di-zhang-fdu/OlympiadExtended huggingface dataset:
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.
Context question:
1a Write down the resonance condition for the absorption of the photon.
Context answer:
\boxed{$\omega_{L}\left(1+\frac{v}{C}\right)$}


Question: 1b Write down the momentum $p_{a t}$ of the atom after absorption, as seen in the laboratory.

Solution:
$p_{a t}=p-\hbar q \approx m v-\frac{\hbar \omega_{L}}{c}$

Final answer: $m v-\frac{\hbar \omega_{L}}{c}$
Subfield: Modern Physics

WHY ARE STARS SO LARGE?

The stars are spheres of hot gas. Most of them shine because they are fusing hydrogen into helium in their central parts. In this problem we use concepts of both classical and quantum mechanics, as well as of electrostatics and thermodynamics, to understand why stars have to be big enough to achieve this fusion process and also derive what would be the mass and radius of the smallest star that can fuse hydrogen.

<img_4491>

Figure 1. Our Sun, as most stars, shines as a result of thermonuclear fusion of hydrogen into helium in its central parts.

$$
\begin{aligned}
& \text { USEFUL CONSTANTS } \\
& \text { Gravitational constant }=G=6.7 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{2} \\
& \text { Boltzmann's constant }=k=1.4 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1} \\
& \text { Planck's constant }=h=6.6 \times 10^{-34} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-1} \\
& \text { Mass of the proton }=m_{p}=1.7 \times 10^{-27} \mathrm{~kg} \\
& \text { Mass of the electron }=m_{e}=9.1 \times 10^{-31} \mathrm{~kg} \\
& \text { Unit of electric charge }=q=1.6 \times 10^{-19} \mathrm{C} \\
& \text { Electric constant (vacuum permittivity) }=\varepsilon_{0}=8.9 \times 10^{-12} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2} \\
& \text { Radius of the Sun }=R_{S}=7.0 \times 10^{8} \mathrm{~m} \\
& \text { Mass of the Sun }=M_{S}=2.0 \times 10^{30} \mathrm{~kg}
\end{aligned}
$$



1. A classical estimate of the temperature at the center of the stars. 

Assume that the gas that forms the star is pure ionized hydrogen (electrons and protons in equal amounts), and that it behaves like an ideal gas. From the classical point of view, to fuse two protons, they need to get as close as $10^{-15} \mathrm{~m}$ for the short range strong nuclear force, which is attractive, to become dominant. However, to bring them together they have to overcome first the repulsive action of Coulomb's force. Assume classically that the two protons (taken to be point sources) are moving in an antiparallel way, each with velocity $v_{r m s}$, the root-mean-square (rms) velocity of the protons, in a onedimensional frontal collision.

Question: 1a What has to be the temperature of the gas, $T_{c}$, so that the distance of closest approach of the protons, $d_{c}$, equals $10^{-15} \mathrm{~m}$ ? Give this and all numerical values in this problem up to two significant figures.

Solution:
We equate the initial kinetic energy of the two protons to the electric potential energy at the distance of closest approach:

$2\left(\frac{1}{2} m_{p} v_{r m s}^{2}\right)=\frac{q^{2}}{4 \pi \varepsilon_{0} d_{c}} ;$ and since

$\frac{3}{2} k T_{c}=\frac{1}{2} m_{p} v_{r m s}^{2}$, we obtain

$T_{c}=\frac{q^{2}}{12 \pi \varepsilon_{0} d_{c} k}=5.5 \times 10^{9} \mathrm{~K}$

Final answer: $5.5 \times 10^{9}$ K
Subfield: Modern Physics


Question: 5c Use the $d_{e} \geq \frac{\lambda_{e}}{2^{1 / 2}}$ condition to write down an equation for the radius of <br> the smallest normal star possible. Take the temperature at the center of the <br> star as typical for all the stellar interior.

Solution:
We assume that <br> $d_{e} \geq \frac{\lambda_{e}}{2^{1 / 2}} \cdot$ Since <br> $\lambda_{e}=\frac{h}{m_{e} v_{r m s}(\text { electron })}$, <br> $\frac{3}{2} k T_{c}=\frac{1}{2} m_{e} v_{r m s}^{2}($ electron $)$, <br> $T_{c}=\frac{q^{4} m_{p}}{24 \pi^{2} \varepsilon_{0}^{2} k h^{2}}$, <br> $\frac{M}{R}=\frac{q^{4}}{12 \pi^{2} \varepsilon_{0}^{2} G h^{2}}$, and <br> $d_{e}=\left(\frac{M}{(4 / 3) \pi R^{3} m_{p}}\right)^{-1 / 3}$, <br> we get that <br> $R \geq \frac{\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$

Final answer: $R \geq \frac{\varepsilon_{o}^{1 / 2} h^{2}}{4^{1 / 4} q m_{e}^{3 / 4} m_{p}^{5 / 4} G^{1 / 2}}$
Subfield: Mechanics

Zero-length springs and slinky coils

A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$.

