to the stress tensor for spinor fields, see https://physics.stackexchange.com/q/315830/ – Connor Behan Oct 07 '23 at 03:04
  • Thank you, but this answer has not addressed the issue of "on-shell" Dirac field. My question is not about how to derive the stress-energy tensor for a spinor field, but rather it concerns about the fate of $g_{\mu\nu}\mathcal{L}\delta g^{\mu\nu}$ term if that field is "on-shell". – SCh Oct 07 '23 at 03:22
  • Right. Your question is about the fate of a term that doesn't appear in the proper approach. The derivative acting on a spinor cannot be generalized to curved space using only a metric. – Connor Behan Oct 07 '23 at 04:00
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    Actually scratch that. I think the term shows up in the papers you linked simply because their Lagrangians have a potential. The Phys.SE answer by contrast deals with the minimally coupled Dirac Lagrangian. – Connor Behan Oct 07 '23 at 04:05

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