di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Mechanics

Two Problems in Mechanics

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

Part A.The Hidden Disk

We consider a solid wooden cylinder of radius $r_{1}$ and thickness $h_{1}$. Somewhere inside the wooden cylinder, the wood has been replaced by a metal disk of radius $r_{2}$ and thickness $h_{2}$. The metal disk is placed in such a way that its symmetry axis $B$ is parallel to the symmetry axis $S$ of the wooden cylinder, and is placed at the same distance from the top and bottom face of the wooden cylinder. We denote the distance between $S$ and $B$ by $d$. The density of wood is $\rho_{1}$, the density of the metal is $\rho_{2}>\rho_{1}$. The total mass of the wooden cylinder and the metal disk inside is $M$.

In this task, we place the wooden cylinder on the ground so that it can freely roll to the left and right. See Fig. 1 for a side view and a view from the top of the setup.

The goal of this task is to determine the size and the position of the metal disk.

In what follows, when asked to express the result in terms of known quantities, you may always assume that the following are known:

$$
r_{1}, h_{1}, \rho_{1}, \rho_{2}, M
\tag{1}
$$

The goal is to determine $r_{2}, h_{2}$ and $d$, through indirect measurements.

а)

<img_4399>

b)

<img_4400>

Figure 1: a) side view b) view from above

We denote $b$ as the distance between the centre of mass $C$ of the whole system and the symmetry axis $S$ of the wooden cylinder. In order to determine this distance, we design the following experiment: We place the wooden cylinder on a horizontal base in such a way that it is in a stable equilibrium. Let us now slowly incline the base by an angle $\Theta$ (see Fig. 2). As a result of the static friction, the wooden cylinder can roll freely without sliding. It will roll down the incline a little bit, but then come to rest in a stable equilibrium after rotating by an angle $\phi$ which we measure.


<img_4560>

Figure 2: Cylinder on an inclined base.

Question: A.1 Find an expression for $b$ as a function of the quantities (1), the angle $\phi$ and the $0.8 p t$ tilting angle $\Theta$ of the base.

Solution:
Geometric solution: use that torque with respect to point of contact is $0 \Rightarrow$ center of gravity has to be vertically above point of contact.

$$
\begin{aligned}
& \sin \phi=\frac{D}{b} \\
& \sin \Theta=\frac{D}{r_{1}}
\end{aligned}
$$

Here $D$ may be called another name. Solve this:

$$
\sin \phi=\frac{r_{1}}{b} \sin \Theta \Rightarrow b=\frac{r_{1} \sin \Theta}{\sin \phi}
$$

Final answer: $b=\frac{r_{1} \sin \Theta}{\sin \phi}$
Subfield: Mechanics


Question: A.2 Find the equation of motion for $\varphi$. Express the moment of inertia $I_{S}$ of the system around its symmetry axis $S$ in terms of $T, b$ and the known quantities (1). You may assume that we are only disturbing the equilibrium position by a small amount so that $\varphi$ is always very small.

Solution:
Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion.

Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$.


Correct equation (either energy conservation or torque equation of motion)

$$
T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}}
$$



(Derivation:

$$
\Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi
$$

so that

$$
\omega^{2}=\frac{b g M}{I_{S}}
$$

Final answer: Kinetic energy: $\frac{1}{2} I_{S} \dot{\varphi}^{2}$ and potential energy: $-b M g \cos \varphi$. Total energy is conserved, and differentiation w.r.t. time gives the equation of motion.

Angular equation of motion from torque, $\tau=I_{S} \ddot{\varphi}=-M g b \sin \varphi$.


Correct equation (either energy conservation or torque equation of motion)

$$
T=2 \pi \sqrt{\frac{I_{S}}{M g b}} \Rightarrow I_{S}=\frac{M g b T^{2}}{4 \pi^{2}}
$$



(Derivation:

$$
\Rightarrow \ddot{\varphi}=-\frac{b M g}{I_{S}} \sin \varphi \simeq-\frac{b g M}{I_{S}} \varphi
$$

so that

$$
\omega^{2}=\frac{b g M}{I_{S}}
$$
Subfield: Electromagnetism

Nonlinear Dynamics in Electric Circuits

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

Introduction

Bistable non-linear semiconducting elements (e.g. thyristors) are widely used in electronics as switches and generators of electromagnetic oscillations. The primary field of applications of thyristors is controlling alternating currents in power electronics, for instance rectification of AC current to DC at the megawatt scale. Bistable elements may also serve as model systems for self-organization phenomena in physics (this topic is covered in part B of the problem), biology (see part C) and other fields of modern nonlinear science.

