Does a free falling charge radiate ?

In summary, a falling charge does radiate, though it does so in ways that are different than when a stationary charge is present.
  • #36
Demystifier said:
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.)
 
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  • #37
Bill_K said:
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.
 
  • #38
Bill_K said:
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.
 
  • #39
Demystifier said:
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.
 
  • #40
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 Tab; w/o a timelike Killing vector ka field (e.g. in an expanding universe) we cannot define the 4-vector Ja = Tab kb 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?
 
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  • #41
pervect said:
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. "


pervect said:
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.

pervect said:
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.
 
  • #42
Bill_K said:
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.
 
  • #43
tom.stoer said:
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.
tom.stoer said:
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 Tab; w/o a timelike Killing vector ka field (e.g. in an expanding universe) we cannot define the 4-vector Ja = Tab kb and again we do not have a reasonable definition of the el.-mag. field energy E = ∫ d³x J0
Agreed.
tom.stoer said:
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.
 
  • #44
TrickyDicky said:
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?
 
  • #45
A.T. said:
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.

A.T. said:
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.
 
  • #46
@TrickyDicky regarding #43: fine, thanks
 
  • #47
A.T. said:
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.
 
  • #48
PeterDonis said:
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.
 
  • #49
TrickyDicky said:
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.

TrickyDicky said:
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.
 
  • #50
PeterDonis said:
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 [Broken]

[edit: See also this thread: https://www.physicsforums.com/showthread.php?t=498039]
 
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  • #51
PAllen said:
geodesic motion is exactly true in GR only in the limit of test particles of zero mass and size.

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.)
 
  • #52
PeterDonis said:
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.)

I think there is more to it: if there is GW, you have decaying orbits. There is a way to package this as pseudo-geodesic motion for the extreme mass ration case.

However, for a symmetrical two body problem, even with no GW (let alone with), I don't see how to even ask the question of moving on geodesics; no one on the thread I linked thought there was any hope of doing this, esp. Sam Gralla for whom this is a specialty of his.
 
  • #53
PAllen said:
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 [Broken]

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

I thought Gralla and Wald http://arxiv.org/abs/0806.3293 derive the MiSaTaQuWa equations, which are described by the Poisson review as geodesic to second order, so even for a point test particle with mass approaching zero (I guess it can't be zero, otherwise it'll move on a null geodesic?), is the geodesic equation "exact"?
 
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  • #54
PAllen said:
for a symmetrical two body problem, even with no GW (let alone with), I don't see how to even ask the question of moving on geodesics; no one on the thread I linked thought there was any hope of doing this, esp. Sam Gralla for whom this is a specialty of his.

Hm, I'll have to read through that thread in more detail. I would have thought it was simple conceptually: you have a numerical solution that describes a metric and two worldlines. Then you just check whether the worldlines are geodesics of the metric. This may not be simple computationally, but that's what the computations would amount to.
 
  • #55
atyy said:
I thought Gralla and Wald http://arxiv.org/abs/0806.3293 derive the MiSaTaQuWa equations, which are described by the Poisson review as geodesic to second order, so even for a point test particle with mass approaching zero (I guess it can't be zero, otherwise it'll move on a null geodesic?), is the geodesic equation "exact"?

Gralla and Wald show there is a precise way in which mass and size can be taken to a zero limit such that the motion becomes exactly a timelike geodesic. The MiSaTaQuWa equations are the first order correction, which include the possibility for a small body radiating GW, thus affecting its trajectory.
 
  • #56
PeterDonis said:
Hm, I'll have to read through that thread in more detail. I would have thought it was simple conceptually: you have a numerical solution that describes a metric and two worldlines. Then you just check whether the worldlines are geodesics of the metric. This may not be simple computationally, but that's what the computations would amount to.

The metric already encodes the motion of the two bodies. A geodesic of this metric would describe the motion of a test body free falling near the two massive bodies.

