Does a free falling charge radiate ?

[greswd]

It appears paradoxical because to an observer falling with the charge it is as though the charge is at rest and therefore should not radiate.

Also, if we place a charge on a table, shouldn't it radiate as there is a normal reaction force from the table?

 

[Bill_K]

Frequently asked question. See the links to previous discussions at the bottom of the page. Yes it radiates, because a charge is not a point object. The field surrounding it is extended, and even though the center may be following a geodesic, the other parts of the field are not.

 

[TrickyDicky]

a charge is not a point object.

I''m confused, what happened to test charges? and aren't electrons(isn't an electron a charge?) point particles according to QFT-QED?

 

[Bill_K]

The electron itself is a pointlike object. But any charged object is surrounded by a Coulomb field which contains stress-energy and extends to infinity. For a charge falling in the Earth's gravitational field, for example - part of the EM field is on the other side of the planet!

In addition to radiating electromagnetic waves, a falling charge gets distorted by the varying gravitational field as it goes along, and radiates gravitational waves too.

 

[TrickyDicky]

The electron itself is a pointlike object. But any charged object is surrounded by a Coulomb field which contains stress-energy and extends to infinity. For a charge falling in the Earth's gravitational field, for example - part of the EM field is on the other side of the planet!

That's a nice explanation, only problem is that the contradiction between the first sentence and the rest of it is too evident to let it pass. Oh well, that rising damp again, let's pretend it's not there.

 

[Nugatory]

That's a nice explanation, only problem is that the contradiction between the first sentence and the rest of it is too evident to let it pass. Oh well, that rising damp again, let's pretend it's not there.

I must confess that I don't see any contradiction between saying that an electron is a point-like object and saying that its electrical field extends out to infinity.

Indeed, I remember watching Edward Purcell standing in front of a blackboard, describing the physics in exactly those terms, just so that we couldthen consider what would happen when the point particle was instantaneously displaced by a small amount...

 

[WannabeNewton]

Indeed, I remember watching Edward Purcell standing in front of a blackboard, describing the physics in exactly those terms, just so that we couldthen consider what would happen when the point particle was instantaneously displaced by a small amount...

You got to meet Purcell? I'm so jealous...T_T but yes this is also how he talks about it in his text.

 

[WannabeNewton]

By the way, Rindler's text has a small discussion on this caveat but it pretty much just reiterates the point made by Bill already.

 

[TopiRinkinen]

I should disagree. The Coulomb field external to electron does not carry electrical charge.
If you calculate the Gauss integral for the electric field of the electron, such that the electron is not enclosed by integrating surface, the integral is zero Coulombs.

On the other hand, if the accelerating electric field generates EM radiation, then it could be the the explanation.

BR, -Topi

The electron itself is a pointlike object. But any charged object is surrounded by a Coulomb field which contains stress-energy and extends to infinity. For a charge falling in the Earth's gravitational field, for example - part of the EM field is on the other side of the planet!

In addition to radiating electromagnetic waves, a falling charge gets distorted by the varying gravitational field as it goes along, and radiates gravitational waves too.

 

[Bill_K]

The Earth itself is a "freely falling particle", following a geodesic in its orbit around the sun. If the Earth had a net charge (and it may well have!), the problem lies entirely within the classical physics of Newton and Maxwell, and the circular motion of this charge would necessarily produce an outgoing EM wave.

 

[Bill_K]

I should disagree.

The bottom line is, for the reasons I gave, a freely falling charge does not exactly follow a geodesic. Due to its extended size there will be additional forces acting on it, that depend not just on the local gravitational field, but everywhere.

 

[TrickyDicky]

The Coulomb gauge is not Lorentz covariant, why would you use it in a QED context?

 

[Jano L.]

This is notorious question. Part of the reason for disagreements is that the situation is often not specified well enough, leaving the contributors to let their imagination fill in the details.

I propose to focus to the original question which is stated almost well enough:

Also, if we place a charge on a table, shouldn't it radiate as there is a normal reaction force from the table?

Well, the experience says that it will not radiate, besides the thermal radiation and scattered radiation. One can isolate charged object and have it on table indefinitely without any time-dependent fields connected due to force of gravity.

This is the answer just for the situation proposed in the question above. It does not say anything about any other scenario, like what free falling observer sees. That is a different question.

