Does a free falling charge radiate ?

pervect

I suspect this is the most common view, but I've seen enough papers with contradictory views to suggest that one needs to make sure that there is agreement about the definition of "radiating" before one tries to answer the question.

I think it'd be handy to have the Cliff Notes version (a short description of the proposed definition and the conditions required for radiation) but I don't think I've seenone.

pervect

You write:

If we apply this to to an accelerating particle in Minkowskii coordinates, I don't quite understand how you conclude that it radiates.

Whatever the solution is, it must be static in those coordinates, because the space-time is static.

You also write

Can you demonstrate, explicitly, such an effect ("slow falloff) in Rindler coordinates?

To insure coordinate independence, I'd like to see an argument for radiation that applies whichever coordinate system is used. Saying that "fermi coordinates are preferred because they are more physical" is sort of a cop-out. (I'm not sure that you actually said such a thing, I'm tempted to think it after a brief reading of your paper though.)

pervect

Suppose we try to apply this to the simple situation of our sun, in a FRW universe.

It seems like the answer to "does the sun radiate" should be obvioiusly "yes" - but without a timelike Killing vector, or an asymptotically flat space-time, how do we justify something as simple as saying "the sun radiates"?

On one hand, I feel like this may be nitpicking. On the other hand, it's a "nit" that has always bothered me.

PAllen

The earth does not move exactly on a geodesic. No particle of nonzero mass does. Geodesic of what background? For a 'particle' with nonzero mass, the spacetime is dynamically affected by the particle, and there is no background geometry in which to specify the geodesic.

PAllen

I disagree with this, in part. If the earth were a point particle of nonzero mass, to the extent you can model that in the limit, it will not move on a geodesic, exactly. However I agree that this fact is related to GW, which is why a point particle earth of non-zero mass would radiate.

tom.stoer

I would like to explain my ideas proposed a few months ago. First we should make clear whether we talk about pointlike test particles or whether we want to study extended objects. For the latter one it's clear that we expect radiation. For pointlike particles we don't expect radiation b/c of the modified geodesic equation. If we set the field to zero (neglecting self-interaction of the test particle with it's own field) then we find the usual geodesic equation and we expect free fall w/o radiation. The main question is then whether the approximatin of pointlike test particles w/o self-interaction makes sense. In reality we expect deviations from this idealized setup and therefore we expect radiation.

What I still don't like is the definition of radiation using 'energy-loss' or the '1/r behaviour'. The problem is that we can neither measure nor define this energy; b/c it's a Coulomb field an integral over the energy density diverges both at r=0 and for r→∞; b/c we may have arbitrarily curved (expanding) spacetime we cannot define E = ∫d³x T⁰⁰ even for well-behaved T^ab; w/o a timelike Killing vector k_a field (e.g. in an expanding universe) we cannot define the 4-vector J^a = T^ab k_b and again we do not have a reasonable definition of the el.-mag. field energy E = ∫ d³x J⁰

That's why I would like to get rid of the concept of energy and 1/r behavour. My proposal is to study local, coordinate-free effects, i.e. the deviation from geodesic motion. Of course this means that we ask a different question: instead of Does a free falling charge radiate? we ask Are charged particles - pointlike test particles or exended objects - in free-fall, i.e. do they follow geodesics?

Demystifier

It's all about how exactly radiation is DEFINED.
In my definition, motivated by the principle of general covariance, the lack of staticity is not a part of the definition of radiation.

Or let me quote from page 8:
" Now we turn back to the attempt to give an operational definition of radiation at large
distances. In our opinion, the only reason why radiating fields deserve special attention in
physics, is the fact that they fall off much slower than other fields, so their effect is much
stronger at large distances. Actually, the distinction between “radiating” and “nonradiating”
fields is quite artificial; there is only one field, which can be written as a sum of components
that fall off differently at large distances. ... In this sense, we can say that radiation does not depend on the observer. "


Yes, it's trivial. First show that the electromagnetic tensor F has a slow falloff in Minkowski coordinates, and then transform F to F' in Rindler coordinates, by Eq. (9). The transformation coefficients f in (9) depend only on local velocity, not on acceleration, and they cannot transform a slow falloff into a fast falloff.

In Sec. 2 I briefly EXPLAIN why these coordinates are to be interpreted as physical coordinates associated with a given observer. More elaborate explanations can also be found in Refs. [11] and [12], where [12] is the authoritative monograph "Gravitation" by Misner, Thorne, and Wheeler.

But if you wish, you can use any coordinates you want. The point is that the only physical quantity is F which transforms as a local tensor. So if, at a certain point far from the source of F, all components of F are of the order of r^-1, then, at this SAME point, the components of F' will also be of the order of r^-1. It is a trivial consequence of the fact that F transforms to F' as a tensor.

TrickyDicky

Nobody has disputed that.
But you seem to be mixing scenarios in a funny way for your own benefit.

Take the Hulse-Taylor binary, a clear example of gravitational radiation, you have there two "extended" masses orbiting each other that are obviously not following exact geodesics. But it is clear that this is a different example from the earth-sun example you were using because in this case it is alrigight to consider the earth as a test particle that is obviously moving in a geodesic, but you need to consider it an extended object to say it radiated and then you cannot claim it is followin a geodesic because in GR there is no defined center of gravity that you can pinpoint as the one that is following a geodesic .
It should be straight-forward that in the Hulse-Taylor binary you cannot model one body as a test body orbitting a source of gravitation since their masses are of similar order of magnitude. So you must make up your mind if in your example you really want to consider the orbitting bodies as extended objects or as test particles.