Goals

To study instabilities and nontrivial dynamics of circuits including elements with non-linear $I-V$ characteristics. To discover possible applications of such circuits in engineering and in modeling of biological systems.

Part A.Stationary states and instabilities

Fig. 1 shows the so-called S-shaped $I-V$ characteristics of a non-linear element $X$. In the voltage range between $U_{\mathrm{h}}=4.00 \mathrm{~V}$ (the holding voltage) and $U_{\text {th }}=10.0 \mathrm{~V}$ (the threshold voltage) this $I-V$ characteristics is multivalued. For simplicity, the graph on Fig. 1 is chosen to be piece-wise linear (each branch is a segment of a straight line). In particular, the line in the upper branch touches the origin if it is extended. This approximation gives a good description of real thyristors.

<img_4550>

Figure 1: $I-V$ characteristics of the non-linear element $X$.

Question: A.1 Using the graph, determine the resistance $R_{\mathrm{on}}$ of the element $X$ on the upper branch of the $I-V$ characteristics, and $R_{\text {off }}$ on the lower branch, respectively. The middle branch is described by the equation

$$
I=I_{0}-\frac{U}{R_{\text {int }}}
\tag{1}
$$

Find the values of the parameters $I_{0}$ and $R_{\text {int }}$.

Solution:
By looking at the $I-V$ graph, we obtain

$$
\begin{gathered}
R_{\text {off }}=10.0 \Omega, \\
R_{\text {on }}=1.00 \Omega, \\
R_{\text {int }}=2.00 \Omega, \\
I_{0}=6.00 \mathrm{~A} .
\end{gathered}
$$



Final answer: $R_{int}=2.00 $ , $I_{0}=6.00$ $\Omega$, A
Subfield: Electromagnetism


Question: A.2 What are the possible numbers of stationary states that the circuit of Fig. 2 may have for a fixed value of $\mathcal{E}$ and for $R=3.00 \Omega$ ? How does the answer change for $R=1.00 \Omega$ ?

Solution:
Kirchoff law for the circuit ( $U$ is the voltage of the bistable element):

$$
\mathcal{E}=I R+U
$$

This yields

$$
I=\frac{\mathcal{E}-U}{R}
$$

Hence, stationary states of the circuit are intersections of the line defined by this equation and the $I-V$ graph of $X$.

For $R=3.00 \Omega$, one always gets exactly one intersection.

For $R=1.00 \Omega$, one gets 1,2 or 3 intersections depending on the value of $\mathcal{E}$.

Final answer: For $R=3.00 \Omega$, one always gets exactly one intersection.

For $R=1.00 \Omega$, one gets 1,2 or 3 intersections depending on the value of $\mathcal{E}$.
Subfield: Electromagnetism


Question: B.2 Find expressions for the times $t_{1}$ and $t_{2}$ that the system spends on each branch of the $I-V$ graph during the oscillation cycle. Determine their numerical values. Find the numerical value of the oscillation period $T$ assuming that the time needed for jumps between the branches of the $I-V$ graph is negligible.

Solution:
Since the non-linear element is oscillating between the switched on and switched off branches we can put $U_{X}=R_{\mathrm{on} / \text { off }} I_{X}$. On either of the branches, the circuit behaves as a standard RC-circuit with conductance $C$ and resistance $R_{\mathrm{on} / \mathrm{off}} R /\left(R_{\mathrm{on} / \mathrm{off}}+R\right)$ (the resistor and the element $X$ being connected in parallel). Another way to express it is to



write the Kirchhoff law for the switched on and switched off branches

$$
R_{\text {on } / \mathrm{off}} R C \frac{d I_{X}}{d t}=\mathcal{E}-\left(R_{\mathrm{on} / \mathrm{off}}+R\right) I_{X}
$$

