Writing Non-Coordinate Basis Continuity Eqs: $\nabla_a T^{ab}$

[pervect]

The continuity equations are $\nabla_a T^{ab} = 0$. In a coordinate basis, we can write out the resulting differential equations as:

$\nabla_a T^{ab} = \partial_a T^{ab} + \Gamma^a{}_{ac}T^{cb} + \Gamma^b{}_{ac}T^{ac}$

(modulo possible typos, though I tried to be careful-ish).

What do we do in a non-coordinate basis, $T^{\hat{a}\hat{b}}$ where we might have the ricci rotation coefficients $\omega^a{}_{bc}$ (or perhaps $\omega_{abc}$ if that's simpler), rather than the Christoffel symbols?

Right now, I'd just convert back to a coordinate basis, take the covariant deriatiave in that coordinate basis, then reconvert back to the non-coordinate basis, but I'd like to see if there is a more direct way.

 

[Mentz114]

The continuity equations are $\nabla_a T^{ab} = 0$. In a coordinate basis, we can write out the resulting differential equations as:

$\nabla_a T^{ab} = \partial_a T^{ab} + \Gamma^a{}_{ac}T^{cb} + \Gamma^b{}_{ac}T^{ac}$

(modulo possible typos, though I tried to be careful-ish).

What do we do in a non-coordinate basis, $T^{\hat{a}\hat{b}}$ where we might have the ricci rotation coefficients $\omega^a{}_{bc}$ (or perhaps $\omega_{abc}$ if that's simpler), rather than the Christoffel symbols?

Right now, I'd just convert back to a coordinate basis, take the covariant deriatiave in that coordinate basis, then reconvert back to the non-coordinate basis, but I'd like to see if there is a more direct way.

In section 11.9 of the text cited below, Hamilton defines the tetrad covariant derivative which I think is what you are looking for.

Good luck calculating the RRCs !

General Relativity, Black Holes, and Cosmology
Andrew J. S. Hamilton

available here http://jila.colorado.edu/~ajsh/astr5770_14/grbook.pdf

 

[WannabeNewton]

You just compute the structure constants $C^{\hat{\alpha}}{}{}_{\hat{\beta}\hat{\gamma}}$ of the tetrad and use the formula for the connection coefficients in terms of the structure constants: $\Gamma^{\hat{\alpha}}_{\hat{\beta}\hat{\gamma}} = \frac{1}{2}g^{\hat{\alpha}\hat{\delta}}(C_{\hat{\beta}\hat{\delta}\hat{\gamma}} + C_{\hat{\gamma}\hat{\delta}\hat{\beta}} - C_{\hat{\beta}\hat{\gamma}\hat{\delta}})+ \frac{1}{2}g^{\hat{\alpha}\hat{\delta}}(g_{\hat{\beta}\hat{\delta}, \hat{\gamma}} + g_{\hat{\gamma}\hat{\delta}, \hat{\beta}} - g_{\hat{\beta}\hat{\gamma}, \hat{\delta}})$.

For a local Lorentz frame the second term vanishes.

 

[martinbn]

Presumably the connection forms had been calculated at this point and the gammas are just the coefficients of the connection forms expressed in terms of the 1-form basis
$\omega^\hat\alpha_\hat\beta = \Gamma^\hat\alpha_{\hat\beta\hat\gamma}\omega^\hat\gamma$