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Thread: gravitational waves, really never detected yet?

  1. #1 gravitational waves, really never detected yet? 
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    It is said that gravitational waves haven't been detected. I don't understand that.

    Gravitometers and other instruments are very very sensible. They can detect minutes variation in gravitation.
    Among other things, they can detect variation in gravity over a period of time. Thats a gravitational wave. Less spectacular, tides can be seen as the result of gravitational waves from the moon/sun. These gravitational waves are of low frequency, but are gravitational waves none the less.

    Producing gravitational waves is trivial, you just wave your hand. Detecting is a bit harder however. If you move enough mass, slowly enough for a gravitometer to detect, then here it is. You produced gravitational waves, and you detected them. That mass doesn't need to be very large to be detected, it's doable at a reasonable budget. Gravitometers are sensible enough, that they can tell if there is snow or not on the roof. See, you don't need to move a mountain...

    All the gravitational waves i'm talking about are low frequency. Nothing spectacular, you'll not use that as a weapon or anything..

    What they really mean by never having detected gravitational waves, is about gravitational waves from distant stars. But i don't get why there's this difference in definitions...
    Actually, i suspect we could produce our selves detectable gravitational waves at higher frequency, if we put in enough money.... I don't get why this is not attempted.....

    So what am i missing??? O_o


    Last edited by Quantum immortal; February 9th, 2014 at 11:44 AM. Reason: messed up title
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    The gravitational waves they are looking for are the ones predicted by Einsteins theory of Relativity. They should be traveling at the speed of light. If they could be found it would be another bit of evidence that Einsteins theories are correct and might lead to more discoveries.
    If they are not found it leaves one of Einstein's predictions unverified which opens the possibility it might be false, that gravity might not be just a distortion of the spacetime.

    If Einstein was actually proved wrong on this it would not likely cause relativity to be abandoned, but it would indicate a new model for gravity might be needed. It might even lead to a new cosmological model.


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    whats the difference with what i describe???
    If you apply GR at the above cases that i describe, you don't get gravitational waves???
    What i describe above isn't going at the speed of light???
    If they aren't gravitational waves, what are they?

    to produce a wave in any field, you just need to accelerate a charge.
    If you accelerate energy, you should produce a gravitational wave.
    We can produce low frequency gravitational waves and detect them at reasonable cost.
    Even the tides are a "gravitational wave" effect, its just not common to see them that way.

    you have an hyperconducting hyper sensitive gravitometer. You move a few tons of mater neer it. It detects its field. You move the mater away, it detects a smaller field. So at the point of the gravitometer, the gravitational field varied. How do you call that? Its not a very low frequency gravitational wave? I just see an arbitrary distinction here.
    Depending how much money you have, you can increase the frequency...
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  5. #4  
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    Quote Originally Posted by Quantum immortal View Post
    It is said that gravitational waves haven't been detected. I don't understand that.

    Gravitometers and other instruments are very very sensible. They can detect minutes variation in gravitation.
    "Very very" is meaningless. "Minute" is meaningless. In instrumentation, we use actual quantities because we must. What is the required sensitivity, compared with the achieved sensitivity?

    You are making a broad qualitative argument, when what is needed is a quantitative assessment.

    So what am i missing??? O_o
    See above.

    If you are interested in the actual answer, do a quick google search for gravitational wave detection. Also see "Hulse-Taylor pulsar" for a description of the first quantitative data in support of gravitational waves.
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  6. #5  
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    Quote Originally Posted by Quantum immortal View Post
    variation in gravity over a period of time. Thats a gravitational wave.
    Actually, it's not. What you describe does not have the characteristic of a radiation, the notion that it takes with it something from the radiating system. Gravitational radiation decreases with distance at a slower rate than ordinary gravitation, and by the peeling theorem will be the only form of gravitation far from the source. If one considers the electromagnetic analogy, then oscillating a charge will produce a variation over time of the electromagnetic field. This electromagnetic field will have two parts: a part that decreases with that is essentially the electromagnetic field associated with the charge itself, and a part that decreases with that is the electromagnetic radiation. Because the energy of the electromagnetic field is proportional to the square of the electromagnetic field, it is the energy flux of the electromagnetic field that decreases with and thus conserved.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    This has got me thinking, so I have had a quick look on Wiki. I came across absolute gravimeters and relative gravimeters. For instance the absolute gravimeters, work by directly measuring the acceleration of a mass during free fall in a vacuum, when the accelerometer is rigidly attached to the ground. I therefore conclude that the device is not detecting gravitational waves ( I could be wrong ). Will the dudes in the lab coats please shed some more light on the questions raised by the OP.

