Evaluating the Ricci tensor effectively

If given a metric of the form $$ds^2=\alpha^2(dr^2+r^2d\theta^2)$$ where $\alpha=\alpha(r)$, then can one immediately conclude that $$R_{\theta\theta}=r^2R_{rr}$$ where $R_{ab}$ is the Ricci tensor, without doing any explicit calculations? I can show that this is true by the long-winded way of computing both explicitly, but it would seem that there may be a more elegant way?

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More on evaluating Ricci tensor: physics.stackexchange.com/q/62907/2451 –  Qmechanic May 11 '13 at 0:18
@Qmechanic: Thanks for the link :) However, I'm afraid I still haven't been able to see if the above mentioned is possible... –  val May 11 '13 at 9:42
Apart from the sufficient answer by Qmechanic, this calculation isn't long-winded at all. The Riemann tensor only has 1 independent component in 2d! –  Vibert May 11 '13 at 10:21

1 Answer

I'm not sure what OP exactly is requesting, but OP's equation follows e.g. from the general fact that for an arbitrary 2D surface, the Ricci tensor

$$R_{\mu\nu} ~\propto~g_{\mu\nu}$$

is always proportional to the metric tensor $g_{\mu\nu}$. This is basically a consequence of that in 2D the Riemann curvature tensor is complete determined by the scalar curvature.

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Sorry for the novice and off-topic question, but what OP exactly stands for? I'm not familiar with this abbreviature yet. –  firtree May 11 '13 at 11:37
OP=Original Poster, in this case val. –  Qmechanic May 11 '13 at 11:42
Thanks, @Qmechanic! –  val May 13 '13 at 16:00