Thanks to answers a previous question, I have realised the difference between helicity, a non-Lorentz-invariant quantity, and the Lorentz invariant chirality.

Let me summarise what I understand, before asking questions on what I do not understand.

Photons are massless, so helicity coincide with chirality. There are two eigenstates, but since they both have mass zero, any photon can be either an eigenstate (circularly polarised, left or right) or a mixture (for instance, linearly polarised). The chirality/helicity state is Lorentz invariant, since a photon cannot be “overtaken”, but can be modified, for instance by going through a “half-wave plate”, that inverts the sign of the chirality/helicity, or a “quarter-wave plate” that can change a circularly polarised photon into a linearly polarised one and vice-versa.

For charged massive leptons (electron, $\mu$,$\tau$) both eigenstates of chirality exist, and with the same rest mass. So while such a lepton, when created by weak interaction, has left chirality, in general such lepton will be a mixed state of chirality. And also mixed state of helicity that does not even have to be equal to chirality. Still, from a remarks by rob and ACuriousmind, when the energy of the electron (created by weak interaction) is much higher than its rest mass, chirality and helicity would tend to both be mostly left-handed. Please, rob and ACuriousmind, correct me if I misunderstood you. So unless I am seriously mistaken, though very energetic chiral-left-handed, mostly helicity-right-handed electrons do certainly exist (it suffices to “overrun” chiral-left, mostly helicity-left-handed electrons, at a speed very, very close to $c$), and though being chiral-left-handed, they do in principle interact through weak interaction, they do so with less efficiency than chiral-left, mostly helicity-left-handed electrons of the same energy.

Isn’it true ? When chiral asymmetry for weak interaction was initially discovered, wasn’t it the helicity, rather than the chirality, of the electrons partaking to the reaction that blew the whistle ?

In the original SM, neutrinos are massless, have only left-handed chirality and helicity (anti-neutrinos having right-handed ones).

In a answer by ACuriousmind I found the sentence

For massless fermions, the evolution equations that couple the right-handed part of a massive Dirac spinor to its left-handed part decouple, meaning a massless Dirac spinor is equivalently a theory to two uncoupled Weyl spinors.

But this is moot since we know neutrinos have mass. Still, though one can indeed "overrun" chiral-left-handed, mostly helicity-left-handed neutrinos and in that frame they would be chiral-left, mostly helicity-right-handed ones, being in principle able to interact through weak interaction, the neutrinos that are created in high energy collisions have always left handed helicity. Chiral-left-handed, right-handed helicity ones are never spontaneously created by weak interaction. Or I seriously misunderstood something.

Now I get into uncharted (for me) territory. I’m proceeding with care.

One thing is clear, is that, contrary to the case of charged leptons (electron, $\mu$, $\tau$), there are no right-handed chirality neutrinos with the same rest mass as left-handed chirality ones. And possibly not at all, in the minimal extension of the original SM to admit neutrino masses.

But the fact I just referred to :

For massless fermions, the evolution equations that couple the right-handed part of a massive Dirac spinor to its left-handed part decouple,

(therefore a single chirality is possible, as is the case in the original SM) is not true for massive particles, even very light ones, is it ?

But if right-handed chirality neutrinos do not exist at all, how can the equations for left-handed chirality "not decouple” from nothing ? It is a bit weird ! A double negative should be a positive, but a positive that does not exist at all… ???

If right-handed chirality neutrinos do exist (but for different, higher masses), then the equations for left-handed chirality neutrinos could couple, but only hardly so, with something that exists.

The admixture of right-handed chirality neutrinos into the propagating "mostly left-handed chirality neutrinos", with different masses, might have a very small coefficient and also oscillate rapidly, about the way the admixture of $\tau$-flavor neutrinos into the lightest propagating (mostly electron-flavor) neutrinos is very small and rapidly oscillating.

Unless I am mistaken, this is what was found in the BEST experiment. Isn’t it so ?

So my questions are

A) Does what I say in the preceding paragraph ( from “Now I get into…” ) make sense, especially my point of needing something to "almost but not quite decouple from" ?

