# Tag Info

4

PDG is discussing charged current interactions, $\nu N \to \mu^- X$ and $\bar\nu N \to \mu^+ X$. These are not charge conjugate processes. The neutron is $udd$. With the neutrino, $\nu$, we need a $W^+$ for the charged current interaction, $$\nu \to W^+ \mu^-$$ We then need $$W^+ d \to u$$ Note that the neutron contains two down quarks. The case of the ...

3

You are perfectly right: neutrinos hold the promise of providing a window that gives us views much deeper into the big bang than the window conventionally provided by photons. The Hubble Space Telescope gives us snapshots of galaxies in a universe that is only 600 million years old. Although this feat is brought to the wider public as big news, the Hubble ...

1

Detecting cosmic neutrino background (~1.95K) is extremely difficult (compared with cosmic microwave background) and never performed directly so far. That's because neutrinos interacts with matter very weakly, unlike photons. We have to build very large detectors. (But if they behave like photons we can't use them to observe early universe.) I also found ...

2

The actual masses are accessible in theory, but not from mixing measurements. Cosmological measurements could give us a useable handle on the sum of the masses (though until we settle the hierarchy questions this may not provide a unique answer), or the combination of a much better model of supernovae plus a precision measurement of the differences in ...

1

We have, at this time, no tools capable of detecting neutrinos at the very low energies of the cosmic neutrino background, and if such tools existed they would have to contend with numerous backgrounds making the experiment ferociously difficult.

0

Unfortunately, this result was later found to be caused by faulty electronics, according to the CERN press release.

3

It really goes deeper than just a theoretical demand on a particular domain. The Hamiltonian for any system must be unitary, because that preserves the total probability at one. This is important because if I start with some state and let it evolve for a while the system must afterwards exist in some state which means that the sum of the probabilities taken ...

2

In this context, the right-handed neutrino is a singlet under the Standard Model gauge groups. Only the right-handed neutrino is allowed a Majorana mass. The left-handed term is not gauge invariant. If the SM and right-handed neutrino fields were embedded into a Grand Unified gauge group, the right-hand term would break that symmetry. It is expected ...

1

I will offer two reasons. First, unitarity of mixing matrices insures that probabilities sum to one. The probability of an oscillating neutrino having electron, muon or tau flavour should equal one. Second, because the neutrino mass matrix is Hermitian it is diagonalised by a unitary matrix.

4

It's a theoretical demand : $$\begin{pmatrix} \nu_{e}\\ \nu_{\mu}\\ \nu_{\tau} \end{pmatrix} = \begin{pmatrix} U_{e1} & U_{e2} & U_{e3} \\ U_{\mu1} & U_{\mu2} & U_{\mu3} \\ U_{\tau1} & U_{\tau2} & U_{\tau3} \end{pmatrix} \begin{pmatrix} \nu_{1}\\ \nu_{2}\\ \nu_{3} \end{pmatrix}$$ You know that all states ...

0

There is a lot to be said about Majorana and Dirac neutrino masses, but I will try to address only the specific point you raise. When you write a Majorana mass term, for example, $$\mathcal{L}_{\text{Majorana}} = -\frac{m_M}{2} \left[\bar\nu_L(\nu_L)^C + \overline{\nu^C_L}\nu_L\right]$$ you are not assuming that $(\nu_L)^C = \nu_R$, but you can write that ...

Top 50 recent answers are included