    Gravimeter - Wikipedia, the free encyclopedia

    The dudes in the lab coats turned up whilst I was typing.
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  8. #7  
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    Quote Originally Posted by Quantum immortal View Post
    whats the difference with what i describe???
    If you apply GR at the above cases that i describe, you don't get gravitational waves???
    What i describe above isn't going at the speed of light???
    If they aren't gravitational waves, what are they?

    to produce a wave in any field, you just need to accelerate a charge.
    If you accelerate energy, you should produce a gravitational wave.
    We can produce low frequency gravitational waves and detect them at reasonable cost.
    Even the tides are a "gravitational wave" effect, its just not common to see them that way.

    you have an hyperconducting hyper sensitive gravitometer. You move a few tons of mater neer it. It detects its field. You move the mater away, it detects a smaller field. So at the point of the gravitometer, the gravitational field varied. How do you call that? Its not a very low frequency gravitational wave? I just see an arbitrary distinction here.
    Depending how much money you have, you can increase the frequency...

    The thing is that such variations in local field strength also occurs with Newtonian gravity, but Newtonian gravity doesn't allow for gravity waves since gravity would act instantaneously. With GR, It is the interaction speed limit that allows for gravity wave formation. Also, in GR, gravity waves can actually carry energy away from a system (Again something that doesn't happen with Newtonian Gravity.) Thus the measurements you mention are not evidence for gravitational waves.
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  9. #8  
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    Quote Originally Posted by Quantum immortal View Post
    whats the difference with what i describe???
    If you apply GR at the above cases that i describe, you don't get gravitational waves???
    What i describe above isn't going at the speed of light???
    If they aren't gravitational waves, what are they?,,,,,.
    Oh sorry.
    The point is they are trying to prove the gravity waves move at the speed of light, almost impossible to do locally.

    However it seems like it is almost impossible to do anyhow, oh well.
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    isn't the distinction between 1/R and 1/R^2 a bit artificial. Both are waves...

    The gravitomer is influenced my matter that is adjacent to it. The definition of the direction of the ground changes a bit.

    I checked once, with c=l*f
    for a reasonable wave length, we would need a ridiculously high frequency....
    looks hopeless

    I understood the points, but i still think that they shouldn't say that we haven't detected gravitational waves.
    Rather, they haven't studied gravitational waves in detail.

    I guess its splitting hair, depending how exactly you define a wave and detecting a wave....
    If the "other" wave like stuff are not waves, what are they then??? I call them waves...
    For me, any acceleration produces waves. If the field varies locally, then work is performed, hence energy was transmitted.
    The speed can't be greater then the speed of light, or relativity would be violated... i take that for granted.... Not too cranky...
    I would call that direct detection, albeit, not a very precise one....

    anyway, its not important, its just terminology...
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  11. #10  
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    Quote Originally Posted by Quantum immortal View Post
    isn't the distinction between 1/R and 1/R^2 a bit artificial.
    No, it's not. Indeed, consideration of the distance law has recently led me to reconsider what the covariant divergence of the Weyl conformal tensor actually means. Because the Weyl conformal tensor, which represents the tidal effect due to external gravitation, decreases with from a static source, its zero covariant divergence in the absence of energy-momentum no longer makes sense to me. However, the generalised Stokes' theorem only applies to vector fields (which the Weyl conformal tensor field is not).

    Putting this in more familiar terms, if one has a vector field representing a flux through a closed surface, then conservation of that vector field implies that the strength of that vector field is inversely proportional to the total area of that surface. Because the surface area of sphere is , this implies a law for a conserved vector field. If one has an energy flux through a surface, then its conservation implies a law, and therefore a law for the corresponding field strength (because the energy is proportional to the square of the field strength). A law for the field strength implies a law for the energy which indicates that there is no energy being radiated away from the source.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    its zero covariant divergence in the absence of energy-momentum no longer makes sense to me
    How come ? The covariant divergence of the Weyl tensor is proportional to the Cotton tensor, which is in itself an exterior derivative of a linear combination of the Ricci tensor and Ricci scalar. In the absence of sources, this must vanish by definition. Or am I missing something here ?
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    what i meant originally

