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Conserved Noether charges for Lorentz symmetry of the action
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[QUOTE="strangerep, post: 5458381, member: 70760"] I suspect you're missing something here. I get: $$\partial_{\mu}(\mathcal{J}^{\mu})^{\rho\sigma}~=~ -(\partial_{\mu}T^{\mu\rho})x^{\sigma} - T^{\mu\rho} \delta^\sigma_\mu + (\partial_{\mu}T^{\mu\sigma})x^{\rho} + T^{\mu\sigma} \delta^\sigma_\mu .$$ The derivatives vanish as you said, but that still leaves ##-T^{\sigma\rho} + T^{\rho\sigma},## which only vanishes if ##T## is symmetric. Usually, that requires an additional argument, wherein one adds a particular total derivative to ##T## which preserves the conservation rule, but also guarantees that its integrals over all space are unchanged. (This might also be the source of your uncertainty about "total angular momentum", which involves such an integral.) Can you access a copy of Greiner & Reinhardt? They explain quite a bit more detail than what's in Tong's notes -- far more than I can reasonably reproduce in a PF post. [/QUOTE]
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Conserved Noether charges for Lorentz symmetry of the action
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