Question about neutrino oscillations

McLaren Rulez
Messages
289
Reaction score
3
Hi.

I'm a college undergrad (junior year, so basic knowledge of QM but not much else) and I'm reading up on neutrino oscillations. I have a few questions.

For neutrinos, which is more fundamental: The mass eigenstates or the flavour eigenstates? In this paper http://www2.warwick.ac.uk/fac/sci/physics/teach/module_home/px435/bkayser.pdf the author says, "Suppose that there are N physical neutrinos (mass eigenstates), \nu_{m}..." So do the mass eigenstates represent physical particles which come together in different linear combinations to produce flavour? I am a bit confused about this issue.

Also, in the same paper, the author mentions in the first page, on the right side column: In the standard treatment it is supposed that we have a beam of neutrinos all having a common fixed momentum, p_{\mu}. So when he talks about momentum, is he saying that all the different mass eigenstates have the same momentum? If so, why is it valid to assume that they all have a fixed common momentum? Why can't different mass eigenstates have different momenta?

Thank you very much for your help.
 
Physics news on Phys.org
I'm afraid you may not like this answer; but, which set of states are more fundamental depends on what you mean by "fundamental." If you're asking about which set of state are more reasonable to think of as the "actual physical states," the best answer is the mass eigenstates; but, if you're asking which should show up in the simplest (or, pretty much equivalently, most symmetric) description of the theory, the answer is the flavor states.
 
Thank you for the reply, Parlyne.

Regarding the second part, do you know why it is ok to think of a beam of mass eigenstates all having the same momentum as the standard treatment for neutrino oscillations? I also see some other sources where they treat the energy of all the eigenstates as equal instead. Why are we allowed to do this?

Thank you.
 
McLaren Rulez said:
why it is ok to think of a beam of mass eigenstates all having the same momentum as the standard treatment for neutrino oscillations?

The neutrino masses are assumed to be much smaller (on the order of 1 eV/c^2) than the neutrino energies in oscillation situations: on the order of 1 MeV for solar neutrinos, and ranging up to the GeV range for accelerator experiments. In these sitations the energy equals the momentum, for all practical purposes, and changing the mass has practically no effect on the momentum, for a given energy.

If the neutrino masses were comparable to other particle masses, then this approximation wouldn't work; but in that case we would have observed the effects of neutrino mass a long time ago in other ways than neutrino oscillations!
 
jtbell said:
The neutrino masses are assumed to be much smaller (on the order of 1 eV/c^2) than the neutrino energies in oscillation situations: on the order of 1 MeV for solar neutrinos, and ranging up to the GeV range for accelerator experiments. In these sitations the energy equals the momentum, for all practical purposes, and changing the mass has practically no effect on the momentum, for a given energy.

If the neutrino masses were comparable to other particle masses, then this approximation wouldn't work; but in that case we would have observed the effects of neutrino mass a long time ago in other ways than neutrino oscillations!

So at what distance from the center of the sun can we expect the solar neutrinos to change into a different neutrino, if that makes sense? Or in other words, how many oscillations from the sun to the Earth?
 
Spinnor said:
So at what distance from the center of the sun can we expect the solar neutrinos to change into a different neutrino, if that makes sense? Or in other words, how many oscillations from the sun to the Earth?

From:

http://en.wikipedia.org/wiki/Neutrino_oscillation


"Oscillation distances, L, in modern experiments are on the order of kilometers"
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
4
Views
2K
Replies
11
Views
2K
Replies
5
Views
2K
Replies
6
Views
2K
Replies
1
Views
1K
Replies
87
Views
18K
Replies
3
Views
2K
Back
Top