    If you move mass at one place, and then remove it. The field at that place varied. Hence there is work. Hence there's energy that got emited. This movement produced a wave of very low frequency. A frequency low enough, so that a gravitometer can follow the variation in the field. For discussion sake, lets say it has a period of 1 hour, or lets say a whole day.
    If this is not a wave, then what is it then? A non-wave variation of the field?
    I define a wave, as any temporal variation of a field.
    I think we just differ in some definiions, not in substance.
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    its zero covariant divergence in the absence of energy-momentum no longer makes sense to me
    How come ? The covariant divergence of the Weyl tensor is proportional to the Cotton tensor, which is in itself an exterior derivative of a linear combination of the Ricci tensor and Ricci scalar. In the absence of sources, this must vanish by definition. Or am I missing something here ?
    Obviously, the mathematical identities themselves aren't in question, but it's the dependency of the Weyl tensor that bothers me. Superficially, this dependency conflicts with the vacuum conservation of the Weyl tensor. Thus, it requires a deeper analysis to resolve the apparent conflict.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    Superficially, this dependency conflicts with the vacuum conservation of the Weyl tensor.
    Could you elaborate on this ? I don't immediately see the conflict.
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    Superficially, this dependency conflicts with the vacuum conservation of the Weyl tensor.
    Could you elaborate on this ? I don't immediately see the conflict.
    A conserved vector field associated with a point source has a dependence because the surface area of a sphere is and the divergence theorem. Although the Weyl tensor is not a vector field and the generalised Stokes' theorem does not apply, it is still odd to me that the Weyl tensor, though conserved, has a dependence rather than a dependence. This leads me to question what the covariant divergence of the Weyl tensor actually means. I had always assumed that it means that the Weyl tensor is conserved, but now I'm not quite so sure (or at least its conservation is less obvious). The main significance of this is being able to derive a covariant expression for gravitational radiation that is distinct from ordinary gravitation. I am also interested in deriving a covariant expression for electromagnetic radiation that does not involve solving the equations. I believe it can be done using an approach related to the above considerations. However, a similar approach for gravitational radiation doesn't appear viable. I should remark however that the work I've done on this is only preliminary.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    I would like to prove that a rotating rigid dumbbell does NOT emit gravitational radiation. It is commonly accepted (including the Wikipedia article on gravitational radiation) that a rotating rigid dumbbell does emit gravitational radiation, but I believe this is incorrect on the basis that mere rotation of a stationary object will not produce gravitational radiation. Unfortunately, my argument is not entirely convincing (even to myself).
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    I would like to prove that a rotating rigid dumbbell does NOT emit gravitational radiation. It is commonly accepted (including the Wikipedia article on gravitational radiation) that a rotating rigid dumbbell does emit gravitational radiation, but I believe this is incorrect on the basis that mere rotation of a stationary object will not produce gravitational radiation. Unfortunately, my argument is not entirely convincing (even to myself).
    But if a rotating dumbbell doesn't produce gravitaional radiation wouldn't that imply that rotating star systems don't either?
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    Quote Originally Posted by dan hunter View Post
    Quote Originally Posted by KJW View Post
    I would like to prove that a rotating rigid dumbbell does NOT emit gravitational radiation. It is commonly accepted (including the Wikipedia article on gravitational radiation) that a rotating rigid dumbbell does emit gravitational radiation, but I believe this is incorrect on the basis that mere rotation of a stationary object will not produce gravitational radiation. Unfortunately, my argument is not entirely convincing (even to myself).
    But if a rotating dumbbell doesn't produce gravitaional radiation wouldn't that imply that rotating star systems don't either?
    No, because a rotating star system also has orbital decay where the objects move closer to each other. I believe that it is this radial motion that produces the gravitational radiation, not the rotation.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    I should remark that there is no cause and effect relationship between gravitational radiation and orbital decay in the sense that one is the cause and the other is the effect. Both gravitational radiation and orbital decay are concomitant. If you somehow prevent one, you'll also prevent the other. With this in mind, consider two equal masses in an otherwise circular orbit around each other. Suppose a precisely fitting rod is placed between the two masses. Because the rod causes the masses to deviate from the geodesic motion that leads to orbital decay, there will be a compressive force on each end of the rod, and therefore a force on the two masses that keeps them in precise circular motion. But because there is no longer any orbital decay, there is no longer any gravitational radiation.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    This leads me to question what the covariant divergence of the Weyl tensor actually means.
    I always equated the covariant divergence of the Weyl tensor with the density of local sources of gravitational radiation. In vacuum solutions, it is thus immediately obvious that this must vanish ( only remote sources ), or else it wouldn't be a vacuum. Am I seeing this in too simplistic a way ?

