Particles, Particles, Particles — Question 3

Question 3

A simplified ionization signal can leave a charge-speed ambiguity

Over a restricted speed interval a detector response is modeled as $S=Cz^2/\beta^2$, where ze is charge and C is a calibration constant. Track A has z=1, $\beta=.400$; track B has z=2, $\beta=.800$.

1. Compare their ionization signals.

2. If both tracks also have equal measured rigidity $p/|q|$, infer their momentum and rest-mass ratios.

Question 3 - Solution

A simplified ionization signal can leave a charge-speed ambiguity

(a) Method and calculation

$$\boxed{S_A/C=1/.400^2=6.25,\quad S_B/C=4/.800^2=6.25}.$$$$p_B/(2e)=p_A/e\Rightarrow\boxed{p_B/p_A=2}.$$

(b) Interpretation and further calculation

$$m=\frac{p}{\gamma\beta c}\Rightarrow\frac{m_B}{m_A}=2\frac{\gamma_A\beta_A}{\gamma_B\beta_B}.$$$$\boxed{m_B/m_A=\sqrt{\frac{1-.800^2}{1-.400^2}}=.654654}.$$

Physical check

The response law is explicitly local, not a full stopping-power formula. Equal ionization and rigidity need not imply equal mass if charge is unknown.