- #1

roomzeig

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[itex]\mathcal{L} = \frac{1}{4}F_{\mu\nu}F_{\mu\nu}, \quad F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu[/itex]

For arbitrary translational symmetries, the Noether conserved current evaluates to:

[itex]T^{\mu \nu}=-F^{\mu \sigma}\eta^{\nu\lambda} \partial_\lambda A_\sigma + \tfrac{1}{4} \eta^{\mu\nu} F_{\alpha \beta} F^{\alpha \beta}[/itex]

which is almost but not equal to the classical enm stress energy tensor:

[itex]\hat T^{\mu \nu}=-F^{\mu \sigma}\eta^{\nu\lambda} F_{\lambda \sigma} + \tfrac{1}{4} \eta^{\mu\nu} F_{\alpha \beta} F^{\alpha \beta}[/itex]

with the difference being a total divergence:

[itex]K^{\mu\nu} \equiv \hat T^{\mu\nu} - T^{\mu\nu} = \partial_\sigma \left( F^{\mu \sigma} A^{\nu} \right) [/itex]

I understand that the difference does not change global conservation etc, but why is it that the Noether procedure does not produce a *symmetric* classical stress energy tensor? is there a physical meaning of K^{\mu\nu}? if \hat T^{\mu nu} is interpreted, on the grounds of ENM, as stress-energy, what is T^{\mu\nu}?

Thanks.