# Covariant derivative in a basis

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.

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,

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

• 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 Jun 2 '19 at 12:53
• @MaxtronMoon please see the edits above. Hopefully, this helps. Jun 2 '19 at 13:18