    I would like to prove that a rotating rigid dumbbell does NOT emit gravitational radiation.
    Wouldn't that depend on the axis of rotation ? If that axis coincides with the axis of symmetry of the body, then all higher multipole moments vanish, and hence it cannot be a source of gravitational radiation. If it rotates around some other axis however, the quadrupole tensor is not identically zero, so it should be a source of radiation.

    But because there is no longer any orbital decay, there is no longer any gravitational radiation.
    I am not convinced. Have you performed a calculation for this ?
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    This leads me to question what the covariant divergence of the Weyl tensor actually means.
    I always equated the covariant divergence of the Weyl tensor with the density of local sources of gravitational radiation. In vacuum solutions, it is thus immediately obvious that this must vanish ( only remote sources ), or else it wouldn't be a vacuum. Am I seeing this in too simplistic a way ?
    No, that's pretty much how I see it also, except for the following correction:

    "I always equated the covariant divergence of the Weyl tensor with the density of local sources of gravitational radiation."

    The Weyl tensor represents all external gravitation, not just gravitational radiation.

    I suppose that what I'm looking for is something that covariantly represents gravitation and for which the generalised Stoke's theorem can be applied.


    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    I would like to prove that a rotating rigid dumbbell does NOT emit gravitational radiation.
    Wouldn't that depend on the axis of rotation ? If that axis coincides with the axis of symmetry of the body, then all higher multipole moments vanish, and hence it cannot be a source of gravitational radiation. If it rotates around some other axis however, the quadrupole tensor is not identically zero, so it should be a source of radiation.
    I'm referring to the axis for which it is assumed there is gravitational radiation. More generally, I am saying that a rotating rigid object of any shape won't emit gravitational radiation. It is naturally accepted that such an object won't radiate if it is not rotating. Why should rotation make any difference?


    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    But because there is no longer any orbital decay, there is no longer any gravitational radiation.
    I am not convinced. Have you performed a calculation for this ?
    No, I haven't. Hence my desire for a proof. But my argument is based on symmetry considerations. If there is no orbital decay, then what from the constrained two mass system is radiated away with the gravitational radiation? The only parameter that can change is the rate of rotation.
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    Maybe I am stupid, but if the dumbbell is rotating around its centre of mass it is in balance, ie symetric. Even if the ends of it are different sizes it is still symetric. It stays symetric no matter what plane it is rotating in so long as it is rotating around its center of mass. right?

    Then there would be no reason for the gravity to change because the sum of the moments would cancel, at least at any reasonable distance from it.

    I have no idea how you would force a binary star system to rotate around anything but its centre of mass.

    Edit: I guess I should be using the word Barycentre instead of saying centre of mass.
    Last edited by dan hunter; February 12th, 2014 at 04:09 AM. Reason: deleting excess verbiage
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    Quote Originally Posted by KJW View Post
    The Weyl tensor represents all external gravitation, not just gravitational radiation.
    Ok, agreed

    I suppose that what I'm looking for is something that covariantly represents gravitation and for which the generalised Stoke's theorem can be applied.
    So you are basically looking for a differential form of some kind that somehow represents gravitation, since that is what Stokes applies to. Interesting. Let me think about that.

    Why should rotation make any difference?
    Is the reduced quadrupole moment not dependent on the rotational motion ? Maybe I am missing something here.
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    Quote Originally Posted by dan hunter View Post
    Maybe I am stupid, but if the dumbbell is rotating around its centre of mass it is in balance, ie symetric. Even if the ends of it are different sizes it is still symetric. It stays symetric no mattter what plane it is rotating in so long as it is rotating around its center of mass. right?

    Then there would be no reason for the gravity to change because the sum of the moments would cancel, at least at any reasonable distance from it.

    I have no idea how you would force a binary star system to rotate around anything but its centre of mass.

