User:RandomPoint~enwiki/Cartan Approach

The Cartan Approach

An alternative derivation of the Schwarzschild solution exists where curvature is computed using exterior differential forms. This is also known as the Cartan approach. We will use the following form of the metric in this derivation:

$ds^{2}=-e^{2\alpha (r)}dt^{2}+e^{2\beta (r)}dr^{2}+r^{2}d\Omega ^{2}\,$

where

$d\Omega ^{2}=d\theta ^{2}+\sin ^{2}\theta d\phi ^{2}\,$

The procedure is then as follows, and is generally applicable.

Orthonormal basis

We start by introducing an orthonormal basis:

$ds^{2}=-({\underline {d\omega }}^{t})^{2}+({\underline {d\omega }}^{r})^{2}+({\underline {d\omega }}^{\theta })^{2}+({\underline {d\omega }}^{\phi })^{2}\,$

We then get that

${\underline {\omega }}^{t}=e^{\alpha (r)}{\underline {d}}t,\quad {\underline {\omega }}^{r}=e^{\beta (r)}{\underline {d}}r,\quad {\underline {\omega }}^{\theta }=r{\underline {d}}\theta ,\quad {\underline {\omega }}^{\phi }=r\sin \theta {\underline {d}}\phi \,$

Connection forms

Next step is to compute the connection forms by applying Cartan's 1. structure equations:

${\underline {d\omega }}^{\mu }=-{\underline {\Omega }}_{\;\;\nu }^{\mu }\wedge {\underline {\omega }}^{\nu }$

This gives

${\underline {d\omega }}^{t}=e^{\alpha }\alpha '{\underline {d}}r\wedge {\underline {d}}t=-e^{-\beta }\alpha '{\underline {\omega }}^{t}\wedge {\underline {\omega }}^{r}=-{\underline {\Omega }}_{\;\;r}^{t}\wedge {\underline {\omega }}^{r}$

and thus

${\underline {\Omega }}_{\;\;r}^{t}=e^{-\beta }\alpha '{\underline {\omega }}^{t}+f_{1}{\underline {\omega }}^{r}$

Note that the last term will vanish when we insert this back into Cartan's 1. structure equation.

Determining the f-functions

Next we must determine the f-functions in the expressions for the connection forms. This is done by applying the anti-symmetric properties

${\underline {\Omega }}_{\mu \nu }=-{\underline {\Omega }}_{\nu \mu }\,$

Using the form from the last paragraph as example we get when raising indices that

${\underline {\Omega }}_{\;\;r}^{t}={\underline {\Omega }}_{\;\;t}^{r}$

When explicitly taking the exterior derivative of ${\underline {\omega }}^{r}\,$ we get that this is 0. Comparing this to Cartan's 1. structure equation we see that we can write

${\underline {\Omega }}_{\;\;\nu }^{r}=h{\underline {\omega }}^{\nu }\,$

which gives for $\nu =t\,$ the connection form ${\underline {\Omega }}_{\;\;t}^{r}$ which we set equal to ${\underline {\Omega }}_{\;\;r}^{t}$ and get that

${\underline {\Omega }}_{\;\;t}^{r}={\underline {\Omega }}_{\;\;r}^{t}=e^{-\beta }\alpha '{\underline {\omega }}^{t}$