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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?

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Try this: blogs.uslhc.us/… – Marek Jan 15 '11 at 15:59

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.

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I don't see anything in this answer that explains how conservation of mass is conserved. I bet its done in the math, but honestly that math is incomprehensible to me (and I got as far as linear algebra in college). Please define your variables, otherwise your answer will only make sense to people who already know the answer. – B T Dec 23 '15 at 21:43
    
The problem is not undefined variables, but a notation unfamiliar to you. I wrote this answer assuming that the reader knew enough QM to be familiar with Dirac's bra-ket notation. The good news is that this notation is just another way of writing linear algebra, so you could presumable bone up on it quickly. However, the is no expectation that mass is conserved: rather energy (including the mass term) and momentum are conserved. – dmckee Dec 23 '15 at 21:59
    
An understanding of bra-ket notation doesn't help if you don't know what the variables mean. – B T Dec 23 '15 at 22:17

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

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If I understand correctly, neutrinos all have a "mass-basis" which is essentially the state of the neutrino that includes its mass and probabilities for what flavor the neutrino interacts as. Neutrinos are created with a particular initial flavor which has a probability of having one of 3 non-oscillating mass-bases ("bay-sees"?), conventionally called v1, v2, v3:

  • ν1 has about 2/3 chance of being detected as an electron-neutrino and 1/6 each of being detected as muon- or tau- neutrinos
  • ν2 has about equal chance of each (tho not quite exactly equal)
  • ν3 has mostly an even chance of between muon- and tau- neutrinos, and a tiny chance of being detected as an electron-neutrino

Saying for example that an electron-neutrino has a particular mass, is misleading. Instead it seems that a neutrino created as an electron-neutrino has a probability of having each mass-basis, and that mass-basis determines how the neutrino will oscillate its flavors.

So a v1 neutrino stays a v1 neutrino the whole way through and energy is indeed conserved. Saying that it keeps the same mass-basis all the way through, though, doesn't mean we can necessarily determine what that mass is. Its oscillation behavior affects how they interact with things like detectors, sometimes passing right through electron-neutrino detectors because it switched to one of the other two flavors while in detection range of the detector.

Read this for a good non-mathematical explanation: http://www.quantumdiaries.org/2010/08/02/solar-neutrinos-astronaut-ice-cream-and-flavor-physics/ . Thanks to Marek for putting that link up there as a comment!

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Your first sentence is wrong. Neutrinos are created in flavor states, not mass states. – dmckee Dec 23 '15 at 22:22
    
It seems like they are created with both a mass state and a flavor state. Is that not correct? – B T Dec 23 '15 at 22:40
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They are created in a flavor state which is a linear combination of mass states. – dmckee Dec 23 '15 at 23:00
    
There is a difference in saying that you don't know what mass state it has and it not having a definite mass state. Conservation of energy holds, and that requires it to have a definite mass state on creation. The mass state of a neutrino does not oscillate. – B T Dec 23 '15 at 23:34
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This is quantum mechanics and the weak interaction term does not commute with the free Hamiltonian, the flavor states no more have a defined mass that a particle with a precisely know position has a well defined momentum. The mechanism is essential the same as the Heisenberg Uncertainty Principle. – dmckee Dec 24 '15 at 0:29

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