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Gravitons have gravity?

  1. Aug 2, 2005 #1
    hi
    i'm in a dielema.
    i dont know if it is the right place to post the thread.
    if every particle is destined by the universe to have some gravitionrl force then does graviton itself have gravity. there is no point of it not having if it is a particle as treaded by quantum physics
    nabodit
    :confused:
     
  2. jcsd
  3. Aug 2, 2005 #2

    DaveC426913

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    I believe that gravitons have no mass, thus no gravity.
     
  4. Aug 2, 2005 #3

    selfAdjoint

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    But photons, which have no mass, do gravitate, as Eddington's observation of the eclipse (claimed to have) showed. Not jus mass, but all forms of momentum and energy gravitate under the Einstein equations, and the graviton is supposed to couple to matter in imitation of those equations.
     
  5. Aug 3, 2005 #4
    if u look at photons u can say that it has created a disortion in space time as it is equivalent to matter*c^2. now there is no point on graviton having no curvature in the fabric causing its own gravity.
     
  6. Aug 3, 2005 #5

    ZapperZ

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    Er... PHOTONS created a distortion in space time? Which nutty part of GR says that?!

    Zz.
     
  7. Aug 3, 2005 #6

    Stingray

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    That nutty thing known as the Einstein equation ;). An electromagnetic field has a nonzero stress-energy tensor, so it does (in theory) 'gravitate.' By the way, the eclipse observations did not demonstrate this effect. They only showed that light is affected by gravity - not that it generates its own field.

    Anyway, the original question cannot really be answered because there is no well-understood, accepted theory involving gravitons. Now, gravitons were originally hypothesized because the linearized Einstein equation looked a lot like a 'spin-2 field equation.' There is actually a self-consistent (as far as I know) quantum field theory that reduces to the linearized Einstein equation in the classical limit. The analogs of photons in this theory are called gravitons. Unfortunately, it has very little to do with the full Einstein equation, which is presumably the only thing that matters to nature.

    Since GR is a very nonlinear theory, I guess you could say that gravitons must gravitate, but this really shouldn't be taken too seriously. For example, if you send two gravitational wave packets at each other, the result won't be their sum when they collide. Because of this, you might imagine that gravitons do not propagate 'freely' when other gravitons are around (even if all the matter fields vanish). This interaction could labelled as 'gravitons generating gravity,' but you should take that with a very large grain of salt.
     
  8. Aug 3, 2005 #7

    ZapperZ

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    Ah, but the "stress-energy tensor" is created by a more general form, and a SUBSTANTIAL part of it is due to mass. This part has been verified. The "photon created distortion in space time" part hasn't and, frankly, I don't see us detecting even a glimpse of that within a few lifetimes. This probably won't be an issue if we're in String Theory, but we're not.

    I, on the other hand, suspect a misunderstanding of some basic idea that caused that statement.

    Zz.
     
  9. Aug 3, 2005 #8

    Stingray

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    What do you mean by a "more general form?" Anyway, radiation-generated gravity is generally considered to have a large effect in cosmology, and this has a direct influence on (for example) CMB observations. I'm not going to say that that interpretation is unique, but it is certainly a non-negligible component in the standard cosmological model.

    I agree with you that detecting more direct effects would be extremely difficult.
     
  10. Aug 3, 2005 #9

    Nereid

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    [controverial mode ON]TeV photons have been detected (albeit indirectly); PeV ones are (widely?) expected to be generated in the 'local' universe, and may too be detected within a decade; the gravitational effect of baryonic mass, travelling non-relativistically, has been detected, for masses as low as a microgram (and lower?); so 'all' we need to do is find a way measure the gravitational pull of a photon that isn't too much different, in magnitude, from that of a small lump of steel :biggrin:[/controversial mode]
     
  11. Aug 3, 2005 #10

    ZapperZ

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    Would you bet your house we can detect this within your lifetime?

    :)

    Zz.
     
  12. Aug 3, 2005 #11
    when matter is equivalent to energy then the net effect of gravity by matter is equal to that by energy.
    -(reference)= faster than the speed of light by joao
     
  13. Aug 3, 2005 #12

    JesseM

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    But you asked "which nutty part of GR says that", not "which experiment has shown that". From a theoretical point of view, GR does predict that electromagnetic fields contribute to the curvature of spacetime, even if this prediction hasn't been experimentally verified.
     
  14. Aug 3, 2005 #13
    All non-zero mass-energy density gravitates.

    This includes photons and gravitons.

    It even includes mechanical potential energy. A wound-up spring
    gravitates more than when it is loose.
     
  15. Aug 3, 2005 #14

    Stingray

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    Well, I'm pretty sure that a concentrated electromagnetic wave packet does not gravitate in any remotely Newtonian way. I don't off-hand know if it would be that easy to detect.

