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


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.


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):



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.

  • 2
    $\begingroup$ Your teacher's formula looks right. Care to show your work? $\endgroup$ – Michael Brown Mar 11 '13 at 5:42
  • 1
    $\begingroup$ Your latest expression $Ruu$ has wrong indices. When summed over, $ij$ should be left, not $il$. $\endgroup$ – Luboš Motl Mar 11 '13 at 5:52
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    $\begingroup$ All of your manipulations look correct. Keep in mind that the final answer has orthonormal-frame indices, not coordinate indices. (I usually use $\hat{\ }$ to denote frame indices just to avoid the perennial confusion.) Are you sure the answer you are comparing to has the correct indices? If not then the conversion will involve the tetrads and this will bring in the factors you mention. $\endgroup$ – Michael Brown Mar 11 '13 at 7:04
  • $\begingroup$ I'm pretty sure this is not the right answer because the solution shows something completely different. I thought maybe the formula that I was given might not have been correct... $\endgroup$ – Atreyu Mar 11 '13 at 7:41

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

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  • 3
    $\begingroup$ This happens to everyone (me included). Like I said, it's helpful to introduce some notation just to clearly seperate the two types of indices. :) $\endgroup$ – Michael Brown Mar 12 '13 at 5:22
  • $\begingroup$ @MichaelBrown: Please could you show me how the exterior derivative and wedge product is taken, I am new to this notation, and not being able to reproduce the calculations of the OP. If I am not wrong $(dX)^a_{\mu \nu}=\partial_{\mu}X_{\nu}^a-\partial_{\nu}X_{\mu^a}$. Then how do I get for e.g. $du^\phi=-\frac{1}{r}[e^{-b}u^\phi\wedge u^r+\cot \theta u^\phi\wedge u^\theta]$. What does $u^\phi\wedge u^\theta$ mean, and why are there negative exponents, and I have no idea how to get these exterior derivatives. $\endgroup$ – user7757 Jun 3 '13 at 1:33

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