Scalar curvature of metric? The metric components in a two-dimensional spacetime are given in terms of the coordinates $(t, x)$ by$$ds^2 = f(x)\,dt^2 + dx^2.$$What is the scalar curvature, $R$, of this metric?
 A: Define the orthonormal basis
$$ e^0 = f \, dt, \quad e^1 = dx. $$
Then
$$ de^0 = \frac{f'}{f} e^1 \wedge e^0, \quad de^1 = 0. $$
Cartan's first structural equation,
$$ de^i = -\omega^i{}_j \wedge e^j, $$
implies that
$$ \omega^0{}_1 = \frac{f'}{f} e^0, $$
which by antisymmetry is effectively the only nonzero curvature form in two dimensions.
The second structural equation is
$$ \Theta^i{}_j = \frac{1}{2} R^i{}_{jkl} e^i \wedge e^j = d\omega^i{}_j + \omega^i{}_k \wedge \omega^k{}_j, $$
and the second term on the right-hand side vanishes in two dimensions because there aren't enough dimensions for the indices $i,j,k$ to be distinct.
Then,
$$ \Theta^0{}_1 = d\left( f' dt \right) = f'' dx \wedge dt = \frac{f''}{f} e^1 \wedge e^0, $$
so
$$ R^0{}_{110} = \frac{f''}{f}, $$
or
$$ R^1{}_{010}=-\frac{f''}{f} $$
in the more useful form, and the Ricci tensor is thus
$$ R_{11}=R_{22}=-\frac{f''}{f}, $$
and summing gives
$$ R=\frac{2f''}{f}. $$
This is by far the quickest way I know to calculate curvatures, especially in two dimensions when the extra wedge term is not present.
