Does a free falling charge radiate ?


by greswd
Tags: charge, falling, free, radiate
PAllen
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#55
Feb20-13, 05:29 PM
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Quote Quote by atyy View Post
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.
PAllen
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Feb20-13, 05:32 PM
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Quote Quote by PeterDonis View Post
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.
PeterDonis
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Feb20-13, 05:33 PM
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Quote Quote by PeterDonis View Post
I'll have to read through that thread in more detail.
I see that Sam Gralla made this comment:

Quote Quote by Sam Gralla View Post
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?
atyy
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Feb20-13, 05:43 PM
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Quote Quote by PAllen View Post
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 )
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Feb20-13, 05:45 PM
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Quote Quote by PeterDonis View Post
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.
PAllen
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Feb20-13, 05:54 PM
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Quote Quote by atyy View Post
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.
atyy
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Feb20-13, 06:02 PM
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Quote Quote by PAllen View Post
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.
PeterDonis
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Feb20-13, 06:12 PM
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Quote Quote by PAllen View Post
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.
PAllen
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Feb20-13, 06:26 PM
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Quote Quote by PeterDonis View Post
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.
PAllen
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Feb20-13, 06:28 PM
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Quote Quote by atyy View Post
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.
atyy
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Feb20-13, 06:38 PM
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Quote Quote by PAllen View Post
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.
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Feb20-13, 06:44 PM
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Quote Quote by atyy View Post
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.
atyy
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Feb20-13, 07:03 PM
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Quote Quote by PAllen View Post
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.
atyy
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Feb20-13, 08:06 PM
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Quote Quote by PAllen View Post
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.
PAllen
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Feb20-13, 09:04 PM
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Quote Quote by atyy View Post
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).
PAllen
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Feb20-13, 09:17 PM
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Quote Quote by PeterDonis View Post
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 vaccuum 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).
TrickyDicky
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#71
Feb21-13, 04:39 AM
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Quote Quote by PeterDonis View Post
I see that Sam Gralla made this comment:



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?
I don't think that is the sense of Gralla's comment. Post-Newtonian methods give valid appproximations for weak-field situations, you might try and use it just to approximate the orbit of 2 bodies when their distance is very large wrt their radii and call them geodesic orbits, but that won't get you a realistic approximation for the gravitational radiation of that system, for that you need strong-field numerical relativity.
Ultimately this might be a definitional problem, but if you want to call geodesic motion to all orbiting bodies, extended or test particles, regardless of the intensity of the radiation (gravitational or EM) they are emitting you basically are saying that all test particles and extended bodies worldlines following some kind of orbit no matter how unstable are following geodesic motion wich I don't think it's true.
I used to also think that extended objects in orbit followed geodesics but was convinced here at PF that this would make gravitational radiation a superfluous notion since if orbiting bodies emitting radiation, regardless of the intensity(strong-field case) didn't see affected their geodesic motion, first:what could actually ever affect a geodesic path? and second: how do we expect that radiation to affect distance bodies detectors if it isn't capable to alter the geodesic path of the emitting body in the least(as long as we still consider third Newton's law as valid of course).
TrickyDicky
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#72
Feb21-13, 05:00 AM
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This post from Wald's collaborator Sam Gralla might be relevant here:


http://physicsforums.com/showpost.ph...30&postcount=9


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