Whats the quickest way to compute the Ricci tensor? I have been going through exam papers and often they ask us to calculate ricci tensor components and affine connections from a given metric.
They seem to take far too long for the time you are supposed to spend on it.  Is there a quick way of evaluating these?
 A: I'll assume you want a pen-and-paper method for the Levi-Cvita connection & curvature and that, as it came from an exam question, the metric is fairly simple, with few nonzero components etc.
The way I compute the connection coefficients starts with the following functional of a path:
$$
  S[x(\tau)] = \frac{1}{2}\int g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu d\tau.
$$
This functional is extremised by geodesics, which satisfy:
$$
  \ddot{x}^\lambda + \Gamma^\lambda _{\mu\nu} \dot{x}^\mu \dot{x}^\nu = 0.
$$
where dots indicate derivatives wrt. $\tau$.
The process is then:

*

*Write out the functional above explicitly.

*Find the Euler-Lagrange equations.

*Solve for $\ddot{x}^\lambda$.

*Read off the $\Gamma^\lambda_{\mu\nu}$, being careful with factors of 2 in the off-diagonal parts over $\mu,\nu$.

Here's one quarter of an example, the Schwarzschild metric:
\begin{align}
 ds^2 &= -\left(1 - \frac{r_s}{r}\right) dt^2 + \left(1 - \frac{r_s}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2, \\
S &= \frac{1}{2} \int -\left(1 - \frac{r_s}{r}\right) \dot{t}^2 + \left(1 - \frac{r_s}{r}\right)^{-1} \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2\theta \dot{\phi}^2 \  d\tau.
\end{align}
The Euler-Lagrange equation for $t$ is
\begin{align}
  \frac{d}{d\tau} \left[ -\left(1 - \frac{r_s}{r}\right) \dot{t} \right] &= 0, \\
  \implies \ddot{t} + \frac{r_s}{r^2} \left(1 - \frac{r_s}{r}\right)^{-1} \dot{t} \dot{r} &= 0, \\
\implies \Gamma^t_{tr} = \Gamma^t_{rt} &= \frac{r_s}{2r^2} \left(1 - \frac{r_s}{r}\right)^{-1},
\end{align}
and all other $\Gamma^t_{\mu\nu} = 0$. I'll leave the other three quarters of this example for you.
I've been told that the following method for the curvature calculation is the fastest, though I haven't used it much. It's based on the spin-connection approach. See appendix of Carroll, "Spacetime and Geometry". Using the differential form notation lets you combine several equations into one.
First, let's pick a convenient vielbein and its inverse:
\begin{align}
 \mathbf{e}^0 &= \left(1 - \frac{r_s}{r}\right)^{1/2} \mathbf{d}t, &
 \mathbf{e}^1 &= \left(1 - \frac{r_s}{r}\right)^{-1/2} \mathbf{d}r, &
 \mathbf{e}^2 &= r \mathbf{d}\theta, &
 \mathbf{e}^3 &= r \sin\theta \mathbf{d}\phi. \\
 \mathbf{e}_0 &= \left(1 - \frac{r_s}{r}\right)^{-1/2} \frac{\partial}{\partial t}, &
 \mathbf{e}_1 &= \left(1 - \frac{r_s}{r}\right)^{1/2} \frac{\partial}{\partial r}, &
 \mathbf{e}_2 &= \frac{1}{r} \frac{\partial}{\partial \theta}, &
 \mathbf{e}_3 &= \frac{1}{r \sin\theta} \frac{\partial}{\partial \phi}.
\end{align}
For the Christoffel connection the zero torsion condition is simple, $\Gamma^\lambda_{\mu\nu} = \Gamma^\lambda_{\nu\mu}$, and metric compatibility is the condition used to compute it. For the spin connection it's the other way round. Metric compatibility is $\boldsymbol{\omega}^{ab} = -\boldsymbol{\omega}^{ba}$. The zero torsion condition is $\mathbf{d}\mathbf{e}^a + \boldsymbol{\omega}^a{}_b \wedge \mathbf{e}^b = 0$. For $a=0$:
\begin{align}
 \mathbf{d}\mathbf{e}^0 &= \frac{r_s}{2r^2} \left(1 - \frac{r_s}{r}\right)^{-1/2} \mathbf{d}r \wedge \mathbf{d}t \\
   &= - \frac{r_s}{2r^2} \mathbf{d}t \wedge \mathbf{e}^1, \\
\implies \boldsymbol{\omega}^0{}_1 = \boldsymbol{\omega}^1{}_0 &= \frac{r_s}{2r^2} \mathbf{d}t.
\end{align}
We used $\boldsymbol{\omega}^0{}_0 = 0$ to derive the last line.
The curvature 2-form and the Ricci tensor are given by
$$
  \mathbf{R}^a{}_b = \mathbf{d} \boldsymbol{\omega}^a{}_b + \boldsymbol{\omega}^a{}_c \wedge  \boldsymbol{\omega}^c{}_b,
\qquad
  R_{\mu\nu} = R^a{}_{b \lambda \nu} e^b{}_\mu e_a{}^\lambda.
$$