The time constant of the circuit is

$$
\frac{R_{\text {on } / \text { off }} R}{R_{\text {on } / \text { off }}+R} C
$$

If the branch in question (switched on or switched off) extended indefinitely, after a long time the system would have landed in a stationary state with the voltage

$$
U_{\text {on } / \text { off }}=\frac{R_{\text {on } / \text { off }}}{R_{\text {on } / \text { off }}+R} \mathcal{E}
$$

Then, the time dependence of the voltage drop on the non-linear element is a sum of the constant term $U_{\text {on/off }}$ and of the exponentially decaying term:

$$
U_{X}(t)=\frac{R_{\text {on } / \text { off }}}{R_{\text {on } / \text { off }}+R} \mathcal{E}+\left(U_{\text {on } / \text { off }}-\frac{R_{\text {on } / \text { off }}}{R_{\text {on } / \text { off }}+R} \mathcal{E}\right) e^{-\frac{R_{\text {on } / \text { off }}+R}{R_{\text {on } / \text { off }}^{R C}} t}
$$

Time spent by the system on the switched on branch during one cycle:

$$
t_{\mathrm{on}}=\frac{R_{\mathrm{on}} R}{R_{\mathrm{on}}+R} C \log \left(\frac{U_{\mathrm{th}}-U_{\mathrm{on}}}{U_{\mathrm{h}}-U_{\mathrm{on}}}\right)=2.41 \cdot 10^{-6} \mathrm{~s},
$$


Time spent by the system on the switched off branch during one cycle:

$$
t_{\mathrm{off}}=\frac{R_{\mathrm{off}} R}{R_{\mathrm{off}}+R} C \log \left(\frac{U_{\mathrm{off}}-U_{\mathrm{h}}}{U_{\mathrm{off}}-U_{\mathrm{th}}}\right)=3.71 \cdot 10^{-6} s
$$

The total period of oscillations:

$$
T=t_{\mathrm{on}}+t_{\mathrm{off}}=6.12 \cdot 10^{-6} \mathrm{~s}
$$

Final answer: $t_{\mathrm{on}}=2.41 \cdot 10^{-6} \mathrm{~s}$, $t_{\mathrm{off}}=3.71 \cdot 10^{-6} \mathrm{~s}$,$T=6.12 \cdot 10^{-6} \mathrm{~s}$
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.

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.

Solution:
Conservation of energy:

$$
m_{p} \cdot c^{2}+V \cdot e=m_{p} \cdot c^{2} \cdot \gamma=\frac{m_{p} \cdot c^{2}}{\sqrt{1-v^{2} / c^{2}}}
$$

Penalties

$$
\text { No or incorrect total energy }
$$

Solve for velocity:

$$
v=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}}
$$

Final answer: $v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$
Subfield: Modern Physics


Question: 3.1

Assume that the ion which has just one electron left the shell.

$\mathrm{A}^{(\mathrm{Z}-1)+}$ is in the ground state.

In the lowest energy state, the square of the average distance of the electron from the nucleus or $\mathrm{r}^{2}$ with components along $\mathrm{x}-\mathrm{y}$ - and $\mathrm{z}$-axis being $(\Delta \mathrm{x})^{2},(\Delta \mathrm{y})^{2}$ and $(\Delta \mathrm{z})^{2}$ respectively and $\mathrm{r}_{\mathrm{0}}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2} \quad$ and also the square of the average momentum by $\mathrm{p}_{\mathrm{o}}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$, whereas $\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}}, \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}}$ and $\Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$.

Write inequality involving $\left(p_{\mathrm{0}}\right)^{2} \cdot\left(r_{\mathrm{0}}\right)^{2}$ in a complete form.