Maybe if you could point to some reference on this? I remain very interested whether there is some way to 'rescue' EP and some form of geodesic motion for similar mass two body problem - but gave up on it when last studying this issue.
 
  • #57
PeterDonis said:
I'll have to read through that thread in more detail.

I see that Sam Gralla made this comment:

Sam Gralla said:
The basic difficulty when you ask about geodesic motion of a body is "geodesic motion in what metric"? For the reasons you identify, it can't be the exact metric. So, approximation is always involved when you talk about geodesic motion (or, in my opinion, the assignment of a "center of mass" worldline to a body in any circumstances).

So you won't make much progress with the exact two-body problem, but there are some limits that you can consider. If one body is much smaller than the other, then the ehlers-geroch theorem as well as all the self-force stuff in Poisson's review will apply. If the bodies are widely separated, then you can use post-Newtonian techniques. Other than that, I think you're stuck with exact solutions (and no notion of CM worldline). Luckily numerical relativity has let us explore these solutions lately, so the two-body problem is pretty well under control.

I think that the binary pulsar case would be an example of the bolded phrase above; the neutron stars in binary pulsars are very far apart compared to their individual sizes. So in that case one might be able to derive post-Newtonian analytic expressions for the metric and its geodesics, if I'm reading him right.

However, he also mentions numerical solutions at the end, which makes me wonder: do numerical solutions not give enough information to even apply the test I described?
 
  • #58
PAllen said:
Gralla and Wald show there is a precise way in which mass and size can be taken to a zero limit such that the motion becomes exactly a timelike geodesic. The MiSaTaQuWa equations are the first order correction, which include the possibility for a small body radiating GW, thus affecting its trajectory.

They seem to need M≠0 to get the exact geodesic for λ=0. (eq 49 in http://arxiv.org/abs/0806.3293 )
 
  • #59
PeterDonis said:
I
However, he also mentions numerical solutions at the end, which makes me wonder: do numerical solutions not give enough information to even apply the test I described?

The question is, what is the test? A geodesic of the numerical metric would represent a test body motion in the spacetime of the two massive bodies, not the motion of of the massive bodies. This is where it all breaks down - the absence of any concept of background metric to define geodesics, or reference for perturbative analysis.
 
  • #60
atyy said:
They seem to need M≠0 to get the exact geodesic for λ=0. (eq 49 in http://arxiv.org/abs/0806.3293 )

That section is describing zero and first order approximation to geodesic. Then, the error terms vanish as M->zero, without reaching it. For small M, the path is almost independent of M; the convergence of these paths as error terms go to zero (an infinitesimal mass particle), the path becomes exact geodesic.
 
  • #61
PAllen said:
That section is describing zero and first order approximation to geodesic. Then, the error terms vanish as M->zero, without reaching it. For small M, the path is almost independent of M; the convergence of these paths as error terms go to zero (an infinitesimal mass particle), the path becomes exact geodesic.

But aren't the errors parameterized by λ, not M?

Intuitively, I'd expect that for non-zero mass, but the taking the test body approximation (body is not a source), then we get an exact timelike geodesic.

Then if we allow backreaction (body is a source), then we get an approximate timelike geodesic.
 
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  • #62
PAllen said:
The question is, what is the test? A geodesic of the numerical metric would represent a test body motion in the spacetime of the two massive bodies, not the motion of of the massive bodies.

I would expect it to represent both, at least at the level of approximation I think is being used for the binary pulsar modeling. AFAIK they are not modeling the pulsars' internal structure; they are just treating them as spherically symmetric objects each with a given mass and radius, where the radius is much smaller than the distance between them. The motion of each body is represented by the motion of the geometric center of each sphere; that's the motion that I believe works out to a geodesic of the overall metric.
 
  • #63
PeterDonis said:
I would expect it to represent both, at least at the level of approximation I think is being used for the binary pulsar modeling. AFAIK they are not modeling the pulsars' internal structure; they are just treating them as spherically symmetric objects each with a given mass and radius, where the radius is much smaller than the distance between them. The motion of each body is represented by the motion of the geometric center of each sphere; that's the motion that I believe works out to a geodesic of the overall metric.