B) If the BEST experiment (and other, similar ones) do prove unambiguously the existence of one sterile neutrino, is this a strong theoretical argument that implies that two more sterile neutrinos (right-handed chirality) must exist, so there is one for each "family”, because all three families of left-handed chirality neutrinos need some right-handed chirality neutrinos to not quite decouple from, since they are very light but not massless ?

  • 1
    $\begingroup$ For examples of decays where the helicity-to-chirality relationship is "wrong," look into the charged weak decays of spin-zero particles, such as $\pi^+\to\mu^+\nu$ or $\pi^+\to\rm e^+\nu$. To conserve angular momentum, the back-to-back decay products must have the same helicity. But to couple to the charged current, the matter and antimatter lepton must have opposite chirality. This is why $\pi\to\rm e\nu$ is suppressed, even though it's energetically favorable: the relativistic $\rm e$ must have the "wrong" chirality relative to its spin, and the overlap is small. $\endgroup$
    – rob
    Jul 22, 2022 at 5:05
  • $\begingroup$ OK, so it does happen but it is suppressed in the case of the positron. And it is the positron, with the largest mass, which has the wrong helicity, not the neutrino. This is coherent with my understanding. The anti-$\mu$ is so close to the $\pi$ in mass, there is no problem at all. $\endgroup$
    – Alfred
    Jul 22, 2022 at 6:07
  • $\begingroup$ Rather than think the suppression is “the positron, not the neutrino,” think instead that that the wrong-helicity overlap goes like some power of $m_e/E$, and that the corresponding overlap for a Dirac-mass neutrino would be correspondingly smaller. $\endgroup$
    – rob
    Jul 22, 2022 at 12:02
  • 1
    $\begingroup$ @rob What you think I wrote is not what I meant. I meant "suppression in the positron case" as opposed to "no, or weak suppression in the 𝜇+ case". And suppression not too bad even in the "electron case" because the electron is not so very light so the wrong helicity goes to it. A wrong helicity on the neutrino would cause a much much worse suppression. I totally agreed with your last remark but expressed myself in an ambiguous way. $\endgroup$
    – Alfred
    Jul 22, 2022 at 14:47

1 Answer 1


It may help to think of the left-circularly polarized and right-circularly polarized modes of a fermion as two completely different particles, because in many ways they are.

Each of the SM fermion fields is only half of what one normally thinks of as a field: it has one circular polarization but not its mirror image. What we call the electron consists of waves in two of these fields, which are coupled together. This is normally hidden in Feynman diagrams, but you could imagine it like this:

    |   |
    *   *   ...
   / \ / \ /
...   *   *
      |   |

where the *s are interaction vertices, the |s are virtual Higgs bosons somehow donated by the vacuum (out of scope for this question), the /s are one of the two chiral fields, and the \s are the other chiral field. I'm suggesting that you think of / and \ as two different particle types that are forever oscillating into each other, rather than as an $e^-$ going in two different directions. The W bosons only couple to one of the two, not because the weak force cares about chirality but because only one of the two has SU(2) charge. They really are different.

The electron mass is the strength of this coupling. Note that there is only one mass, not two. The two chiral fields can't have different masses. They don't have mass at all. It's the coupling that has (or is) a mass.

In the original Standard Model, there are (by one count) fifteen chiral fermion fields per generation, so they can't all be paired up in this way. The odd one out is the neutrino. So it's not really the case that the neutrino's mass is zero: rather, the whole concept of mass doesn't exist for the neutrino, because there is no pair of fields to couple in the first place.

The simplest way to accommodate neutrino mass in the Standard Model (at least simplest in the sense of being the most like the other fermions) is to add a sixteenth fermion field which pairs up like the others.