Solution:
3.1

$\mathrm{r}_{0}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}+(\Delta \mathrm{z})^{2}$

$\mathrm{p}_{0}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{z}}\right)^{2}$

since

$\Delta \mathrm{p}_{\mathrm{x}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{x}} \quad \Delta \mathrm{p}_{\mathrm{y}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{y}} \quad \Delta \mathrm{p}_{\mathrm{z}} \geq \frac{\hbar}{2 \cdot \Delta \mathrm{z}}$

gives

$\mathrm{p}_{0}^{2} \geq \frac{\hbar^{2}}{4} \cdot\left[\frac{1}{(\Delta \mathrm{x})^{2}}+\frac{1}{(\Delta \mathrm{y})^{2}}+\frac{1}{(\Delta \mathrm{z})^{2}}\right]$

and

$(\Delta \mathrm{x})^{2}=(\Delta \mathrm{y})^{2}=(\Delta \mathrm{z})^{2}=\frac{\mathrm{r}_{0}^{2}}{3}$

thus

$$
\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}
$$

Final answer: $\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2} \geq \frac{9}{4} \cdot \hbar^{2}$
Subfield: Modern Physics


Question: 3.2 

The ion represented by $\mathrm{A}^{(\mathrm{Z}-1)+}$ may capture an additional electron and consequently emits a photon.

Write down an equation which is to be used for calculation the frequency of an emitted photon.

Solution:
3.2

$\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture

$\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing

$\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing

$\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon

conservation of energy:

$\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$

where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively.

conservation of momentum:

$m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$

where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.

Final answer: 3.2

$\left|\vec{v}_{\mathrm{e}}\right| \ldots . .$. speed of the external electron before the capture

$\left|\vec{V}_{i}\right| \ldots . .$. speed of $A^{(Z-1)+}$ before capturing

$\left|\vec{V}_{\mathrm{f}}\right| \ldots . .$. speed of $\mathrm{A}^{(\mathrm{Z}-1)+}$ after capturing

$\mathrm{E}_{\mathrm{n}}=\mathrm{h} . v \quad \ldots . .$. energy of the emitted photon

conservation of energy:

$\frac{1}{2} \cdot m_{e} \cdot V_{e}^{2}+\frac{1}{2} \cdot\left(M+m_{e}\right) \cdot V_{i}^{2}+E\left[A^{(Z-1)+}\right]=\frac{1}{2} \cdot\left(M+2 \cdot m_{e}\right) \cdot V_{f}^{2}+E\left[A^{(Z-2)+}\right]$

where $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-1)+}\right)$ and $\mathrm{E}\left[\mathrm{A}^{(\mathrm{Z}-2)+}\right]$ denotes the energy of the electron in the outermost shell of ions $\mathrm{A}^{(\mathrm{Z}-1)+}$ and $\mathrm{A}^{(\mathrm{Z}-2)+}$ respectively.

conservation of momentum:

$m_{e} \cdot \vec{v}_{e}+(M+m) \cdot \vec{v}_{i}=\left(M+2 \cdot m_{e}\right) \cdot \vec{v}_{f}+\frac{h \cdot v}{c} \cdot \overrightarrow{1}$

where $\overrightarrow{1}$ is the unit vector pointing in the direction of the motion of the emitted photon.
Subfield: Mechanics

EVOLUTION OF THE EARTH-MOON SYSTEM 

Scientists can determine the distance Earth-Moon with great precision. They achieve this by bouncing a laser beam on special mirrors deposited on the Moon's surface by astronauts in 1969, and measuring the round travel time of the light (see Figure 1).

<img_4459>

Figure 1. A laser beam sent from an observatory is used to measure accurately the distance between the Earth and the Moon.

With these observations, they have directly measured that the Moon is slowly receding from the Earth. That is, the Earth-Moon distance is increasing with time. This is happening because due to tidal torques the Earth is transferring angular momentum to the Moon, see Figure 2. In this problem you will derive the basic parameters of the phenomenon.



<img_4430>

Figure 2. The Moon's gravity produces tidal deformations or "bulges" in the Earth. Because of the Earth's rotation, the line that goes through the bulges is not aligned with the line between the Earth and the Moon. This misalignment produces a torque that transfers angular momentum from the Earth's rotation to the Moon's translation. The drawing is not to scale.

