- #1

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## Main Question or Discussion Point

How do you prove that the energy-momentum tensor is divergence-free?

∂μTμν=0

∂μTμν=0

- Thread starter ClaraOxford
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- #1

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How do you prove that the energy-momentum tensor is divergence-free?

∂μTμν=0

∂μTμν=0

- #2

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∂[itex]_{\mu}[/itex]T[itex]^{\mu\nu}[/itex]=0

T[itex]^{\mu\nu}[/itex]=F[itex]^{\mu\alpha}[/itex]F[itex]^{\nu}[/itex][itex]_{\alpha}[/itex]-1/4F[itex]^{\alpha\beta}[/itex]F[itex]_{\alpha\beta}[/itex][itex]\eta[/itex][itex]^{\mu\nu}[/itex]

I don't know whether to use Lagrangian variables or the Einstein tensor or if there's a simpler way to just expand the tensor and work it out?

- #3

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[tex]

\partial_\nu F^{\mu \nu} = J^\mu, \partial_{\mu} F^{\nu \rho} + \partial_{\nu} F^{\rho \mu} + \partial_{\rho} F^{\mu \nu} = 0, \; F^{\mu \nu} = -F^{\nu \mu}

[/tex]

- #4

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It won't be divergence-free if you use those equations. Instead use the vacuum Maxwell equations (above with J=0). Alternatively use the above to find the divergence to equal [tex]F_{ab}J^b[/tex] (up to sign).

[tex]

\partial_\nu F^{\mu \nu} = J^\mu, \partial_{\mu} F^{\nu \rho} + \partial_{\nu} F^{\rho \mu} + \partial_{\rho} F^{\mu \nu} = 0, \; F^{\mu \nu} = -F^{\nu \mu}

[/tex]

- #5

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Ah, of course. There is work done on charges by the electromagnetic field. The above energy gives the 4-Lorentz force per unit volume.It won't be divergence-free if you use those equations. Instead use the vacuum Maxwell equations (above with J=0). Alternatively use the above to find the divergence to equal [tex]F_{ab}J^b[/tex] (up to sign).

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