 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/section-a/5/solution

For a proper lifetime inferred from a delayed decay flash, the measured interval ends when the light returns to the detector, not when the muon decays. Put passage at

(t, x) = (0, 0)

in the detector frame. If decay occurs at

t_{d}

, its distance is

v t_{d}

, and the backwards photon takes another

v t_{d} / c

to arrive. Therefore

t_{record} = t_{d} (1 + v / c), t_{d} = \frac{10 ns}{1 + 1/2} = \frac{20}{3} ns .

The muon's proper elapsed time is

Δ τ = t_{d} 1 - v^{2} / c^{2} = 10 ns \frac{1 - v / c}{1 + v / c} = \frac{10}{3} ns \approx 5.77 ns .

Applying time dilation directly to the 10 ns reception interval would transform the wrong pair of events.

 New to topics? Read the docs here!