weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: 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,

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

Answer: $A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$
Problem: A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.

Solution: Energy (3) of the question sheet can be interpreted as describing a system of a mass $\mu$ in a circular orbit with angular velocity $\Omega$, radius $L$, around a mass $M$ (at rest). Equating the gravitational acceleration to the centripetal acceleration:

$$
G \frac{M}{L^{2}}=\Omega^{2} L
\tag{8}
$$

This is indeed Kepler's third law (for circular orbits). Then, from 7,

$$
E=-\frac{1}{2} G \frac{M \mu}{L}
\tag{9}
$$

Answer: $\beta=-\frac{1}{2}$
Problem: B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain:

$$
\mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
\tag{9}
$$

What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.]

Solution: First take the derivatives:

$$
\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}=4 \Omega^{3} \mu L^{2}\left(\begin{array}{ccc}
\sin 2 \Omega t & -\cos 2 \Omega t & 0 \\
-\cos 2 \Omega t & -\sin 2 \Omega t & 0 \\
0 & 0 & 0
\end{array}\right)
\tag{13}
$$

Then perform the sum:

$$
\frac{\mathrm{d} E}{\mathrm{~d} t}=\frac{G}{5 c^{5}}\left(4 \Omega^{3} \mu L^{2}\right)^{2}\left[2 \sin ^{2}(2 \Omega t)+2 \cos ^{2}(2 \Omega t)\right]=\frac{32}{5} \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6} .
\tag{14}
$$

Answer: $\xi=\frac{32}{5}$
Problem: B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form

$$
\left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5}
\tag{10}
$$

where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name "chirp" is inspired by the high pitch sound (increasing frequency) produced by small birds.]

Solution: Now we assume a sequency of Keplerian orbits, with decreasing energy, which is being taken from the system by the GWs.

First, from (9), differentiating with respect to time,

$$
\frac{\mathrm{d} E}{\mathrm{~d} t}=\frac{G M \mu}{2 L^{2}} \frac{\mathrm{d} L}{\mathrm{~d} t}
\tag{15}
$$

Since this loss of energy is due to GWs, we equate it with (minus) the luminosity of GWs, given by (14)

$$
\frac{G M \mu}{2 L^{2}} \frac{\mathrm{d} L}{\mathrm{~d} t}=-\frac{32}{5} \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
\tag{16}
$$

We can eliminate the $L$ and $\mathrm{d} L / \mathrm{d} t$ dependence in this equation in terms of $\Omega$ and $\mathrm{d} \Omega / \mathrm{d} t$, by using Kepler's third law (8), which relates:

$$
L^{3}=G \frac{M}{\Omega^{2}}, \quad \frac{\mathrm{d} L}{\mathrm{~d} t}=-\frac{2}{3} \frac{L}{\Omega} \frac{\mathrm{d} \Omega}{\mathrm{d} t}
\tag{17}
$$



Substituting in (16), we obtain:

$$
\left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=\left(\frac{96}{5}\right)^{3} \frac{\Omega^{11}}{c^{15}} G^{5} \mu^{3} M^{2} \equiv\left(\frac{96}{5}\right)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5}
\tag{18}
$$

Answer: $M_{\mathrm{c}}=(\mu^{3} M^{2})^{1 / 5}$
Problem: B.4 Using the information provided above, relate the orbital angular velocity $\Omega$ with the GW frequency $f_{\mathrm{GW}}$. Knowing that, for any smooth function $F(t)$ and $a \neq 1$,

$$
\frac{\mathrm{d} F(t)}{\mathrm{d} t}=\chi F(t)^{a} \quad \Rightarrow \quad F(t)^{1-a}=\chi(1-a)\left(t-t_{0}\right)
\tag{11}
$$

where $\chi$ is a constant and $t_{0}$ is an integration constant, show that (10) implies that the GW frequency is

$$
f_{\mathrm{GW}}^{-8 / 3}=8 \pi^{8 / 3} \xi\left(\frac{G M_{c}}{c^{3}}\right)^{(2 / 3)+p}\left(t_{0}-t\right)^{2-p}
\tag{12}
$$

and determine the constant $p$.

