It is a fact that two real (massless) photons, gluons, or gravitons can't react by means of their virtual counterparts (for example, two external photons that interact via one of these massless internal virtual mediators giving two external real photons) represented by a Feynman diagram with real external massless photons, etc. and internal virtual photons, gluons, or gravitons. I think this also holds for two real (massive) $W$ and $Z$ particles and internal massless virtual photons, gluons, or gravitons.

Why is this so? And if not can someone give an example that they can?

I think though that two external photons (or gravitons) can react by means of a virtual internal graviton Or am I wrong here?

  • $\begingroup$ Are you asking why a real photon can't turn into a virtual one? That is just energy-momentum conservation. $\endgroup$ – knzhou Jun 12 at 11:39
  • 1
    $\begingroup$ Sort of odd to start a question with "it's a fact that" followed by non-facts which can easily be corrected by looking at some standard source such as Wikipedia. $\endgroup$ – Andrew Steane Jun 12 at 17:13
  • 1
    $\begingroup$ @descheleschilder There is no photon-photon-photon vertex in QED or in the Standard Model. $\endgroup$ – G. Smith Jun 12 at 19:19
  • 2
    $\begingroup$ Massless vector fields can couple to themselves. This is what gluons do. Photons don’t. Mathematically, this has to do with the gauge symmetry of the strong interaction being nonabelian and the gauge symmetry of electromagnetism being abelian. Questions about why nature is like this are beyond the current ability of physics to answer, other than to say if it were not like this then you would probably not be here asking these questions. $\endgroup$ – G. Smith Jun 12 at 21:15
  • 1
    $\begingroup$ Also, multiverse proponents would say that the reason “why” there is no photon-photon-photon vertex is because of the way that the six extra dimensions happened to randomly compactify into a Calabi-Yau space in our particular universe. $\endgroup$ – G. Smith Jun 12 at 21:22

Two real gravitons can scatter by exchanging a virtual graviton, when one quantizes small perturbations in General Relativity. So your “fact” is not a fact.

See section 4.3 of this paper for the Feynman diagrams and a calculation of the scattering amplitude. (Note: I think this calculation was first done by deWitt, but his landmark 1967 paper is behind a paywall. It is now common enough to just be an exercise for a masters thesis!)

The reason this can happen for gravitons and not photons is that gravity is nonlinear while electromagnetism is linear. In terms of Feynman vertices, there are vertices where three graviton lines or four graviton lines can interact, but no vertices where just photon lines can interact. The only vertex in QED is one where two charged-particle lines and one photon line meet.

QCD is also nonlinear and has three- and four-gluon vertices, similar to those for gravitons. But color confinement means that we don’t observe real gluons.

Of course, we don’t observe real gravitons either. But this is more for technological reasons (much too difficult, because gravity is so weak) than for theoretical ones. It’s true that the quantized GR that predicts gravitons is not a consistent theory, but string theory also predicts gravitons. I don’t know whether graviton-graviton scattering via virtual gravitons has been calculated in string theory.

Finally, string theory may not be correct, but most physicists assume that gravity must be some kind of quantum field just like all other fundamental fields. So everyone expects that whatever theory of quantum gravity emerges as the winner is going to have gravitons, is going to be nonlinear, and is going to reduce to quantized GR at low energies.

  • $\begingroup$ I know all Feynman diagrams in QED are built up out of the basic vertex of two charged particles and one photon (which obviously can't exist on its own), but I was just running off of the mind! Everything's clear now, but there is one thing I don't understand: why can't two external photons don't interact by the exchange of a virtual graviton? $\endgroup$ – descheleschilder Jun 12 at 17:37
  • $\begingroup$ Two real photons can exchange a virtual graviton. Light gravitates! There is a photon-photon-graviton vertex. This scattering amplitude has been calculated. Since every kind of particle carries energy and momentum, and energy and momentum are what cause gravity in GR, every kind of particle, massless or massive, has a vertex with a graviton. $\endgroup$ – G. Smith Jun 12 at 19:23
  • $\begingroup$ That's what I asked in the question! $\endgroup$ – descheleschilder Jun 12 at 21:04
  • 1
    $\begingroup$ I see that now. I originally thought you wanted real photons interacting via virtual photons, or real gluons interacting via virtual gluons, or real gravitons interacting via virtual gravitons. Sorry for not reading carefully. $\endgroup$ – G. Smith Jun 12 at 21:19

Actually many of these interactions do happen. It has nothing to do with being real or virtual. The 3 gluon vertex is well established. The process $e^+e^- \to W^+W^-$, seen at LEP2, involves real Ws and virtual photons and/or Zs.

Some processes do not occur. Photons do not interact with photons. Photons, W and Z bosons do not interact with gluons. The reason why can be given in two equivalently valid ways:

1) Photons have no electric (or strong) charge so they do not interact with photons (or gluons). The W and Z have no strong charge so do not interact with gluons. Gluons have no electric charge so do not interact with photons.

2) There is no term in the Lagrangian which would give a corresponding vertex that could appear in a Feynman diagram

  • $\begingroup$ The point is that I'm asking if reactions can occur between two real massless photons, photons, or (hypothetical) gravitons by means of the virtual versions of these particles (for example the reaction between two real photons by means of a virtual graviton). The three gluons in the vertex you mention are all three virtual particles, as are all gluons. The decay of the positron and electron in a $W^{+}$ and a $W^-$ doesn't conform to the reactions in my question. $\endgroup$ – descheleschilder Jun 12 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.