<img_4535>

Figure 1: the relation between the force $F$ and the spring length $L$

A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as "slinky" may be (but not necessarily) such a ZLS.

Part A: Statics

Question: A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring?

Solution:
The force $F$ causes the spring to change its length from $L_{0}$ to $L$. Since equal parts of the spring are extended to equal lengths, we get: $\frac{\Delta y}{\Delta l}=\frac{L}{L_{0}} \rightarrow \Delta y=\frac{L}{L_{0}} \Delta l$.

Since $L=\max \left\{\frac{F}{k}, L_{0}\right\}$, we get $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$. From this result we see that any piece of length $\Delta l$ the spring behaves as a ZLS with spring constant $k^{*}=k \frac{L_{0}}{\Delta l}$.



Final answer: $\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$
Subfield: Mechanics

Zero-length springs and slinky coils

A zero effective length spring (ZLS) is a spring for which the force is proportional to the spring's length, $F=k L$ for $L>L_{0}$ where $L_{0}$ is the minimal length of the spring as well as its unstretched length. Figure 1 shows the relation between the force $F$ and the spring length $L$ for a ZLS, where the slope of the line is the spring constant $k$.

<img_4535>

Figure 1: the relation between the force $F$ and the spring length $L$

A ZLS is useful in seismography and allows very accurate measurement of changes in the gravitational acceleration $g$. Here, we shall consider a homogenous ZLS, whose weight $M g$ exceeds $k L_{0}$. We define a corresponding dimensionless ratio, $\alpha=k L_{0} / M g<1$ to characterize the relative softness of the spring. The toy known as "slinky" may be (but not necessarily) such a ZLS.

Part A: Statics
Context question:
A.1 Consider a segment of length $\Delta \ell$ of the unstretched ZLS spring which is then stretched by a force $F$, under weightless conditions. What is the length $\Delta y$ of this segment as a function of $F, \Delta \ell$ and the parameters of the spring?
Context answer:
$\Delta y=\max \left\{\frac{F}{k L_{0}} \Delta l, \Delta l\right\}$ , $k^{*}=k \frac{L_{0}}{\Delta l}$


Question: A.2 For a segment of length $\Delta \ell$, calculate the work $\Delta W$ required to stretch it from its original length $\Delta \ell$ to a length $\Delta y$.

Solution:
Let us compute the work of the force. From Task A.1: $d W=F(x) d x=\frac{k L_{0}}{\Delta l} x d x$.

Hence, $\Delta W=\int_{\Delta l}^{\Delta y} \frac{k L_{0}}{\Delta l} x d x=\left.\frac{k L_{0}}{\Delta l} \frac{x^{2}}{2}\right|_{\Delta l} ^{\Delta y}=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$.

Final answer: $\Delta W=\frac{k L_{0}}{2 \Delta l}\left(\Delta y^{2}-\Delta l^{2}\right)$
Subfield: Optics

Thermoacoustic Engine

A thermoacoustic engine is a device that converts heat into acoustic power, or sound waves - a form of mechanical work. Like many other heat machines, it can be operated in reverse to become a refrigerator, using sound to pump heat from a cold to a hot reservoir. The high operating frequencies reduce heat conduction and eliminate the need for any working chamber confinement. Unlike many other engine types, the thermoacoustic engine has no moving parts except the working fluid itself.

The efficiencies of thermoacoustic machines are typically lower than other engine types, but they have advantages in set up and maintenance costs. This creates opportunities for renewable energy applications, such as solar-thermal power plants and utilization of waste heat. Our analysis will focus on the creation of acoustic energy within the system, ignoring the extraction or conversion for powering external devices.

Part A: Sound wave in a closed tube

Consider a thermally insulating tube of length $L$ and cross-sectional area $S$, whose axis lies along the $x$ direction. The two ends of the tube are located at $x=0$ and $x=L$. The tube is filled with an ideal gas and is sealed on both ends. At equilibrium, the gas has temperature $T_{0}$, pressure $p_{0}$ and mass density $\rho_{0}$. Assume that viscosity can be ignored and that the gas motion is only in the $x$ direction. The gas properties are uniform in the perpendicular $y$ and $z$ directions.

<img_4558>

Figure 1

Question: A.1 When a standing sound wave forms, the gas elements oscillate in the $x$ direction with angular frequency $\omega$. The amplitude of the oscillations depends on each element's equilibrium position $x$ along the tube. The longitudinal displacement of each gas element from its equilibrium position $x$ is given by

$$
u(x, t)=a \sin (k x) \cos (\omega t)=u_{1}(x) \cos (\omega t)
\tag{1}
$$

(please note the $u$ here describes the displacement of a gas element)

where $a \ll L$ is a positive constant, $k=2 \pi / \lambda$ is the wavenumber and $\lambda$ is the wavelength. What is the maximum possible wavelength $\lambda_{\max }$ in this system?

