If a neutrino has mass then it travels less than the speed of light. Suppose I boost myself to the rest frame; i.e. bring it to rest in the laboratory. Now if it oscillates between different states and masses sitting there, where does the oscillating excess/loss energy and mass reside? In some internal state? I have a very limited knowledge of QM but in SR it would seem strange that "mass" would move into another place. A pointer to where I should start studying would (I hope) be sufficient.

  • $\begingroup$ Neutrinos propagate in their mass eigenstate so they don't change mass as they oscillate. $\endgroup$ Commented Jan 1, 2014 at 19:21
  • $\begingroup$ @BrandonEnright Well, they propagate in a superposition of mass eigenstates. $\endgroup$ Commented Jan 1, 2014 at 21:43
  • $\begingroup$ @dmckee I thought a mass eigenstate was a superposition of flavor eigenstates. I guess I still don't understand the mixing. I really need to learn the math :-/ $\endgroup$ Commented Jan 1, 2014 at 22:19
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    $\begingroup$ @BrandonEnright It works both ways. Both the mass states and the flavor states constitute a basis for the space. $\endgroup$ Commented Jan 1, 2014 at 22:57

2 Answers 2


When you say "Suppose I boost myself to the rest frame" you have in effect asserted that you are looking at a particular mass state rather than at the superposition of mass-states that arises from allowing a flavor state to propagate freely.

This is related to the question of how neutrinos can oscillate though the lepton flavors have differing masses (also examined in "Neutrino Oscillations and Conservation of Momentum").

Boris Kayser's explanation of the latter problem has the flavor of consistent histories to me, though I suppose that you can also understand it in therms of the uncertainty principle.

As an experimental limitation, say that you have achieved the rest frame of a neutrino. How will you know? You'd have to measure it's interaction with something, but the cross-sections are vanishing even by the understanding of neutrino physics and the energies involved are likely much less than 1 eV. It is very challenging even in principle.

  • $\begingroup$ Experimental difficulties don't stop theory :) and I was trying to understand the present day theory. Although I think present day theory will be supplemented in the long run; I have no idea what that will be so I try to stick with present ideas. Your right about not being able to measure without instantiating the state/particle. Thank you for the references. I obviously haven't done enough internet searching. $\endgroup$
    – rrogers
    Commented Jun 17, 2016 at 21:49

OK, let's accept your conceit.

You ran after $\nu_1$, the lightest neutrino, for the sake of argument, and jumped on it. You ran real fast, as your Lorentz factor γ is several millions, (In natural units $\hbar=c=1$, in the lab frame, $E=p+m_1^2/(2p)+...$, not yours). You watch what is happening around you. You do not oscillate to anybody: You are a non-interacting plane wave. You will not change your energy or momentum--how could you? Near you, there are two other plane waves/neutrinos that you started your trip with, with different masses, energies and momenta, but only by a little!

They are related to you, as they were produced by the same pion decay, in your parent superposition $\nu_\mu$, a fictitious convenience superposition state helping with the accounting. Since they have slightly different masses, $m_2, m_3$, they are zipping by at slightly different energies and momenta, which they took from the pion at the moment of their birth, (e.g. as the e or μ in a K decay take different energies and momenta away, but since their masses are so enormously different, one would not bother superposing them!) Now there are sticklers who would arrange some uncertainty in the time you got emitted, or the position your parent decay occurred, to artificially arrange for a common momentum, or energy, or speed with your 2 fellow travelers. But you are going so fast, it hardly matters, to this order of the calculation.

The crucial thing, however, is that you and your two partners' waves will evince "beats" about the common average frequency and momentum, without any interaction, beyond the mere superposition--Fourier analysis: the envelope of the average wavetrain will show some oscillation with the distance of your trip. That is differences in your formation with your two partners: you'll just keep your steady course, and not exchange any momentum or energy or anything else.

And then you get detected: you deposit your energy and momentum to the observing atom. And so do your two partners. Nobody observed you during your trip, so you were virtual (!) but, still, since the trip was so long and the time of it likewise (yes! in our units) you were all but on-shell, close to the energy-momentum relation dictated by your mass. The shape of the group wave envelope, so the altered relationships with your partners will dictate the type of fictitious interaction combination ("flavor eigenstate") likely to hit the detector, but, in an oscillation experiment the parameters are chosen so it may be different than your birth flavor eigenstate. The atom that was hit, did not know what hit it, even though it would get a different energy and momentum from each of you three (by a little... so little it is hardly accounted for...) it registers being hit by a given flavor eigenstate formation of p~E, but it does not make much sense to ascribe a mass to that formation; only the component mass eigenstates, you and your 2 partners, had a mass.

The effect is kinematic, linear Fourier analysis beats, very similar to the musical phenomenon, except now it is relativistic.

The straightforward math for all this in in the PDG summary, eqns (14.9-15).

  • $\begingroup$ Thank you for your analysis, and that is exactly the question/experiment I was envisioning. I wish I had had the knowledge to phrase it so well. One small thing is: with different masses they will deposit different momentum, but with elastic scattering I presume the momentum will be distributed according to typical energy/mass/momentum equations. And similar calculations will hold for inelastic scattering. Thank you for the reference. I seem to have missed it or my copy of PDG doesn't have such detail. $\endgroup$
    – rrogers
    Commented Jun 15, 2016 at 22:02
  • $\begingroup$ Indeed, each of the 3 neutrino mass eigenstates will behave ever so slightly differently kinematically, but the differences are too small to asses....and the exact production momenta and energies were never determined to parts per million. You might see that effect, though, at the spectrum endpoints of the KATRIN experiment low energy beta-decay. $\endgroup$ Commented Jun 15, 2016 at 22:36

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