Frames, Tetrads and GR

Question

Given a general metric, $g_{ab}$ I can select an orthonormal basis $\omega^{a}$ such that,

$$g_{ab} = \eta_{ab}\omega^a \otimes \omega^b$$

where $\eta_{ab}$ = $\mathrm{diag}(1,-1,-1,-1).$ We may conveniently compute the spin connection and curvature form by employing Cartan's equations. The problem I have lies in getting back to the coordinate basis. I know the general formula,

$$R^{\mu}_{\nu \lambda \tau} = (\omega^{-1})^{\mu}_a \, \omega^b_\nu \, \omega^c_\lambda \, \omega^d_\tau \, R^{a}_{bcd}$$

where the l.h.s. $R^{\mu}_{\nu \lambda \tau}$ is in the coordinate basis. The objects $\omega^b$ are familiar, they're just the orthonormal basis, so what is the object $\omega^b_\nu$ (with the extra index)? The $(\omega^{-1})^{\mu}_a$ are the inverse vielbeins? How are these obtained?

I've visited several sources, including Wikipedia, but it's still not 100% clear. I'd appreciate any clarification, especially a small explicit example if possible.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user JamalS

Comments

Related question: physics.stackexchange.com/q/102722/2451

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Qmechanic

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@Qmechanic I noticed that question from another user, but it wasn't entirely helpful as I still don't know what the $\omega^b_\mu$'s are, or the $e^\mu_a$.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user JamalS

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The notation seems to be bad and idiosyncratic. What you call $\omega^{a}$ other sources call $e^{a}={e^{a}}_{\mu}\,\mathrm{d}x^{\mu}$. The ${\omega^{a}}_{b}$ are the spin connections, and can be obtained via Cartan's equations.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Alex Nelson

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@AlexNelson: Is the relation between the Riemann tensors of differing basis correct, if interpreted using your notation?

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user JamalS

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@Qmechanic: Thank you for editing my post; it looks like my misunderstanding arose due to differing notation/terminology.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user JamalS

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For a good, free, online reference, perhaps the TangentBundle's article on the spin connection would help; and $\mathrm{d}{\omega^{a}}_{b}+{\omega^{a}}_{c}\wedge{\omega^{c}}_{b} = {R^{a}}_{bfg}e^{f}e^{g}$ is the relationship between the Riemann tensor and the spin connection IIRC...possibly up to some factor of 1/2, depending on reference...

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Alex Nelson

Answer 1

The point is that, with your second equation, you are dealing with in a local coordinate patch, say an open set $U\subset M$ equipped with coordinates $x^\mu \equiv x^1,x^2,x^3,x^4$. Therefore, if $p\in U$, you can handle two bases of the tangent space $T_pM$. One is made of (pseudo)orthonormal vectors $e_a$, $a=1,2,3,4$ and the other is the one associated with the coordinates $\frac{\partial }{\partial x^\mu}|_p$, $\mu= 1,2,3,4$. The metric at $p$ reads: $$g_p = \eta_{ab} \omega^a \otimes \omega^b$$ where, by definition, the co-vectors $\omega^a \in T^*_pM$ (defining another pseudo-orthonormal tetrad but in the cotangent space at $p$) satisfy $$\omega^a (e_b) = \delta^a_b\:\:.\qquad(0)$$

Correspondingly you have: $$\omega^a = \omega^a_\mu dx^\mu|_p \:, \qquad (1)$$ and $\Omega :=[\omega^a_\mu]$ is a $4\times 4$ invertible matrix. Invertible because it is the transformation matrix between two bases of the same vector space ($T_p^*M$). Similarly $$e_a = e_a^\mu \frac{\partial}{\partial x^\mu}|_p\:,\qquad (2)$$ where $E:= [e^a_\mu]$ is a $4\times 4$ invertible matrix. It is an elementary exercise to prove that, in view of (0): $$E= \Omega^{-1t}\:,\qquad (3)$$ so you can equivalently write $$e_a^\mu = (\omega^{-1})_a^\mu$$ where the transposition operation in (3) is now apparent from the fact that we have swapped the positions of Greek and Latin indices (compare with (1)).

An identity as yours: $$R^{\mu}_{\nu \lambda \tau} = (\omega^{-1})^{\mu}_a \, \omega^b_\nu \, \omega^c_\lambda \, \omega^d_\tau \, R^{a}_{bcd}$$ is understood as a trivial change of basis relying upon (1) and (2). It could equivalently be written down as: $$R^{\mu}_{\nu \lambda \tau} = e^{\mu}_a \, \omega^b_\nu \, \omega^c_\lambda \, \omega^d_\tau \, R^{a}_{bcd}\:.$$

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Valter Moretti

Comments

Thank you for your detailed answer, +1.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user JamalS

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@V. Moretti. Fyi you can use the command \tag{<your label>} to label equations.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user joshphysics

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thanks, I will do.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Valter Moretti

Answer 2

They are your tetrad. If we take, say, Schwarzschild spacetime in Schwarzschild coordinates, one choice for the $e_{\mu}{}^{I}$ is:

$e_{t}{}^{t} = \sqrt{1-\frac{2M}{r}}$

$e_{r}{}^{r} = \frac{1}{\sqrt{1-\frac{2M}{r}}}$

$e_{\theta}{}^{\theta} = r$

$e_{\phi}{}^{\phi} = r\sin\theta$

with all others zero. The key point to notice here is that $g_{ab} = e_{a}{}^{I}e_{b}{}^{J}\eta_{IJ}$. It is often convention to leave off the internal indices, but I would recommend leaving them in until it's obvious what you're doing. I think you're confused by your first equation.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Jerry Schirmer