    Edit: I guess I should be using the word Barycentre instead of saying centre of mass.
    There is no doubt that for an observer that is in the non-rotating frame of reference and at rest relative to the barycentre, the gravitation varies with time. This is not what gravitational radiation is. Gravitational radiation is a form of radiation that carries gravitational "energy" away from the radiating system. Thus, it is not sufficient that the gravitation varies with time. An observer who circles a non-rotating dumbbell will also observe time-varying gravitation, but there is no gravitational radiation from the non-rotating dumbbell.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    I suppose that what I'm looking for is something that covariantly represents gravitation and for which the generalised Stoke's theorem can be applied.
    So you are basically looking for a differential form of some kind that somehow represents gravitation, since that is what Stokes applies to. Interesting. Let me think about that.
    A tensor that has recently attracted my attention is the Lanczos tensor, which seems as if it may be a representation of the gravitational field itself rather than its tidal effect.


    Why should rotation make any difference?
    Is the reduced quadrupole moment not dependent on the rotational motion ? Maybe I am missing something here.[/QUOTE]

    My view is that the standard derivation of gravitational radiation is only be an approximation and not fully covariant. I prefer to consider the problem from a different perspective.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post
    A tensor that has recently attracted my attention is the Lanczos tensor
    Yes, I've heard of that one too. Apparently this can be considered the gauge field that generates curvature in the gauge field formulation of GR. I find it rather odd though that this is a rank-3 tensor, I wonder what the physical significance of that is.

    My view is that the standard derivation of gravitational radiation is only be an approximation and not fully covariant.
    But there are exact solution also, like the Bondi-Pirani wave : http://www.itp.kit.edu/~schreck/gene...lane_waves.pdf
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    Have you guys looked through the Living Reviews in Relativity Journal at Max-Planck-Institut für Gravitationsphysik?
    http://relativity.livingreviews.org/...s/subject.html

    They have published a few papers that seem related to what you are talking about.
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    A tensor that has recently attracted my attention is the Lanczos tensor
    Yes, I've heard of that one too. Apparently this can be considered the gauge field that generates curvature in the gauge field formulation of GR. I find it rather odd though that this is a rank-3 tensor, I wonder what the physical significance of that is.

    My view is that the standard derivation of gravitational radiation is only be an approximation and not fully covariant.
    But there are exact solution also, like the Bondi-Pirani wave : http://www.itp.kit.edu/~schreck/general_relativity_seminar/Gravitational_waves_in_general_relativity_exact_pl ane_waves.pdf
    It seems like a lot is needed to keep spacetime flat when it could be as simple as a latitude that needs to be maintained each moment in a three dimensional rotating collapsing sphere one after another. The need to create a sandwich wave supports this as two things are happening at once. Weight is a force that needs gravity to exist.
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    Quote Originally Posted by Markus Hanke View Post
    But there are exact solution also, like the Bondi-Pirani wave : http://www.itp.kit.edu/~schreck/gene...lane_waves.pdf
    Interesting. Thanks.

    However, from the brief look I had, the article deals only with a particular type of gravitational wave itself, and does not address what does or does not emit gravitational radiation in general. I feel it is important to take a fresh approach and not rely on calculations that may not correctly apply to the situation under consideration.
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    The general assumption seems to be that because there is a time-varying gravitational field associated with an rotating rigid irregular object, that this means that such an object is emitting gravitational radiation. But gravitational radiation is a form of radiation that takes something away from the emitting source. Therefore, the emitting source must change as it is emitting gravitational radiation. But the only change that can occur is a change in the rate of rotation. So the question becomes: Can the emission of gravitational radiation simply change the rate of rotation?

    It is said that when a binary system undergoes orbital decay due to the emission of gravitational radiation, that radiation carries away angular momentum from the binary system. But in this case, the moment of inertia is changing, and therefore the change in angular momentum is less straightforward than a simple change in the rate of rotation, where one can directly appeal to rotational symmetry of the laws of physics and Noether's theorem.
    Last edited by KJW; February 12th, 2014 at 06:42 AM.
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    Quote Originally Posted by KJW View Post
    The general assumption seems to be that because there is a time-varying gravitational field associated with an rotating rigid irregular object, that this means that such an object is emitting gravitational radiation. But gravitational radiation is a form of radiation that takes something away from the emitting source. Therefore, the emitting source must change as it is emitting gravitational radiation. But the only change that can occur is a change in the rate of rotation. So the question becomes: Can the emission of gravitational radiation simply change the rate of rotation?

    It is said that when a binary system undergoes orbital decay due to the emission of gravitational radiation, that radiation carries away angular momentum from the binary system. But in this case, the moment of inertia is changing, and therefore the change in angular momentum is less straightforward than a simple change in the rate of rotation, where one can directly appeal to rotational symmetry of the laws of physics and Noether's theorem.
    Hi KJW. Just popping in. I see you are dealing with your old gem of a thought experiment again in relation to that rigid dumbell.

