Suppose that instead of doublets consisting of lepton/neutrino flavor eigenstates, the sterile neutrino is a pure singlet with zero flavor. Then, since there would be no partner lepton for it to tunnel into upon switching up the weak interaction, it seems that the sterile neutrino, if it has any interaction at all, must interact exclusively through other interactions.

If there are four fundamental interactions, and nature doesn't necessarily generate a new interaction from a new symmetry, but sometimes merely augments the existing interactions with new dynamic symmetries, then it becomes plausible that; because the EM interaction is ruled out due to its connection to the weak sector; because gravity is far too weak to have any influence at microscopic scales, that sterile particles are strongly interacting.

If so, it could be the case that such an augmentation is uniquely a property of the strong interaction because there would be no reason to believe such an augmentation exists universally--hence; there are no selectrons, etc.

Phenomenologically, super-symmetry would never be any other than an approximate dynmanic symmetry since a sterile particle would eventually decay into a familiar quark/gluon combination.

Interestingly, it can be argued that the dynamic supersymmetry can persist over over cosmological time scales, while the the decay of the usual quark/gluon composites into sterile particles is much (much) slower than their decay by the usual decay mechanisms --thus explaining why sterile particles haven't been seen in earth based accelerator/detector experiments


Your question is rather confusingly worded. You're using many big words, and not quite in the right context. However, your question can be interpreted in three ways.

First, can a sterile neutrino participate in the strong interaction? No. The point of a sterile neutrino is to account for the known neutrino masses using the seesaw mechanism: if two species of neutrino are allowed to mix, then it's easy to get one very light one and one very heavy one. But you can't have this mixing if the sterile neutrinos are color charged and the known ones aren't.

Second, can some "sterile" particle participate in the strong interaction? No. The definition of the word sterile is "not participating in the electroweak or strong interactions". That's important because it means sterile particles are hard to see directly. If you allow them to participate in the strongest interaction there is, they're not sterile.

Third, could there be a new particle that only participates in the strong interaction? Of course. People have made thousands of guesses as to what new particles could be out there, and I'm sure in at least a few hundred, there are particles that only interact strongly. For example, supersymmetry predicts such a particle, the gluino. The main issue is that any such particle must either be quite heavy or interact quite weakly, or else it would have already been seen. If you posit that the interaction is weak, you need to explain why it is. For example, the axion is a light strongly-interacting particle, but it only interacts via nonperturbative effects.

  • $\begingroup$ The point of my remarks was to argue that super-symmetry might not be a universal property of all physical interactions, but only a dynamical symmetry existing within the strong interaction. If so, such particles would only be sterile in the sense that they only interact strongly and would not be the same kind of sterile particle usually considered. and EM interactions $\endgroup$ – H. Cooper Oct 8 '18 at 16:24
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    $\begingroup$ @H.Cooper I don't think it makes sense to say supersymmetry is "a dynamical symmetry existing within the strong interaction". You're welcome to ask another question, though you should provide more technical details to explain what you mean. $\endgroup$ – knzhou Oct 8 '18 at 16:25
  • $\begingroup$ If the strong interaction is the only interaction with super-symmetry, it be only approximate because its particle content, would eventually decay into the usual composites. Independent of interaction strength, those decays would tend to be blocked by energy conversation since the decay products would be much heavier the than the initial sterile particles in their CM, while at the same time, decay of conventional composites into sterile particles would, for a decay into neutrinos, be blocked by phase space factors consisting of the ratio of the sterile mass to the quark mass. $\endgroup$ – H. Cooper Oct 8 '18 at 17:19

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