Atmospheric electron neutrinos

[Zuzana]

Hi,

I would like to ask question about atmospheric electron neutrinos.
It is known that atmospheric electron neutrinos originate from the decay of muon in the atmosphere, but we can also calculate that muon with energy more than 10 GeV is able to penetrate about 100 km, so it does not decay and no electron neutrino is produced. It means that electron neutrinos are produced from muons with energy < ~10 GeV and thus electron neutrino posses energies in this range (< ~ 10 GeV). Does it mean that we cannot detect atmospheric electron neutrinos with higher energies?

Thanks for the reply.

 

[Astronuc]

so it does not decay and no electron neutrino is produced.

How does that follow from

we can also calculate that muon with energy more than 10 GeV is able to penetrate about 100 km,

?

 

[Zuzana]

Atmospheric muon with energy > 10 GeV can reach the Earth and can be detected, so electron neutrino won't be detected, since the muon did not decay in detector. Does it mean that only electron neutrinos, which come from decay of muon (energy < 10 GeV) can be detected at the Earth?

 

[Orodruin]

Atmospheric muon with energy > 10 GeV can reach the Earth and can be detected

Huh? The atmospheric muon being detected or not is irrelevant. It is however true that high-energy muons will start reaching the Earth. Those will then be slowed down by scattering and decay with a lower energy than their original energy. This phenomenon is well known and does result in a predominantly muon neutrino flux at high energies. At low energies the muon-to-electron neutrino ratio is about 2. At higher energies it is predominantly muons from meson decays.

since the muon did not decay in detector

It has nothing to do with the muon being detected or not. It has to do with the muons reaching the Earth's surface and slowing down or not.

Does it mean that only electron neutrinos, which come from decay of muon (energy < 10 GeV) can be detected at the Earth?

No. Even at higher energies there will be some muons that decay. Decay is not a deterministic process. It becomes a game of rates whether the resulting higher energy neutrinos can be detected or not.

 

[mfb]

Muons can decay at any point in time, as mentioned. There is also the angle to consider: If you look closer to the horizon the muons had more time to decay before slowing down in the ground.

 

[Zuzana]

Huh? The atmospheric muon being detected or not is irrelevant. It is however true that high-energy muons will start reaching the Earth. Those will then be slowed down by scattering and decay with a lower energy than their original energy. This phenomenon is well known and does result in a predominantly muon neutrino flux at high energies. At low energies the muon-to-electron neutrino ratio is about 2. At higher energies it is predominantly muons from meson decays.It has nothing to do with the muon being detected or not. It has to do with the muons reaching the Earth's surface and slowing down or not.No. Even at higher energies there will be some muons that decay. Decay is not a deterministic process. It becomes a game of rates whether the resulting higher energy neutrinos can be detected or not.

Thank you very much for the explanation.

 

[Vanadium 50]

Muons can decay at any point in time, as mentioned.

indeed, the most probable decay time is t = 0.

 

[Orodruin]

indeed, the most probable decay time is t = 0.

No time like the present …

 

[vanhees71]

The survival probability is $P_{\text{survival}}(t)=\exp(-\Gamma t)$, given that the particle exists at time $t=0$. Correspondingly the probability for being decayed at time $t$ is $P_{\text{decayed}}=1-\exp(-\Gamma t)$. It's not so clear, what you mean by the "most probable decay time". The average lifetime is $1/\Gamma$.

 

[Orodruin]

The survival probability is $P_{\text{survival}}(t)=\exp(-\Gamma t)$, given that the particle exists at time $t=0$. Correspondingly the probability for being decayed at time $t$ is $P_{\text{decayed}}=1-\exp(-\Gamma t)$. It's not so clear, what you mean by the "most probable decay time". The average lifetime is $1/\Gamma$.

Those are the cdf for decay and 1-cdf for decay. For any given time, the pdf of the time of decay given it has not yet decayed is the largest at that time. Given that the particle exists at t=0, the most probable time of decay -as given by the pdf- is just after that.

 

[vanhees71]

Ok, then you simply define it as the maximum of the probability density for decay, i.e., \dot{P}_{\text{dec}}=\Gamma \exp(-\Gamma t)$, and then it's of course at$t=0##.

 

[Orodruin]

Ok, then you simply define it as the maximum of the probability density for decay, i.e., \dot{P}_{\text{dec}}=\Gamma \exp(-\Gamma t)$, and then it's of course at$t=0##.

Well, it was v50 that said it, but that was my interpretation at least