# Divergence of Energy-momentum Tensor

## Main Question or Discussion Point

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

∂μTμν=0

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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?

use the fact that:
$$\partial_\nu F^{\mu \nu} = J^\mu, \partial_{\mu} F^{\nu \rho} + \partial_{\nu} F^{\rho \mu} + \partial_{\rho} F^{\mu \nu} = 0, \; F^{\mu \nu} = -F^{\nu \mu}$$

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).

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.