Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Since the total mass-energy for the neutrino presumably does not change when a neutrino changes lepton flavor, though the mass is different, what compensates for the gain or loss of mass? Does the propagation speed of the neutrino change?

share|improve this question
1  
Try this: blogs.uslhc.us/… –  Marek Jan 15 '11 at 15:59
add comment

2 Answers

There are a couple of misconceptions here.

  1. The flavor states are not mass states. That is, the electron neutrino does not have a mass $m_{\nu_e}$ and the muon neutrino a mass of $m_{\nu_\mu}$. Rather, there are two different basis' in which to examine the neutrino. So a neutrino known to be $l$ flavored, is a mixture of mass states (numbered) like $$ |l> = \sum_{i=1}^3 U_{li}^* |i> $$ where $|l>$ is a flavor state (for $l = e, \mu, \tau$); $|i>$ is a mass state; and $U$ is the unitary mixing matrix.

    Neutrinos interact in the flavor basis, but must propagate in the mass basis, so a neutrino mixing experiment probes the probability of detecting a neutrino in state $\beta$ and it was created in state $\alpha$ $$ P_{\alpha,\beta} = \left| <\beta|\alpha(t)> \right|^2 = \left| <\beta|Ue^{iEt}U^*|\alpha>\right|^2 $$ which is a pretty complicated expression in the full three-flavor analysis.

  2. Nor is it a single mass state which propagates, all three do using the usual $e^{iEt}$ propagator, where $E = \sqrt{m_i^2 - p^2}$, which is where the mixing enters, because this reduces to oscillating trig functions.

    The expression above becomes $$ P_{\alpha,\beta} = \left| \sum_i \sum_j U^*_{\alpha,i}U_{\beta,j} \sin \left(2X_{i,j} \right) <\beta|\alpha> \right|^2 $$ where $X_{i,j} = \frac{m_i^2 - m_j^2}{4E}L$, which can be reduced further because $<\beta|\alpha> = \delta_{\alpha\beta}$ and by using some trig identities, but all that is left as an exercise.

Finally, note that all the neutrinos we can interact with have energies measures in MeVs or GeVs, and all the mass states are understood to be less than 1 eV, so all neutrinos are ultra-relativistic: they move a the speed of light for nearly all practical purposes. (The exception here is the hope of comparing the arrival time of the neutrino and light wave-fronts from distant supernovae.

If this were not the case, you would expect to the the probability distribution for a initially well defined neutrino pulse to differentiate by mass state as a function of time, with the leading edge being composed of the lightest state (i.e $m_1$ [$m_3$] if the normal [inverted] hierarchy obtains), and the trailing edge of the heaviest state ($m_3$ [$m_2$]). But those states would still mix to all flavors, it just that the mixture would be time dependant. I've been informed of a more rigorous way to treat this part of the problem. Overview at http://physics.stackexchange.com/a/21382/520.

share|improve this answer
add comment

The reason neutrino oscillations are confusing to those students who think carefully about them is partially because of the history of how the neutrinos were discovered.

Originally, it was thought that the neutrinos were massless and so the flavor eigenstates were the only states that existed. Then the neutrinos were named electron neutrino $\nu_e$, muon neutrino $\nu_\mu$ and tau neutrino $\nu_\tau$. But these were not the mass eigenstates. We usually call the mass eigenstates $\nu_1$, $\nu_2$, $\nu_3$.

So rather than think of the situation as one involving the transmission of a single neutrino of a known (or unknown) mass, think of the situation as one involving three Feynman diagrams involving three different neutrinos $\nu_1$, $\nu_2$ and $\nu_3$. Each diagram contributes a complex number to the amplitude. By the rules of quantum mechanics, the three diagrams interfere.

Looked at this way, the mystery of neutrino oscillation becomes simply an interference that you're already familiar with. You'd have had the same type of interference if there were three possible energies of photons emitted.

For a reference to this way of looking at things, see slide 18 and following in Smirnov's presentation: http://physics.ipm.ac.ir/conferences/lhp06/notes/smirnov1.pdf

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.