1. Conservation of Angular Momentum. 

Let $L_{1}$ be the present total angular momentum of the Earth-Moon system. Now, make the following assumptions: i) $L_{1}$ is the sum of the rotation of the Earth around its axis and the translation of the Moon in its orbit around the Earth only. ii) The Moon's orbit is circular and the Moon can be taken as a point. iii) The Earth's axis of rotation and the Moon's axis of revolution are parallel. iv) To simplify the calculations, we take the motion to be around the center of the Earth and not the center of mass. Throughout the problem, all moments of inertia, torques and angular momenta are defined around the axis of the Earth. v) Ignore the influence of the Sun.

Question: 1a Write down the equation for the present total angular momentum of the Earth-Moon system. Set this equation in terms of $I_{E}$, the moment of inertia of the Earth; $a_{E 1}$, the present angular frequency of the Earth's rotation; $I_{M 1}$, the present moment of inertia of the Moon with respect to the Earth's axis; and $a_{M 1}$, the present angular frequency of the Moon's orbit.

Solution:
$L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$

Final answer: $L_{1}=I_{E} \omega_{E 1}+I_{M 1} \omega_{M 1}$
Subfield: Mechanics


Question: 2h Find the final angular frequency $\mathrm{a}_{2}$ in s $^{-1}$, as well as the final duration of the day in units of present days.

Solution:
$\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days

Final answer: $\omega_{2}=1.6 \times 10^{-6} \mathrm{~s}^{-1}$, that is, a period of 46 days
Subfield: Modern Physics

THEORETICAL PROBLEM 2 

DOPPLER LASER COOLING AND OPTICAL MOLASSES

The purpose of this problem is to develop a simple theory to understand the so-called "laser cooling" and "optical molasses" phenomena. This refers to the cooling of a beam of neutral atoms, typically alkaline, by counterpropagating laser beams with the same frequency. This is part of the Physics Nobel Prize awarded to S. Chu, P. Phillips and C. Cohen-Tannoudji in 1997.

<img_4354>

The image above shows sodium atoms (the bright spot in the center) trapped at the intersection of three orthogonal pairs of opposing laser beams. The trapping region is called "optical molasses" because the dissipative optical force resembles the viscous drag on a body moving through molasses.

In this problem you will analyze the basic phenomenon of the interaction between a photon incident on an atom and the basis of the dissipative mechanism in one dimension.

PART I: BASICS OF LASER COOLING

Consider an atom of mass $m$ moving in the $+x$ direction with velocity $v$. For simplicity, we shall consider the problem to be one-dimensional, namely, we shall ignore the $y$ and $z$ directions (see figure 1). The atom has two internal energy levels. The energy of the lowest state is considered to be zero and the energy of the excited state to be $\hbar a_{0}$, where $\hbar=h / 2 \pi$. The atom is initially in the lowest state. A laser beam with frequency $a_{L}$ in the laboratory is directed in the $-x$ direction and it is incident on the atom. Quantum mechanically the laser is composed of a large number of photons, each with energy $\hbar \boldsymbol{c}_{L}$ and momentum $-\hbar q$. A photon can be absorbed by the atom and later spontaneously emitted; this emission can occur with equal probabilities along the $+x$ and $-x$ directions. Since the atom moves at non-relativistic speeds, $v / c \ll 1$ (with $c$ the speed of light) keep terms up to first order in this quantity only. Consider also $\hbar q / m v \ll 1$, namely, that the momentum of the atom is much larger than the



momentum of a single photon. In writing your answers, keep only corrections linear in either of the above quantities.

<img_4516>

internal energy levels of the atom

<img_4343>

Fig. 1 Sketch of an atom of mass $m$ with velocity $v$ in the $+x$ direction, colliding with a photon with energy $\hbar c_{L}$ and momentum $-\hbar q$. The atom has two internal states with energy difference $\hbar \boldsymbol{c}_{0}$.

Assume that the laser frequency $a_{L}$ is tuned such that, as seen by the moving atom, it is in resonance with the internal transition of the atom. Answer the following questions:

1. Absorption.

Question: 1a Write down the resonance condition for the absorption of the photon.

Solution:
$\omega_{0} \approx \omega_{L}\left(1+\frac{v}{C}\right)$

Final answer: $\omega_{L}\left(1+\frac{v}{C}\right)$