Why Do Neutrinos and Anti-Neutrinos Oscillate Differently?

  • Thread starter Thread starter sanman
  • Start date Start date
  • Tags Tags
    Neutrinos
Physics news on Phys.org
They don't. The article was just wrong, and the reporter didn't know what he was talking about.
 
Why do neutrinos and anti-neutrinos oscillate differently?
Whether they do or not, and if they do, why, is a long-standing unresolved issue. The LSND neutrino oscillation experiment back in the 1990's seemed to see an excess oscillation from anti-muon neutrino to anti-electron neutrino. More than was expected, and difficult to explain. The MiniBoone experiment which ran from 2002 to 2005 looked at muon neutrino to electron neutrino, and to everyone's relief did not see an excess. But then they switched to using antineutrinos, and got a result confirming the one from LSND! This seemed to imply that neutrinos and antineutrinos behave differently.

See here and here for a summary.
 
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