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How to show that massless particles do not carry charges from QFT's point of view?


marked as duplicate by Community Aug 14 '15 at 16:26

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    $\begingroup$ Gluons are massless and carry color charge. Before spontaneous symmetry breaking the electroweak gauge bosons are massless and have electroweak charge. $\endgroup$ – Robin Ekman Aug 14 '15 at 16:06
  • $\begingroup$ Oh. thank you very much. Can massless particles carry electric charges? $\endgroup$ – Xiaoyi Jing Aug 14 '15 at 16:08
  • $\begingroup$ It depends on what you mean by electric charge. If you mean the unbroken part of the electroweak symmetnry group, then if you omit the Higgs-fermion interaction $\mathcal L_y$ from the Weinberg-Salam-Glashow model the fermions are still massless after spontaneous symmetry breaking and still interact with the massless gauge boson, i.e., have electric charge. I have to admit I'm not sure if their masslessness is protected during renormalization, because the symmetry that prevented mass terms in the first place is now broken. If by electric charge you mean any $U(1)$ theory, then you could, $\endgroup$ – Robin Ekman Aug 14 '15 at 16:25
  • $\begingroup$ I think, write down a chiral $U(1)$ gauge theory with massless fermions. $\endgroup$ – Robin Ekman Aug 14 '15 at 16:26
  • $\begingroup$ Would you elaborate for chiral U(1) gauge theory? $\endgroup$ – Xiaoyi Jing Aug 14 '15 at 16:35

Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism.

The reason may be explained in different ways. One of them is using "running couplings". Renormalization implies that the electromagnetic coupling constant "runs". The dimensionless value of $e$, while naively constant in classical physics, has a subtle dependence on $\log E$ according to quantum mechanics thanks to the shielding by polarized particle-antiparticle pairs.

In the real world, the running begins at low energies comparable to the electron (or positron) mass, the lightest charged particle. At energy scales below the electron mass, there is basically no running anymore. If the electron mass were sent to zero, the coupling would keep on running to arbitrarily low energies. The coupling is weakening at lower energies. It would be zero in classical physics if a massless particle species existed.

It's hard to decide whether such a situation would be inconsistent. It would surely be different than it is in our world where the fine-structure constant has a finite, well-defined low-energy limit, $1/137.036\dots$.

The world with massless charged particles would either be inconsistent; or had a vanishing coefficient in front of Coulomb's law for macroscopic low-energy objects (which would be enough to say that we shouldn't call it "electromagnetism" anymore); or would demand an infinite magnitude of the coupling for any finite energy scales which is probably inconsistent.

One might worry that the arguments based on running are unreliable. There is a more direct, experimental way to approach the question: We know that massless particles don't exist because they would be easily pair-produced at colliders – and pretty much in any process – because the probability of their pair-production is governed by the (known to be finite) fine-structure constant and can't be suppressed, and there can't be a shortage of energy, either (because they're massless).

Concerning the "loopholes" that are important in the theoretical literature, supersymmetric field theories are full of massless charged particles. But whenever they occur, the behavior of the forces at long distances is very different from what we know in the real world. These $U(1)$ forces are typically a part of a non-Abelian symmetry that is either confining like QCD; or is conformal (scale-invariant) when the negative contribution to the running from gauge bosons is cancelled by some other charged particles.

There is nothing inconsistent about these theories – supersymmetric theories like that are more consistent, not less consistent – but the actual physics in that world would not allow the macroscopic charged static objects we know from the first encounter with the Coulomb law simply because all massive charged particles would be more or less guaranteed to be unstable. Mathematically, these theories are theories with massless charged particles. But physically, the phenomena in such Universes wouldn't follow the history of electromagnetism started by the Coulomb law etc., so from an empirical viewpoint, the corresponding $U(1)$ interaction shouldn't be considered a special case of "our" electromagnetism.

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    $\begingroup$ Thank you so much professor, Luboš Motl. I am very surprised that you would answer my question. I should collect all your answers and comments in stack exchange and overflow because your explanations are always inspiring and illuminating. $\endgroup$ – Xiaoyi Jing Aug 14 '15 at 16:33

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