[Physics] Calculating Riemann Tensor Using Tetrad Formalism

Question

I was trying to calculate the Riemann Tensor for a spherically symmetric metric:

$ds^2=e^{2a(r)}dt^2-[e^{2b(r)}dr^2+r^2d\Omega^2]$

I chose the to use the tetrad basis:

$u^t=e^{a(r)}dt;\, u^r=e^{b(r)}dr;\, u^\phi=r\sin\theta d\phi; \, u^\theta=rd\theta$

Using the torsion free condition with the spin connection $\omega^a{}_{b}\wedge u^b=-du^a$ I was able to find the non-zero spin connections.

In class my teacher presented the formula:

$\Omega^i{}_{j}=d\omega^i{}_j+\omega^i{}_k\wedge \omega^k{}_j=\frac{1}{2}R^i{}_j{}_k{}_l\,u^k\wedge u^l$

But this can't be right since I calculate with this:

$\Omega^t{}_\phi=-\frac{1}{r}a_r \,e^{-2b}u^t\wedge u^\phi \implies R^t{}_\phi{}_\phi{}_t=-\frac{1}{r}a'e^{-2b}$

The real answer involves a factor of $e^a$, $\sin \theta$ and no $\frac{1}{r}$ term.

Any help is appreciated.

Edit:

Here is some of my work:

$du^t=-a_re^{-b}u^t\wedge u^r$

$du^\phi=-\frac{1}{r}[e^{-b}u^\phi\wedge u^r+\cot \theta u^\phi\wedge u^\theta]$

(I will not show the calculations for the rest)

From no-torsion equation, we get 2 out of 4 spin connections (the rest require the two missing exterior derivatives that I have not shown in this post):

$\omega^t{}_r=a_re^{-b}u^t$

$\omega^\phi{}_r=\frac{1}{r}e^{-b}u^\phi$

Then $\Omega^t{}_\phi$ is as shown above. Explicitly:

$\Omega^t{}_\phi=d\omega^t{}_\phi+\omega^t{}_r\wedge \omega^r{}_\phi+\omega^t{}_\theta\wedge \omega^\theta{}_\phi$

where the first and last terms are 0 since $\omega^t{}_\phi$ and $\omega^t_\theta$ are 0.

Answer 1

Turns out Michael Brown was right after all. The calculations are correct for the curvature terms in the tetrad basis.