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When a muon decays from rest, typically what fraction of the energy is carried off by the electron? I tried looking into some papers, but I wasn't sure how to interpret the graphs they displayed.

Sesquipedal
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2 Answers2

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There are two neutrinos involved, so there is a spectrum. Both the maximum and the minimum are easy to get.

The most energetic case involves the two neutrinos being emitted in the same direction and the electron recoiling. By treating the neutrinos as massless (an acceptable approximation) we get

$$ pc + \sqrt{m_e^2 c^4 + p^2 c^2} = m_\mu c^2 \,,$$

which can be solved for $p$ and the electron energy found directly.

The minimum would be $\frac{m_e}{m_\mu}$. The two neutrinos are emitted back-to-back and the electron is left at rest.

dmckee --- ex-moderator kitten
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  • Thanks very much. Is there a way to estimate the probability associated with a given outcome (i.e. electron momentum)? – Sesquipedal Dec 11 '13 at 00:59
  • Certainly. To a first approximation it is proportional to the total phasespace. See my answers http://physics.stackexchange.com/a/11671/520, http://physics.stackexchange.com/a/31517/520 and all the usual texts. – dmckee --- ex-moderator kitten Dec 11 '13 at 02:44
  • I should clarify: ultimately what I'm interested in is how much of a muon's mass energy will end up being distributed thermally if the muon decays within a material. I assume that the energy carried off by the neutrinos essentially just disappears for this purpose, which is why I'm interested in the typical electron energy. Naively I would guess that this would be about 1/3 or 1/2 of the muon's rest energy, but the pitfall I want to avoid is a situation where the electron has a high probability of emerging almost stationary. Unfortunately, my background is insufficient to compute this. – Sesquipedal Dec 11 '13 at 03:58
  • A Monte Carlo integral is almost certainly easier that a closed form calculation. I'd throw the neutrino energies and angles randomly and weight the event by the total phase-space. Quick and dirty but on modern machines more than fast enough. – dmckee --- ex-moderator kitten Dec 11 '13 at 04:03
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A (somewhat idealized plot) is shown on page 6 of this paper. In a three body decay, the energy peaks close to the maximum due to phase space. The maximum is $53$ MeV and the peak looks to be around $45-48$ MeV

Ross Millikan
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