Quantifying an Off-axis Neutrinos (probability and intensity)

Rodger
Messages
4
Reaction score
0
Quantifying an "Off-axis" Neutrinos (probability and intensity)

Considering the modelling of a high energy proton beam neutrino experiment. I have questions concerning the scattering of neutrinos from the axis of the proton beam.

I understand that a muon beam (derived from a proton beam) incident on a target causes the muons to decay in part into (muon) neutrinos which are scattered from the initial axis of the muon and proton beams.

I can quantify the energies of the muons and neutrinos with respect to the angle that they scatter in.


1) How might I go about quantifying the probability that neutrinos are scattered at a given angle from the incident axis?

2) How might I then go about quantifying the intensity of the scattered neutrinos at a given angle?
 
Physics news on Phys.org
I never saw numbers for that, but I would expect that neutrino scattering is negligible.

The flight direction of muons can deviate from the proton direction (probably complicated to model), and the neutrino emission does not have to happen exactly in the flight direction of the muons: go to the muon rest frame, find the angular distribution (uniform? Or do we have some spin to consider?) and the energy distribution (3-body decay), transform back to the lab frame.
 
  • Like
Likes 1 person
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}...
Back
Top