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Post a Comment On: Backreaction

"Anti-Gravitation"

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Anonymous Garrett said...

Anti-gravitation -- fun to think about! Here's a problem I see with it:

All stuff in GR with no other forces acting on it travels along geodesics. Even postulated anti-gravitating stuff is going to have to travel along geodesics in GR -- what else could it do? You could try saying it's going to run backwards on a geodesic, but that's still a geodesic. So the conclusion I come to is all stuff has to travel along geodesics, and behave like normal gravitating matter. Unless there are other forces at work. Which means no anti-gravity, at least in GR.

4:23 AM, April 20, 2006

Blogger Bee said...

Dear garrett,

you have asked exactly the right question! Here is the answer: in a curved space, anti-gravitating matter does not move on geodesics. Geodesics are defined through the covariant derivative. The derivative depends on the transformation property of the quantity to be differentiated.

When you drop the assumption that the gravitational energy (momentum) has to be identical to the kinetic energy (momentum), it turns out that there are two quantities which you could want to preserve under parallel transport.

The one gives the usual geodesics, the other a different - but well defined - curve (in flat space, both agree). This becomes possible because the anti-gravitating fields do not transform as 'usual' tensor fields under general diffeomorphism. Nevertheless, their transformation behaviour is known from the representation it belongs to, and it can be used to define an appropriate covariant derivative.

This is not so different from gauge-theories. E.g. in an electromagnetic field, a positron moves on a different curve than an electron does. It has a different coupling to the gauge-field, which shows up in the covariant derivative.

In the Newtonian limit, you can interpret the covariant derivative of the particle as the conservation of total energy. A usual particle falls down on earth, thereby it gains kinetic energy and lowers its gravitational energy. An anti-gravitating particle would do the same when falling 'up'. Or, with sufficient initial momentum directed towards earth, it would fall down, but thereby loose kinetic energy and gain gravitational energy.

Best,

B.

2:51 PM, April 20, 2006

Anonymous Garrett said...

Ahh, so you have to give up the nice action for a free particle -- it has to have an action other than the proper time of the path. The new free particle action must explicitly include the connection, just like for a charged particle interacting with other gauge forces. And you have to give up the equivalence principle.

Inertial mass no longer must equal gravitational mass...

Hey, you know what? This stuff sounds exactly like the teleparallel formulation of gravity. That's effectively equivalent to GR, but allows for particle equations of motion with unequal gravitational and inertial mass. Are you familiar with that?

1:48 PM, April 21, 2006

Blogger Bee said...

Hi garrett,

you are absolutely right, I have to give up the action one commonly uses to derive the geodesic equation as the curve of a particle.

However, also in usual GR, to get the curve of the particle one does not need to postulate an extra-action for the particle. Though not widely done in textbooks, it is possible to derive the curve directly from Einsteins's field equations, using the energy momentum tensor of a pointlike particle. If you use this approach you find two possibilities for the curve, depending on the structure of the energy momentum tensor - it is either that of the usual particle, or that of the the anti-gravitational particle. This corresponds to either parallel transporting the 'usual' or the 'new' kinetic momentum.

I stumbled across teleparallel gravity several times, but never looked really into it. Thanks for the remark, I will have a closer look.

Best,

B.

2:01 PM, April 21, 2006

Anonymous garrett said...

It makes sense that you could get the motion from the energy-momentum tensor. And you could probably get the particle action by contracting that tensor with g. (I like actions, can you tell?)

I'd probably get a better idea of what you're doing if I had your paper available, which I don't. :( I'll have access to the UCSD library in a couple of months though, and I'll look at it then.

The basic idea of teleparallel gravity is to treat gravity as a force field in flat Minkowski space. So, there's a metric (or rather a vierbein) and a connection, but its curvature is zero. The torsion of the connection, though, is not zero, and the dynamics are in there. It's a pretty theory, and agrees with GR predictions. But I (and most) still like good 'ol geometric GR -- but I'm always open to whatever works best in the big picture.

Here's an intro to teleparallel GR that includes the equation of motion for free particles with different inertial and gravitational mass:
http://arxiv.org/abs/gr-qc/0410042

Probably worth a quick look from you just to see if you spot similarities.

3:07 PM, April 21, 2006

Anonymous Chris said...

SHIT happens!

interesting idea, keep us updated.

11:13 PM, April 21, 2006

Blogger Bee said...

Hi garrett,

I had a brief look at the paper about teleparralel (TP) gravity. I can't say I really understand it, but it seems to me there is no relation to my model. The only similairity is that in the TP it is possible to have an inertial mass other than the graviational. However, the price to pay for this seems to me quite high.

In the TP, the ratio between gravitational and inertial mass is a continuos parameter, connected to a continuous deformation of the Geodesic motion. (Not sure whether the deformed curves still have a geometrical meaning).

In my model, the ratio is discrete, it is either +1 or -1, it is more a charge symmetry than a deformation. The crucial point, which does not appear in the TP, is the modified transformation behavior of the new fields. The cause for the non-standard-geodesic motion in the TP is a completely different one.

BTW, I read yesterday the Physics Today from April and found there is a letter (from R.E.Becker) with an comment on TP, as well as the corresponding answer from S. Weinberg, in case you are interested.

Best,

B.

9:19 PM, April 24, 2006

Anonymous Garrett said...

Hi Sabine,
I have no great love for teleparallel GR. (I'm flexible about whether torsion exists or not, but I don't think it should exist at the expense of curvature!) I just thought it might be relevant, since it's compatible with having two different connections and corresponding different mass "charges." But I guess not.

I'm going to go post a question about your paper on PF, where I can write math that doesn't make me barf.

2:24 AM, April 25, 2006

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