Solution:
The boundary conditions are: $u(0, t)=u(L, t)=0$. As a result, $\sin \left(\frac{2 \pi}{\lambda} L\right)=0$, so we get $\lambda_{\max }=2 L$.

Final answer: $\lambda_{\max }=2 L$
Subfield: Optics


Question: A.6 For the purpose of this task only, we assume a weak thermal interaction between the tube and the gas. As a result, the standing sound wave remains almost unchanged, but the gas can exchange a small amount of heat with the tube. The heating due to viscosity can be neglected.

For each of the points in Figure 2 (A, C at the edges of the tube, B at the center) state whether the temperature of the tube at that point will increase, decrease or remain the same over a long time.

![](https://cdn.mathpix.com/cropped/2023_12_21_2fffc9c901c17aa00e2eg-1.jpg?height=191&width=826&top_left_y=2417&top_left_x=615)

Figure 2

Solution:
The movement of the gas parcels inside the tube conveys heat along its boundary. To determine the direction of the convection, we combining the result of Task A. 5 and the expression (1) for $u(x, t)$. We see that when $0<x<\frac{L}{2}$, the gas is colder when the displacement $u(x, t)$ is positive. Likewise, when $\frac{L}{2}<x<L$, the gas is colder when the displacement $u(x, t)$ is negative. Hence, heat flows into the gas near the point $B$, cooling it down, and out of the gas near the points $A$ and $C$, heating them up.

Final answer: 证明题
Subfield: Thermodynamics

Consider two liquids A and B insoluble in each other. The pressures $p_{i}(i=\mathrm{A}$ or $\mathrm{B})$ of their saturated vapors obey, to a good approximation, the formula:

$$
\ln \left(p_{i} / p_{o}\right)=\frac{\alpha_{i}}{T}+\beta_{i} ; \quad i=\mathrm{A} \text { or } \mathrm{B}
$$

where $p_{o}$ denotes the normal atmospheric pressure, $T$ - the absolute temperature of the vapor, and $\alpha_{i}$ and $\beta_{i}$ ( $i=\mathrm{A}$ or $\left.\mathrm{B}\right)$ - certain constants depending on the liquid. (The symbol $\ln$ denotes the natural logarithm, i.e. logarithm with base $e=2.7182818 \ldots)$

The values of the ratio $p_{i} / p_{0}$ for the liquids $\mathrm{A}$ and $\mathrm{B}$ at the temperature $40^{\circ} \mathrm{C}$ and $90^{\circ} \mathrm{C}$ are given in Tab. 1.1.

Table 1.1

| $t\left[{ }^{\circ} \mathrm{C}\right]$ | $i=\mathrm{A}$ | $p_{i} / p_{0}$ |
| :---: | :---: | :---: |
|  | 0.284 | 0.07278 |
| 40 | 1.476 | 0.6918 |
| 90 |  |  |

The errors of these values are negligible.

Question: A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$.

Solution:
PART A

The liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid $i(i-\mathrm{A}$ or $\mathrm{B}$ ), one should determine such a temperature $T_{b i}$ (or $t_{b i}$ ) for which $p_{i} / p_{0}=1$.

Then $\ln \left(p_{i} / p_{0}\right)=0$, and we have:

$$
T_{b i}=-\frac{\alpha_{i}}{\beta_{i}}
$$



The coefficients $\alpha_{i}$ and $\beta_{i}$ are not given explicitly. However, they can be calculated from the formula given in the text of the problem. For this purpose one should make use of the numerical data given in the Tab. 1.1.

For the liquid A, we have:

$$
\begin{aligned}
& \ln 0.284=\frac{\alpha_{A}}{(40+273.15) \mathrm{K}}+\beta_{A}, \\
& \ln 1.476=\frac{\alpha_{A}}{(90+273.15) \mathrm{K}}+\beta_{A} .
\end{aligned}
$$

After subtraction of these equations, we get:

$$
\begin{gathered}
\ln 0.284-\ln 1.476=\alpha_{A}\left(\frac{1}{40+273.15}-\frac{1}{90+273.15}\right) \mathrm{K}^{-1} . \\
\alpha_{A}=\frac{\ln \frac{0.284}{1.476}}{\frac{1}{40+273.15}-\frac{1}{90+273.15}} \mathrm{~K} \approx-3748.49 \mathrm{~K}
\end{gathered}
$$

Hence,

$$
\beta_{A}=\ln 0.284-\frac{\alpha_{A}}{(40+273.15) \mathrm{K}} \approx 10.711
$$

