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.
  • #71
PeterDonis said:
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 which 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).
 
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  • #73
PAllen said:
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).

Yes, the physical argument is good. I'm guess it is right that the massless limit (λ→0) is an exact timelike geodesic, and it is also right that this limit depends on M≠0, where M is the ADM mass for a scaled observer at any λ > 0, so that the massless case "remembers" that it has come as the self-similar limit of a massive body. So the massless point particle on an exact timelike geodesic exists, but cannot exist on its own, and must come from a limit of massive bodies.
 
  • #74
atyy said:
Yes, the physical argument is good. I'm guess it is right that the massless limit (λ→0) is an exact timelike geodesic, and it is also right that this limit depends on M≠0, where M is the ADM mass for a scaled observer at any λ > 0, so that the massless case "remembers" that it has come as the self-similar limit of a massive body. So basically, the massless case cannot exist on its own, and must come from a limit of massive bodies.

That's the way I interpret it. Especially given that we know that over 50 orders of magnitude (Jupiter around sun, to atom around sun), the path is essentially the same. The limit is a continuation of this process. Call it the 'infinitesimal limit of mass and size' rather than a mass of zero.
 
  • #75
PAllen said:
That's the way I interpret it. Especially given that we know that over 50 orders of magnitude (Jupiter around sun, to atom around sun), the path is essentially the same. The limit is a continuation of this process. Call it the 'infinitesimal limit of mass and size' rather than a mass of zero.

That paper is like magic. First he says point particles don't make sense, then he shows they magically pop out when you consider a family of massive bodies. The appearance of the ADM mass is the most amazing thing.
 
  • #76
One thing that prevents any possibility of such a limiting process approaching a null geodesic is that there is no such thing, in GR, of an 'almost null' timelike path, in any coordinate independent sense. So you must get a limiting timelike path, as the physics suggests.
 
  • #77
PAllen said:
One thing that prevents any possibility of such a limiting process approaching a null geodesic is that there is no such thing, in GR, of an 'almost null' timelike path, in any coordinate independent sense. So you must get a limiting timelike path, as the physics suggests.

But shouldn't be an analogous argument for null geodesics, since those are what test light rays move on?
 
  • #78
TrickyDicky said:
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.

This makes sense, and I agree it might indeed be what Gralla was trying to say. Basically this would mean that, since the orbital period of the binary pulsar is much shorter than the characteristic time scale for gravitational radiation from that system, the post-Newtonian method would give approximate geodesic orbits for time scales short compared to the radiation time scale, but if you tried to model the long-term behavior of the system that way you would have to adjust the orbital parameters every so often as gravitational radiation was emitted to keep the approximation close enough. To actually predict the long-term changes in the orbital parameters, you would need to do the strong field numerical simulation.

TrickyDicky said:
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 which I don't think it's true.

I'm not trying to say this. I agree there would be little point in the concept of a geodesic if it didn't give you a way to pick out some meaningful subset of all possible worldlines.

TrickyDicky said:
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).

Geodesic paths can "change" and still be geodesic paths because of changes in the metric that defines what a geodesic path is. Consider the detector scenario; suppose we have a GW detector that uses interferometry, like LIGO or LISA. When a gravitational wave passes through the detector, it shows interference fringes; but the individual mirrors that reflect the laser light that shows the fringes (because of small changes in the proper length between the mirrors) are in free fall the whole time. They are following geodesics, but geodesics of a time-varying metric.

Similarly, the two neutron stars in the binary pulsar could be following geodesics, but still have their orbital parameters change, because the metric is changing. In fact, that's probably the wrong way to think about it, though, because the changes in the orbital parameters, at least to a first approximation, *are* the changes in the metric. The stars themselves don't change, considered in isolation; what changes is their relationship. The overall metric of the system as a whole includes the relationship between the stars, so if that changes, the metric changes, even if each star remains exactly the same internally. This doesn't mean the stars don't travel on geodesics; it means that there is a single self-consistent solution realized by Nature (which we can only approximate at our current level of knowledge) that has each star (more precisely, each star's center of mass) traveling on a geodesic of the full, time-dependent metric that is ultimately due to the two stars acting together as sources.

But *why* would the orbital parameters change, if the stars themselves are not changing internally? AFAIK the answer to this involves the light-speed time delay in the propagation of gravity, as outlined, for example, in this paper by Carlip:

http://arxiv.org/abs/gr-qc/9909087

Of course it's possible that the full, self-consistent solution realized by Nature does not have the stars traveling on exact geodesics of the full, time-dependent metric; that's what Gralla seems to think, for example. We won't know for sure until we can construct such solutions and make more precise observations. But I don't think we can rule out the possibility that, at least for systems like the binary pulsar, gravitational waves can be emitted without requiring any deviation from geodesic motion to explain them.
 
