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Do all massless particles (e.g. photon, graviton, gluon) necessarily have the same speed $c$?

+ 6 like - 0 dislike
5013 views

I suppose there was a discussion already on speed-of-gravity-and-speed-of-light.

But I silly wonder whether all the massless mediators of four fundamental forces, i.e.

Graviton: $g_{\mu\nu}$ (gravity)

Photon $\gamma$: $A_\mu$ (electromagnetism)

Gluons: $A_\mu^a$ (strong interactions)

Necessarily travel at the same speed? Is there a no-go theorem or theoretical proof to say that it is impossible to have these three mediators have different speeds?

Or does QCD confinement makes the story of gluons any different from gravitons and photons?

[ps. excluded massive $Z^{0}$ and $W^{\pm}$ bosons (weak interactions)]

Another way to say this: Speed of photon, graviton, gluon all equal to $c$? or Whether all massless particles necessarily have the same speed?

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Idear
asked Dec 17, 2013 in Theoretical Physics by wonderich (1,400 points) [ no revision ]
retagged Jun 4, 2014
Isn't it because of the geometry of spacetime?

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user hwlau
Can we have bimetric gravity with, say, photons and gravitons coupled to different metrics?

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user user23660
@hwlau: can you be more explicit? Actually now I wonder whether one scheme is that the SUSY unification on gauge couplings of U(1)xSU(2)xSU(3) and gravity at the GUT scale 10^16 GeV, implies that the mediators of all forces needs to have the same origin, so the speed of massless particles needs to be the same? What may be a theoretical constraint? or a no-go theorem?

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Idear
ps. If $c_{gravity}, c_{photon}$ and $c_{gluon}$ are different, we can imagine there are more fundamental constants, and HEP people certainly hate to see this happens.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Idear
There was a theory which predicted different limit speeds for each particle, but I can find the reference.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user jinawee

4 Answers

+ 6 like - 0 dislike

Another way to say this: Speed of photon, graviton, gluon all equal to c? or Whether all massless particles necessarily have the same speed?

You must not have been introduced to the concept of a virtual particle:

In physics, a virtual particle is a transient fluctuation that exhibits many of the characteristics of an ordinary particle, but that exists for a limited time. The concept of virtual particles arises in perturbation theory of quantum field theory where interactions between ordinary particles are described in terms of exchanges of virtual particles. Any process involving virtual particles admits a schematic representation known as a Feynman diagram, in which virtual particles are represented by internal lines.

A virtual particle is an internal line in a Feynman diagram which represents the propagator mathematics that has to be substituted to get the integral necessary for computing measurable quantities . Virtual particles have the quantum numbers of their homonymous ( having the same name) particles except not the mass. The mass is off shell.

So it is a general rule that massless particles travel at the velocity of light, but only when in external lines in Feynman diagrams. This is true for photons, and we thought it was true for neutrinos but were proven wrong with neutrino oscillations.

Gluons on the other hand we only find within a nucleus and these are by definition internal lines in Feynman diagrams and therefore are not constrained to have a mass of 0, even though in the theory they are supposed to. In the asymptotically free case, at very high energies they should display a mass of zero.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user anna v
answered Dec 17, 2013 by anna v (1,875 points) [ no revision ]
Most voted comments show all comments
The path integral of physical fields and virtual particles...I feel like it's a bit like we consider the exponetial function $\mathrm e^x$, make a tailor expansion around $\pi$ up to second order and absorb an overall factor $\mathrm e^{-\pi}/2$ because it's not physically detactable. We end up with $\mathrm e^{x}\ \overset{\text{sort of}}{=}\ 2+\pi^2-2\pi(1+x)+x(2+x)$. We absorb the constant and reparametrize $ex\equiv x+1$ to emphasise the beautiful symmetry of our system $-2\pi\ ex+(ex-1)(ex+1)$. We call $ex$ the "expandion" and it will provide great counterexamples for other somethingions.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user NikolajK
I think Nick's comment is a serious concern. At least in QED it seems like this perturbative calculation makes sense (weak coupling), at least for a while. However, the series is asymptotic. In QCD we have an even bigger problem with strong coupling. So while the mathematics is useful and suggests that there are virtual particles, I think the onotlogical status of such objects hasn't really been established. Are they physical or just an artifact of the perturbative technique?

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Kevin Driscoll
@KevinDriscoll Well, when really kicked out of the nucleus gluons do make jets so their reality seems to me to be backed by experiment.cerncourier.com/cws/article/cern/29201

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user anna v
@NickKidman I think the difference with your example lies in the quantum numbers . The virtual particles are not arbitrary functions, they carry the quantum number conservations and the pole/propagetor that represents them has the on mass shell value except that in the internal lines things go imaginary.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user anna v
@annav: Are you implying the expandions are arbitary? While $x$ might seem like a random variable, only the expandions let us explore the quadrahedron density $\mathcal L_*[ex]:=(ex-1)(ex+1)$. Note that this purely algebraic (=geometric) object is invariant under $ex\mapsto -\mathrm{wau}\cdot ex$, (Ref 17). Indeed, the symmetry group is isomorphic to the Galois group which emerges when you pass from the unique Archimedean complete totally ordered field to its algebraic closure, (Ref 32).