Solution: Angular and cycle frequencies are related as $\Omega=2 \pi f$. From the information provided above: GWs have a frequency which is twice as large as the orbital frequency, we have

$$
\frac{\Omega}{2 \pi}=\frac{f_{\mathrm{GW}}}{2}
\tag{19}
$$

Formula (10) of the question sheet has the form

$$
\frac{\mathrm{d} \Omega}{\mathrm{d} t}=\chi \Omega^{11 / 3}, \quad \chi \equiv \frac{96}{5} \frac{\left(G M_{\mathrm{c}}\right)^{5 / 3}}{c^{5}} .
\tag{20}
$$

Thus, from (11) of the question sheet

$$
\Omega(t)^{-8 / 3}=\frac{8}{3} \chi\left(t_{0}-t\right),
\tag{21}
$$

or, using (20) and the definition of $\chi$ gives

$$
f_{\mathrm{GW}}^{-8 / 3}(t)=\frac{(8 \pi)^{8 / 3}}{5}\left(\frac{G M_{\mathrm{c}}}{c^{3}}\right)^{5 / 3}\left(t_{0}-t\right)
\tag{22}
$$

Answer: p=1
Problem: A.1 Derive an expression for the cyclotron radius, $r$, of the circular trajectory of an electron acted upon by a magnetic force perpendicular to its velocity, and express that radius as a function of its kinetic energy, $K$; charge modulus, $e$; mass, $m$; and magnetic field, $B$. Assume that the electron is a non-relativistic classical particle.

Solution: The magnetic force is the centripetal force:

$$
m \frac{v^{2}}{r}=e v B \Rightarrow r=\frac{m v}{e B}
$$

First express the velocity in terms of the kinetic energy,

$$
K=\frac{1}{2} m v^{2} \Rightarrow v=\sqrt{\frac{2 K}{m}},
$$

and then insert it in the expression above for the radius to get

Answer: $r=\frac{\sqrt{2 K m}}{e B}$
Problem: A.2 Calculate the minimum value of the momentum of an electron that allows it to escape the inner part of the detector in the radial direction. The inner part of the detector has a cylindrical shape with a radius of 1.1 meters, and the electron is produced in the collision point exactly in the center of the cylinder. Express your answer in $\mathrm{MeV} / c$.

Solution: The radius of the circular motion of a charged particle in the presence of a uniform magnetic field is given by,

$$
r=\frac{m v}{e B}
$$

This formula is valid in the relativistic scenario if the mass correction, $m \rightarrow \gamma m$ is included:

$$
r=\frac{\gamma m v}{e B}=\frac{p}{e B} \Rightarrow p=r e B
$$

Note that the radius of the circular motion is half the radius of the inner part of the detector. One obtains $\left[1 \mathrm{MeV} / c=5.34 \times 10^{-22} \mathrm{~m} \mathrm{~kg} \mathrm{~s}^{-1}\right.$ 

Answer: $p=330$
Problem: A.3 A particle is called ultrarelativistic when its speed is very close to the speed of light. For an ultrarelativistic particle the emitted power can be expressed as:

$$
P=\xi \frac{e^{4}}{\epsilon_{0} m^{k} c^{n}} E^{2} B^{2}
$$

where $\xi$ is a real number, $n, k$ are integers, $E$ is the energy of the charged particle and $B$ is the magnetic field. Find $\xi, n$ and $k$.

Solution: The acceleration for the particle is $a=\frac{e v B}{\gamma m} \sim \frac{e c B}{\gamma m}$, in the ultrarelativistic limit. Then,

$$
P=\frac{e^{4} c^{2} \gamma^{4} B^{2}}{6 \pi \epsilon_{0} c^{3} \gamma^{2} m^{2}}=\frac{e^{4} \gamma^{2} c^{4} B^{2}}{6 \pi \epsilon_{0} c^{5} m^{2}}
$$

Since $E=\gamma m c^{2}$ we can obtain $\gamma^{2} c^{4}=\frac{E^{2}}{m^{2}}$ and, finally,



$$
P=\frac{e^{4}}{6 \pi \epsilon_{0} m^{4} c^{5}} E^{2} B^{2}
$$

Answer: $\xi=\frac{1}{6 \pi}, n=5$ , $k=4$
Problem: A.4 In the ultrarelativistic limit, the energy of the electron as a function of time is:

$$
E(t)=\frac{E_{0}}{1+\alpha E_{0} t},
$$

where $E_{0}$ is the initial energy of the electron. Find $\alpha$ as a function of $e, c, B, \epsilon_{0}$ and $m$.

