Question: Čerenkov radiation is emitted by a high energy charged particle which moves through a medium with a velocity greater than the velocity of electromagnetic wave propagation in the medium.

A Čerenkov radiation particle detector is made by fitting a long pipe of one atmosphere, $20^{\circ} \mathrm{C}$ hydrogen gas with an optical system capable of detecting the emitted light and of measuring the angle of emission $\theta$ to an accuracy of $\delta \theta=10^{-3}$ radian. A beam of charged particles with momentum of $100 \mathrm{GeV} / \mathrm{c}$ are passed through the counter. Since the momentum is known, the measurement of the Čerenkov angle is, in effect, a measurement of the particle rest mass $m_{0}$. For particles with $m_{0}$ near $1 \mathrm{GeV} / \mathrm{c}^{2}$, and to first order in small quantities, what is the fractional error (i.e., $\delta m_{0} / m_{0}$ ) in the determination of $m_{0}$ with the Čerenkov counter?
Answer: As shown in Fig. 5.13, the radiation emitted by the charge at $Q^{\prime}$ at time $t^{\prime}$ arrives at $P$ at time $t$ when the charge is at Q. As the radiation propagates at the speed $c / n$ and the particle has speed $v$ where $v>c / n$, we have

$$
\mathrm{Q}^{\prime} \mathrm{P}=\frac{c}{n}\left(t-t^{\prime}\right), \quad \mathrm{Q}^{\prime} \mathrm{Q}=v\left(t-t^{\prime}\right),
$$

or

$$
\frac{\mathrm{Q}^{\prime} \mathrm{P}}{\mathrm{Q}^{\prime} \mathrm{Q}}=\cos \theta=\frac{c}{v n}=\frac{1}{\beta n},
$$

where $\beta=\frac{v}{c}$. At all the points intermediate between $Q^{\prime}$ and $Q$ the radiation emitted will arrive at the line QP at time $t$. Hence QP forms the wavefront of all radiation emitted prior to $t$.

Fig. $5.13$

As $|\cos \theta| \leq 1$, we require $\beta \geq \frac{1}{n}$ for emission of Čerenkov radiation. Hence we require

$$
\gamma=\frac{1}{\sqrt{1-\beta^{2}}} \geq\left(1-\frac{1}{n^{2}}\right)^{-\frac{1}{2}}=\frac{n}{\sqrt{n^{2}-1}} .
$$

Thus the particle must have a kinetic energy greater than

$$
\begin{aligned}
T &=(\gamma-1) m_{0} c^{2} \\
&=\left[\frac{n}{\sqrt{(n+1)(n-1)}}-1\right] m_{0} c^{2} \\
& \approx\left(\frac{1}{\sqrt{2 \times 1.35 \times 10^{-4}}}-1\right) \times 0.5 \\
& \approx 29.93 \mathrm{MeV} .
\end{aligned}
$$For a relativistic particle of momentum $P \gg m_{0} c$,

$$
\gamma=\frac{E}{m_{0} c^{2}}=\frac{\sqrt{P^{2} c^{2}+m_{0}^{2} c^{4}}}{m_{0} c^{2}} \approx \frac{P}{m_{0} c} .
$$

With $P$ fixed we have

$$
d \gamma=-\frac{P}{c} \cdot \frac{d m_{0}}{m_{0}^{2}} .
$$

Now $\beta=\frac{\gamma \beta}{\gamma}=\sqrt{\frac{\gamma^{2}-1}{\gamma^{2}}}=\sqrt{1-\frac{1}{\gamma^{2}}} \approx 1-\frac{1}{2 \gamma^{2}}$ for $\gamma \gg 1$, so that

$$
d \beta=\frac{d \gamma}{\gamma^{3}} .
$$

For the Čerenkov radiation emitted by the particle, we have

$$
\cos \theta=\frac{1}{\beta n},
$$

or

$$
d \beta=n \beta^{2} \sin \theta d \theta .
$$

Combining the above we have

$$
\left|\frac{d m_{0}}{m_{0}}\right|=\frac{m_{0} c}{P} d \gamma \approx \frac{d \gamma}{\gamma}=\gamma^{2} \beta=n \beta^{2} \gamma^{2} \sin \theta d \theta=\beta \gamma^{2} \tan \theta d \theta .
$$

With $\gamma=\frac{P c}{m_{0} c^{2}}=\frac{100}{1}=100, n=1+1.35 \times 10^{-4}$, we have

$$
\beta \approx 1-\frac{1}{2 \times 10^{4}}=1-5 \times 10^{-5},
$$

$\cos \theta=\frac{1}{\beta n}=\left(1-5 \times 10^{-5}\right)^{-1}\left(1+1.35 \times 10^{-4}\right)^{-1} \approx 1-8.5 \times 10^{-5}$, 

$$
\begin{aligned}
\tan \theta=\sqrt{\frac{1}{\cos ^{2} \theta}-1} &=\sqrt{\left(1-8.5 \times 10^{-5}\right)^{-2}-1} \\
& \approx \sqrt{1.7 \times 10^{-4}} \approx 1.3 \times 10^{-2},
\end{aligned}
$$

and hence

$$
\left|\frac{d m_{0}}{m_{0}}\right|=\left(1-5 \times 10^{-5}\right) \times 10^{4} \times 1.3 \times 10^{-2} \times 10^{-3}=0.13
$$