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:

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

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:

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

This gives

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

and thus

$$\underline{\Omega}^t_{\;\;r} = e^{-\beta}\alpha'\underline{\omega}^t+f_1\underline{\omega}^r$$

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

$$\underline{\Omega}^t_{\;\;r} = \underline{\Omega}^r_{\;\;t}$$

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

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

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

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

Curvature forms
We now determine the curvature forms by applying Cartan's 2. structure equation:

$$\underline{R}^{\mu}_{\;\;\nu} = \underline{d\Omega}^{\mu}_{\;\;\nu}+\underline{\Omega}^{\mu}_{\;\;\alpha}\wedge \underline{\Omega}^{\alpha}_{\;\;\nu}$$

This is straight forward and gives, for example

$$\underline{R}^{t}_{\;\;r} = -e^{-2\beta}(\alpha''+\alpha'^{2}-\alpha'\beta')\underline{\omega}^t\wedge\underline{\omega}^r$$

Riemann's curvature tensor
The components of Riemann's curvature tensor are found from the relation

$$\underline{R}^{\mu}_{\;\;\nu} = \frac{1}{2}R^{\mu}_{\;\;\nu\alpha\beta}\underline{\omega}^{\alpha}\wedge\underline{\omega}^{\beta} = R^{\mu}_{\;\;\nu\left|\alpha\beta\right |}\underline{\omega}^{\alpha}\wedge\underline{\omega}^{\beta}$$

This is read directly out from the expressions for the curvature forms, for example

$$R^{t}_{\;\;rtr} = e^{-2\beta}(\alpha''+\alpha'^2-\alpha'\beta')$$

Ricci's curvature tensor
This is found by the contraction

$$R_{\mu\nu} = R^{\alpha}_{\;\;\mu\alpha\nu}$$

Ricci's curvature scalar
Another contraction gives this one

$$R = R^{\mu}_{\;\;\mu}$$

Components of the Einstein tensor
We can now finally compute the components of the Einstein tensor which are given by

$$E_{\mu\nu} = R_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}R$$

where $$\eta = diag(-1,1,1,1)\,$$. For this case the field equations say that all the components of the Einstein tensor are 0. This is here the same as saying that the components of Ricci's curvature tensor are 0, but because this need not apply in general, we've taken it all the way to the Einstein tensor.

The solution
By solving the equations we get from computing the components of the Einstein tensor and setting them equal to 0, the resulting metric becomes

$$ds^2 = -\left(1-\frac{K}{r} \right)dt^2+\left(1-\frac{K}{r} \right)^{-1}dr^2+r^2d\Omega^2$$

where K is a constant, and we see that if K = 0 we get the back the Minkowski metric. This constant is determined from the weak field approximation and turns out to be Schwarzschild's radius.