    I have to agree with your very valid viewpoint taken from the frame of reference of an observer who is stationary with respect to the barycentre of the rotating dumbell and the observer who is circling the stationary dumbell. The symmetry must be observed between these two respective frames of reference. As you have mentioned it is the radial component that seems to be the nub of the problem. I was just thinking that if we just seperate out this radial component for the moment and look at it closely then the problem seems to be one relating to the notion that 'perfectly rigid bodies are forbidden by relativity' which got me looking at born rigidity. I am wondering whether the Herglotz-Noether theorem may help you. This paper seems to have an interesting contemporary take on it.

    Cheers :-))
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  33. #32  
    KJW
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    Quote Originally Posted by Implicate Order View Post
    the problem seems to be one relating to the notion that 'perfectly rigid bodies are forbidden by relativity' which got me looking at born rigidity.
    On the other forum when this issue was raised, my reply was that even if one removes the condition of perfect rigidity, realistic rigidity would limit the gravitational radiation produced such that in the limit of perfect rigidity, there would be no gravitational radiation. The real issue is whether or not rotation itself is responsible for gravitational radiation, particularly since it intuitively seems as if it would be.
    Implicate Order likes this.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by Markus Hanke View Post
    Quote Originally Posted by KJW View Post
    But because there is no longer any orbital decay, there is no longer any gravitational radiation.
    I am not convinced. Have you performed a calculation for this ?
    Some time ago in another thread, I mentioned that gravitational radiation does not contain energy-momentum because the contraction of the Weyl tensor is zero. You asked me to provide an explanation for the decay of a binary orbit. This carried with it the presumption that the orbital decay represents a loss of energy-momentum, but does it? What is actually being lost is gravitational potential. In other words, gravitational radiation is the radiation of pure gravitation, and the radiating source is also losing a corresponding amount of pure gravitation. Thus, in the case of the rotating rigid dumbbell, because there are no gravitational changes occurring (considered in the rotating frame of reference), there can be no gravitational radiation.

    Let's consider what happens in the Newtonian theory. If I drop a mass from the top of a tall building, as the mass falls, the kinetic energy increases by the amount that the gravitational potential energy decreases. This conservation of energy is responsible for stable orbits, and there is no gravitational radiation. But in fact, if I drop a mass from the top of a tall building, the speed of the ball will not quite match the decrease in potential energy. In other words, the loss of gravitational potential energy exceeds the gain in kinetic energy, and this discrepancy is the gravitational radiation emitted, which in Newtonian theory would be a form of energy-momentum. It is worth noting that the Newtonian gravitational field is perfectly linear, so that as masses move relative to each other, the change in the total gravitational field does not produce gravitational radiation. Thus it would appear that gravitational radiation is due to the non-linearity of spacetime curvature in general relativity.
    There are no paradoxes in relativity, just people's misunderstandings of it.
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    Quote Originally Posted by KJW View Post

    Some time ago in another thread, I mentioned that gravitational radiation does not contain energy-momentum because the contraction of the Weyl tensor is zero. You asked me to provide an explanation for the decay of a binary orbit. This carried with it the presumption that the orbital decay represents a loss of energy-momentum, but does it? What is actually being lost is gravitational potential. In other words, gravitational radiation is the radiation of pure gravitation, and the radiating source is also losing a corresponding amount of pure gravitation. Thus, in the case of the rotating rigid dumbbell, because there are no gravitational changes occurring (considered in the rotating frame of reference), there can be no gravitational radiation.

    Let's consider what happens in the Newtonian theory. If I drop a mass from the top of a tall building, as the mass falls, the kinetic energy increases by the amount that the gravitational potential energy decreases. This conservation of energy is responsible for stable orbits, and there is no gravitational radiation. But in fact, if I drop a mass from the top of a tall building, the speed of the ball will not quite match the decrease in potential energy. In other words, the loss of gravitational potential energy exceeds the gain in kinetic energy, and this discrepancy is the gravitational radiation emitted, which in Newtonian theory would be a form of energy-momentum. It is worth noting that the Newtonian gravitational field is perfectly linear, so that as masses move relative to each other, the change in the total gravitational field does not produce gravitational radiation. Thus it would appear that gravitational radiation is due to the non-linearity of spacetime curvature in general relativity.
    Interesting This is something I need to think about first.
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