    The much more serious experimental concern would be that any piece of apparatus near such a photon would be completely overwhelmed by electromagnetic forces - unlike with a lump of steel.

    Antiphon, what's mass-energy? That doesn't mean anything outside of a particular physical theory. Each of them define it differently, and it doesn't work very well at all in exact GR, let alone whatever the theory of quantum gravity ends up as.
     
  16. Aug 3, 2005 #15

    pervect

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    Eotovos type experiments done on Aluminum and gold also help confirm that electromagnetism contributes to the stress energy tensor. The electrostatic binding energy in these two elements are considerably different, .1% vs .4% of the total mass - but we can see that both materials fall at the same rate, nonetheless.
     
  17. Aug 3, 2005 #16
    I mean it in the context of GR where rest mass is related to rest energy
    by [tex] E = m_0c^2 [/tex]. If an electron-positron pair anhilate one another,
    the resulting gamma rays produce the equivalent gravitation.

    But it's important to note that even the classical notions of kintetic and potential
    energy are included in GR as source terms of gravitation. Spinning flywheels
    gravitate more than stationary ones, the coiled sping I mentioned above, etc.


    I'm not sure what you mean by this.
     
    Last edited: Aug 3, 2005
  18. Aug 3, 2005 #17

    Stingray

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    That's special relativity. There are no good definitions of rest mass or rest energy in GR, except in very special cases (or in some approximations).

    Electron-positron annihilations (a quantum mechanical process) are not technically handled in classical GR. So you'd have to go to quantum field theory in curved spacetime to discuss that, which I doubt you really want to do :). The concept of 'particle' actually loses its meaning, among other things.
     
  19. Aug 3, 2005 #18

    Nereid

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    Well, maybe.

    I suspect detection of the 'gravity' of photons won't come from a lab setup that resembles anything anyone reading this post would have familiarity with. For example, detection of TeV photons is, at present, via air showers - the cherenkov footprint of a photon can be distinguished from that of proton/electron/etc. Perhaps there's a signature in the CMBR that can be interpreted as photon's gravity? Perhaps some aspect of a certain class of GRBs will betray this GR aspect of photons?
    Me, no (and saying any more would give you far too many clues as to my age!).

    However, underestimating the ingenuity of experimentalists is all too easy to do; if detecting this effect were to become one of the hottest things in physics, period, I'd not be at all surprised if some group came up with a breath-takingly clever way to get a fix on it. :smile:
     
  20. Aug 3, 2005 #19
    I'll take your word for this, but I didn't think [tex] E=mc^2[/tex] had
    trouble holding true in GR. Nevertheless, the equations of GR have
    any energy or matter distribution as a gravitational source.

    Although particle creation isn't handled directly by GR, I don't think this
    would change the status of the gravitational field implied by the conservation
    of energy.
     
  21. Aug 3, 2005 #20

    ZapperZ

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    If you are going to play by the speculation rule of joao, then I should be able to also. For example, if the Higgs really is the mass field, then it doesn't couple to photons (nor gluons). So there IS a difference between "mass" and "energy" even though one can be CONVERTED to another.

    Zz.
     
  22. Aug 3, 2005 #21

    pervect

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    Just some quick notes on some semantic issues. "Mass-energy" is a term used in MTW's gravitation. It can be considered to be the same as energy.

    Non-relativistically, energy has an extra degree of freedom in that one can add any scalar to the value of energy (as in Newtonian mechanics).

    Relativistically we require E^2-p^2 = m^2, so we don't have this extra degree of freedom.

    MTW calls energy with the requirement that the zero point be set such that E^2-p^2=m^2 "mass-energy".

    GR does have a good concept of non-gravitational energy density (even strong field GR has this concept), this is the stress-energy tensor (or more specifically, the T_00 component of this tensor). It may be puzzling as to how GR can have a good notion of energy density, without always having a well-defined value of energy.

    One way of describing the basic issue with energy in GR is that one cannot simply find the total energy of a system by adding together (integrating) the energy densities due to the curvature of space-time. Another way of describing this difficulty is that the difficulty lies in finding the "energy" in a gravitational field. Note that the "energy" in the stress-energy tensor does not include any gravitational sources of energy.

    Hopefully these semantic issues (the definition of "mass-energy" ala MTW, and the distinction between energy and energy density) will help clarify some of the points being made.
     
  23. Aug 3, 2005 #22

    Nereid

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    Please keep a vigilant eye pervect; it is so extraordinarily to trip up over these terms. :bugeye:

    Yet a timely, simple 'let's sing from the same hymn sheet' can avoid pages of emotion-soaked rhetoric which, in the end, amounts to little more than 'violent agreement'.
     