Solution: The power emitted by the particle is given by,

$$
P=-\frac{\mathrm{d} E}{\mathrm{~d} t}=\frac{e^{4}}{6 \pi \epsilon_{0} m^{4} c^{5}} E^{2} B^{2}
$$

The energy of the particle as a function of time can be calculated from

$$
\int_{E_{0}}^{E(t)} \frac{1}{E^{2}} \mathrm{~d} E=-\int_{0}^{t} \frac{e^{4}}{6 \pi \epsilon_{0} m^{4} c^{5}} B^{2} \mathrm{~d} t
$$

where $E(0)=E_{0}$. This leads to,

$$
\frac{1}{E(t)}-\frac{1}{E_{0}}=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} t \quad \Rightarrow \quad E(t)=\frac{E_{0}}{1+\alpha E_{0} t},
$$

Answer: $\alpha=\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}}$
Problem: A.5 Consider an electron produced at the collision point along the radial direction with an energy of $100 \mathrm{GeV}$. Estimate the amount of energy that is lost due to synchrotron radiation until the electron escapes the inner part of the detector? Express your answer in MeV.

Solution: If the initial energy of the electron is $100 \mathrm{GeV}$, the radius of curvature is extremely large ( $r=\frac{E}{e B c} \approx 167 \mathrm{~m}$ ). Therefore, in approximation, one can consider the electron is moving in the inner part of the ATLAS detector along a straight line. The time of flight of the electron is $t=R / c$, where $R=1.1 \mathrm{~m}$ is the radius of the inner part of the detector. The total energy lost due to synchrotron radiation is,

$$
\Delta E=E(R / c)-E_{0}=\frac{E_{0}}{1+\alpha E_{0} \frac{R}{c}}-E_{0} \approx-\alpha E_{0}^{2} \frac{R}{c}
$$

Answer: -56
Problem: A.6 Find an expression for the cyclotron frequency of the electron as a function of time in the ultrarelativistic limit.

Solution: In the ultrarelativistic limit, $v \approx c$ and $E \approx p c$. The cyclotron frequency is,

$$
\omega(t)=\frac{c}{r(t)}=\frac{e c B}{p(t)}=\frac{e c^{2} B}{E(t)}
$$

Answer: $\omega(t)=\frac{e c^{2} B}{E_{0}}(1+\frac{e^{4} B^{2}}{6 \pi \epsilon_{0} m^{4} c^{5}} E_{0} t)$
Problem: a. Calculate the distance $s$ that the spring is stretched beyond its rest length when the box is just about to start moving.

Solution: This is when the spring force equals the maximal static friction, $k s=\mu_{0} m g$, so $s=\mu_{0} m g / k$.

Answer: $s=\mu_{0} m g / k$
Problem: b. Let the box start at $x=0$, and let $t=0$ be the time the box first starts moving. Find the acceleration of the box in terms of $x, t, v_{0}, s$, and the other parameters, while the box is moving.

Solution: The net stretching of the spring is $s+v_{0} t-x$, leading to a rightward force $k s$. When the box is moving, it is always moving to the right, so the kinetic friction force $\mu m g$ is always in the leftward direction, which means



$$

m a=k\left(s+v_{0} t-x\right)-\mu m g

$$



which means



$$

a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g .

$$

Answer: $a=\frac{k}{m}\left(s+v_{0} t-x\right)-\mu g$
Problem: c. Find the time $t_{0}$ when the box stops for the first time.

Solution: Taking the derivative, the velocity of the box is



$$

v=v_{0}(1-\cos \omega t)+(1-r) s \omega \sin \omega t .

$$



The box stops when this is equal to zero for the first time. There are several ways to







evaluate this condition. First, we can use half-angle identities to find



$$

0=2 v_{0} \sin ^{2} \frac{\omega t}{2}+2(1-r) s \omega \sin \frac{\omega t}{2} \cos \frac{\omega t}{2}

$$



As a result, the box stops when



$$

\tan \frac{\omega t}{2}=-\frac{(1-r) s \omega}{v_{0}}

$$



Using a basic property of the tangent function,



$$

\tan \left(\pi-\frac{\omega t}{2}\right)=\frac{(1-r) s \omega}{v_{0}}

$$



Solving for $t$, we conclude that



$$

t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}

$$



Note that we cancelled a factor of $\sin (\omega t / 2)$, which has a zero at $t=2 \pi / \omega$. However, this is a larger time than the one we just found, so it is irrelevant.