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  • #79
atyy said:
But shouldn't be an analogous argument for null geodesics, since those are what test light rays move on?

I would think it should be possible to make a similar argument for light. Start with some SET representing a wave packet in some background geometry. Theoretically it acts as a source and can produce GW, thus not following a geodesic exactly. Then, in some limiting process to zero size and energy, it does exactly follow a null geodesic. Not saying I have any idea how to pull this off. Note, that there are null paths that are not geodesics. This would presumably take the limit of nearly geodesic null paths to the geodesic limit.
 
  • #80
PAllen said:
I would think it should be possible to make a similar argument for light. Start with some SET representing a wave packet in some background geometry. Theoretically it acts as a source and can produce GW, thus not following a geodesic exactly. Then, in some limiting process to zero size and energy, it does exactly follow a null geodesic. Not saying I have any idea how to pull this off. Note, that there are null paths that are not geodesics. This would presumably take the limit of nearly geodesic null paths to the geodesic limit.

It's easy. Just kidnap Sam Gralla ...
 
  • #81
PAllen said:
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.

I'm really skeptical this can be done in any meaningful sense. Consider a fluid body with internal flows and eddies. Then geodesics of the metric from the SET would seem to represent the local flow at a point. You might then have many different convoluted geodesics. If you then try a limiting process like Wald & Gralla, here you have to keep actual (not scaled) mass constant as you shrink the bodies - at least if you want your limiting case to include GW. But then you get singularities ...
If you try an averaging approach, how do you make it precise? Average of a bunch of convoluted geodesics? If the approach only applies to sufficiently simple SET, it isn't much of aresult.
 
  • #82
PeterDonis said:
Of course it's possible that the full, self-consistent solution realized by Nature does not have the stars traveling on exact geodesics of the full, time-dependent metric; that's what Gralla seems to think, for example. We won't know for sure until we can construct such solutions and make more precise observations. But I don't think we can rule out the possibility that, at least for systems like the binary pulsar, gravitational waves can be emitted without requiring any deviation from geodesic motion to explain them.

I haven't seen you address the issue of geodesics of what, exactly? Clearly, if you had a complete solution including matter and vaccuum, representing co-orbiting stars, geodesics of the vacuum metric would represent test particle paths for particles under the action of the stars and their GW. Geodesics in the matter region are more complex, but I would think they represent matter flow lines which could be all over the place for a general SET. I am completely failing to see how to state the proposition that the star is following a geodesic.

And if I imagine limiting this to zero sized stars, at every stage, vacuum geodesics near a star represent a test particle orbit, while the star itself becomes singular.
 
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  • #83
PAllen said:
I haven't seen you address the issue of geodesics of what, exactly?

Consider a simple example: a spherically symmetric gravitating body like a neutron star. The worldline of the center of mass of the star (which is also its geometric center, and the spatial origin r = 0 under the usual Schwarzschild-type coordinates) will be a geodesic; just ask yourself what its proper acceleration is. If it's nonzero, then it picks out some preferred direction in space (the direction of the acceleration), which breaks spherical symmetry. So whatever the full metric in the matter region is, the star's CoM follows a geodesic of it.

It is true that, since the star has finite size, no portion of the star other than its CoM will travel along a geodesic; again, that's obvious just by considering that the proper acceleration of any piece of matter at r > 0 must be nonzero. So there are certainly pieces of matter present that are traveling on non-geodesic worldlines. But if we don't need to worry about the star's internal structure, we can ignore all that, and just treat the motion of the star's CoM as the motion of the star.

What I'm hypothesizing is that a similar dodge will work in a case like the binary pulsar: we can ignore the internal structure of the two neutron stars and treat the motion of each star's CoM as the motion of the star itself. Then the question becomes: what is the proper acceleration of each star's CoM? I'm hypothesizing that it's still zero; the only thing that I can see that would make it nonzero is that curvature effects would cause a net force on the star as a whole, something like "spacetime swimming":

http://dspace.mit.edu/handle/1721.1/6706

In the binary pulsar case I would expect any effect of this type to be too small to matter because the separation between the stars is so much larger than their sizes; but I admit I have not tried to do any calculation along these lines.
 
  • #84
Well, for starters, there are big problems defining COM in GR. In any case, I assume you no longer would claim you know that this will work.