Unfortunately for the simplicity of the theory, there is an additional allowed coupling that is unique to this sixteenth field (the reason being that it uniquely has no gauge charges). This is the so-called Majorana coupling or Majorana mass term. As a result, the neutrinos in this model do have two masses per generation, while the other fermions have only one. However, the different masses still aren't associated with left and right polarization. They aren't even equal to the two mass coefficients (Higgs and Majorana), but are derived from them, per the seesaw mechanism. In any case, if you somehow slow a neutrino to a stop, or just boost to its rest frame, it isn't half heavy and half light. It's just light. The other mass is the mass of a different oscillation mode of the same pair of fields. That's the sterile neutrino–though the term "sterile neutrino" also seems to be used in other ways.

You could also set the Majorana coupling to zero and set the Higgs coupling to the observed neutrino mass, making the neutrinos just like the other particles, but there is no known theoretical reason why the Majorana coupling should be zero, so it's more likely that it isn't.

very energetic chiral-left-handed, mostly helicity-right-handed electrons

This seems backward to me. An electron can't be only left-chiral, because you can't turn off the Higgs interaction, but it can have just one helicity (with respect to any particular inertial frame). Internal electron edges in Feynman diagrams could be purely left-chiral if Feynman diagrams were drawn that way.

For massless fermions, the evolution equations that couple the right-handed part of a massive Dirac spinor to its left-handed part decouple, meaning a massless Dirac spinor is equivalently a theory to two uncoupled Weyl spinors.

That statement makes sense if you start from the Dirac equation, which describes two mirror-image chiral fields coupled together. As you take the coupling constant (mass) to zero, it becomes a theory of two independent fields. But no such mirror-image fields have been found in the real world.

  • $\begingroup$ Your answer is not useful. You speak of "one neutrino" instead of three. The point of my question is, if BEST proves the existence of ONE sterile neutrino, does it imply that three ones exist ? Besides, contrary to your claim, the different mass are definitely related to different chiralities. The light one is left-chiral, and IF the other, heavier one really exists, it is sterile, right-chiral. And finally, you do claim that to accommodate SM to neutrino mass, one adds one field, but this is not generally accepted. Before BEST and even now, not everyone agrees that sterile neutrinos exist $\endgroup$
    – Alfred
    Jul 22, 2022 at 8:22
  • 2
    $\begingroup$ @Alfred, the answer (v1) talks about fifteen fermion fields per generation, and presumably adds a sterile sixteenth field to each generation. However, in the absence of experimental data, aesthetically unpleasant models are also on the table, such as having the number of “sterile flavors” be more or fewer than the three interacting flavors. It is possible that the one of the known neutrino mass states is exactly massless. It is possible there is a fourth generation whose members, including the neutrino, are all very massive. When you go beyond the data, speculation becomes unconstrained. $\endgroup$
    – rob
    Jul 22, 2022 at 12:32
  • $\begingroup$ @rod Before BEST, there were zero data on sterile neutrinos. Lots of people do not think that they exist at all. But how can the left-chiral massive (if very light) neutrino "almost decouple" from nothing ? Well, if there really was no evidence at all for a sterile neutrino, maybe look at weird scenarios. But with at least serious evidence (not yet, maybe, BEST results need independent confirmation, but I hope there will be) then it seems to me that the most natural solution is one sterile neutrino per family. And if the one found by BEST is much too light for Dark Matter, the next one ? $\endgroup$
    – Alfred
    Jul 22, 2022 at 14:04
  • $\begingroup$ @rob You see, I am not convinced by the MOND scenario of gravitation. I like GR, I'd like to find a likely candidate for Dark Matter. SUSY candidates don't seem to work. But with the lightest sterile neutrino around a few eVs, either the second one or the third one might have the correct mass and correct lifetime to explain Dark Matter and not produce too much radiation, the absence of which would kill this scenario. I am not a particle physics expert. I am interested in Cosmology (as a hobby, my specialty in Physics being Mathematical Physics with little or no connection to the real world) $\endgroup$
    – Alfred
    Jul 22, 2022 at 14:13
  • $\begingroup$ @Alfred, that’s not correct. There were hints about sterile or missing neutrinos in LSND, in the low-statistics region of MiniBoone, and in a number of other neutrino results, but the community was rightly skeptical. For massive neutrinos as a component of dark matter, see this post and links therein. $\endgroup$
    – rob
    Jul 22, 2022 at 14:17

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