0
$\begingroup$

What happens in terms of neutrino oscillation if the distance is extremely high (much more than Sun-Earth distance) : is oscillation lost ? Why ?

Is there decoherence of oscillation ? What does this mean ?

Are astrophysics neutrinos (=very far distance) considered for neutrino oscillation ? If yes, how could it be if the oscillation is lost ?

Such figure makes believe that for extremely long distance, the oscillation would not be observable... (?)

http://hep.bu.edu/~superk/osc.html

enter image description here

$\endgroup$
1
$\begingroup$

We refer to it as decoherence. Flavor changing effects are still seen though. The effect is that

$$P(\nu_\alpha\to\nu_\beta)=\sum_{i=1}^3|U_{\alpha i}|^2|U_{\beta i}|^2$$

Another way to write this that may be more useful is:

$$P(\nu_\alpha\to\nu_\beta)=\sum_{i=1}^3P_{\alpha i}P_{i\beta}$$

So for example if you knew the flavors produced at the source you can calculate what they are at the Earth. As an example, if the flavor ratio at the source is $(\frac13,\frac23,0)$ (that is, no $\nu_\tau$'s, and twice as many $\nu_\mu$'s as $\nu_e$'s) given the measured values of the lepton mixing matrix, we end up with approximately $(\frac13,\frac13,\frac13)$ (that is, the same amount of each flavor). Note that the fact that they all end up as nearly the same in this example is a coincidence.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thank you @jazzwhiz. But I thought that having decoherence was meaning "no more having quantum effect", that is "no more having oscillation" ? I also thought (without proof) that for a large distance enough from a given source of neutrino, whater is the flavour, we would end up with the same fraction for all flavours. Is there something in the spirit of that ? $\endgroup$ – Mathieu Krisztian Dec 23 '19 at 18:38
  • 1
    $\begingroup$ The decoherence here is due to a loss of phase constancy. If two neutrinos have even slightly different oscillation lengths, say to ie slightly different energies, after enough distance their oscillations mood longer line up and the signal goes away. $\endgroup$ – Bob Jacobsen Dec 24 '19 at 2:46
  • 1
    $\begingroup$ @MathieuKrisztian I changed the wording. I should have said "flavor changing" not "oscillations." Sometimes they are (incorrectly) used interchangeably. $\endgroup$ – jazzwhiz Dec 24 '19 at 19:24
  • 1
    $\begingroup$ @MathieuKrisztian re: astrophysical sources: we do. There are many papers about this looking at IceCube data. It is even harder than already hard terrestrial or solar experiments because we know neither the initial flavor ratio nor the initial flux. $\endgroup$ – jazzwhiz Dec 24 '19 at 19:25
  • 1
    $\begingroup$ Oscillations do not happen on those distances, only flavor changing: the "effect associated with the mixing" as @MathieuKrisztian mentioned. There are several reasons why there are no oscillations, but they are all so strong that only one is sufficient. It is true that the mass eigenstates will decohere from each other, but there is also the fact that we will never know the distance to the source to a precision smaller than a wavelength, so we're in the oscillation averaged regime anyway, which is the same as the states having decohered. $\endgroup$ – jazzwhiz Dec 25 '19 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.