Another way to arrive at the answer is to rewrite the original condition as



$$

1=\cos \omega t-\tan \alpha \sin \omega t

$$



Squaring both sides and using some trigonometric identities gives



$$

\tan \omega t=-\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}

$$



This can then be further simplified using the tangent half-angle identity, upon which we recover the same result as above.

Answer: $t_{0}=\frac{2 \pi-2 \alpha}{\omega}, \quad \alpha=\tan ^{-1} \frac{(1-r) s \omega}{v_{0}}$
Problem: e. After the box stops, how long will it stay at rest before starting to move again?

Solution: Using a result we found in part (d), the stretch is



$$

\Delta \ell=r s+\frac{v_{0}}{\omega \cos \alpha} \sin \left(\omega t_{0}+\alpha\right)

$$



when the box stops. Plugging in the value of $t_{0}$ found in part (c),



$$

\Delta \ell=r s+\frac{v_{0}}{\omega \cos \alpha} \sin (2 \pi-\alpha)=r s-\frac{v_{0}}{\omega} \tan \alpha=(2 r-1) s

$$



The box starts to move again when the stretch becomes $s$, so the time is



$$

\frac{s-(2 r-1) s}{v_{0}}=\frac{2(1-r) s}{v_{0}}

$$



The pattern of motion investigated in this problem is known as "stick-slip" and occurs in many practical contexts.

Answer: $\frac{2(1-r) s}{v_{0}}$
Problem: a. If no point on the sphere's surface can travel faster than the speed of light (in the frame of the sphere's center of mass), what is the maximum magnetic moment that the sphere can have? You may use the integral:



$$

\int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3}

$$

Solution: A point on the sphere's equator moves at a speed $\omega R$, where $\omega$ is the angular velocity of rotation. Setting $\omega R=c$ gives $\omega=c / R$.



The spinning sphere can be thought of as a stack of infinitesimal current loops, all of which have a magnetic moment pointing in the same direction. Consider making a thin, circular slice of the sphere's surface, corresponding to polar angles in the range $(\theta, \theta+d \theta)$. This slice has a radius $R \sin \theta$, so that the surface area of the slice is



$$

d s=2 \pi R \sin \theta R d \theta

$$



The charge of the slice is



$$

d Q=-\frac{q d s}{4 \pi R^{2}}=-\frac{q \sin \theta}{2}

$$



Since the charge $d Q$ moves around the rotation axis one time per period $T=2 \pi / \omega$, the corresponding current is



$$

d I=\frac{d Q}{T}=-\frac{\omega q \sin \theta}{4 \pi}

$$



The magnitude of the magnetic moment of this slice is



$$

d \mu=\pi(R \sin \theta)^{2}|d I|=\frac{1}{4} q \omega R^{2} \sin ^{3} \theta d \theta

$$



Using the provided integral, the total magnetic moment is



$$

\mu=\int_{0}^{\pi} \frac{1}{4} q \omega R^{2} \sin ^{3} \theta d \theta=\frac{1}{3} q c R

$$



If you weren't able to do this, you could also have given the answer $\mu \sim q c R$, which can be derived by dimensional analysis, for partial credit.

Answer: $\mu=q c R / 3$
Problem: b. The electron's magnetic moment is known to be $\mu=q \hbar / 2 m$, where $\hbar$ is the reduced Planck constant. In this model, what is the minimum possible radius of the electron? Express your answer in terms of $m$ and fundamental constants.

Solution: Since the magnetic moment is fixed, and we want the radius to be small, we want the electron to be spinning as fast as possible. Thus, the magnetic moment has the value found in part (a), and equating this to the known value gives



$$

R=\frac{3}{2} \frac{\hbar}{m c}

$$



Again, you can get $R \sim \hbar / m c$ by dimensional analysis.

