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Are all particles subject to gravity?

  1. May 1, 2015 #1
    Theoretically are all particles subject to gravity. Or is there any particle which need not be affected by gravity.
     
  2. jcsd
  3. May 1, 2015 #2
    All objects that have energy do interact gravitationally... Since E=sqrt(m^2+p^2), there are no observable particles with 0 energy. So all particles (including massless) are subject to gravity.
     
  4. May 1, 2015 #3
    Does that mean gravitons attract all other particles?
     
  5. May 1, 2015 #4

    mfb

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    If gravitons exist (we don't know), they would have self-interactions. I would not call this "gravitons attract all other particles", however, and this is a tiny effect. Gravitons (again, if they exist) mainly transmit the gravitational interaction between other objects.
     
  6. May 1, 2015 #5
    Ok. Gravitons are exceptional particles which are said to be those who create (functioning) gravity But I can't agree completely to you. Because as you said particles subject to gravity, it means particles are subjected to gravitons, right? :confused:
     
  7. May 1, 2015 #6

    mfb

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    Gravitons would interact with all particles, similar to the way photons interact with charged particles, sure.
     
  8. Jan 27, 2016 #7
    what does the phrase "subject to gravity" mean exactly?
     
  9. Jan 27, 2016 #8

    ohwilleke

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    "All particles couple to graviton with a strength proportional to their mass-energy" would be the more conventional way in Standard Model terminology of expressing what you mean. I am not sure, theoretically, if a virtual particle couples to gravitons, although I think that the answer is still yes. In particular, it is conventional to define a graviton such that it couples to other gravitons just as it couples to other particles in much the same manner that gluons can couple to gluons as well as quarks.

    This feature of a graviton, which makes it non-renormalizable, is the principal reason that the quantum gravity is a problem that has defied theorists for many decades, and QCD calculations remain accurate only to about 1% precision, while QED calculations (whose carrier boson the photon does not have a self-interaction) are the most precise calculations in all of physics (that don't have exact solutions).

    I am not 100% convinced that this graviton formulation applied with all of the other rules of the SM fundamental particles is really exactly equivalent to GR in the classical limit.

    For example, in ordinary classical GR, the energy of the gravitational field is treated differently than all other kinds of matter and energy and cannot be localized, only evaluated at a global level, and that has some pretty important consequences. In any graviton based theory, in contrast, it is always true that the energy of the gravitational field can be localized and there is no obvious reason to exclude it from the stress-energy tensor as one does in the case of classical gravitational fields in GR. It is not at all obvious in my mind that this significant difference in the treatment of the self-interactions of the gravitational field between classical GR and generically, any quantum gravity theory that involves spin-2 zero mass gravitons, actually has no phenomenological consequences and is rigorously equivalent to GR in the classical limit. It might be true, but I have not see a proof of that which is sufficiently rigorous to make me comfortable that this is really the case. (Of course, if there is a difference, this would be a way to experimentally test a quantum gravity theory, but first you have to build one that works and that you can do calculations with, which no one has managed to do yet.)

    Generally speaking, of course, as expected and intended, quantum gravity theories also don't give rise to true mathematical singularities in the strong field limit of black holes and the Big Bang, but these deviations from classical GR are at the fringe of the theory's zone of applicability so an exact match to classical GR is not expected there. But, the locality issue also conceivably leads to subtle deviations from classical GR in weak gravitational fields.

    One of the motivations behind loop quantum gravity type theories relative to graviton based stringy theories of quantum gravity, is that in LQG, gravity continues to operate via space-time rather than something moving within space-time, which may, in principal at least, not give rise to the localized gravitational field distinction from GR which is present in graviton based theories.
     
  10. Jan 28, 2016 #9

    vanhees71

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    Gravity is coupling universaly to the energy-momentum tensor of all kinds of matter and radiation, not only to mass-energy!
     
  11. Jan 28, 2016 #10

    ohwilleke

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    I'm confused. What in the energy-momentum tensor isn't ultimately mass or energy or both? Momentum implies kinetic energy. Radiation implies energy. I can't imagine a kind of matter or radiation that is not mass or energy.

    To be clear, the term "mass-energy" in the context of gravitation generally means energy of all kinds plus mass multiplied by c^2 (pressure terms in the energy-momentum tensor are also a form of energy). I've never seen the term "mass-energy" used in the context of gravitation to mean energy attributable to mass only.
     
  12. Jan 29, 2016 #11

    vanhees71

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    Usually what's meant when somebody talks about "mass-energy" it's the rest energy of a body, ##E_0=m c^2##, where ##m## is the invariant mass of the body, but it's the full energy-momentum tensor coupling to the gravitational field in the Einstein-Hilbert action and not only the parts from the mass terms in the Lagrangian.
     
  13. Jan 29, 2016 #12
    Yes, and there's several different ways of saying this. Because of diffeomorphism invariance, yes, all particles must either be coupled to gravity or else live on a completely different manifold (i.e. not exist in our manifold). In terms of Weinberg's soft-graviton theorem, the equivalence principle must hold and so all particles must couple to gravity equivalently (Source gravity through their stress-energy tensor).
     
  14. Feb 1, 2016 #13

    ohwilleke

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    This isn't the usage I've seen (I would say what you are calling "mass-energy" is what I would call simply a coupling to "mass"), but at any rate it is clear that we mean the same thing even if we are assigning different technical definitions to the terms we are using.
     
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