 

[tom.stoer]

How should we define 'radiation' in curved and non-stationary spacetimes?
- non-geodesics trajectories
- non-conservation of energy along a trajectory
- 1/r behaviour in the Coulomb potential for large r
- ...

 

[TrickyDicky]

What I'd like to get right is if from Bill's reply we must infer that point charges don't exist.

 

[Bill_K]

How should we define 'radiation' in curved and non-stationary spacetimes?
- non-geodesics trajectories
- non-conservation of energy along a trajectory
- 1/r behaviour in the [STRIKE]Coulomb potential[/STRIKE] field for large r

The last one is correct. Radiation is defined as the presence of a 1/r field at future null infinity, in asymptotically flat coordinates. For GR this issue was resolved back in the 60's by Bondi, Newman, Penrose, et al, where "field" means Riemann tensor.

 

[tom.stoer]

Bill, we had this discussion a couple of times and my answer was always that this definition does not work in not-asymptotically flat spacetimes, therefore one should look for a local definition; this is a general idea in GR: replace global definitions by local ones, look at all horizon discussions where one tries to get rid of null-infinity

(of course "field" Is correct and "potential" was nonsense, I am sorry for the confusion)

So why not using a comparison of trajectories of a non-charged and a charged particle? Of course this does not answer the radiation question directly, but it turns it round: we do no longer ask whether free falling particles radiate, but whether charged particles are in free fall according to the equivalence principle.

In parallel we should address the question whether (why) charged particles which are not in free fall do or don't radiate, i.e. particles which are stationary in a gravitational field, e.g. at fixed radius in a lab on the earth.

 

[atyy]

In parallel we should address the question whether (why) charged particles which are not in free fall do or don't radiate, i.e. particles which are stationary in a gravitational field, e.g. at fixed radius in a lab on the earth.

http://arxiv.org/abs/physics/9910019 suggested this answer. I believe the paper is serious, except for the claim that the principle of equivalence is validated, since I'm sure they know it's not applicable to this situation.

 

[TrickyDicky]

Bill, we had this discussion a couple of times and my answer was always that this definition does not work in not-asymptotically flat spacetimes,

This was my concern from the start of this thread, why mix notions from static time-independent scenarios (Coulomb fields, 1/r...) in which there would seem no radiation is even possible in principle with a question that requires moving charges-time dependent scenario?
IMHO it can only contribute to confuse even more the issue and the OP.

 

[Bill_K]

Bill, we had this discussion a couple of times and my answer was always that this definition does not work in not-asymptotically flat spacetimes, therefore one should look for a local definition

Tom, the understanding of radiation is rather firmly established - it's a global phenomenon, not a local one, in which a bounded system irreversibly loses energy to infinity. In the framework of GR one could perhaps extend the analysis from asymptotically flat spacetimes to de Sitter, or some other open cosmology.

But the radiation concept is not limited to GR. It's a basic feature of electromagnetism, as well as mechanical systems, such as elastic media. There are several reasons we treat it asymptotically.

One is simplicity - radiation exhibits common features at infinity which are far simpler than the details of what is happening at the source. For example, radiation may be conveniently described in terms of time-varying multipole moments, and knowing only these you know the energy loss.

A second reason is that some source motions transfer energy without producing radiation. Energy may be transferred from one part of the source to another "inductively", e.g. a pair of orbiting planets in Newtonian gravity, which constantly exchange energy and momentum through the inductive zone, which is 1/r² rather than 1/r. Likewise many electromagnetic examples.

Or even such a simple system as a pair of pendulums, coupled to each other, and also coupled to an infinitely long spring. You imagine there could be a local definition of radiation? As one of the two pendulums loses amplitude, it would be impossible to tell from its (local) motion alone whether the energy is being transferred (temporarily) to the other, or (permanently) lost to infinity. The answer must necessarily involve an analysis of the entire system, not just the one pendulum. That is, it must be global.

 

[Bill_K]

Don't believe everything you read on arXiv! The paper cited by atyy claims that a motionless particle radiates, the necessary energy coming about because, although the charge itself remains stationary, its surrounding electric field sags somewhat due to gravity, and goes on indefinitely sagging more and more!