Answer: $R=\frac{3}{2} \frac{\hbar}{m c}$
Problem: c. Assuming the radius is the value you found in part (b), how much energy is stored in the electric field of the electron? Express your answer in terms of $E_{0}=m c^{2}$ and the fine structure constant,



$$

\alpha=\frac{q^{2}}{4 \pi \epsilon_{0} \hbar c} \approx \frac{1}{137}

$$

Solution: For a collection of charges, the total energy stored in the electrostatic field is



$$

U_{E}=\frac{1}{2} \sum_{i} q_{i} V_{i}

$$



where $V_{i}$ is the electric potential at $q_{i}$. In this case, the total charge is $q$, and all of the charge is at potential $q / 4 \pi \epsilon_{0} R$, so



$$

U_{E}=\frac{q^{2}}{8 \pi \epsilon_{0} R}

$$



Using the result of part (b),



$$

U_{E}=\frac{1}{3} \alpha E_{0}

$$



Note that you can't get this answer by dimensional analysis alone, since $\alpha$ is dimensionless. (However, if you found $R$ by dimensional analysis, and additionally reasoned that $U_{E}$ could
depend only on $q, \epsilon_{0}$, and $R$, then you could derive $U_{E} \sim \alpha E_{0}$, for partial credit.)

Answer: $U_{E}=\frac{1}{3} \alpha E_{0}$
Problem: d. Roughly estimate the total energy stored in the magnetic field of the electron, in terms of $E_{0}$ and $\alpha$. (Hint: one way to do this is to suppose the magnetic field has roughly constant magnitude inside the sphere and is negligible outside of it, then estimate the field inside the sphere.)

Solution: Following the hint, we can estimate



$$

U_{B} \sim \frac{B_{0}^{2}}{2 \mu_{0}}\left(\frac{4}{3} \pi R^{3}\right)

$$



where $B_{0}$ is the typical magnetic field inside the sphere. Actually finding the value of $B_{0}$ would require doing some complicated integrals. To get a rough estimate, note that if we replaced the sphere with a ring of charge, then at the center of the ring,



$$

B_{0} \sim \frac{\mu_{0} I}{R} \sim \frac{\mu_{0} q c}{R^{2}}

$$



Thus, we have



$$

U_{B} \sim \frac{1}{\mu_{0}}\left(\frac{\mu_{0} q c}{R^{2}}\right)^{2} R^{3} \sim \frac{\mu_{0} q^{2} c^{2}}{R} \sim \frac{\mu_{0} m c^{3} q^{2}}{\hbar}

$$



To get this in terms of the fine structure constant, we use $c^{2}=1 / \mu_{0} \epsilon_{0}$, giving



$$

U_{B} \sim m c^{2} \frac{q^{2}}{\epsilon_{0} \hbar c} \sim \alpha E_{0}

$$



An even faster way to get this result is to note that in general, the energy stored in magnetic fields tends to be a factor of order $(v / c)^{2}$ smaller than the energy stored in electric fields, where $v$ is the speed of the charge. In this problem the charge is all moving relativistically, so we must have $U_{B} \sim U_{E}$.

Answer: $\alpha E_{0}$
Problem: i. What is the minimum heat extraction $-d Q_{2, \text { min }}$ required by the laws of thermodynamics to heat up the room by $d Q_{1}$ ?

Solution: The second law of thermodynamics implies that, no matter what you do, you must have $\mathrm{d} S_{\text {universe }} \geq 0$, and if your process is to be as efficient as possible, we can assume it is reversible, so



$$

\mathrm{d} S_{\text {universe; reversible }}=0

$$



If we do extract any work while allowing heat to transfer between reservoirs, we will later use that work to transfer more heat. So in the entire process, there are only heat transfers, and by conservation of energy,



$$

\mathrm{d} Q_{0}+\mathrm{d} Q_{1}+\mathrm{d} Q_{2}=0

$$



The entropy change associated with each reversible heat transfer is $d S=d Q / T$, so our assumption of zero entropy production becomes



$$

\frac{\mathrm{d} Q_{0}}{T_{0}}+\frac{\mathrm{d} Q_{1}}{T_{1}}+\frac{\mathrm{d} Q_{2}}{T_{2}}=0

$$



By combining these equations, we can eliminate $d Q_{0}$ and solve for $d Q_{2}$, giving



$$

-\mathrm{d} Q_{2 ; \min }=\frac{T_{2}}{T_{1}} \frac{T_{1}-T_{0}}{T_{2}-T_{0}} \mathrm{~d} Q_{1}

$$



For the provided numbers, this happens to be about $0.11 d Q_{1}$. That is, a heat pump can be much more efficient than direct heating. This problem was inspired by Jaynes, E. T, "Note on thermal heating efficiency.", American Journal of Physics 71.2 (2003): 180-182. (You can also solve the problem by considering an explicit procedure using Carnot engines. But since Carnot engines are reversible, all such procedures will just give the same answer.)

