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Are there any massless (zero invariant mass) particles carrying electric charge?

If not, why not? Do we expect to see any or are they a theoretical impossibility?

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  • $\begingroup$ So far, there are not: Only photon and gluon are massless. Neutrinos aren't massless, but even if they were, they have no charge. As for the possibility: I don't know. But I believe it couldn't have a "rest" charge in that case, because for charge density there's the same relativistic increase as for mass. $\endgroup$ – Lagerbaer Apr 2 '11 at 4:00
  • $\begingroup$ @Lagerbaer Charge is invariant, only the charge density transforms relativistically. $\endgroup$ – Paracosmiste Jun 29 '13 at 7:07
  • $\begingroup$ I think all answers below are not quite correct: massless charged particle cannot exists because they are unstable against creation from the vaccum, had they existed our world would immediately blow up $\endgroup$ – John Feb 12 '14 at 18:22
  • $\begingroup$ Related question: Why are there no elementary charged, spin-zero particles? $\endgroup$ – BMS Sep 12 '14 at 20:20
  • $\begingroup$ Of possible interest: (Classical) Electrodynamics of massless charged particles $\endgroup$ – Luke Burns Nov 16 '16 at 14:53
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There's no problem in writing down a theory that contains massless charged particles. Simple $\mathcal{L} = \partial_{\mu} \phi \partial^{\mu} \phi^*$ for a complex field $\phi$ will do the job. You might run into problems with renormalization but I don't want to get into that here (mostly because there are better people here who can fill in the details if necessary).

Disregarding theory, those particles would be easy to observe assuming their high enough density. Also, as you probably know, particles in Standard Model compulsorily decay (sooner or later) into lighter particles as long as conservation laws (such as electric charge conservation law) are satisfied. So assuming massless charged particles exist would immediately make all the charged matter (in particular electrons) unstable unless those new particles differed in some other quantum numbers.

Now, if you didn't mention electric charge in particular, the answer would be simpler as we have massless (color-)charged gluons in our models. So it's definitely nothing strange to consider massless charged particles. It's up to you whether you consider electric charge more important than color charge.

Another take on this issue is that Standard Model particles (and in particular charged ones) were massless before electrosymmetric breaking (at least disregarding other mechanisms of mass generation). So at some point in the past, this was actually quite common.

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  • $\begingroup$ So you say that, although not theoretically impossible, there shouldn't be such a particle given our observations? $\endgroup$ – Eelvex Apr 2 '11 at 4:25
  • $\begingroup$ @Eelvex: it depends on your definition of theoretically impossible. But yeah, they are basically ruled out by experiment because a world with charged massless particles would be very different from ours. $\endgroup$ – Marek Apr 2 '11 at 9:34
  • $\begingroup$ Like flying elephants: theoretically possible but surely non existent :) $\endgroup$ – Eelvex Apr 2 '11 at 10:05
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    $\begingroup$ @Eelvex: but unlike flying penguins :) youtube.com/watch?v=9dfWzp7rYR4 $\endgroup$ – Marek Apr 2 '11 at 10:07
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    $\begingroup$ To be precise, you need to turn your partial derivatives into covariant derivatives to minimally couple the scalar the field to the photon field: $\mathcal{L} = D_\mu\phi^\ast D^\mu\phi$ for $D_\mu = \partial_\mu + ie\hat{Q}A_\mu$. From here, note that the photon-loop diagram would give a mass renormalization. Unless there is some symmetry which protects/prevents the renormalization of the $\phi$ field's mass, there is no reason to assume that this bare Lagrangian should give physically massless particles! $\endgroup$ – josh May 16 '12 at 4:08
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Massless charged particles can't exist in Nature because they would be easily produced by the colliders, and they haven't been. Such a production would simply arise from the Feynman diagram with an intermediate photon that "splits" into the new charged massless particle and its antiparticle. The cross section of this process would be calculable, and not small in any way.

Also, the fine-structure constant $\alpha=1/137.036$, one expressing the strength of the electrostatic interactions in the natural units, is not a real constant. It's running. However, it's only running at energy scales such that there exist lighter charged particles. In Nature, it means that the constant is only running above the mass of the electron or positron - the lightest charged particles.

If there were massless charged particles, the electron and positron would become unstable - one problem - and the fine-structure constant would run to $\alpha=0$ at very long distances - another problem, and it obviously doesn't. So massless charged particles are theoretically impossible in our world - assuming that we empirically know some things such as the fact that there is a limiting Coulomb force at long distances.

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  • $\begingroup$ Using that logic, does that mean the weak force constant should run all the way to the neutrino mass, since the neutrinos have weak charge? $\endgroup$ – John Apr 2 '11 at 8:21
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    $\begingroup$ @John: for weak interactions you need to consider weak isospin (because the group is SU(2) and acts on doublets). Because it has to be conserved you always need to include some massive particles in your weak diagrams too. In your particular case, you'd have e.g. $W^- \to e^- \bar{\nu}$. $\endgroup$ – Marek Apr 2 '11 at 10:21
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    $\begingroup$ Dear John, the $SU(2)\times U(1)$ symmetry is broken at the electroweak scale, 246 GeV or so, which means that the corresponding potential isn't just $g^2/r$, the Coulomb Ansatz, but $g^2 \exp(-vr)/r$: it is exponentially decreasing at distances longer than the W-boson Compton wavelength. This "classical" exponential decrease is far more important than some logarithmic corrections from the running. Your question effectively assumes that the potential is $g(r)^2/r$ even at long distances which is surely wrong. But yes, neutrino loops of course make some impact on all processes for $E>m_\mu$. $\endgroup$ – Luboš Motl Apr 2 '11 at 17:34
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    $\begingroup$ What does it mean when you say: "the fine-structure constant is running" - and then "above the mass of the electron"? $\endgroup$ – Gerard Apr 2 '11 at 20:26
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    $\begingroup$ You don't have to appeal to collider experiments. If there were massless charged particles, there would be obvious effects that we would notice in everyday life. We'd see pair production when visible-light photons interacted with matter. $\endgroup$ – Ben Crowell Aug 11 '11 at 14:33
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The following is just my attempt to rewrite part of Lubos Motl's argument in a way that may be more understandable to those of us who are not technically fluent in field theory. I think this is equivalent to what he said in terms of running coupling constants. I would be grateful for feedback as to whether this is right.