Indeed, depending on how the charge is supported its field may sag, but an equilibrium is eventually reached where the distortion of the field is enough to resist more sagging. So there is not an indefinite source of energy to feed the radiation.

 

[tom.stoer]

Bill, regarding radiation you are right; the only question is whether this is the appriate question; I think that the main is geodesic motion = free fall.

 

[atyy]

Don't believe everything you read on arXiv! The paper cited by atyy claims that a motionless particle radiates, the necessary energy coming about because, although the charge itself remains stationary, its surrounding electric field sags somewhat due to gravity, and goes on indefinitely sagging more and more!

Indeed, depending on how the charge is supported its field may sag, but an equilibrium is eventually reached where the distortion of the field is enough to resist more sagging. So there is not an indefinite source of energy to feed the radiation.

Thanks for the warning! I'd long been wondering whether that article was correct, and hoped to get your opinion!

 

[Demystifier]

Even though I agree one shouldn't trust everything on arXiv, someone might want to see my opinion on that stuff, written in an arXiv paper:
http://arxiv.org/abs/gr-qc/9909035

For those who do not have time to read all this, let me just write down the main conclusions (for classical theory):
1. Radiation does not depend on the observer.
2. A charge radiates if and only if it does not move along a geodesic.

 

[Bill_K]

2. A charge radiates if and only if it does not move along a geodesic.

I gave what I thought was a valid counterexample of that. The Earth moves along a geodesic in its orbit about the sun. It surely radiates gravitational waves, and if it should carry a slight nonzero charge it will radiate electromagnetic waves also.

 

[TrickyDicky]

I gave what I thought was a valid counterexample of that. The Earth moves along a geodesic in its orbit about the sun. It surely radiates gravitational waves, and if it should carry a slight nonzero charge it will radiate electromagnetic waves also.

If it radiates gravitational waves it cannot be moving along a geodesic.

 

[TrickyDicky]

More specifically if you consider the Earth an idealized test particle orbiting the sun in a perfect geodesic it is obvious that it cannot radiate gravitational waves (nor EM ones following the correct point 2. by Demystifier).

If you consider it a real body with real mass and net charge, you run into problems with GR (two body problem etc) and if you were to claim that there is a point about the Earth's core that can be considered to be following an exact geodesic you stumble on the difficult problem that in GR there is no defined center of gravity as there is in Newtonian mechanics, so you can't really claim the Earth as a whole is following an exact geodesic path unless you follow the usual idealization of the Earth as a test body.

 

[Demystifier]

I agree with TrickyDicky. If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves. But it is an extended object consisting of many particles between which other (non-gravitational) forces act, so that individual particles do not move along a geodesic.

In my arXiv paper I explain that a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.

 

[atyy]

I agree with TrickyDicky. If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves. But it is an extended object consisting of many particles between which other (non-gravitational) forces act, so that individual particles do not move along a geodesic.

In my arXiv paper I explain that a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.

http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html (after Eq 19.84) discusses conditions under which one can use two complementary pictures (approximate, but very good) in which a point mass radiates and moves on a geodesic.

"It should be noted that Eq. (19.84) is formally equivalent to the statement that the point particle moves on a geodesic in a spacetime ... This elegant interpretation of the MiSaTaQuWa equations was proposed in 2003 by Steven Detweiler and Bernard F. Whiting [53]. Quinn and Wald [151] have shown that under some conditions, the total work done by the gravitational self-force is equal to the energy radiated (in gravitational waves) by the particle."

 

[TrickyDicky]

http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html (after Eq 19.84) discusses conditions under which one can use two complementary pictures (approximate, but very good) in which a point mass radiates and moves on a geodesic.

"It should be noted that Eq. (19.84) is formally equivalent to the statement that the point particle moves on a geodesic in a spacetime ... This elegant interpretation of the MiSaTaQuWa equations was proposed in 2003 by Steven Detweiler and Bernard F. Whiting [53]. Quinn and Wald [151] have shown that under some conditions, the total work done by the gravitational self-force is equal to the energy radiated (in gravitational waves) by the particle."

