+ 2 like - 0 dislike
78 views

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

recategorized Apr 3, 2015
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
@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
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
@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
@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
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

+ 3 like - 0 dislike

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
answered Mar 21, 2014 by (2,025 points)

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user JamalS
@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
thanks, I will do.

This post imported from StackExchange Physics at 2015-03-04 16:09 (UTC), posted by SE-user Valter Moretti
+ 0 like - 0 dislike

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
answered Mar 20, 2014 by (130 points)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.