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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$\begingroup$ Related question: Why are there no elementary charged, spin-zero particles? $\endgroup$– BMSCommented Sep 12, 2014 at 20:20
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$\begingroup$ Of possible interest: (Classical) Electrodynamics of massless charged particles $\endgroup$– Luke BurnsCommented Nov 16, 2016 at 14:53
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8 Answers
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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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1$\begingroup$ So you say that, although not theoretically impossible, there shouldn't be such a particle given our observations? $\endgroup$– EelvexCommented Apr 2, 2011 at 4:25
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$\begingroup$ @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. $\endgroup$– MarekCommented Apr 2, 2011 at 9:34
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4$\begingroup$ @Eelvex: but unlike flying penguins :) youtube.com/watch?v=9dfWzp7rYR4 $\endgroup$– MarekCommented Apr 2, 2011 at 10:07
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7$\begingroup$ 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! $\endgroup$– joshCommented May 16, 2012 at 4:08
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1$\begingroup$ Wouldn't the energy of the electrostatic field be the mass an "otherwise" massless particle would have? (I.e, a "massless charged particle" is a contradictio in adjecto, an impossible contradiction by definition.) I actually always wondered whether the electron's rest mass is simply the sum of its electrostatic and weak fields. (There is certainly no "ball" "having" a charge, for example; the electron, like all other "particles", is the field.) $\endgroup$ Commented Sep 15, 2020 at 12:33
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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.
answered Apr 2, 2011 at 5:43
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1$\begingroup$ 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? $\endgroup$– JohnCommented Apr 2, 2011 at 8:21
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1$\begingroup$ @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. $W^- \to e^- \bar{\nu}$. $\endgroup$– MarekCommented Apr 2, 2011 at 10:21
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3$\begingroup$ 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 $E>m_\mu$. $\endgroup$ Commented Apr 2, 2011 at 17:34
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9$\begingroup$ What does it mean when you say: "the fine-structure constant is running" - and then "above the mass of the electron"? $\endgroup$– GerardCommented Apr 2, 2011 at 20:26
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2$\begingroup$ 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. $\endgroup$– user4552Commented Aug 11, 2011 at 14:33
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Lubos' answer is good for the present day universe. Here I reply to the question made duplicate "Can a massless object have a charge associated with it? [duplicate]".
In the standard model of particle physics, before symmetry breaking in the cosmological model, all symmetries exist and all the particles with their quantum numbers exist. The gauge bosons are zero mass before symmetry breaking, and also all the fermions were zero mass before. So if we include the time of the universe in the question the answer is yes. Experiments have not reached the energies and conditions necessary to reproduce the conditions before symmetry breaking, so the answer for present day particles is given by Lubos' answer.
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There are no massless particles with no electric charge. All the fermions have mass and the leptons that are not neutrinos have electric charge. The quarks also have electric charge. The Bosons, W+ and W- have mass and are charged. So as far as we know all particles that have charge have a reasonable amount of mass. However the particle with a charge and has the lowest mass, is the electron(and the positron).
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$\begingroup$ I suspect that this answer contains a typo. The photon is a massless particle with no electric charge. Gluons are also massless, with no electric charge, although they do carry a color charge. $\endgroup$– PM 2RingCommented Jul 10, 2020 at 1:35
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Since a particle only can be said to exist if it can express its existence, its properties in interactions if it has energy and localized energy is a source of gravity and we define mass as something which exerts and feels gravity, then there cannot exist massless particles. (That is, if we define a particle as an entity which at all times has a well-defined position -which a massless particle like a photon hasn't as from its own point of view, its transmission takes no time at all.)
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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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2$\begingroup$ Yet photons feel the force of gravity; $F=ma$ isn't always the case. $\endgroup$ Commented Jan 14, 2015 at 0:12
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$\begingroup$ Classical $F=ma$ equation doesn't work for massless particles — they are ultrarelativistic in all cases. $\endgroup$– RuslanCommented Mar 18, 2015 at 12:02
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$\begingroup$ But this answer is smart and funny argument :) $\endgroup$– kakazCommented Jun 10, 2019 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. $\endgroup$– EelvexCommented Feb 12, 2014 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.. $\endgroup$ Commented Feb 12, 2014 at 17:46
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4$\begingroup$ Newton's law should be $F = dp/dt$ here, not $F = ma$. $\endgroup$ Commented Nov 16, 2016 at 13:58
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If you think mathematically then if you divide Coulomb's force law by mass of the charge, then we get it's acceleration. If there exist a massless charge then for $M=0$, then. $\dfrac{KQ_1Q_2}{R^2}$. $M$ will give infinite acceleration that will lead to infinite velocity!!
IMPOSSIBLE NA!! interesting
