A Pause for Reflection — Question 2

Section 44-9 · Physics pages 7454–7455

Question 2 A two-body decay can reconstruct its parent without knowing the boost Two massless daughters have laboratory energies 4 and 9 GeV with opening angle 60 degrees. Use c=1. 4 GeV 9 GeV 60◦ Schematic; labels, not drawing lengths, specify the data. (a) Calculate their parent invariant mass and laboratory Lorentz factor. (b) Find their common energy in the parent rest frame and check why laboratory energies can differ.
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Question 2 - Solution A two-body decay can reconstruct its parent without knowing the boost 4 GeV 9 GeV 60◦ (a) Method and calculation M2 = 2E1E2(1 −cos θ) = 2(4)(9)(1 −.5) = 36. M = 6 GeV, γ = (4 + 9)/6 = 13/6 . (b) Interpretation and further calculation E∗ 1 = E∗ 2 = M/2 = 3 GeV . E1,2 = γE∗(1 ± β cos θ∗), E1 + E2 = γM . Physical check Invariant mass is independent of the parent motion, whereas individual daughter energies are not.

Question 2

A two-body decay can reconstruct its parent without knowing the boost

Two massless daughters have laboratory energies 4 and 9 GeV with opening angle 60 degrees. Use c=1.

Diagram shown in the original SVG worksheet.

  1. Calculate their parent invariant mass and laboratory Lorentz factor.

  2. Find their common energy in the parent rest frame and check why laboratory energies can differ.

Question 2 - Solution

A two-body decay can reconstruct its parent without knowing the boost

Diagram shown in the original SVG worksheet.

(a) Method and calculation

M2=2E1E2(1−cos⁡θ)=2(4)(9)(1−.5)=36.M^2=2E_1E_2(1-\cos\theta)=2(4)(9)(1-.5)=36.M=6GeV,γ=(4+9)/6=13/6.\boxed{M=6\ {\mathrm{GeV}},\quad\gamma=(4+9)/6=13/6}.

(b) Interpretation and further calculation

E1*=E2*=M/2=3GeV.\boxed{E_1^*=E_2^*=M/2=3\ {\mathrm{GeV}}}.E1,2=γE*(1±βcos⁡θ*),E1+E2=γM.E_{1,2}=\gamma E^*(1\pm\beta\cos\theta^*),\quad\boxed{E_1+E_2=\gamma M}.

Physical check

Invariant mass is independent of the parent motion, whereas individual daughter energies are not.