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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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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. –  Lagerbaer Apr 2 '11 at 4:00
@Lagerbaer Charge is invariant, only the charge density transforms relativistically. –  metacompactness Jun 29 '13 at 7:07
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 –  John Feb 12 at 18:22
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7 Answers

up vote 10 down vote accepted

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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So you say that, although not theoretically impossible, there shouldn't be such a particle given our observations? –  Eelvex Apr 2 '11 at 4:25
@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. –  Marek Apr 2 '11 at 9:34
Like flying elephants: theoretically possible but surely non existent :) –  Eelvex Apr 2 '11 at 10:05
@Eelvex: but unlike flying penguins :) youtube.com/watch?v=9dfWzp7rYR4 –  Marek Apr 2 '11 at 10:07
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! –  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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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? –  John Apr 2 '11 at 8:21
@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. $W^- \to e^- \bar{\nu}$. –  Marek Apr 2 '11 at 10:21
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 $E>m_\mu$. –  Luboš Motl Apr 2 '11 at 17:34
BTW preemptively, you could also ask about the running of the $SU(3)$ QCD coupling. Indeed, this $SU(3)$ is not broken - but rather confined - which means that it's running to arbitrarily long distances, not only the Compton wavelength of the lightest quarks. In some sense, the potential energy goes like $|r|$ for long distances (although it gets converted to quark-antiquark pairs if $|r|$ is too large). –  Luboš Motl Apr 2 '11 at 17:37
What does it mean when you say: "the fine-structure constant is running" - and then "above the mass of the electron"? –  Gerard Apr 2 '11 at 20:26
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None found so far - but they may be possible apparently (don't understand how!)


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Thanks for the link. The paper says that there seems to be some lower limit on the mass of a charged particle. –  Eelvex Apr 2 '11 at 4:24
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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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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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Since 'zero speed' is not invariant, any massless particle must move at c. –  Eelvex Feb 12 at 13:12
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.. –  user34793 Feb 12 at 17:46
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I retract the following answer (see comments): emphasized textA massless charged particle would violate special relativity. It could be accelerated by an electric field beyond the speed of light.

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-1, you can't change invariant mass of particles by accelerating them. Consequently, you can't accelerate massless particles. –  Marek Apr 2 '11 at 18:15
@Marek then how could a massless charged particle exist if it couldn't experience the electric field? –  jaskey13 Apr 2 '11 at 18:18
@Marek meaning isn't it a contradiction to call such a particle charged if it cannot be accelerated by the electric field? –  jaskey13 Apr 2 '11 at 18:27
Experience of the field (which we call interaction) has nothing to do with invariant masses (unless you are talking about renormalization). By the way, are you at least familiar with terms such as off-shell momenta? And with QCD and gluons (which are massless charged particles that do experience strong field)? –  Marek Apr 2 '11 at 18:28
There is no contradiction. You are just confused because you are trying to apply classical intuition to elementary particles interaction. First of all, there is no acceleration. Interactions are described by Feynman diagrams where few particles can exchange momenta (or more generally, some particles are annihilated and others created). The individual particles there have no identity and it makes no sense asking about their acceleration because there is just no way to measure velocity of an individual particle during a prolonged interval. –  Marek Apr 2 '11 at 18:33
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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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