TrickyDicky

Hi Tom, I assure you I wrote my previous post before reading this this, and we're saying basically the same thing.
Agreed.
This was the sense of my first question in this thread.

A.T.

Is gravitational radiation created due to the fact that parts of extended masses don't follow geodesics? What if the two big masses where just clouds of particles, that create gravity but do not otherwise interact with each other. So each particle follows a geodesic. Would that system create gravitational radiation?

TrickyDicky

Well this would be putting it in maybe too simplified terms. Gravitational radiation is a prediction of GR, and asserting that only test particles in GR can be claimed to follow exact geodesics is a consequence of a limitation of GR as a theory. The absence of practical solutions to the EFE that can handle the SET of test particles, or in this case the n-body problem.

The bolded part looks like a description of a clump of Dark matter, answering this would be entering into (probably wild) especulation. But it is an interesting question.

tom.stoer

@TrickyDicky regarding #43: fine, thanks

PeterDonis

AFAIK the numerical models of binary pulsar systems that make predictions about the orbital parameters compute geodesic paths for the pulsars (at least to the level of approximation of the models). Since these models match the experimental data very well, that would indicate that a system of two neutron stars following geodesics, at least to a good approximation, can create gravitational radiation.

TrickyDicky

Numerical (nonperturbative) methods in relativity are simply strong-field approximations that benefit of the tremendous calculational power of computers to calculate a good approximation of the gravitational waves radiated by these systems i.e. in the case of black hole binaries orbits. The method of calculation is independent of whether one labels the orbital paths in the problem as geodesic or non geodesic paths, it is just a crunching numbers method that depends on the initial data fed to the computer.

To decide whether what they are computing are geodesic paths or not it would be wise to look at the geodesic definitions, see for intance Wikipedia page on "Geodesics in General relativity":

"True geodesic motion is an idealization where one assumes the existence of test particles. Although in many cases real matter and energy can be approximated as test particles, situations arise where their appreciable mass (or equivalent thereof) can affect the background gravitational field in which they reside.

This creates problems when performing an exact theoretical description of a gravitational system (for example, in accurately describing the motion of two stars in a binary star system). This leads one to consider the problem of determining to what extent any situation approximates true geodesic motion. In qualitative terms, the problem is solved: the smaller the gravitational field produced by an object compared to the gravitational field it lives in (for example, the Earth's field is tiny in comparison with the Sun's), the closer this object's motion will be geodesic."

If one then considers that "in metric theories of gravitation, particularly general relativity, a test particle is an idealized model of a small object whose mass is so small that it does not appreciably disturb the ambient gravitational field." And remembers that gravitational radiation is a gravitational field disturbance that in the case of binary neutron stars or black holes should be not negligible I think there's not really much more to discuss.

PeterDonis

Yes; you look at the definition of a geodesic, then you look at the paths that are computed and the metric that is computed, and you determine whether the paths that are computed are geodesics (to within the given approximation) of the metric that is computed. They are.

Saying that the two neutron stars in the binary pulsar system both follow geodesics of the metric due to their combined masses and configurations does not require claiming that the two neutron stars are test particles. Obviously they aren't, not just because their motion generates non-negligible disturbances to the field, but because they generate the entire field to begin with.

The fact remains that (AFAIK) the trajectories computed for each neutron star turn out to be geodesics of the metric that is computed based on their masses and configurations, and this remains true as the system emits gravitational waves (meaning the metric changes with time). And this, as I said, is good evidence that gravitational waves can be generated by the geodesic motion of objects.

PAllen

I have always read the opposite - that geodesic motion is exactly true in GR only in the limit of test particles of zero mass and size. There is a slight generalization for the case of inspiral with extreme mass ratio; however none is known for similar mass co-orbit.

For the general topic, a paper that points to much of the history of results, along with yet more rigorous derivation of geodesic motion for test particles (only):

http://arxiv.org/abs/1002.5045 [see esp. section II, where the mass as well as size are required to approach zero
to derive exact geodesic motion] Also: http://arxiv.org/abs/0907.0414

For the slight generalization only (and not exactly) true for extreme mass ratio, see the discussion of the The Detweiler–Whiting Axiom in section 24.1 of:

http://relativity.livingreviews.org/.../fulltext.html

[edit: See also this thread: http://physicsforums.com/showthread.php?t=498039]

PeterDonis

As far as an analytical proof goes, yes, I believe you are right. But when we actually look at how objects of finite mass and size, like planets and stars and binary pulsars, move, we find that they appear to move on geodesics to the level of approximation we can measure, even though we can't analytically derive that result.

It's possible that the neutron stars in the binary pulsar are not moving on *exact* geodesics; of course we can't tell for sure because we can't make precise direct measurements of the system. But it does seem like they move on curves that are close enough to geodesics that we can't tell the difference with our current measurements, even though the two neutron stars are certainly not test objects. (At least, that's my understanding of the current models.)