Subfield: Modern Physics

Large Hadron Collider

Please read the general instructions in the separate envelope before you start this problem.

In this task, the physics of the particle accelerator LHC (Large Hadron Collider) at CERN is discussed. CERN is the world's largest particle physics laboratory. Its main goal is to get insight into the fundamental laws of nature. Two beams of particles are accelerated to high energies, guided around the accelerator ring by a strong magnetic field and then made to collide with each other. The protons are not spread uniformly around the circumference of the accelerator, but they are clustered in so-called bunches. The resulting particles generated by collisions are observed with large detectors. Some parameters of the LHC can be found in table 1 .

| LHC ring |  |
| :--- | :--- |
| Circumference of ring | $26659 \mathrm{~m}$ |
| Number of bunches per proton beam | 2808 |
| Number of protons per bunch | $1.15 \times 10^{11}$ |
| Proton beams |  |
| Energy of protons | $7.00 \mathrm{TeV}$ |
| Centre of mass energy | $14.0 \mathrm{TeV}$ |

Table 1: Typical numerical values of relevant LHC parameters.

Particle physicists use convenient units for the energy, momentum and mass: The energy is measured in electron volts $[\mathrm{eV}]$. By definition, $1 \mathrm{eV}$ is the amount of energy gained by a particle with elementary charge, e, moved through a potential difference of one volt $\left(1 \mathrm{eV}=1.602 \cdot 10^{-19} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\right)$.

The momentum is measured in units of $\mathrm{eV} / c$ and the mass in units of $\mathrm{eV} / c^{2}$, where $c$ is the speed of light in vacuum. Since $1 \mathrm{eV}$ is a very small quantity of energy, particle physicists often use $\mathrm{MeV}\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right)$, $\mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)$ or $\mathrm{TeV}\left(1 \mathrm{TeV}=10^{12} \mathrm{eV}\right)$.

Part A deals with the acceleration of protons or electrons. Part B is concerned with the identification of particles produced in the collisions at CERN.

Part A.LHC accelerator

Acceleration:

Assume that the protons have been accelerated by a voltage $V$ such that their velocity is very close to the speed of light and neglect any energy loss due to radiation or collisions with other particles.
Context question:
A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.
Context answer:
\boxed{$v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$}


Extra Supplementary Reading Materials:

A design for a future experiment at CERN plans to use the protons from the LHC and to collide them with electrons which have an energy of $60.0 \mathrm{GeV}$.
Context question:
A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.
Context answer:
\boxed{$\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$}


Extra Supplementary Reading Materials:

We now return to the protons in the LHC. Assume that the beam pipe has a circular shape.

Question: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.

Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.

Solution:
Balance of forces:

$$
\frac{\gamma \cdot m_{p} \cdot v^{2}}{r}=\frac{m_{p} \cdot v^{2}}{r \cdot \sqrt{1-\frac{v^{2}}{c^{2}}}}=e \cdot v \cdot B
$$

In case of a mistake, partial points can be given for intermediate steps (up to max 0.2). Examples:

$$
\begin{array}{ll}
\text { Example: } & \text { Lorentz force } \\
\text { Example: } & \frac{\gamma \cdot m_{p} \cdot v^{2}}{r}
\end{array}
$$

Energy:

$$
E=(\gamma-1) \cdot m_{p} \cdot c^{2} \simeq \gamma \cdot m_{p} \cdot c^{2} \rightarrow \gamma=\frac{E}{m_{p} c^{2}}
$$

Therefore:

$$
\frac{E \cdot v}{c^{2} \cdot r}=e \cdot B
$$

With

$$
v \simeq c \text { and } r=\frac{L}{2 \pi}
$$

follows:

$$
\rightarrow B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}
$$

Solution:

$$
B=5.50 \mathrm{~T}
$$

Final answer: $B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$ ,$\mathrm{~T}$