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I always thought the non-linearity of Einsteins field equations implies that there should be direct graviton-graviton interactions. But I stumbled upon Wikipedia which argues:

If gravitons exist, then, like photons and unlike gluons, gravitons do not interact with other particles of their kind. That is, gravitons carry the force of gravitation but are not affected by it. This is apparent due to gravity being the only thing which escapes from black holes, in addition to having an infinite range and traveling in straight lines, similarly to electromagnetism.

Is Wikipedia correct? If not, why not? And what then are the arguments that there must be graviton-graviton interactions?

(As of this question being asked, the above paragraph has been removed from Wikipedia.)

doppelgreener
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asmaier
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    Well, it seems like the problem's been solved. The Wikipedia article no longer contains the quoted paragraph. – tparker Nov 21 '16 at 04:12
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    In QFT, everything interacts with everything else :) Even photons interact with one another by means of virtual leptons (electrons). However, gravity is a nonlinear theory and gravitons must have self-interaction vertices in Feynman rules. That is, of course, if there is a way to make sense of perturbative quantum gravity, which there isn't. – Prof. Legolasov Nov 21 '16 at 10:17
  • this is relevant https://arxiv.org/abs/0901.4005 – anna v Feb 04 '17 at 05:26
  • Related/worth checking: https://physics.stackexchange.com/q/772144/226902 and https://physics.stackexchange.com/a/772150/226902 – Quillo Jul 19 '23 at 12:54

2 Answers2

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I'm pretty sure that you are right and Wikipedia is wrong. In the linearized gravity approximation at weak curvature, you ignore the gravitational self-back-reaction, but in general gravitons carry energy (as evidenced by the work done by gravitational waves on the LIGO detectors) and therefore contribute to the stress-energy tensor of general relativity, therefore sourcing more gravitons. Also, some quick Googling finds lots of references to multiple-graviton vertices in effective quantum gravity field theories, whereas the Wikipedia article paragraph you quote has no references.

The issue of how gravitons can "escape" from a black hole without needing to travel faster than light is discussed at How does gravity escape a black hole?. The short answer is that gravitons can't escape from a black hole, but that's okay because they only carry information about gravitational radiation (which also can't escape from inside a black hole), not about static gravitational fields.

Community
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tparker
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    +1: The wikipedia paragraph is definitely wrong. Gravitons have self-interactions in any theory which reduces to Einstein gravity at low energies; the Einstein Hilbert action is not quadratic. – user1504 Nov 21 '16 at 00:32
  • Agree Wikipedia is conceptually wrong. Physicists who are not general relativists and use the linear approximations for gravitational radiation see it as simply a linear perturbation, and so treat it like photons. That has to be taken with lots of doubt for any arguments beyond the linear approximations. There are calculations of how gravitational waves scatter from or get 'absorbed' or in fact 'enhanced' by black holes and neutron stars. None of the treatments in this question and answers deal with any more than the linear approximations. – Bob Bee Nov 21 '16 at 07:40
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    I'm not a physicist, but wouldn't it be correct to deduce that (1) gravitons have energy (just like photons), therefore (2) they have mass, therefore (3) they must also be affected by other masses? I'm not claiming any mathematical understanding, but I feel like that shouldn't be necessary here...? – user541686 Nov 21 '16 at 07:40
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    @Merhdad. Absolutely right. They have energy, so they are part of what creates spacetime curvature, and they are affected by it (since they have to travel in their geodesics). – Bob Bee Nov 21 '16 at 07:48
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    @Mehrdad you are right, but not absolutely. Gravitons have energy, but they have no rest mass. Fortunately it is energy, not mass (more precisely, the stress-energy tensor) which acts as a source field for General Relativistic gravity. – Prof. Legolasov Nov 21 '16 at 10:20
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    @SolenodonParadoxus: Yeah, I didn't mean rest mass -- photons don't have them either, right? But great, thanks! – user541686 Nov 21 '16 at 10:26
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    @Mehrdad yes, I see. Its just that the concept of relativistic mass (dynamic mass, etc.) is old-fashioned and nowadays the term "mass" refers to "rest mass". Instead we talk of energy depending on the velocity. – Prof. Legolasov Nov 21 '16 at 10:29
  • @SolenodonParadoxus: Ahh I see, didn't know it's become old-fashioned :-) cool! – user541686 Nov 21 '16 at 10:30
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So in quantum field theory, the gluon is an operator which changes the color charge of a field. Since the gluon field itself carries color charge, the gluon-gluon interaction has the same strength as the gluon-quark interaction. Furthermore, since the QCD coupling constant is $\alpha_S \approx 0.1$, Feynman diagrams with virtual QCD particles in loops contribute with roughly the same strength as one-gluon exchange. The inability to ignore higher-order corrections is why we call QCD a "non-perturbative" theory.