  24. Aug 3, 2005 #23

    Stingray

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    My point is that the only well-known, and general concept of energy conservation in GR that is useful in this context is [tex]\nabla_{a} T^{ab}=0[/tex]. I have no idea what that has to do with electron-positron annihilation, or the gravitational field of such a process. You're vastly oversimplifying things by using language unnatural to the theory.

    Actually, there are some constructions that allow things like energy and mass to be defined nicely for finite bodies, but these are extremely difficult to define and interpret. They also don't necessarily satisfy the one's intuitive expectations for things like "mass" or "momentum."

    It is also not rigorous to talk about point particles in GR (meaning things with 1+1 dimensional stress-energy tensors) if you want to go there. There are no solutions to Einstein's equation with such objects.
     
  25. Aug 3, 2005 #24

    pervect

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    This is an interesting question to which I don't have an answer, though I can speculate a bit.

    The first idea that occurs is to take the limit of a massive body moving by at a velocity v with a constant total energy E, the limit as v approaches 'c' (shirking the the invariant mass of the body to keep E constant).

    In aid of this approach, I recently worked out what tidal force a body experiences when a massive body moves by at a velocity of 'v' at the point of "closest approach".

    Unfortunately, the specified limit doesn't exist. The tidal force approaches infinity, because the tidal force in the radial direction is proportional to (2m/r^3)*(1+v^2/2)/(1-v^2)

    To me, this suggests that looking at the Aichelberg-Sexyl solution might be productive, and that the "gravitational force" in this case may be an impulse function.

    http://www.arxiv.org/abs/gr-qc/0110032

     
    Last edited: Aug 3, 2005
  26. Aug 4, 2005 #25

    pervect

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    You definitely have to be careful here. Take a look at

    http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

    for some of the complexities of energy in full GR. You probably don't want to go around saying that GR "incorporates" classical notion of potential energy, especially gravitational potential energy as people will assume you are talking about generalized GR including cosmology, in which case the above is definitely not true.

    Nonetheless, most of what you say above is reasonably close to being true in the weak-field limit of GR. In the weak-field limit, you can find the total energy (and mass, if you divide by c^2) of a *closed* system by integrating the sum of (energy density + pressure_x + pressure_y + pressure_z). This will be approximately the energy of the system - the error should be somewhere around the amount by which g_00 differs from 1 in the system under consideration. The smaller the system, the better the apprxomiation.

    Objects as large as the size of our sun are still reasonably good as far as this apprxoimation goes. (The metric coefficients differs from Mikowskain flatness only in parts per million at the surface of the sun - good but but not really great as far as approximations go). When you get to cosmological scales, though, you definitely need to avoid this approximation.

    Now we get to a second point:

    The pressure terms will cancel out in most static systems, leaving you with the intergal of the energy density, which is in the weak-field limit equal to the energy.

    While the pressure terms cancel out in _most_ systems, I'm not sure if they cancel out in rotating systems or not. One of these days I'm going to take a closer look at the rotating flywheel question, as it comes up a lot. However, even the sci.physics.faq "wimps out" on calculating the mass of a rotating flywheel.

    A third point should probably be mentioned:

    Finding the mass of a non-closed system is liable to give one confusing or contradictory results. I'm not sure if there is a rigorous defintion of mass that applies well to a non-closed system - the formulas I wrote above for finding the "weak-field" mass of a system probably won't work for a non-closed system, reason enough IMO to avoid non-closed systems.

    A closed system would consist of (for example) a stretched spring and the bar that keeps it stretched. An open system would be a stretched spring that does not include whatever is holding it in the stretched position.

    [add]
    I thought of something else I should add. Having all the metric coefficients close to 1 also implies that one has a low velocity relative to the center of mass frame of the system under consideration. (The definition of "low' depends on the accuracy desired). High velocites will cause off-diagonal terms in the metric, and if they get too large the approximation starts to fail.

    Reading what I wrote earlier, it's not really accurate to say that just g_00 must be close to 1 - all the metric coefficients have to be nearly Minkowskian, which means that all the on-diagonal metric coefficients must be near unity, and all the off-diagonal coefficients must be nearly zero.

    The details of the justification for this approximation go like this: the nearly Minkowskian metric has a nearly timelike killing vector, thus one is in an approximately stationary space-time. This means that one can use, to a high degree of accuracy, the formulas for mass that apply to stationary space-times. There are both surface-intergal and volume intergal forms of this formula, the volume intergal forms give the result I mentioned above, that one integrates the sum of the energy and the three pressure terms over the volume. I can quote the detailed formula from Wald used if anyone is interested (one might also be able to find referenes to said formula in some of my previous posts).
     
    Last edited: Aug 4, 2005
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