We had heated debates about this in the past, a distinction was considered important between the linearized gravity approximation and full non-linear GR, your example belongs to the former case.
In any case gravitational radiation has certain idiosyncratic properties for instance relating to the SET and energy of gravity that have generated lots of threads here and that make it a world of its own. If anyone wants to claim that objects following geodesics(and thus subjected only to the gravitational interaction) radiate just like the ones that don't , it's fine (but it kind of makes the distinction between geodesic motion and non geodesic motion as based in whether one can measure the acceleration with an accelerometer moot).
But EM radiation stress energy doesn't have the problems of gravitational SET(or absence of), and there is nothing in that article that allows a test particle with charge following a geodesic to radiate.

 

[Bill_K]

If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves.

That noise you just heard was my jaw hitting the floor. Needless to say (I hope it is needless to say!) it is a basic fact in general relativity that two massive particles orbiting each other do radiate gravitational waves.

a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.

This is reversing cause and effect. The backreaction from the emitted radiation may well make the particle deviate from a geodesic, but that is a result of the radiation, not a cause of it.

 

[Demystifier]

Needless to say (I hope it is needless to say!) it is a basic fact in general relativity that two massive particles orbiting each other do radiate gravitational waves.

That's true. But our real issue here is electromagnetic radiation of a test charge in a fixed gravitational background.

The backreaction from the emitted radiation may well make the particle deviate from a geodesic, but that is a result of the radiation, not a cause of it.

It is a sort of a chicken-or-egg dilemma. There is a self-reaction involved here, so I think it is both a cause and a result.

 

[A.T.]

This is reversing cause and effect. The backreaction from the emitted radiation may well make the particle deviate from a geodesic, but that is a result of the radiation, not a cause of it.

So what if the charged object is forced to move on a geodesic, by a rail or something. Can it radiate if the rail goes around a big mass?

 

[Bill_K]

Yes, the lesson of the 60's tells us that the gravitational radiation emitted by a source depends only on its mass multipole moments and not on the details of its composition or the nature of the forces holding it together, or whether the internal fields are weak or strong. In lowest order the radiated power is proportional to the third time derivative of the quadrupole moment. And these are completely well-defined quantities.

Just as we learned that in basic electromagnetism it's the second time derivative of the electric dipole moment that comes into play. And GR has not repealed this fact.

 

[pervect]

Tom, the understanding of radiation is rather firmly established - it's a global phenomenon, not a local one, in which a bounded system irreversibly loses energy to infinity.

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]

I agree with TrickyDicky. If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves. But it is an extended object consisting of many particles between which other (non-gravitational) forces act, so that individual particles do not move along a geodesic.

In my arXiv paper I explain that a charged particle in curved spacetime also does NOT move along a geodesic, and therefore radiates.

You write:

On the other hand, if the charge accelerates,
then, even in the small neighborhood, Eqs. (11) no longer look like the Maxwell equations in Minkowski spacetime. This gives rise to a more complicated solution, which includes the terms proportional to r−1.

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

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. If one knows the distance of the charge that produced the electromagnetic field Fμ
ext and measures the intensity of its effects described
by (12), then one can determine whether this effect is “large” or “small”, i.e., whether the charge radiates or not.

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]

Tom, the understanding of radiation is rather firmly established - it's a global phenomenon, not a local one, in which a bounded system irreversibly loses energy to infinity.

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]

I gave what I thought was a valid counterexample of that. The Earth moves along a geodesic in its orbit about the sun. It surely radiates gravitational waves, and if it should carry a slight nonzero charge it will radiate electromagnetic waves also.

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 agree with TrickyDicky. If Earth was a point particle, it would indeed move along a geodesic and would not radiate gravitational waves. But it is an extended object consisting of many particles between which other (non-gravitational) forces act, so that individual particles do not move along a geodesic.
.

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]

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.

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. "

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

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.

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.)

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]

two massive particles orbiting each other do radiate gravitational waves.

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]

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.

Hi Tom, I assure you I wrote my previous post before reading this this, and we're saying basically the same thing.

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

Agreed.

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?

This was the sense of my first question in this thread.

 

[A.T.]

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.

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]

Is gravitational radiation created due to the fact that parts of extended masses don't follow geodesics?

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.

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?

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]

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?

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]

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).

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]

To decide whether what they are computing are geodesic paths or not it would be wise to look at the geodesic definitions

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.

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.

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]

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.

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/Articles/lrr-2011-7/fulltext.html

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