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 $a=2.7 \times 10^{-4} \mathrm{~atm}^{-1}$

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.

Solution:
For the Cherenkov effect to occur it is necessary that $n>\frac{C}{v}$, that is $n_{\min }=\frac{C}{v}$.

Putting $\zeta=n-1=2.7 \times 10^{-4} P$, we get

$$
\zeta_{\min }=2.7 \times 10^{-4} P_{\min }=\frac{C}{v}-1=\frac{1}{\beta}-1
\tag{1}
$$

Because

$$
\frac{M c^{2}}{p c}=\frac{M c}{p}=\frac{M c}{\frac{M v}{\sqrt{1-\beta^{2}}}}=\frac{\sqrt{1-\beta^{2}}}{\beta}=K
\tag{2}
$$

then $K=0.094 ; 0.05 ; 0.014$ for proton, kaon and pion, respectively.

From (2) we can express $\beta$ through $K$ as

$$
\beta=\frac{1}{\sqrt{1+K^{2}}}
\tag{3}
$$

Since $K^{2}<<1$ for all three kinds of particles we can neglect the terms of order higher than 2 in $K$. We get

$$
1-\beta=1-\frac{1}{\sqrt{1+K^{2}}} \approx \frac{1}{2} K^{2}=\frac{1}{2}\left(\frac{M c}{p}\right)^{2}
\tag{3a}
$$
$$
\frac{1}{\beta}-1=\sqrt{1+K^{2}}-1 \approx \frac{1}{2} K^{2}=\frac{1}{2}\left(\frac{M c}{p}\right)^{2}
\tag{3b}
$$

Putting (3b) into (1), we obtain



$$
P_{\min }=\frac{1}{2.7 \times 10^{-4}} \times \frac{1}{2} K^{2}
\tag{4}
$$

We get the following numerical values of the minimal pressure:

$$
\begin{array}{ll}
P_{\min }=16 \mathrm{~atm} & \text { for protons, } \\
P_{\min }=4.6 \mathrm{~atm} & \text { for kaons, } \\
P_{\min }=0.36 \mathrm{~atm} & \text { for pions. }
\end{array}
$$



Final answer: $16, 4.6, 0.36$ \mathrm{~atm}