Divergence of Energy-momentum Tensor

[ClaraOxford]

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

∂μTμν=0

 

[ClaraOxford]

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]

use the fact that:
<br /> \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}<br />

 

[Sam Gralla]

use the fact that:
<br /> \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}<br />

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]

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.