# Computing the Einstein tensor for a spherically symmetrical metric using the tetrad formalism

I am having some trouble understanding how to use the tetrad formalism. I will start with what I have so far, my question will be after that.

I begin with the metric

$$\text{d}s^2 = e^{2a} \text{ d}t^2 - e^{2b} \text{ d}r^2 - r^2 \text{ d} \theta^2 - r^2 \sin^2 \theta \text{ d} \phi^2$$

Where $a$ and $b$ are functions of the coordinate $r$ only. Now I want to find the basis one-forms $\theta^{a}$ such that

$$\text{d}s^2 = \eta_{ab} \theta^{a} \otimes \theta^{b}$$

where $\eta_{ab} = \operatorname{diag} (+1, -1, -1, -1)$. So

$$\theta^{t} = e^{a} \text{ d} t\\ \theta^{r} = e^{b} \text{ d} r\\ \theta^{\theta} = r \text{ d} \theta \\ \theta^{\phi} = r \sin \theta \text{ d} \phi$$

Now I take the exterior derivatives of each basis one-form

\begin{align} \text{d} \theta^{t} &= a' e^{a} \text{ d} r \wedge \text{ d} t \\ \text{d} \theta^{r} &= b' e^{b} \text{ d} r \wedge \text{ d} r = 0 \\ \text{d} \theta^{\theta} &= \text{ d} r \wedge \text{ d} \theta \\ \text{d} \theta^{\phi} &= \sin \theta \text{ d} r \wedge \text{ d} \phi + r \cos \theta \text{ d} \theta \wedge \text{ d} \phi \end{align}

Up to here, I am pretty sure my results are correct. Where I am confused is when it comes to computing the spin connection one-forms from the no-torsion condition

$$\omega^{a}_{\phantom{a} b} \wedge \theta^{b} = - \text{d} \theta^{a}$$

How can I use this to find the connection one-forms? My naive guess is that, for example,

$$\omega^{t}_{\phantom{a} b} \wedge \theta^{b} = - \text{d} \theta^{t} = - a' e^{a} \text{ d} r \wedge \text{ d} t$$

so that this picks out the term on the left side of the form $\text{ d} r \wedge \text{ d} t$ so that

$$e^{a} \omega^{t}_{\phantom{a} t} = - a' e^{a} \text{ d} r \implies \omega^{t}_{\phantom{a} t} = - a' \text{ d} r$$

But the wedge product is anticommutative, so this should also say that

$$\omega^{t}_{\phantom{a} b} \wedge \theta^{b} = a' e^{a} \text{ d} t \wedge \text{ d} r$$

Am I reading this incorrectly? Is there some symmetry property that I'm missing?

• multiply the no-torsion condition to the left against a one-form with a new labeled index $\theta^c$ and use the properties of the wedge product – lurscher Mar 18 '14 at 21:36
• A general tip: when computing the exterior derivatives of your basis, or any quantity, re-express it in terms of the basis. It simplifies using Cartan's equations. – user32361 Mar 18 '14 at 21:46