Last night I did find an old write up by Synge of the following:

Under very broad assumptions, given a matter world tube with sharp boundary, and assuming no non-gravitational radiation, and an exterior vacuum metric region with no assumptions made (e.g. could be non static; no asymptotic flatness assumed), then there exists in the matter world tube a locus of points of no proper acceleration that form a continuous curve*. However, he claimed (without showing it, only by reference to ancient literature he was borrowing the treatment from) that such a locus not only need not be the path of any matter, it could be spacelike! (Even though every piece of matter is following a timelike path).

* For a non-spinning body, it is necessary to assume things about pressure that seem physically plausible, to get this result. For a spinning body, he had to assume a limit on amount of spin. Without assuming these, it did not follow that there was a locus of no-acceleration at all.

So, I think you have a lot of work to make your argument really convincing.
 
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  • #85
PeterDonis said:
Geodesic paths can "change" and still be geodesic paths because of changes in the metric that defines what a geodesic path is.
But we don't have any such solution of the EFE and it's hard to see that coming soon. This IMO would get us close to the no useful distinction between timelike geodesics and timelike worldlines scenario.

PeterDonis said:
Consider the detector scenario; suppose we have a GW detector that uses interferometry, like LIGO or LISA. When a gravitational wave passes through the detector, it shows interference fringes; but the individual mirrors that reflect the laser light that shows the fringes (because of small changes in the proper length between the mirrors) are in free fall the whole time.

Interferometry for GW detection either ground-based or space-based measures motion between separated 'free' or test masses. That motion is obviously a deviation of the expected free fall motion of the masses that gets reflected by the mirrors, otherwise nothing is measured, that deviation is the deviation from geodesic motion of the emitter.

If GW are disturbances of the gravitational field i.e propagation of changes in spacetime curvature, it is hard to see how it would be possible for them to be emitted and detected maintaining geodesic motion of the body that emits them or the detector that receives them as long as we think of geodesics as the straightest spacetime lines in a curved manifold.
 
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  • #86
PAllen said:
Well, for starters, there are big problems defining COM in GR.

In the general case, yes. I don't mean to say that the sort of thing I'm talking about will work in the general case. I'm trying to see if something like the binary pulsar can be treated as close enough to an idealized special case where the big problems don't arise.

PAllen said:
In any case, I assume you no longer would claim you know that this will work.

I didn't mean to claim that in the first place. Sorry if that wasn't clear.

PAllen said:
such a locus not only need not be the path of any matter, it could be spacelike! (Even though every piece of matter is following a timelike path).

Hm, interesting. I don't suppose anything like this is available online? :wink:
 
  • #87
TrickyDicky said:
Interferometry for GW detection either ground-based or space-based measures motion between separated 'free' or test masses. That motion is obviously a deviation of the expected free fall motion of the masses

"Expected" in the absence of GWs.

TrickyDicky said:
that deviation is the deviation from geodesic motion of the emitter.

Not necessarily. That's the question under discussion.

TrickyDicky said:
If GW are disturbances of the gravitational field i.e propagation of changes in spacetime curvature, it is hard to see how it would be possible for them to be emitted and detected maintaining geodesic motion of the body that emits them or the detector that receives them as long as we think of geodesics as the straightest spacetime lines in a curved manifold.

I think you're failing to consider that the spacetime curvature of the manifold is time-dependent; a time-dependent metric means that a geodesic curve won't look "straight" in the sense you would expect it to based on looking at geodesics in time-independent metrics. If GWs are present then the metric has bumps and wiggles in it; that's what the GWs *are*. So geodesics of such a metric will also have bumps and wiggles, which will be followed by objects lying in the path of the GWs. AFAIK the fact that the motion of the mirrors in an interferometer-type GW detector is geodesic is well-established; I'm pretty sure Kip Thorne goes into this in Black Holes and Time Warps, for example, at least at a lay person's level.
 
  • #88
PeterDonis said:
Hm, interesting. I don't suppose anything like this is available online? :wink:

Not that I could find.

I did find a way to visualize this surprising claim:

Imagine there is fluid wave motion inside the body. Then the locus no acceleration could reflect that some 'particle' in the wave has no proper acceleration, while a nearby particle in a slightly different phase of the wave is the 'next' particle with no proper acceleration. Then the locus of no acceleration represents something more like a phase propagation than a material propagation. I can imagine it in a spacelike zigzag through the world tube.

Perhaps under much more restrictive assumptions about the SET, you could get a nicer result. But, again, I've looked and not found any sign of such claim in the literature (but I don't have access to a university library, and don't claim to any great searching skills).

