# Einstein-Cartan theory

Tags:
1. Oct 3, 2014

### michael879

I've been reading up on EC theory, and the basic premise and the math behind it are all very straightforward. What I'm a little confused about is more the intuitive side of the theory, and I'm sure it stems from a very poor intuitive understanding of chirality (I do have some intuitive understanding from the research I've done, and know all about the math behind it, but the whole thing is still pretty foreign to me on any intuitive level).

Anyway, I was wondering if someone could explain the difference between chirality and classical angular momentum in the EC framework (no quantum mechanics, chiral fields/particles do have classical descriptions). The way I see it, a chiral geodesic is a geodesic capable of describing the rotation of point particles along it. So the extra d.o.f. in the SET/metric store this extra information. With non-chiral point particles (or fields) these rotations aren't observed because the particles don't change state under rotations (i.e. a 2pi rotation leaves them unchanged)

The problem I'm having is that all the angular momentum of a system is stored in the SET, which in normal GR is symmetric. So a symmetric SET is perfectly capable of describing rotating objects, and yet for some reason you need the 6 extra d.o.f. of an asymmetric tensor to model chirality! Clearly if you take the limit as R goes to 0 of some rotating ball with radius R, the originally symmetric SET will remain symmetric. However this ball will now be a point particle with angular momentum right? I think this entire issue might just boil down to the difference between SU(2) rotations and SO(3) rotations, but then I remember that GR can describe effects similar to EC theory. For example, a ball with finite radius will rotate as it follows the geodesic of its c.o.m, and all of this can be predicted with symmetric SETs/metrics!

Also a related question: I'm aware that Kerr-Newman black holes have a lot of similarities to fundamental particles (g-factor, no hair, radius, etc.), but do they also share the same transformation properties? i.e. Do Kerr-Newman black holes transform as SU(2) under rotations? I'm thinking it could be possible, if you use the fully extended space-time with the r < 0 universe, as it kind of reminds me of the plate trick demonstration

2. Oct 8, 2014

### Greg Bernhardt

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Oct 16, 2014

### michael879

Not really, except that the visualizations of chirality make it a lot clearer to me now. A chiral geodesic doesn't simply rotate, it drags "something" along with it in order to make it a spin group rotation (rather than SO(3)). The way I see it, the ONLY way to see chiral effects in our world is to rotate an object with something non-rigid attached to it. So I might be wrong in extending this to EC theory, but I would expect that the chiral geodesics twist space-time as they propagate. This is very different than a classical test particle which, by definition, doesn't frame-drag.

This does bring me back to my second question though (if nobody wants to try to answer the first). Given the frame dragging effect of Kerr black holes, could they be modeled as chiral objects?? Just the frame dragging alone would cause it to twist space-time in a way that takes a 4pi rotation to arrive back at the original state! On the other hand, if these are chiral objects, how can the SET be symmetric?

4. Oct 17, 2014

### Staff: Mentor

I'm not sure whether Kerr black holes really can be modeled as chiral objects; but since the Kerr geometry is a vacuum solution, the "SET" is symmetric in the vacuous sense (pun intended) that it's identically zero. I guess the question that raises is whether a chiral object can be "made" out of vacuum spacetime curvature.

5. Sep 23, 2015

### michael879

sorry to resurrect this, but I was going through my old posts and stumbled across it. I never really got an answer to the question I was asking, but Peter made a good point about the SET being 0 for any vacuum solution. However, in principle the SET should diverge at the origin if there is matter there right? Does anyone know a way to represent the KN SET as some limit as r->0? Its easy with spin-less black holes, since composite objects are also described by the Schwarzschild metric outside, it's straightforward to take the limit as r->0. However, rotating stars aren't described by the KN metric outside, so I don't think they would work

*edit* yea, so basically my question has become: what composite object is described by the KN metric outside of itself?

6. Sep 23, 2015

### Staff: Mentor

No--at least, not if I'm understanding your question correctly. In the solution for a static, non-rotating object that you describe, where the metric is Schwarzschild in the vacuum region exterior to the object, the SET is nonzero inside the object, but is perfectly finite everywhere; nothing diverges at $r = 0$.

Do you mean, take the limit as the radial coordinate $R$ of the surface of the object goes to zero? You can't do that, because as soon as $R$ becomes less than $9/8$ of the Schwarzschild radius $2M$, there are no static solutions. So you can't view the Schwarzschild solution with vacuum everywhere as a limit of composite static solutions with Schwarzschild exterior regions down to some radius $R$, as $R$ approaches zero.

7. Sep 24, 2015

### michael879

Sorry, I may have not explained that well. I meant for the black hole solutions the SET diverges at r=0 (or is undefined, depending on your definition). This is what you would expect by taking the limit of a static object as r->0

Ok that's a good point... Is there any way to discuss the SET of a black hole at R=0 then? Clearly it's divergent, and you just showed it can't be expressed as a limit R→0 for the usual black hole that contains non-stationary regions. What about a naked singularity though?

8. Sep 24, 2015

### Staff: Mentor

For black hole solutions, the SET is zero everywhere. They're vacuum solutions. What diverges at $r = 0$ (which, technically, is not part of the manifold) is the Kretschmann curvature invariant, which can be nonzero even if the SET is zero (since it includes Weyl curvature).

I'm still unclear on how you would take this limit. If the mass is held constant, the object cannot remain static for arbitrarily small $r$, for the reason I gave before--there are no static solutions for fixed $M$ and arbitrarily small $r$. If the mass is decreased to allow the solution to remain static for arbitrarily small $r$, then $M$ must go to zero as $r$ goes to zero, and you no longer have a divergence; the limit in that case is just Minkowski spacetime.

No, it isn't; it's zero by definition. See above.

I'm not sure if solutions with a naked singularity in the vacuum everywhere case (e.g., Reissner-Nordstrom for $Q > M$) can be the exterior geometry of a static object. I don't think I've seen this case discussed anywhere.