In a couple weeks, I will conduct a lab experiment where I measure the lifetime of the muons from the secondary cosmic radiation. For that, we have two detectors above each other, one will give a start signal, the other will give a stop signal, assuming the muon came to rest in the second detector and decayed.

There is a preparation question that says:

Muons as well as anti muons arrive at the surface of earth. Which nuclear process is possible for muons but not for anti muons? How does this qualitatively affect the measured lifetime?

As far as I know, the decay of the muon goes like this: $$ \mu^- \to \nu_\mu + \bar \nu_e + \mathrm e^-$$ And for the anti muon, it goes like this: $$ \mu^+ \to \bar \nu_\mu + \nu_e + \mathrm e^+$$

The Wikipedia page says:

The mean lifetime of the (positive) muon is 2.1969811±0.0000022 µs.[1] The equality of the muon and antimuon lifetimes has been established to better than one part in 104.

What is that difference that they are asking about in the lab course manual?


If you consider a muon decay in vacuum then there is no difference between the lifetimes of muons and antimuons.

However a muon can interact with a proton via the weak force to form a neutron, while an antimuon cannot. Since the air is full of protons this means the muon lifetime in air is slightly shorter than the antimuon lifetime. If the air were full of antiprotons instead of protons the effect would be the other way round.

  • $\begingroup$ Is that $\mu + \mathrm p \to \mathrm n + \bar\nu_\mu + \nu_\text e$? $\endgroup$ – Martin Ueding Jun 22 '14 at 10:58
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    $\begingroup$ @queueoverflow Almost, but that doesn't conserve lepton flavor. It'd be $\mu^- + \mathrm p \to \mathrm n + \nu_\mu$. $\endgroup$ – rob Jun 22 '14 at 11:32
  • $\begingroup$ I see that when I use $\mathrm W^- \to \mu^- + \bar\nu_\mu$ and isolate $\mu^-$ on one side. Then it is $\mu^- + \mathrm p \to \mathrm W^- + \nu_\mu + \mathrm p \to \mathrm n + \nu_\mu$. Okay. $\endgroup$ – Martin Ueding Jun 22 '14 at 11:40
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    $\begingroup$ @queueoverflow: it's the muonic version of electron capture. $\endgroup$ – John Rennie Jun 22 '14 at 11:42

The flux of negative muons at the surface of the earth can be smaller, than the flux of positive muons, because of muon capture by proton (which typically follows after the formation of an hydrogen-like atom with the $\mu^-$ in place of the electron). Such a reduction of the flux, if not taken into account properly, could be misinterpreted as shorter life time compared to the one of $\mu^+$. See e.g. This link http://www2.fisica.unlp.edu.ar/~veiga/experiments.html


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