[In particular, I did a lot of searching on 'generalized equivalence principle' and 'Detweiler-Whiting' to see if there even any proposals that these could be generalized. I found none. The implications of some writers was clearly that this could only be expected for the extreme mass ratio case covered by the MiSaTaQua equation. ]
 
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  • #89
PeterDonis said:
I think you're failing to consider that the spacetime curvature of the manifold is time-dependent; a time-dependent metric means that a geodesic curve won't look "straight" in the sense you would expect it to based on looking at geodesics in time-independent metrics.
I have no problem with the geodesics in a time-dep. metric. Think of the FRW metric. I just said we don't have such a metric for a spacetime compatible with GWs.



If GWs are present then the metric has bumps and wiggles in it; that's what the GWs *are*. So geodesics of such a metric will also have bumps and wiggles, which will be followed by objects lying in the path of the GWs.
Well the existence of such metrics as solutions of the EFE is what's being discussed.

AFAIK the fact that the motion of the mirrors in an interferometer-type GW detector is geodesic is well-established;
Yes, but that motion has to be disturbed to register a GW.
 
  • #91
atyy said:
http://arxiv.org/abs/1206.6538 has interesting comments about GW from binaries.

Great find! This did not exist when I last looked (2011). It will take much time to digest...
 
  • #92
TrickyDicky said:
we don't have such a metric for a spacetime compatible with GWs.

Perhaps not for sources, but we do for detectors like LIGO and LISA. By the time a GW reaches such a detector, it's weak enough to be treated by linearized GR. MTW has at least one whole chapter on this IIRC (my copy isn't handy right now to check).

TrickyDicky said:
Yes, but that motion has to be disturbed to register a GW.

You're missing the point: the geodesic motion *is* the "disturbance". Saying that the mirrors follow geodesics that have bumps and wiggles in them, and saying that the mirrors are disturbed by the GW, are different ways of saying the same thing. There is no "disturbance" over and above the behavior of the geodesics due to the time-dependence of the metric.
 
  • #93
PeterDonis said:
Perhaps not for sources, but we do for detectors like LIGO and LISA. By the time a GW reaches such a detector, it's weak enough to be treated by linearized GR. MTW has at least one whole chapter on this IIRC (my copy isn't handy right now to check).
I thought the thread's discussion (though switching back and forth from EM source to gravitational source here and there) was more centered in the sources.

PeterDonis said:
You're missing the point: the geodesic motion *is* the "disturbance". Saying that the mirrors follow geodesics that have bumps and wiggles in them, and saying that the mirrors are disturbed by the GW, are different ways of saying the same thing. There is no "disturbance" over and above the behavior of the geodesics due to the time-dependence of the metric.

This point seems worth being missed because IMO it contradicts basic concepts in GR. For instance I guess this quote from wikipedia "Spacetime" must not be righ to you:
"The concept of geodesics becomes critical in general relativity, since geodesic motion may be thought of as "pure motion" (inertial motion) in spacetime, that is, free from any external influences."
I consider GWs an external influence, don't you?

Maybe it would be interesting here to consider absolute gravitometers that are a type of accelerometers that work by directly measuring the acceleration of a mass during free fall in a vacuum that includes a retroreflector and a Michelson interferometer so interferometry is also used. I don't think this is a very different mechanism ultimately (obviously the details are very different, the gravitometer is attached to the ground for one but so are all ground-based GW detectors) from that used in LIGO to detect GWs, the detector is ultimately a very specialized , very sensitive type of accelerometer.
I've read you in several threads defining geodesic motion as that which reads no acceleration in an accelerometer. But now you define geodesics as something that includes exactly the type of "bumps and wiggles" disturbances that an accelerometer should measure.
That is odd. I mean you don't bring up the time-dependence of the metric when talking about the absolute acceleration notion in GR.
 
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  • #94
TrickyDicky said:
"The concept of geodesics becomes critical in general relativity, since geodesic motion may be thought of as "pure motion" (inertial motion) in spacetime, that is, free from any external influences."
I consider GWs an external influence, don't you?

This is a good example of why Wikipedia is not to be trusted. By "external influences", they really mean non-gravitational external influences. GWs are not a non-gravitational external influence. They are fluctuations in spacetime curvature.

TrickyDicky said:
the detector is ultimately a very specialized , very sensitive type of accelerometer.

No, it isn't. It doesn't measure proper acceleration; it measures fluctuations in spacetime curvature, which manifest as fluctuations in the proper distance between two endpoints that are moving on geodesics.

TrickyDicky said:
I mean you don't bring up the time-dependence of the metric when talking about the absolute acceleration notion in GR.

Because proper acceleration is curvature of the *path*, not spacetime; it is a different thing from curvature of spacetime, time-varying or otherwise. Here we're talking about curvature of spacetime; the paths of the mirrors in the GW detector are straight, but they're straight in a manifold that's curved, and whose curvature varies with time.
 
