Subfield: Optics

CHERENKOV LIGHT AND RING IMAGING COUNTER 

Light propagates in vacuum with the speed $c$. There is no particle which moves with a speed higher than $c$. However, it is possible that in a transparent medium a particle moves with a speed $v$ higher than the speed of the light in the same medium $\frac{c}{n}$, where $n$ is the refraction index of the medium. Experiment (Cherenkov, 1934) and theory (Tamm and Frank, 1937) showed that a charged particle, moving with a speed $v$ in a transparent medium with refractive index $n$ such that $v>\frac{C}{n}$, radiates light, called Cherenkov light, in directions forming with the trajectory an angle

$$
\theta=\arccos \frac{1}{\beta n}
\tag{1}
$$

where $\beta=\frac{v}{c}$.

<img_4441>

1. To establish this fact, consider a particle moving at constant velocity $v>\frac{C}{n}$ on a straight line. It passes $\mathrm{A}$ at time 0 and $\mathrm{B}$ at time $t_{1}$. As the problem is symmetric with respect to rotations around $\mathrm{AB}$, it is sufficient to consider light rays in a plane containing $\mathrm{AB}$.

At any point $\mathrm{C}$ between $\mathrm{A}$ and $\mathrm{B}$, the particle emits a spherical light wave, which propagates with velocity $\frac{c}{n}$. We define the wave front at a given time $t$ as the envelope of all these spheres at this time.
Context question:
1.1. Determine the wave front at time $t_{1}$ and draw its intersection with a plane containing the trajectory of the particle.
Context answer:
<img_4447>


Context question:
1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$.
Context answer:
\boxed{$\varphi=\arcsin \frac{1}{\beta n}$}


Context question:
2. Let us consider a beam of particles moving with velocity $v>\frac{C}{n}$, such that the angle $\theta$ is small, along a straight line IS. The beam crosses a concave spherical mirror of focal length $f$ and center $\mathrm{C}$, at point $\mathrm{S}$. SC makes with SI a small angle $\alpha$ (see the figure in the Answer Sheet). The particle beam creates a ring image in the focal plane of the mirror.


Explain why with the help of a sketch illustrating this fact. Give the position of the center $\mathrm{O}$ and the radius $r$ of the ring image.

This set up is used in ring imaging Cherenkov counters (RICH) and the medium which the particle traverses is called the radiator.

Note: in all questions of the present problem, terms of second order and higher in $\alpha$ and $\theta$ will be neglected.
Context answer:
<img_4436>


Extra Supplementary Reading Materials:

3. A beam of particles of known momentum $p=10.0 \mathrm{GeV} / \mathrm{c}$ consists of three types of particles: protons, kaons and pions, with rest mass $M_{\mathrm{p}}=0.94 \mathrm{GeV} / c^{2}$, $M_{\kappa}=0.50 \mathrm{GeV} / c^{2}$ and $M_{\pi}=0.14 \mathrm{GeV} / c^{2}$, respectively. Remember that $p c$ and $M c^{2}$ have the dimension of an energy, and $1 \mathrm{eV}$ is the energy acquired by an electron after being accelerated by a voltage $1 \mathrm{~V}$, and $1 \mathrm{GeV}=10^{9} \mathrm{eV}, 1 \mathrm{MeV}=10^{6} \mathrm{eV}$.

The particle beam traverses an air medium (the radiator) under the pressure $P$. The refraction index of air depends on the air pressure $P$ according to the relation $n=1+a P$ where $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$
Context question:
3.1. Calculate for each of the three particle types the minimal value $P_{\min }$ of the air pressure such that they emit Cherenkov light.
Context answer:
$$
\begin{array}{ll}
P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\
P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\
P_{\min }=0.36 \mathrm{~atm} & \text { for pions. }
\end{array}
$$

Question: 3.2. Calculate the pressure $P_{\frac{1}{2}}$ such that the ring image of kaons has a radius equal to one half of that corresponding to pions. Calculate the values of $\theta_{\kappa}$ and $\theta_{\pi}$ in this case.

Is it possible to observe the ring image of protons under this pressure?

Solution:
For $\theta_{\pi}=2 \theta_{\kappa} \quad$ we have

$$
\cos \theta_{\pi}=\cos 2 \theta_{\kappa}=2 \cos ^{2} \theta_{\kappa}-1
\tag{5}
$$

We denote

$$
\varepsilon=1-\beta=1-\frac{1}{\sqrt{1+K^{2}}} \approx \frac{1}{2} K^{2}
\tag{6}
$$

From (5) we obtain

$$
\frac{1}{\beta_{\pi} n}=\frac{2}{\beta_{\kappa}^{2} n^{2}}-1
\tag{7}
$$

Substituting $\beta=1-\varepsilon$ and $n=1+\zeta$ into (7), we get approximately:

$$
\begin{aligned}
& \zeta_{\frac{1}{2}}=\frac{4 \varepsilon_{\kappa}-\varepsilon_{\pi}}{3}=\frac{1}{6}\left(4 K_{\kappa}^{2}-K_{\pi}^{2}\right)=\frac{1}{6}\left[4 .(0.05)^{2}-(0.014)^{2}\right], \\
& P_{\frac{1}{2}}=\frac{1}{2.7 \times 10^{-4}} \zeta_{\frac{1}{2}}=6 \mathrm{~atm} .
\end{aligned}
$$

The corresponding value of refraction index is $n=1.00162$. We get:

$$
\theta_{\kappa}=1.6^{\circ} ; \quad \theta_{\pi}=2 \theta_{\kappa}=3.2^{\circ} \text {. }
$$

We do not observe the ring image of protons since

$$
P_{\frac{1}{2}}=6 \mathrm{~atm}<16 \mathrm{~atm}=P_{\min } \text { for protons. }
$$



Final answer: $P_{\frac{1}{2}}=6 $ , $\theta_{\kappa}=1.6$ , $\theta_{\pi}=3.2$ $\mathrm{~atm}$, $^{\circ}$, $^{\circ}$