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ClaraOxford
Oct16-11, 05:23 AM
How do you prove that the energy-momentum tensor is divergence-free?

∂μTμν=0

ClaraOxford
Oct16-11, 05:42 AM
I mean

∂_{\mu}T^{\mu\nu}=0

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


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?

Dickfore
Oct16-11, 05:44 AM
use the fact that:

\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}

Sam Gralla
Oct17-11, 05:09 PM
use the fact that:

\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}


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 F_{ab}J^b (up to sign).

Dickfore
Oct17-11, 10:59 PM
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 F_{ab}J^b (up to sign).

Ah, of course. There is work done on charges by the electromagnetic field. The above energy gives the 4-Lorentz force per unit volume.