  • #95
PeterDonis said:
No, it isn't. It doesn't measure proper acceleration; it measures fluctuations in spacetime curvature, which manifest as fluctuations in the proper distance between two endpoints that are moving on geodesics.
It isn't exactly a regular accelerometer as I said, in fact rather than proper acceleration what GW detectors measure are variations ("fluctuations") in proper acceleration, much like gravity gradiometers do.
Even though WP is not the most reliable source this seems accurate: "Pairs of accelerometers extended over a region of space can be used to detect differences (gradients) in the proper accelerations of frames of references associated with those points. These devices are called gravity gradiometers, as they measure gradients in the gravitational field. Such pairs of accelerometers in theory may also be able to detect gravitational waves."
So of course they are not exactly the same thing but theoretically when used in groups to detect variations of proper acceleration they share a basically similar mechanism.

Also in a curved spacetime, there may be more than one geodesic between two events, so the proper length between the endpoints is not uniquely defined, and if it is not uniquely defined I wonder the sense of measuring its "fluctuations", if you are really right about GW detectors measuring varaitions in proper length, with respect to what?


PeterDonis said:
Because proper acceleration is curvature of the *path*, not spacetime; it is a different thing from curvature of spacetime, time-varying or otherwise. Here we're talking about curvature of spacetime; the paths of the mirrors in the GW detector are straight, but they're straight in a manifold that's curved, and whose curvature varies with time.
See above.
Can you define what you call path? , you said that the paths of the mirrors were geodesics, I agree they are, until they are modified by the gravitational wave, something has to trigger the motion of the test masses in order to then be registered by interferometry, no?
I'm yet to understand how a mass can be made to change its state of motion without a proper acceleration being involved.
 
  • #96
The GW detection involves the geodesic deviation equation of neighboring geodesics and is therefore related directly to the space - time curvature (the time dependent perturbations when we are talking about GW waves) as can be seen in the equation. Proper acceleration (as measured by an accelerometer) is related to a single wordline. This is exactly what PeterDonis has said already.
 
  • #97
WannabeNewton said:
The GW detection involves the geodesic deviation equation of neighboring geodesics and is therefore related directly to the space - time curvature (the time dependent perturbations when we are talking about GW waves) as can be seen in the equation. Proper acceleration (as measured by an accelerometer) is related to a single wordline. This is exactly what PeterDonis has said already.
Hi WN, I made clear that difference when I spoke of the gravity gradiometer that in fact involves several accelerometers for neighbouring worldlines.
 
  • #98
TrickyDicky said:
Hi WN, I made clear that difference when I spoke of the gravity gradiometer that in fact involves several accelerometers for neighbouring worldlines.
Sorry I typed up my response and left it alone for a bit before submitting so I didn't get to see your response before then.
 
  • #99
No problem.
 
  • #100
WannabeNewton said:
The GW detection involves the geodesic deviation equation of neighboring geodesics and is therefore related directly to the space - time curvature (the time dependent perturbations when we are talking about GW waves) as can be seen in the equation.
This may be just nit-picking but I always thought geodesic deviation to be caused by tidal forces is there a straight forward way to relate GWs and tidal forces?
 
  • #101
TrickyDicky said:
This may be just nit-picking but I always thought geodesic deviation to be caused by tidal forces is there a straight forward way to relate GWs and tidal forces?
http://arxiv.org/pdf/gr-qc/9712019.pdf
Go to, in particular, page 159 out of 238 in the pdf itself (not page 159 in the notes).
 
  • #102
WannabeNewton said:
http://arxiv.org/pdf/gr-qc/9712019.pdf
Go to, in particular, page 159 out of 238 in the pdf itself (not page 159 in the notes).

Thanks, pal. I had read those notes long ago but I think I skipped some bits.
But yes the bottom line is that gravitational waves induce a form of tidal effect on the test masses of the modified Michelson interferometer that is used in modern GWs detectors.

I think I'll start a new thread on GW detection, tidal forces and accelerometers in order not to go so much OT here.
 
  • #103
TrickyDicky said:
I think I'll start a new thread on GW detection, tidal forces and accelerometers in order not to go so much OT here.
Yeah that would be cool. It is quite instructive to take a ring of test particles as your family of closely separated geodesics and see how the gravitational waves expand and shear them. There is a close relationship between the Weyl tensor as a contributor to shear and the Ricci tensor as a contributor to expansion, as codified by the Raychaudhuri equation (however they are coupled so it isn't exactly totally independent)
 
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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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