2
$\begingroup$

I'm struggling with Anthony Zee's chapter on differential forms in Einstein Gravity in a Nutshell, page 600. He asks us to prove that $$\omega= \Lambda \omega' \Lambda^{-1} - (d\Lambda) \Lambda^{-1}$$ using $$e=\Lambda e'$$ and $$ de+ \omega e= 0.$$ I'm not sure what $ \Lambda $ is exactly. He describes it as not being a coordinate transform but merely a local Lorentz transform (or in euclidean space a rotation) of the orthonormal frame. I originally thought it was Infinitismal transform but now I'm thinking I can put in any Lorentz Transform. Everytime I start the calculation I get lost somewhere in the notation and I'm not sure if I'm even starting from the correct transform anyway.

$\endgroup$
  • 1
    $\begingroup$ Make $\Lambda$ a function of $x$, use the Leibniz rule for differentiation after substituting for $e$ in the parallel transport equation and you should be able to get there. $\endgroup$ – Void Jun 19 '15 at 9:32
  • $\begingroup$ How would I make it a function of x? Maybe I should just do it for the rotation case first then move to lorentz. $\endgroup$ – mathmath12 Jun 19 '15 at 9:41
  • 2
    $\begingroup$ Simply: $e^\mu(x) = \Lambda^\mu_\nu(x) e'^\nu(x) $. You do not really have to think of $\Lambda$ as anything else than a non-degenerate matrix which varies from point to point, that's the beauty of diffeomorphism invariance of general relativity. $\endgroup$ – Void Jun 19 '15 at 9:52
  • $\begingroup$ Ok. I will try thst $\endgroup$ – mathmath12 Jun 19 '15 at 10:01
  • $\begingroup$ Related: physics.stackexchange.com/q/225413/2451 $\endgroup$ – Qmechanic Nov 8 '15 at 19:34
2
$\begingroup$

Well, this is linked to what the cotetrad $e_\mu^I$ is.

It is customary to present the cotetrad as a diagonalization of the metric and indeed, we have: $$g_{\mu\nu} = e_\mu^I e_\nu^J \eta_{IJ}$$ Note here that the cotetrad has two kind of indices. The greek type corresponds to spacetime coordinates but the latin type indices are indices in the tangent space. So, in a way, the cotetrad is a natural basis for the metric.

The physical idea behind this is actually quite simple: the metric is always locally minkowskian. It takes this form (locally) in a free falling frame of reference. So given a metric, you can assign to each point a free-falling frame (described by the cotetrad) in which the metric is diagonal (and even minkowskian).

To be a bit more precise, the cotetrad is the inverse of the tetrad, usually noted $e_I^\mu$ (note the position of the indices - this is kind of a loose notation but it is standard). The tetrad is the collection of four vectors labelled by $I$ with coordinates labelled by $\mu$. These vectors are orthogonal and normed. One of them is timelike and corresponds to the normed tangent vector along the free-falling trajectory.

But when you do this, you have an arbitrary choice: there are an infinite amount of free-falling frames at a given point and they are linked together by Lorentz transformations. So at each point, you have possible Lorentz freedom. This is your local Lorentz transform: it relates the free-falling frames of reference at each point.

Back to the math now, your Lorentz transform will not act on the coordinates $\mu$, these are fixed here. It will act on the tangent space. So, writing down the indices, you have: $${e'}_\mu^I(x) = \Lambda^I_{~J}(x) e_\mu^J(x)$$ With that you can conclude quite easily since the second equation defines $\omega_{\mu~J}^{~I}(x)$ as a connection.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.