By contrast, the photon couples to electric charge, but is itself electrically neutral. Photon-photon vertices therefore don't appear in the Feynman diagrams that describe electromagnetism. However, photons can interact with virtual particle loops: each photon spends some fraction of its time as a virtual electron-positron pair, and other photons can interact with those virtual charged particles. This is negligible because the electromagnetic coupling constant, $\alpha_\text{EM} \approx 1/137$, is about ten times feebler than for the strong interaction. So we can describe electromagnetism quite well, especially at low energy densities, by considering only one-photon exchange between charged particles and ignoring loop corrections, including photon-photon scattering.

Since the gravitational force between the charged fundamental particles is $\sim 10^{40}$ times weaker than the electric force, any perturbation-theoretical approach to gravity will have totally negligible interactions between gravitons, for the same reason that electromagnetism allows you to neglect interactions between photons. I don't think they're impossible, which seems to be the statement that's bothering you; but I think that they're negligible.

rob
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    The thing is, if you neglect graviton self-interactions, you won't get the right classical limit. They may be negligible, but they appear at tree level. – user1504 Nov 21 '16 at 00:41
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    @user1504 I guess you're right: the gravitons, with nonzero energy density, are "charged" under gravity and so the gluon analogy is more appropriate. However I also had the impression that one-boson exchange between charges gives you the $1/r$ potential of classical gravity. My handwaving explanation is that for any matter density, apart from black holes and possibly neutron stars, the energy density due to matter is so much larger than the energy density due to the gravitational field itself that the gravitational self-interaction isn't important --- that's the "classical limit" as I see it. – rob Nov 21 '16 at 00:51
  • Very interesting to think of quantum gravity as a weak, but still nonperturbative, theory. – rob Nov 21 '16 at 00:53
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    The start of the third paragraph seems to contrast two incomparable (or at least incorrectly compared) things. One is the ratio of the strength of gravity to the strength of electromagnetism. The other is the ratio of the strength of matter-graviton interactions to the strength of graviton-graviton interactions. Sure, gravity may be 40 orders of magnitude weaker, but gravitons may have enhanced interactions with gravitons versus other matter... (They don't, but my point is that the paragraph seems to make an apples-to-oranges comparison.) – Eric Towers Nov 21 '16 at 01:55
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    @Eric I was assuming that (a) the "charge" associated with gravitation is, in some hand-wavy sense, the total energy of a particle, and (b) gravitons, as quanta with energy $\hbar\omega$, generally aren't associated with frequencies high enough that their total energies are comparable to any of the massive particles. – rob Nov 21 '16 at 03:43
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    This is a very insightful answer, and I agree that in practice graviton scattering is probably negligible in most situations. But I think the Wikipedia article is still extremely misleading at the conceptual level, because its statement that "gravitons do not interact with [each] other" is simply wrong - sure, they interact very weakly, but they do interact. – tparker Nov 21 '16 at 04:08
  • Well, it's a wrong assumption or approximation that you can ignore black holes or neutron stars. Maybe black holes are too strange to treat, but two black holes with 65 total solar masses were detected in LIGO producing 3 solar masses in gravitational radiation, about 5%. Not ignorable. Neutron stars may produce something less we know they exist. When gravitational radiation is important is when you have very strong fields. The fact that gravitons can have large wavelengths means they interact with a lot of matter fields at the same time. Not easy to quantum calculate. – Bob Bee Nov 21 '16 at 07:34
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    Importance of higher-order corrections does not make a theory non-perturbative, it just makes it slowly converging. QCD is non-pertubative at low scales because its beta function has a negative sign, and hence the coupling diverges there... such that "higher order" has no physical meaning at all. QCD at a relatively high scale like $M_Z \simeq 91 GeV$, where $\alpha_s \simeq 0.118$, is quite successfully perturbative. And fundamentally the question here is not about whether the coupling is small, but whether the theory is non-abelian, i.e. has tree-level self-interactions... cf. EW theory. – andybuckley Nov 26 '16 at 16:34
  • But, the classical Maxwell theory has non-interacting electromagnetic waves. Einstein theory's classical form has a self-interacting gravitational field. Photon-photon scattering in QED require some mediating particle. An expansion of the the Hilbert action will have graviton-graviton vertices. – Zo the Relativist Jan 20 '22 at 15:15