0
$\begingroup$

Reading through this paper, I saw that the energy momentum conservation:

$$\nabla_\mu T^{\mu\nu}=0$$

can be evaluated as:

$$\partial_t(\sqrt{-g}T^{t}_\nu)=-\partial_i(\sqrt{-g}T^{i}_\nu)+\sqrt{-g}T^{\kappa}_\lambda\Gamma_{\nu\kappa}^{\lambda}$$

Can someone explain how this is done?

I guess this might be a simple question, but not having the experience, I would appreciate any help.

$\endgroup$
0
1
$\begingroup$

Applying the Levi-Civita connection (torsion-free and metric compatible) of a rank 2 covariant tensor in a coordinate patch $(t,x^i)$ has the following evaluation (please comment further if you want a derivation of this law),

$ \nabla_{\alpha}T^{\mu\nu}=\partial_{\alpha}T^{\mu\nu}+\Gamma^{\mu}_{\alpha\beta}T^{\beta\nu}+\Gamma^{\nu}_{\alpha\beta}T^{\mu\beta}$

Contracting on $\alpha$ and $\mu$ gives,

$ \nabla_{\mu}T^{\mu\nu}=\partial_{\mu}T^{\mu\nu}+\Gamma^{\mu}_{\mu\beta}T^{\beta\nu}+\Gamma^{\nu}_{\mu\beta}T^{\mu\beta}=0$

by conservation. Multiplying by $\sqrt{-g}$ gives,

$\partial_{\mu}(\sqrt{-g}T^{\mu\nu})-T^{\mu\nu}\partial_{\mu}(\sqrt{-g})+\sqrt{-g}\Gamma^{\mu}_{\mu\beta}T^{\beta\nu}+\sqrt{-g}\Gamma^{\nu}_{\mu\beta}T^{\mu\beta}=0$

Note that,

$\frac{\partial g}{\partial g_{\alpha\beta}}=gg^{\alpha\beta}\implies \partial_{\mu}(\sqrt{-g})=\frac{\sqrt{-g}}{2}g^{\alpha\beta}\partial_{\mu}g_{\alpha\beta}$

Further,

$\Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}g^{\mu\nu}(\partial_{\alpha}g_{\beta\nu}+\partial_{\beta}g_{\alpha\nu}-\partial_{\nu}g_{\alpha\beta})\implies \Gamma^{\mu}_{\mu\beta}=\frac{1}{2}g^{\mu\nu}(\partial_{\mu}g_{\beta\nu}+\partial_{\beta}g_{\mu\nu}-\partial_{\nu}g_{\mu\beta})=\frac{1}{2}g^{\mu\nu}(\partial_{\beta}g_{\mu\nu})$

by the symmetry of the inverse metric. So,

$\partial_{\mu}(\sqrt{-g})=\sqrt{-g}\Gamma^{\alpha}_{\alpha\mu}$

from the chain rule. Finally,

$\partial_{\mu}(\sqrt{-g}T^{\mu\nu})-\sqrt{-g}T^{\mu\nu}\Gamma^{\alpha}_{\alpha\mu}+\sqrt{-g}\Gamma^{\mu}_{\mu\beta}T^{\beta\nu}+\sqrt{-g}\Gamma^{\nu}_{\mu\beta}T^{\mu\beta}=0$

Hence the result. Note I've been liberal with my use of relabelling dummy indices. Hopefully, this doesn't cause problems.

Edit: elaboration on derivative of the determinant.

We can write the determinant as,

$g=\sum_{\nu=0}^{n-1}g_{\mu\nu}C^{\mu\nu}$

where $(C^{\mu\nu})$ is the cofactor matrix. Hence,

$\frac{\partial g}{\partial g_{\mu\nu}}=C^{\mu\nu}$

Next recall that for an invertible matrix $A$, we have $A^{-1}=\frac{1}{\det A}C^T$, so $C^T=(\det A)A^{-1}$. The inverse metric is symmetric so we have,

$C^{\mu\nu}=g g^{\mu\nu}$.

$\endgroup$
2
  • $\begingroup$ Thank you very much for your service. Could you elaborate more on $\frac{\partial g}{\partial g_{\alpha\beta}}=gg^{\alpha\beta}\implies \partial_{\mu}(\sqrt{-g})=\frac{\sqrt{-g}}{2}g^{\alpha\beta}\partial_{\mu}g_{\alpha\beta}$ On the first derivative and how it implies the latter $\endgroup$ Jun 2 '19 at 12:53
  • $\begingroup$ @MaxtronMoon please see the edits above. Hopefully, this helps. $\endgroup$
    – Sam Colley
    Jun 2 '19 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.