Classically, if we integrate the energy contained in the electric field of a point charge, we get an improper integral that diverges at small distances like $1/r$. Through $E=mc^2$, this would give the electron infinite mass, which is not what we observe. To get the right mass, we can introduce a cut-off in this integral at the classical electron radius $r_e=e^2/m$ (multiplied by a possible coupling constant depending on your system of units). The interpretation of this cut-off is that at short ranges, quantum mechanics modifies the classical picture. (IIRC the semiclassical idea is that the creation of virtual $\text{e}^+\text{e}^-$ pairs makes the vacuum polarizable.)

For a massless particle named foo, the classical foo radius diverges to infinity. This suggests that it never has a classical $1/r^2$ field, no matter how big $r$ is. But this is in contradiction with the way we would normally define what charge is, using Gauss's law.

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  • $\begingroup$ Dear Ben, I am not sure what you're trying to do here. Like all other simple formulae, 1/r^2 is not exact in Nature, but it becomes arbitrarily accurate, by relative error margin, far enough from the charged source. At any rate, I think that these comments don't answer anything about massless charges - which was being asked - or about running couplings - which you claimed to address. $\endgroup$ – Luboš Motl Jan 2 at 10:51
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It would seem that our current understanding of physics would predict that a charged, massless paticle would not be attracted or repelled by any other charged particle because the acceleration caused by a difference in charges is caused by a force, and $F=ma$. If there is a charged, massless particle, it would be able to influence the motion of charged, massful particles without itself being affected, which would violate Newton's third law of motion. This doesn't mean that such a particle couldn't exist, but it seems that it would upset our understanding of physics.

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    $\begingroup$ Yet photons feel the force of gravity; $F=ma$ isn't always the case. $\endgroup$ – HDE 226868 Jan 14 '15 at 0:12
  • $\begingroup$ Classical $F=ma$ equation doesn't work for massless particles — they are ultrarelativistic in all cases. $\endgroup$ – Ruslan Mar 18 '15 at 12:02
  • $\begingroup$ But this answer is smart and funny argument :) $\endgroup$ – kakaz Jun 10 at 12:54
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Suppose such a particle existed. Question is what would happen if it was to enter an electric field? Consider $p$ ($m = 0$, $q > 0$) entering an electric field $E_i$, on a manifold $M (i,j)$ $$F_i = q E_i \; \; \;\text{but} \; \; \; F_i = m a_i$$

It follows that $F_i = 0$ since $m = 0$ meaning either $q = 0$ or $E = 0$, but such is not the case, $F_i$ (electric field ) is not equal $F_i$ (Newton's force)

Consider the same situation, we may write the following

$$F_j = q E_j \; \; \;\text{and} \; \; \; F_i = m a_i $$

Again we note that $F_i = 0$ and $F_j$ doesn't exist in the dimension of $e^i$, but it lies on the same manifold as $F_i$. We may use the matrix $A_i{}^j$ to transform $F_j$ to $F_i$, i.e $F_i = A_i{}^j F_j$, this means $A_i{}^j = 0$. The only way this can be so is if the angle between the two forces $\theta$ is given by: $$ \theta= 0 + k90 $$ where $k = 1,3,5,\ldots,n$. So $A_i{}^j = g^{ik} g_{jk} = \delta^i{}_j = 0$ since $j$ is not equal $i$.

So such a particle would be stationary in our dimension (or it would be whizzing through space at $c$, its speed is indeterminant) but one thing certain it is not bound to our spacetime.

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  • $\begingroup$ Since 'zero speed' is not invariant, any massless particle must move at c. $\endgroup$ – Eelvex Feb 12 '14 at 13:12
  • $\begingroup$ Let us consider the particle's mass to be a function of (theta), In which case F_i = m(theta) a_i = 0 when (theta) = 90,where m(theta) = (rest mass) cos(theta),from this we realize that when theta = 0 ,I.e when it apears to be moving in space we realise F_i = F_j but F_i = (rest mass) a_i which contradicts the special relativIty since its speed would be c,therefore its rest would have to be infinite and when it is whizzing of at speeds greater than c we realize it would have complex.. $\endgroup$ – user34793 Feb 12 '14 at 17:46
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    $\begingroup$ Newton's law should be $F = dp/dt$ here, not $F = ma$. $\endgroup$ – Luke Burns Nov 16 '16 at 13:58
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From $\gamma$ <-> $e^+e^-$ I can consider that the EM field is a charge separation that propagate side by side. We measure a net charge 0 until we make a way to separate the charges in a pair.

There is a possibility that if we have charge after (in $e^+e^-$ pair)
then we must have it also before (in $\gamma$) in such a way we can not detect it. A non detection is not the same as 'non existent'.

AFAIK the standard interpretation is not like this one I post and you must forget this if you are a regular student.

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