Particles, Particles, Particles — Question 2

Question 2

Time of flight separates particles having the same momentum

Two singly charged particles each have $pc=1.00$ GeV and traverse a straight L=3.00 m flight path. Their rest energies are .100 and .800 GeV. Use c=$3.00\times10^8$ m/s.

1. Find each speed and flight time from the relativistic energy relation.

2. Find their arrival-time separation and compare it with a .500 ns timing resolution.

Question 2 - Solution

Time of flight separates particles having the same momentum

(a) Method and calculation

$$\beta_i=\frac{pc}{\sqrt{p^2c^2+m_i^2c^4}},\quad\boxed{\beta_1=.995037,\ \beta_2=.780869}.$$$$t_i=\frac Lc\sqrt{1+(mc/p)^2}\Rightarrow\boxed{t_1=10.0499\,\mathrm{ns},\ t_2=12.8062\,\mathrm{ns}}.$$

(b) Interpretation and further calculation

$$\boxed{\Delta t=2.75637\,\mathrm{ns}}.$$$$\boxed{\Delta t/(.500\,\mathrm{ns})=5.51274}\quad\text{several resolution units}.$$

Physical check

This comparison uses the stated single timing scale; a statistical separation significance would require the two measurement-error distributions.