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Given the Lagrangian density for a real scalar field $\mathcal{L}(\phi, \partial_\mu \phi)$, we obtain from Noether's theorem the canonical stress-energy tensor $$ T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \partial_\nu \phi - g^{\mu \nu} \mathcal{L} $$ Can anyone tell me, or provide a (hopefully detailed) reference, why there is always (I think?) a rank $3$ tensor $X^{\lambda \mu \nu}$ such that $$\widetilde{T}^{\mu \nu} = T^{\mu \nu} + \partial_{\lambda} X^{\lambda \mu \nu}$$ is symmetric? I'm fine with assuming $g$ is the Minkowski metric, but I can only find this done in particular cases of $\mathcal{L}$ and no general argument.

Qmechanic
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Pedro
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    See Wikipedia: Belinfante–Rosenfeld stress–energy tensor – G. Smith Mar 04 '21 at 04:34
  • Here is the important part of that article: “The curious combination of spin-current components required to make $T_B^{\mu\nu}$ symmetric and yet still conserved seems totally ad hoc, but it was shown by both Rosenfeld and Belinfante that the modified tensor is precisely the symmetric Hilbert energy-momentum tensor that acts as the source of gravity in general relativity.” – G. Smith Mar 04 '21 at 04:37
  • For a reference, I know this is discussed in the relevant section of Di Francesco et al.'s conformal field theory book, though only briefly. – Richard Myers Mar 04 '21 at 05:03
  • Possible duplicates: https://physics.stackexchange.com/q/119838/2451 , https://physics.stackexchange.com/q/68564/2451 and links therein. – Qmechanic Mar 04 '21 at 06:02
  • For those interested in a complete reference, see the article 'Stress-Energy-Momentum Tensors and the Belinfante-Rosenfeld Formula' Gotay, M. J. and J. E., Marsden Contemp. Math., (1992) 132, 367-392 – Pedro Mar 05 '21 at 01:25

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