Answer: $\frac{T_{2}}{T_{1}} \frac{T_{1}-T_{0}}{T_{2}-T_{0}} \mathrm{~d} Q_{1}$
Problem: i. As a parcel of air moves upward, it accelerates. Find a rough estimate for the average speed $v_{0}$ during its upward motion.

Solution: The temperature of the air is higher than its surroundings by a fractional amount of order $\Delta T / T$. Thus, by the ideal gas law, the density is lower than its surroundings by a fraction of order $\Delta T / T$, which means the upward acceleration due to the buoyant force is of order $a=g \Delta T / T$. Since this is roughly uniformly accelerated motion, $v_{0}^{2} \propto a h$,







which implies



$$

v_{0} \sim \sqrt{g h \frac{\Delta T}{T_{0}}}

$$



Note that because $d P / d z=-\rho g$ in hydrostatic equilibrium, the pressure of the surrounding air varies between the bottom and top of the container, by a fractional amount of order $\rho_{0} g h / P_{0}$. But since we assumed $\rho_{0} g h / P_{0} \ll \Delta T / T_{0}$, we can neglect this effect.

Answer: $\sqrt{g h \frac{\Delta T}{T_{0}}}$
Problem: ii. In the steady state, warm parcels of air are continuously moving upward from the bottom, and cold parcels of air are continuously moving downward from the top. Find a rough estimate for the net rate of upward energy transfer per area.

Solution: The extra energy carried by a parcel of gas is



$$

n C_{p} \Delta T \sim n R \Delta T \sim P_{0} V \frac{\Delta T}{T_{0}}

$$



where $V$ is the volume of the parcel. The net volume of warm air transported upward per unit time is of order $A v_{0}$, where $A$ is the cross-sectional area of the box. Thus, the average power per area is roughly



$$

P_{0} v_{0} \frac{\Delta T}{T_{0}} \sim P_{0} \sqrt{g h}\left(\frac{\Delta T}{T_{0}}\right)^{3 / 2}

$$



This is a simplified version of the mixing length theory of convection, which is essential for modeling the interiors of stars.

Answer: $P_{0} v_{0} \frac{\Delta T}{T_{0}}$
Problem: a. How much time elapses on Carla's stopwatch with each revolution?

Solution: From time dilation, her clock ticks slower by a factor $\gamma$. Therefore, each revolution takes



$$

\frac{2 \pi R}{\gamma v}=\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}

$$



when measured by Carla's stopwatch.

Answer: $\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$
Problem: b. In the lab frame (the reference frame of the clocks, which are at rest), what is the offset between Clock $A$ and Clock $B$ ?

Solution: Carla's motion is perpendicular to the displacement between Clock $A$ and Clock $B$ when they are synchronized in CIRF. Therefore, the simultaneous synchronization in CIRF is also simultaneous in the lab frame. Thus, the offset is 0 .



To understand why this offset is 0 , you can also imagine placing an lightbulb halfway between the two clocks and having it send a light pulse at some known time. In both Carla's frame and the lab frame, the light pulse reaches the two clocks simultaneously.

Answer: 0
Problem: c. If Carla's stopwatch measures an elapsed time $\tau$, what does Clock A measure in CIRF?

Solution: By symmetry, the speed at which the center clock ticks according to CIRF cannot change. In one revolution, Carla's stopwatch measures $\frac{2 \pi R \sqrt{1-v^{2} / c^{2}}}{v}$, while the center clock measures $\frac{2 \pi R}{v}$. Then,





$$

t_{A}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}

$$

Answer: $t_{A}(\tau)=\frac{\tau}{\sqrt{1-v^{2} / c^{2}}}$