Neutrino's and Cherenkov Radiation

Drakkith
Mentor
Messages
23,175
Reaction score
7,625
Do neutrinos passing through matter emit some kind of Cherenkov radiation like charged particles do?
 
Physics news on Phys.org
No, since they are not electrically charged.
 
But if they hit a nucleon, they can produce a charged lepton from their generation. In that case, the charged lepton produces Cherenkov radiation, since its speed is close to the speed of light. That's how neutrinos are detected.
 
in addition to what haael said: This is a cool experiment being done in the antarctic:

http://en.wikipedia.org/wiki/Antarctic_Muon_And_Neutrino_Detector_Array

(In most cases) A Neutrino strikes a nucleon creating a Muon. This Muon creates the Cherenkov Radiation that we detect( sometimes over a meter long!). These scientists have turned the water beneath the polar ice cap into a neutrino observatory! Also, since there are several optical modules organized in a array, we can detect the motion of the radiation, thus determining the direction the neutrino came from.

A little off track: another cool Neutrino detector is the "Super-Kamiokande"Detector in Japan
http://bionicbong.com/travel/places-in-japan/super-kamiokande-neutrino-detector/
 
Back to the OP's post, I wonder if neutrinos could produce some non-electrical analog to the Cherenkov radiation. I mean, neutrinos are weakly charged. The role of photons in the weak interaction is carried by Z-bosons. If they was some medium that interacts "strongly" with Z-bosons and "weakly" with neutrinos, then neutrinos in that medium would produce a shock wave of Z-bosons, similar to the Cherenkov radiation. Of course, despite the word "strong", this interaction would be extremely week. Maybe it would be possible in some very high density state of matter, like quark-gluon plasma. Z-bosons have mass, so I don't know if the analogy would be exact, but it seems reasonable.
 
That would need all week interactions with the many particles to emit the resulting wave in phase to get a wave front. "In phase" here is stringent as it relates with a wave having the speed of light. As opposed, I believe weak interaction is very slow, so its delay must vary an awful lot, meaning the phase information is lost during the interaction.

Provided what I imagine makes any sense - I'm very far here from anything I believe to know - this would prevent Cerenkov-like (not necessarily optical) radiation through the weak interaction.
 
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}...

Similar threads

Replies
7
Views
4K
Replies
4
Views
1K
Replies
13
Views
4K
Replies
2
Views
1K
Replies
6
Views
1K
Replies
4
Views
13K
Back
Top