Thus, the boiling temperature of the liquid $\mathrm{A}$ is equal to

$$
T_{b A}=3748.49 \mathrm{~K} / 10.711 \approx 349.95 \mathrm{~K} .
$$

In the Celsius scale the boiling temperature of the liquid $\mathrm{A}$ is

$$
t_{b A}=(349.95-273.15)^{\circ} \mathrm{C}=76.80^{\circ} \mathrm{C} \approx 77^{\circ} \mathrm{C} .
$$

For the liquid B, in the same way, we obtain:

$$
\begin{aligned}
\alpha_{B} & \approx-5121.64 \mathrm{~K} \\
\beta_{B} & \approx 13.735 \\
T_{b B} & \approx 372-89 \mathrm{~K} \\
t_{b B} & \approx 99.74^{\circ} \mathrm{C} \approx 100^{\circ} \mathrm{C} .
\end{aligned}
$$

Final answer: $76.8$ , $100$ $^{\circ}\mathrm{C}$
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>


Question: 1.2. Express the angle $\varphi$ between this intersection and the trajectory of the particle in terms of $n$ and $\beta$.

Solution:
<img_4447>

Figure 1

Let us consider a plane containing the particle trajectory. At $t=0$, the particle position is at point $\mathrm{A}$. It reaches point $\mathrm{B}$ at $t=t_{1}$. According to the Huygens principle, at moment $0<t<t_{1}$, the radiation emitted at $\mathrm{A}$ reaches the circle with a radius equal to $\mathrm{AD}$ and the one emitted at $\mathrm{C}$ reaches the circle of radius $\mathrm{CE}$. The radii of the spheres are proportional to the distance of their centre to $\mathrm{B}$ :

$$
\frac{\mathrm{CE}}{\mathrm{CB}}=\frac{c\left(t_{1}-t\right) / n}{v\left(t_{1}-t\right)}=\frac{1}{\beta n}=\mathrm{const}
$$

The spheres are therefore transformed into each other by homothety of vertex B and their envelope is the cone of summit B and half aperture $\varphi=\operatorname{Arcsin} \frac{1}{\beta n}=\frac{\pi}{2}-\theta$, where $\theta$ is the angle made by the light ray $\mathrm{CE}$ with the particle trajectory.

Final answer: $\varphi=\arcsin \frac{1}{\beta n}$
Subfield: Mechanics

LIGO-GW150914

In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.

Part A: Newtonian (conservative) orbits

Question: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,

$$
M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
\tag{!}
$$

The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
\tag{2}
$$

where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.

Solution:
Apply Newton's law to mass $M_{1}$ :

$$
M_{1} \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=G \frac{M_{1} M_{2}}{\left|\vec{r}_{2}-\vec{r}_{1}\right|^{2}} \frac{\vec{r}_{2}-\vec{r}_{1}}{\left|\vec{r}_{2}-\overrightarrow{r_{1}}\right|}
\tag{1}
$$

Use, from eq. (1) of the question sheet

$$
\vec{r}_{2}=-\frac{M_{1}}{M_{2}} \vec{r}_{1}
\tag{2}
$$

in eq. (1) above, to obtain

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\frac{G M_{2}^{3}}{\left(M_{1}+M_{2}\right)^{2} r_{1}^{2}} \frac{\vec{r}_{1}}{r_{1}} .
\tag{3}
$$

Final answer: $n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$
Subfield: Mechanics

LIGO-GW150914

In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.

Part A: Newtonian (conservative) orbits
Context question:
A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,

$$
M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
\tag{!}
$$

The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
\tag{2}
$$

where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
Context answer:
\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}


Question: A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:

$$
E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
\tag{3}
$$

where

$$
\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
\tag{4}
$$

are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.

Solution:
The total energy of the system is the sum of the two kinetic energies plus the gravitational potential energy. For circular motions, the linear velocity of each of the masses reads

$$
\left|\vec{v}_{1}\right|=r_{1} \Omega, \quad\left|\vec{v}_{2}\right|=r_{2} \Omega
\tag{4}
$$

Thus, the total energy is

$$
E=\frac{1}{2}\left(M_{1} r_{1}^{2}+M_{2} r_{2}^{2}\right) \Omega^{2}-\frac{G M_{1} M_{2}}{L}
\tag{5}
$$

Now,

$$
\left(M_{1} r_{1}-M_{2} r_{2}\right)^{2}=0 \quad \Rightarrow \quad M_{1} r_{1}^{2}+M_{2} r_{2}^{2}=\mu L^{2}
\tag{6}
$$

Thus,

$$
E=\frac{1}{2} \mu L^{2} \Omega^{2}-G \frac{M \mu}{L} .
\tag{7}
$$

Final answer: $A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$