The Leptons — Question 3

Section 44-3 · Physics pages 7336–7337

Question 3 The muon-decay endpoint follows from the smallest missing invariant mass A muon at rest decays to an electron and two effectively massless neutrinos. Use mµc2 = 105.66 MeV and mec2 = .511 MeV. Treat the neutrino pair as a composite system of invariant mass Mνν. µ− e− ¯νe + νµ Schematic; labels, not drawing lengths, specify the data. (a) Derive electron total energy in terms of Mνν and find its maximum. (b) Find the maximum electron kinetic energy and explain why the electron generally has a continuous energy spectrum.
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Question 3 - Solution The muon-decay endpoint follows from the smallest missing invariant mass µ− e− ¯νe + νµ (a) Method and calculation M2 ννc4 = (mµc2 −Ee)2 −p2 ec2 = m2 µc4 + m2 ec4 −2mµc2Ee. Ee,max = m2 µ + m2 e 2mµ c2 = 52.83124 MeV (Mνν = 0). (b) Interpretation and further calculation Ke,max = Ee,max −mec2 = 52.32024 MeV . Mνν varies ⇒Ee varies continuously between allowed limits. Physical check The endpoint is a limiting collinear-neutrino configuration. A two-body decay would instead fix the electron energy in the parent rest frame.

Question 3

The muon-decay endpoint follows from the smallest missing invariant mass

A muon at rest decays to an electron and two effectively massless neutrinos. Use mμc2=105.66m_\mu c^2=105.66 MeV and mec2=.511m_ec^2=.511 MeV. Treat the neutrino pair as a composite system of invariant mass MννM_{\nu\nu}.

Diagram shown in the original SVG worksheet.

  1. Derive electron total energy in terms of MννM_{\nu\nu} and find its maximum.

  2. Find the maximum electron kinetic energy and explain why the electron generally has a continuous energy spectrum.

Question 3 - Solution

The muon-decay endpoint follows from the smallest missing invariant mass

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Mνν2c4=(mμc2−Ee)2−pe2c2=mμ2c4+me2c4−2mμc2Ee.M_{\nu\nu}^2c^4=(m_\mu c^2-E_e)^2-p_e^2c^2=m_\mu^2c^4+m_e^2c^4-2m_\mu c^2E_e.Ee,max=mμ2+me22mμc2=52.83124MeV(Mνν=0).\boxed{E_{e,\max}=\frac{m_\mu^2+m_e^2}{2m_\mu}c^2=52.83124\,\mathrm{MeV}}\quad(M_{\nu\nu}=0).

(b) Interpretation and further calculation

Ke,max=Ee,max−mec2=52.32024MeV.\boxed{K_{e,\max}=E_{e,\max}-m_ec^2=52.32024\,\mathrm{MeV}}.Mνν varies⇒Ee varies continuously between allowed limits.M_{\nu\nu}\text{ varies}\Rightarrow E_e\text{ varies continuously between allowed limits}.

Physical check

The endpoint is a limiting collinear-neutrino configuration. A two-body decay would instead fix the electron energy in the parent rest frame.