# Are all massless particles "born" at light speed?

Tags:
1. Nov 14, 2015

### xpell

(I think) I know that massless particles can only exist traveling at c, but I find it somehow counter-intuitive (like many other real things... :D ) Would anyone please be so kind to confirm that, for instance, a gamma photon generated by the radioactive decay of a stationary isotope is already "born" at light speed, with zero "acceleration time"? And, when this gamma ray interacts with matter and produces a pair of massive particles (let's say an electron and a positron), what is the initial speed of these massive particles?

By the way, then a deconfined gluon (yes, I know, this is impossible, but let's say for the sake of the argument) would be instanly "propelled" at c too, wouldn't it?

Thanks for your patience! :)

Last edited: Nov 14, 2015
2. Nov 14, 2015

### Staff: Mentor

It does not have an "acceleration time". Quantum mechanics can make the definition of a speed tricky, but it certainly does not travel at a speed different from the speed of light..
Any value below the speed of light. Also, the speed of massive particles depends on the reference frame.
There is nothing "propelling" it. But here you really cannot ignore quantum mechanics, and "speed" gets ill-defined for realistic gluons.

3. Nov 14, 2015

### xpell

Thank you very much again, Mfb. I suppose (no, sure) I lack some basic knowledge to properly understand many of these things, but I'm a curious person and I love science! :)
I thought so, being massive particles. But I thought too that they would "start" at a speed very close to c, which maybe it's not true, is it?
Relative to a stationary observer measuring the gamma ray and the pair production (I think...)
Yes, yes, I know, it was just a way to say it with my limited knowledge. :)
Please, are you aware of any site where I could learn more about this at an outreach level (more or less...)?

Thank you once more!

4. Nov 14, 2015

### Staff: Mentor

If massive particles are created, their initial speed can have any value from 0 to arbitrarily close to the speed of light - and this speed depends on the reference frame.
Stationary relative to what?
I never collected books and websites with good explanations, but textbooks should cover that.

5. Nov 14, 2015

### xpell

Uh... To the Earth, I suppose. :D I was thinking in a guy studying the process in his lab. Maybe I'm saying stupid things. :)

6. Nov 14, 2015

### Staff: Mentor

Well, Earth is no special reference frame. The samw laws of physics are valid for all observers.

7. Nov 14, 2015

### xpell

My (most possibly wrong) idea was that if a massless particle at the speed of light produces a pair of particles, the initial speed of those particles would be "grazing" the speed of light before losing velocity or doing any other thing, as seen by a person observing the whole process. :)

I suppose I've totally made this up in my mind out of the classical conception that if an object traveling at a speed x breaks apart or suffers any other transformation, the initial speed of the products thus obtained would be x or slightly inferior to x too. But I undertand this is not applicable when a massless particle produces a pair of massive particles?

8. Nov 15, 2015

### vanhees71

This is a very difficult to answer question. One must first remember that what we call "particles" in theory (quantum field theory) are defined as asymptotic free states. What you consider is the creation of massless particles in collisions (e.g., the annihilation process $e^+ + e^- \rightarrow \gamma+\gamma$). What you have are asymptotic free electrons and positrons in the initial state (it's another story what these states truely are; usually we use momentum eigenstates of single particles, but that's not entirely correct as the infrared divergences of QED show). These interact and with some probability, which you can calculate from the formalism (evaluating Feynman diagrams), they get annihilated to form two photons. The photons are again defined as asymptotic free states and since they are massless they obey $E=|\vec{p}|$, but to stress it again, that's only true for the asymptotic free states which have a definite energy and momentum. To get "on-shell", however takes time, the formation time, which in the case of massless particles you can estimate as $\sim 1/E$. Formally you switch on and off the interaction (a la Gell-Mann and Low), i.e., you consider the process as starting in the "infinite past" and ending in the "infinite future". It is not well defined in which sense you can talk about particles/photons in the transient state, where all the quantum fields interact.

9. Nov 15, 2015

### xpell

Heck, I hate not to know enough to propertly understand you here. :( Does this means that in those Feynman diagrams I've been studying, we are not sure if the "gamma ray part" between the electron-positron annihilation and the quark-antiquark (plus gluon?) formation is truly a gamma photon, please?

10. Nov 15, 2015

### Staff: Mentor

No, they start at lower speeds (and keep that). A nucleus at the conversion place has to get some momentum to conserve both energy and momentum at the same time, that's why pair production does not happen in a perfect vacuum. It is more a collision process than a "break up".
All internal lines are virtual particles which can have some odd properties (and things like their speed are not well-defined). All external lines (those that do not both end and begin in the diagram) correspond to real particles - gamma rays produced in electron/positron annihilation are actual gamma rays.

11. Nov 16, 2015

### xpell

Thank you once more, Mfb. :)

12. Nov 17, 2015

### vanhees71

I don't know, which specific diagram you have in mind. Do you mean some $e^+ + e^- \rightarrow \text{hadrons}$ reaction? If so, in leading order you have, e.g., a photon propagtor between the $e^+ e^- \gamma$ vertex and the rest of the diagram. This internal line does, of course, not symbolize a real photon. The momentum running through it is not a light-like vector (if this were so, you'd have trouble with IR divergences, and there is souch trouble in other diagrams, when you look at a $t$ or $u$-channel process like electron-electron scattering). Usually one calls this a "virtual photon", but it's a bit misleading. You should not think of these intermediate states in an S-matrix calculation as particles. It's more like a field, but even this is not completely correct. The only thing, you can really interpret is the S-matrix element itself. Its modulus squared is some transition-probability-density rate, i.e., it gives the probability per unit time and unit volume for a transition from the initial asymptic free particle state (in your case an $e^+$ and an $e^-$) to the final asymptotic free particle state. You cannot say much more at this very general level, and that's so far almost always sufficient to describe the measured cross sections of all kinds of particle experiments up to the highest available energies.

An important example, where this very general picture is not sufficient are neutrino oscillations, where the finite distance of the detectors plays an important role, and the proper treatment has to take into account both the neutrino production and the neutrino detection process in terms of wave packets, but that's another topic.