English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

How many different types of Neutrino were actually assumed to exist if they exist at all?

2006-08-18 01:45:00 · 3 answers · asked by goring 6 in Science & Mathematics Physics

3 answers

Well, at first the neutrino was supposed to be massless, because it is a fermion (spin 1/2) but alwys seemed to be spinning one way and not the other with respect to its linear momentum. That is, it seemed to exist in a helicity eigenstate. You couldn't catch up to it. So it must be travelling at the speed of light.

However: there are three kinds of neutrino, and it seems that these kinds have been able to turn into one another. This is the phenomenon called "neutrino oscillation." Neutrino oscillation seems to require mass--in fact, masses, different for different kinds of neutrinos. So the measurement of the magnitude of neutrino oscillation--in, say, a beam of neutrinos--puts a lower limit on the neutrino masses. By the way, the poster above me is referring to an upper bound.

2006-08-18 03:50:12 · answer #1 · answered by Benjamin N 4 · 1 0

Neutrinos always were believed to have mass but the mass was considered "vanishing." That is, it aproached zero

When a neutron undergoes beta decay it produces a proton, an electron and an anti-nutrion; when proton undergoes beta decay it produces a neurton, a positron, and a neutrion.

Because the total energy of the system must be conserved. We know the rest mass energy of the neutron (M_n * C^2) must equal the mass energy of the proton plus the kinetic energy of the proton, plus the rest mass energy of the electron, plus the kinnetic ernegy of the electron, plus (finally) the total energy of the neutrion as shown below:

Mc^2[Neutron] = Mc^2[Proton] +K[Proton} + Mc^2[electron] + K[electron] + E[neutrino]

Near the far end of the electron spectrum for beta decay is ther region of maximum K[e] and minimu E[n]. Since:

p[n] = sqrt(E[n]^2 - M[n]^2*c^4) / c

M[n] can only approach zero because a zero mass particle cannot be at rest in any lorentz frame. p is determined experimentally, E from the first equation, from that you calculate M.

Current results suggest a lower bound near 30eV.

2006-08-18 03:28:15 · answer #2 · answered by selket 3 · 1 0

I didnt know neutrinos have mass! Guess im behind the times

2006-08-18 01:50:03 · answer #3 · answered by megalomaniac 3 · 0 0

fedest.com, questions and answers