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user NikolajK
Most recent comments show all comments
@annav I wasn't trying to doubt the existence of real gluons. I was questioning whether the virtual particles which we use to draw internal lines in Feynman diagrams have the same ontological status as external lines. To me a 'virtual particle' is an unfortunate term. I think they aren't really particles at all, but just transient bundles of some field which we deem particles thanks to the structure of our perturbative technique.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Kevin Driscoll
@KevinDriscoll Well, these are the tools we have. In the propagator picture , where the virtual particle is a pole when on mass shell, the on mass shell is a limiting condition, there exists a continuity in the formalism, the particle will become "real" because it will not only have the quantum numbers but also the mass, or be very very close to the mass

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user anna v
+ 3 like - 0 dislike

If basic symmetry and homogeneity assumptions about the Universe hold, then yes, all massless real particles (see Anna V's answer for virtual particles must travel at a universal constant $c$, the speed of a massless particle, in all frames of reference.

Given these basic symmetry and homogeneity assumptions, one can derive the possible co-ordinate transformations for the relativity of inertial frames: see the section "From Group Postulates" on the Wikipedia Page "Lorentz Transformation". (Also see my summary here). Galilean relativity is consistent with these assumptions, but not uniquely so: the other possibility is that there is some speed $c$ characterizing relativity such that $c$ is the same when measured from all frames of reference. Time dilation, Lorentz-Fitzgerald contraction and the impossibility of accelerating a massive particle to $c$ are all simple consequences of these other possible relativities.

So now it becomes an experimental question as to which relativity holds: Galilean or Lorentz transformation? And the experiment is answered by testing how speeds transform between inertial frames. Otherwise put, the experimental question is are there any speeds that are the same for all inertial observers?. The question is not about measuring the values of any speed, but rather, how they transform. Now of course we know the answer: the Michelson Morley experiment found such a speed, the speed of light. So there are two conclusions here: (1) Relativity of inertial frames is Lorentzian, not Galilean (which can be thought of as a Lorentz transformation with infinite $c$) and (2) light is a massless particle, because light is observed to go at this speed that transforms in this special way.

Notice that at the outset of this argument we mention nothing about particles or any particular physical phenomenon (even though special relativity's historical roots were in light). It follows that, if $c$ is experimentally observed to be finite (i.e. Galilean relativity does not hold), then the specially invariant speed is unique: it can only be reached by massless particles and there can't be more than one such $c$ - the Lorentz laws are what they are and are the only ones consistent with our initial symmetry and homogeneity assumptions. So if we observed two different speeds transforming like $c$, this would falsify our basic symmetry and homogeneity assumptions about the World. No experiment gives us grounds for doing that.

This is why all massless particles have the same speed $c$.

Incidentally, if we confine massless particles, e.g. put light into a perfectly reflecting box, the box's inertia increases by $E/c^2$, where $E$ is the energy content. This is the mechanism for most of your body's mass: massless gluons are confined and are accelerating backwards and forwards all the time, so they have inertia just as the confined light in a box did. Likewise, an electron can be thought of as comprising two massless particles, tethered together by a coupling term that is the mass of the electron. The Dirac and Maxwell equations can be written in the same form: the left and right hand circularly polarized components of light are uncoupled and therefore travel at $c$, but the massless left and right hand circular components of the electron are tethered together. This begets the phenomenon of the Zitterbewegung - whereby an electron can be construed as observable at any instant in time as traveling at $c$, but it swiftly oscillates back and forth between left and right hand states and is thus confined in one place. Therefore it takes on mass, just as the "tethered" light in the box does.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
answered Dec 17, 2013 by WetSavannaAnimal (485 points) [ no revision ]
+ 3 like - 0 dislike

The straightforward answer to this is that there is only one frame invariant speed in relativity, and if there are two speeds, the other speed has to be relative to an ether. So if there is relativistic symmetry, there is only one speed for massless particles, which is equal to the asymptotic speed of high energy particles. The "no-go" theorem is simply relativity, and it fails exactly when relativistic symmetry fails.

answered Jun 4, 2014 by Ron Maimon (7,535 points) [ no revision ]
+ 1 like - 0 dislike

I thought the same thing for a long time. I wondered why gluons don't fly out of the nucleus at the speed of $c$. The difference is that photons don't interact with other photons and gravitons don't interact with other gravitons. They can move around and pass through each other. On the other hand, gluons do interact with each other.

In fact, gluons form chains/flux tubes which is part of why quarks are confined. Gluons do travel at $c$ but not for very far before they interact with other quarks or gluons, which keeps them from moving any appreciable distance.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Brandon Enright
answered Dec 17, 2013 by Brandon Enright (10 points) [ no revision ]
no. this is not true? gravitons DO interacting with gravitons. gravity are badly non-renormalizable.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Idear
@Idear I'm pretty sure gravitons behave normally in one-loop diagrams and don't interact very strongly with each other when they do interact. The strong force, on the other hand, is many orders of magnitude stronger in interaction.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Brandon Enright
Thanks Brandon. What you address is not exactly what I asked. I wonder whether there is no go theorem for preventing the speeds of massless particles to be different. Perhaps if the speed of one is the largest, say c_1 >c_2 >c_3> .... Then only c_1 can be defined as massless?

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Idear
@Idear graviton-related interaction question physics.stackexchange.com/questions/8112/…

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Brandon Enright
@Idear John Rennie also addresses strong-force propagation velocity via Gluons and Mesons here: physics.stackexchange.com/questions/57137/…

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user Brandon Enright
I think the spirit of the question is to look at the speed a free gluon would propagate. This speed is the speed of light. For an electron say, it doesn't have to travel at the speed of light, and it's mass gives a second speed scale. However a massless particle does not have this second speed scale. Is it possible for a lorentz invariant QFT to have massless particles that don't propagate at the speed of light when interactions are turned off? I would think not because there is no way to get a second speed scale.

This post imported from StackExchange Physics at 2014-06-04 11:34 (UCT), posted by SE-user NowIGetToLearnWhatAHeadIs

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