New answers tagged

2

The left-handed neutrino is a 2-spinor field $\eta_A$, $A=0,1$, and the Majorana mass term is a bilinear, $\Delta L = \pm 2$ term without the complex conjugation in each term, $$ m\cdot \eta_A \eta_B \cdot \epsilon^{AB} + \text{Hermitian conjugate}$$ Note that this Majorana term is the only bilinear term without derivatives that one may construct from a ...


3

Because we have observed processes where the photon number is not conserved. For example positronium can decay into 2 or 3 (or more) photons. This means that it is not possible to assign a global conserved charge to photons. For neutrinos we can assign lepton number and so far we have not observed a process that would violate total lepton number (lepton ...


0

The solution you quote actually doesn't conserve momentum. You can use that $p^2$ is a Lorentz invariant and solve $(p_\nu+p_p)^2=(p_e+p_n)^2$, considering the left hand side in the lab frame and the right hand side in the CM frame. You can find $E_\nu$ and check that now momentum is conserved. Anyway, as mentioned in the previous comment, ...


2

I am very bad at drawingin "paint", but the process can go as antineutrino +neutrino to Z0 , Z0 to s antis (or up antiup), a gluon vertex from the quark to an up antiup (or strange antistrange quark) the parenthesis are the alternate diagram. , So it is not forbidden, it has two weak vertices and so very small cross section.


2

Your first diagram is wrong, since there is no vertex in a Lorentz invariant theory where three fermions and a vector meet. However, I don't see why you say the interaction is forbidden. It would surely be insanely suppressed since amplitudes are extremely weak, but I don't see a problem with the diagram (for instance): Notice that quarks mix, so the ...


0

The problem with this process is, that there is no term in the standard model which allows a coupling of a fermion a W-boson and a quark-antiquark pair. Even with the neutrino having no charge, you have to take this arrow you draw on the fermion line seriously, so the fermion flow is violated at this interaction points. Furthermore you can only couple ...


3

All the neutrino detectors we have or might build will have course angular resolution because they detect the direction of scattered products of neutrino interactions rather than the direction of the neutrinos themselves. Worse, the solar neutrinos are relatively low energy (a few MeV), which means the scattering angles are large. Yes, we could accumulate ...


2

First, let's clear up some terminology: the usual statement "Majorana fermions are their own antiparticles" is correct, but confusing because the words we usually use to describe neutrinos are made for Dirac fermions. If neutrinos had no mass at all, there would be two independent types of neutrino: a left-handed and a right-handed neutrino. These particles ...



Top 50 recent answers are included