2
$\begingroup$

Massless particles, such as photons and gluons, move with the speed of light and consequently experience no time. According to Special Relativity, time stops at the speed of light. As a result, massless particles exist in the same moment of their proper time from the emission to absorption. Therefore massless particles cannot decay or experience any other change. A practical example of this logic is that neutrinos have mass, because massless neutrinos cannot oscillate.

It seems that two massless particles that don't experience time should not be able to interact with each other (at least without the interaction mediated by virtual massive particles). This holds true for photons not interacting with each other (in the first order) resulting in the linearity of the electromagnetic field.

QCD is a non-linear theory with gluons interacting with each other via 3 and 4 vertices. Specifically the 3-gluon vertex (please correct me if I am wrong) under specific conditions can be viewed as a gluon decay into two gluons. However, massless particles cannot decay according to Special Relativity. What is the explanation of why massless gluons can interact with each other and apparently even decay while not experiencing time?

Improve this question
$\endgroup$
4
  • $\begingroup$ Do you have a similar problem with massless electrons exchanging photons? $\endgroup$
    – gj255
    Commented Jun 7, 2018 at 17:28
  • $\begingroup$ @gj255 Massless electrons? $\endgroup$
    – safesphere
    Commented Jun 7, 2018 at 17:29
  • $\begingroup$ Massless QED is another QFT with this feature, but far simpler than QCD. It is not physically relevant (perhaps), but is perfectly conceivable from a theoretical perspective. $\endgroup$
    – gj255
    Commented Jun 7, 2018 at 17:38
  • $\begingroup$ @gj255 I see your point. Yes, I would have a concern of a potential contradiction with relativity. For example, in GR photons must have a gravitational field and thus emit gravitons, but this is prohibited by relativity. So the rigorous calculations show that the gravitons are actually emitted only from the photon emission and absorption points thus resolving the relativity concern. $\endgroup$
    – safesphere
    Commented Jun 7, 2018 at 18:28

2 Answers 2

Reset to default
7
$\begingroup$

That's...not how special relativity works.

Your question seems predicated on the idea that massless particles "experience no time", because special relativity says that "time stops at the speed of light". Neither of these is true, see this answer by Ben Crowell. The point is that the notion of "experiencing time" is ill-defined for particles moving at the speed of light, since there simply is no notion of frame for a hypothetical observer moving at the speed of light (because the formula for Lorentz transformations diverges as $v\to c$, a manifestation of the non-compactness of the Lorentz group). Ill-defined does not mean that "time stops", it means it is not meaningful to try to speak about the "time experienced by an object moving at the speed of light" at all.

As a side remark, a single massless particle is indeed forbidden by relativity to decay into two daughter particles! But not because of any mystical rule imposed by it "not experiencing time", but simply because such a decay cannot fulfill the kinematic constraints of energy and momentum conservation.

Improve this answer
$\endgroup$
10
  • $\begingroup$ I appreciate you taking time to respond, but your post only criticizes the language of my question instead of answering it. "Time stops" vs. "time is ill-defined" is a "tomayto" vs. "tomahto" terminology difference. You quote Ben, but he says, "the proper time s of a particle moving at c is constant, ds=0", exactly the same as mine, "massless particles exist in the same moment of their proper time". Your "kinematic constraints" argument doesn't explain why massless neutrinos cannot oscillate. You are known as an expert here with better answers expected from you without dv questions you answer. $\endgroup$
    – safesphere
    Commented Jun 7, 2018 at 18:15
  • 1
    $\begingroup$ @safesphere No, this is not mere quibbling about words: That the proper time of an object moving at light speed is zero does not imply that it cannot decay, and the way your question is worded gives the impression that you derived that idea from imagining the "experience of time" of such an object. I'm pointing out this doesn't work; if your idea of why massless particles cannot decay is different, you might need to make it more clear. $\endgroup$
    – ACuriousMind ♦
    Commented Jun 7, 2018 at 18:49
  • $\begingroup$ Neutrino oscillation is indeed impossible for massless neutrinos but that's because more generally neutrinos having the same mass don't oscillate; all being massless is merely a special case of that, so that's not a phenomenon intrinsically linked to masslessness, either. $\endgroup$
    – ACuriousMind ♦
    Commented Jun 7, 2018 at 18:49
  • $\begingroup$ A decay is a change, a change requires a passage of proper time. The proper time of massless particles is constant (per Ben C.), so they can't decay. The idea is clear and you got it right, but argue that it is wrong. The problem is that you provide no proof, so your comment is just opinion based. Note that my question did not refer to massless particles experiencing time. This would require a frame and a concept of observer that you correctly objected. Instead my question referred to massless particles not experiencing time and the absence of experience does not involve a frame or observer. $\endgroup$
    – safesphere
    Commented Jun 7, 2018 at 22:19
  • $\begingroup$ @safesphere Why are you so concerned with the proper time of a massless particle if there's no frame in which that proper time actually behaves like a time (i.e. no rest frame)? I would argue that yes, the proper time of a massless particle is constant, but that the quantity labeled "proper time" has no relation to physical time. It's more like a label for a particular light-like trajectory. $\endgroup$
    – probably_someone
    Commented Jun 7, 2018 at 23:46
4
$\begingroup$

In our laboratory experiments gluons are always off mass shell so the problem does not arise as they take virtual masses under the integrals.

Your question would be relevant in the quark gluon plasma at cosmological times, where anyway symmetry is unbroken and all elementary particles in the table have zero mass. Calculations are not done in the rest system of zero mass particles, but in some appropriate observer frame, and interactions take place following the rules of quantum field theory.

After all there are photon photon interactions in our laboratory experiments, higher order, but still there, and gamma gamma colliders are being proposed. They will be studied in the laboratory frame, not in the system of one of the gammas!!

Improve this answer
$\endgroup$
1
  • $\begingroup$ I see, these are virtual gluons that play by different rules. Makes sense. And for real gluons my question makes little sense, because they would hadronize with a mess of jets. Thanks! $\endgroup$
    – safesphere
    Commented Jun 7, 2018 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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