I'd like to see some reference for that. I could find no such thing when I last researched this. I was specifically looking for a way to treat it like that (each body moving on a geodesic fo the total spacetime), but could find nothing.
 
  • #64
atyy said:
But aren't the errors parameterized by λ, not M?

Intuitively, I'd expect that for non-zero mass, but the taking the test body approximation (body is not a source), then we get an exact timelike geodesic.

Then if we allow backreaction (body is a source), then we get an approximate timelike geodesic.

A body not a source is counter-factual unless body has infinitesimal mass. Thus, a no source approximation is a limit of mass approaching zero.
 
  • #65
PAllen said:
A body not a source is counter-factual unless body has infinitesimal mass. Thus, a no source approximation is a limit of mass approaching zero.

I guess he has λ→0, but M≠0. So you are saying mass goes to zero because λ→0, whereas I am saying mass is not zero, because M≠0. I do think λ→0 is kind of a mass→0, so I see your point, but I still don't understand then why M≠0.
 
  • #66
atyy said:
I guess he has λ→0, but M≠0. So you are saying mass goes to zero because λ→0, whereas I am saying mass is not zero, because M≠0. I do think λ→0 is kind of a mass→0, so I see your point, but I still don't understand then why M≠0.

I think the treatment in section II of:

http://arxiv.org/abs/1002.5045

Which is based on Gralla and Wald, is a bit simpler and easier to understand. They make explicit that mass must decrease to zero as λ decreases to zero.
 
  • #67
PAllen said:
I think the treatment in section II of:

http://arxiv.org/abs/1002.5045

Which is based on Gralla and Wald, is a bit simpler and easier to understand. They make explicit that mass must decrease to zero as λ decreases to zero.

Looks like there are 2 masses. He says λ→0 is mass going to zero, but at the end M≠0, which he says is the ADM mass.
 
  • #68
PAllen said:
I think the treatment in section II of:

http://arxiv.org/abs/1002.5045

Which is based on Gralla and Wald, is a bit simpler and easier to understand. They make explicit that mass must decrease to zero as λ decreases to zero.

For λ→0, it's more like a size going to zero. He says in words mass goes to zero, otherwise it's a black hole. But I don't see a problem with the point particle being a black hole, so is there a need to say λ→0 is size and mass going to zero?

It seems conceptually ok to have the "word description" of λ→0 as size going to zero, we allow the point particle to be a black hole, and we end up with non-zero mass M≠0.
 
  • #69
atyy said:
For λ→0, it's more like a size going to zero. He says in words mass goes to zero, otherwise it's a black hole. But I don't see a problem with the point particle being a black hole, so is there a need to say λ→0 is size and mass going to zero?

It seems conceptually ok to have the "word description" of λ→0 as size going to zero, we allow the point particle to be a black hole, and we end up with non-zero mass M≠0.

I think I see the resolution. The metric is scaled during the limiting process, and the ADM mass is a scaled ADM mass. That is, unscaled, the mass goes to zero, but the scaled ADM mass remains constant:

"The results of this section (i.e., the results of sec. IV of [4]) may be summarized as
follows. Consider a one-parameter-family of spacetimes containing a body whose size and
mass decrease to zero, according to the stated assumptions."

"Furthermore, the “particle mass” M is indeed the
ADM mass of the body (as measured in the scaled limit)."

All of this must be true based on my physical argument: you cannot treat a finite mass body as not being a source, no matter how small you make it (without also decreasing its mass).
 
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  • #70
PeterDonis said:
I would expect it to represent both, at least at the level of approximation I think is being used for the binary pulsar modeling. AFAIK they are not modeling the pulsars' internal structure; they are just treating them as spherically symmetric objects each with a given mass and radius, where the radius is much smaller than the distance between them. The motion of each body is represented by the motion of the geometric center of each sphere; that's the motion that I believe works out to a geodesic of the overall metric.

