Scalar curvature of metric?

Question

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?

Answer 1

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.