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?

  • $\begingroup$ 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 $\endgroup$ – lurscher Mar 18 '14 at 21:36
  • $\begingroup$ 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. $\endgroup$ – user32361 Mar 18 '14 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.