Let me try to get at the crux of the matter. There is some exact metric (which we don't know) representing two similar mass orbiting bodies. If we assume they are BH's, what might be presumed to represent a world line trajectory for one is the world line of a singularity - oops, better not go their; that's not in the manifold. Thus, we better not assume BH.

Then, a world line representing a body trajectory is one that is always inside a world tube of non-vanishing SET (call it a matter region). What would need to be shown is that there exists a timelike geodesic in matter region that remains always in the matter region. Then if the matter size is small, this is reasonably a body trajectory. I am unaware of any such result being referred to in the literature. It would be really cool if it were true and someone provided a convincing argument for it.

Then, also, a timelike geodesic in the vacuum region would represent a test body trajectory.

Another take on this would be to imagine co-orbiting spherical shells. Then also what we would like to believe is that some geodesic inside each shell (one with the right initial tangent) is always inside the shell, never hitting the edge. (Note that while for one shell, you have Minkowski space inside, for two shells you do not - there is no such thing as a gravity shield, and one shell influences geometry inside the other shell).
 
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<h2>1. What is free fall?</h2><p>Free fall is the motion of an object under the influence of gravity alone, without any other external forces acting on it. In this state, the object is accelerating towards the ground at a constant rate of 9.8 meters per second squared.</p><h2>2. Why does a free falling charge radiate?</h2><p>A free falling charge radiates because it is undergoing acceleration, which causes it to emit electromagnetic radiation. This is known as synchrotron radiation, and it is a fundamental principle of classical electromagnetism.</p><h2>3. How does the rate of radiation change during free fall?</h2><p>The rate of radiation during free fall increases as the object falls faster. This is because the acceleration of the object increases, causing it to emit more radiation. As the object reaches terminal velocity, the rate of radiation remains constant.</p><h2>4. Does the mass of the falling object affect the amount of radiation emitted?</h2><p>Yes, the mass of the falling object does affect the amount of radiation emitted. According to the Larmor formula, the power of the radiation emitted is directly proportional to the square of the charge and the acceleration of the object, and inversely proportional to the square of the distance from the object. Therefore, a heavier object will emit more radiation than a lighter object with the same charge and acceleration.</p><h2>5. Is the radiation emitted by a free falling charge harmful?</h2><p>The radiation emitted by a free falling charge is not harmful to humans. This type of radiation is in the form of radio waves and is in the non-ionizing part of the electromagnetic spectrum, which means it does not have enough energy to cause damage to cells. However, it is still important to limit exposure to high levels of radiation for safety reasons.</p>

1. What is free fall?

Free fall is the motion of an object under the influence of gravity alone, without any other external forces acting on it. In this state, the object is accelerating towards the ground at a constant rate of 9.8 meters per second squared.

2. Why does a free falling charge radiate?

A free falling charge radiates because it is undergoing acceleration, which causes it to emit electromagnetic radiation. This is known as synchrotron radiation, and it is a fundamental principle of classical electromagnetism.

3. How does the rate of radiation change during free fall?

The rate of radiation during free fall increases as the object falls faster. This is because the acceleration of the object increases, causing it to emit more radiation. As the object reaches terminal velocity, the rate of radiation remains constant.

4. Does the mass of the falling object affect the amount of radiation emitted?

Yes, the mass of the falling object does affect the amount of radiation emitted. According to the Larmor formula, the power of the radiation emitted is directly proportional to the square of the charge and the acceleration of the object, and inversely proportional to the square of the distance from the object. Therefore, a heavier object will emit more radiation than a lighter object with the same charge and acceleration.

5. Is the radiation emitted by a free falling charge harmful?

The radiation emitted by a free falling charge is not harmful to humans. This type of radiation is in the form of radio waves and is in the non-ionizing part of the electromagnetic spectrum, which means it does not have enough energy to cause damage to cells. However, it is still important to limit exposure to high levels of radiation for safety reasons.

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