Surely there must be a better approach than this?

Certainly; it's the Cartan formalism which employs differential forms. Consider your case of a sphere, with a metric tensor,

$$ds^2 = r^2dr^2+r^2\sin^2 \theta \, d\theta^2$$

We can choose an orthonormal basis $e^a$ such that $ds^2 = \eta_{ab}e^{a}e^{b}$, and $e^a = e^a_\mu dx^\mu$. For our case:

$$e^r = r \, dr, \, \, e^\theta = r\sin \theta \, d\theta$$
We now compute the exterior derivative of each, and re-express them in terms of the basis:

$$de^r = 0, \quad de^\theta = \sin \theta \, dr \wedge d\theta = \frac{1}{r^2} \, e^r \wedge e^\theta$$

We can now use the first structure equation,$^{1}$ namely $de^a + \omega^a_{b} \wedge e^b = 0$, to deduce the connection components $\omega^a_b$. With some experience, you'll be able to read them off:

$$\omega^r_r = \omega^\theta_\theta = 0, \quad \omega^\theta_r = \frac{1}{r^2}e^\theta = \frac{1}{r}\sin \theta \, d\theta$$

The second structure equation allows us to directly compute components of the Ricci tensor $R^{a}_b$ in the orthonormal basis, namely, $R^a_b = d\omega^a_b + \omega^a_c \wedge \omega^c_b$. We find that,

$$R^\theta_\theta = R^r_r = 0, \quad R^\theta_r = \frac{1}{r^4} e^\theta \wedge e^r$$

Given that $R^a_{b} = R^a_{bcd} e^c \wedge e^d$, we can compute the terms of the full Riemann tensor. To convert back from our orthonormal basis, we can use the relation

$$R^\lambda_{\mu \nu \sigma} = ( e^{-1})^{\lambda}_a \, R^a_{bcd} \, e^b_\mu \, e^c_\nu \, e^d_\sigma$$

By this method, we find for example, $R^\theta_{r\theta r} = \sin^2 \theta$ as expected for the sphere. Generally, the Cartan formalism is much easier than direct computation, but be careful in particular when,

- Deducing the connections from the first structure equation; it's not always straightforward;
- Converting back to the orthonormal basis.

For an excellent exposition of the method, see the gravitational physics lectures by Professor Ruth Gregory available from the Perimeter Scholars website.

$1$ In fact, the structure equation states $de^a + \omega^a_b + e^b = T^a$, where $T^a$ is the torsion, but in general relativity, we may assume $T^a = 0$. See my own question: Why can we assume torsion is zero in GR?

This post imported from StackExchange Physics at 2016-07-21 06:45 (UTC), posted by SE-user JamalS