di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Modern Physics

## Flashlight



Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame.



<img_4373>


a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.

Question: i. What is the horizontal velocity of the muons in Bob's reference frame?

Solution:
The muons were fired straight down in Alice's frame, so in Bob's frame their horizontal velocity is the same as Alice's, $v=\beta c$.



One way to see this is to imagine Alice was carrying a vertical pole with her. In her reference frame, the muons travel along the length of the pole. This must remain true in any frame, so in Bob's frame the muons must have the same horizontal velocity as Alice.

Final answer: $v=\beta c$
Subfield: Modern Physics

## Flashlight



Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame.



<img_4373>


a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.
Context question:
i. What is the horizontal velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$v=\beta c$}


Question: ii. What is the vertical velocity of the muons in Bob's reference frame?

Solution:
The muons have speed $c$ and horizontal velocity $v=\beta c$, so they have vertical velocity $c \sqrt{1-\beta^{2}}$ by the Pythagorean theorem.

Final answer: $c \sqrt{1-\beta^{2}}$
Subfield: Modern Physics

## Flashlight



Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame.



<img_4373>


a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.
Context question:
i. What is the horizontal velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$v=\beta c$}


Context question:
ii. What is the vertical velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$c \sqrt{1-\beta^{2}}$}


Question: iii. How long does it take the muons to reach the ground in Bob's reference frame?

Solution:
The time is the height divided by the vertical velocity of the muons,



$$

\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}

$$



Alice's velocity $v$ is directed an angle $\theta$ away from Bob. For the rest of the problem, you may additionally express your answers in terms of $\theta$.



<img_4402>

Final answer: $\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$
Subfield: Modern Physics

## Flashlight



Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame.



<img_4373>


a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.
Context question:
i. What is the horizontal velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$v=\beta c$}


Context question:
ii. What is the vertical velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$c \sqrt{1-\beta^{2}}$}


Context question:
iii. How long does it take the muons to reach the ground in Bob's reference frame?
Context answer:
\boxed{$\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$}


Question: b. In Bob's reference frame, how much time is there between when he sees Alice first fire the beam, and when he sees the beam first hit the ground? (Hint: remember to account for the travel time of light to Bob's eyes.)

Solution:
In Bob's frame, during the time the muons take to reach the ground, Alice's car moves a distance $\Delta r=(v \Delta t) \cos \theta$ away from Bob, which means the light from the muons hitting the ground takes an extra time $(\Delta r) / c$ to reach Bob. Thus, the time interval Bob sees, with his eyes, is



$$

\Delta t+\frac{v \cos \theta}{c} \Delta t=\frac{h}{c} \frac{1+\beta \cos \theta}{\sqrt{1-\beta^{2}}}

$$

Final answer: $\frac{h}{c} \frac{1+\beta \cos \theta}{\sqrt{1-\beta^{2}}}$
Subfield: Modern Physics

## Flashlight



Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame.



<img_4373>


a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.
Context question:
i. What is the horizontal velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$v=\beta c$}


Context question:
ii. What is the vertical velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$c \sqrt{1-\beta^{2}}$}


Context question:
iii. How long does it take the muons to reach the ground in Bob's reference frame?
Context answer:
\boxed{$\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$}


Context question:
b. In Bob's reference frame, how much time is there between when he sees Alice first fire the beam, and when he sees the beam first hit the ground? (Hint: remember to account for the travel time of light to Bob's eyes.)
Context answer:
\boxed{$\frac{h}{c} \frac{1+\beta \cos \theta}{\sqrt{1-\beta^{2}}}$}


Question: c. In this part, suppose that $\beta=1 / 2$. Does there exist a value of $\theta$ so that the time it takes the muons to hit the ground in Alice's frame is equal to the time taken according to Bob's eyes, in Bob's frame? If so, find the value of $\theta$ in degrees. If not, briefly explain why not.

Solution:
This will be true if



$$

\frac{h}{c \sqrt{1-\beta^{2}}}(1+\beta \cos \theta)=\frac{h}{c}

$$



Solving for $\theta$, we find



$$

\theta=\cos ^{-1}\left(\sqrt{\beta^{-2}-1}-\beta^{-1}\right) .

$$



This has a solution as long as the argument of the inverse cosine is less than 1 . In the case $\beta=0.5$, it is, and the result is



$$

\theta=105.5^{\circ} \text {. }

$$



That is, this value of $\beta$ is low enough so that the motion of Alice towards Bob can make up for the time dilation effect.

Final answer: $105.5$ $^{\circ}$
Subfield: Modern Physics

## Flashlight



Alice the Mad Scientist, travelling in her flying car at height $h$ above the ground, shoots a beam of muons at the ground. Bob, observing from the ground at distance $R \gg h$ from Alice's car, decides to check some facts about special relativity. Assume the muons travel extremely close to the speed of light in Alice's frame.



<img_4373>


a. Alice's car flies at horizontal speed $v=\beta c$. Alice shoots her muon beam straight down, in her reference frame. Express your answers in terms of $\beta, h, R$ and fundamental constants.
Context question:
i. What is the horizontal velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$v=\beta c$}


Context question:
ii. What is the vertical velocity of the muons in Bob's reference frame?
Context answer:
\boxed{$c \sqrt{1-\beta^{2}}$}


Context question:
iii. How long does it take the muons to reach the ground in Bob's reference frame?
Context answer:
\boxed{$\Delta t=\frac{h}{c \sqrt{1-\beta^{2}}}$}


Context question:
b. In Bob's reference frame, how much time is there between when he sees Alice first fire the beam, and when he sees the beam first hit the ground? (Hint: remember to account for the travel time of light to Bob's eyes.)
Context answer:
\boxed{$\frac{h}{c} \frac{1+\beta \cos \theta}{\sqrt{1-\beta^{2}}}$}


Context question:
c. In this part, suppose that $\beta=1 / 2$. Does there exist a value of $\theta$ so that the time it takes the muons to hit the ground in Alice's frame is equal to the time taken according to Bob's eyes, in Bob's frame? If so, find the value of $\theta$ in degrees. If not, briefly explain why not.
Context answer:
\boxed{$105.5$}


Question: d. Suppose Alice is carrying a radio transmitter set to frequency $f$. To what frequency would Bob have to set his radio receiver in order to receive Alice's transmission?

Solution:
The key point is that all the logic in part (b) still works, if we replace Bob's eyes with the radio receiver. Since the frequency of a wave is the inverse of the time between maxima,



$$

f^{\prime}=\frac{\sqrt{1-\beta^{2}}}{1+\beta \cos \theta} f

$$







This is the two-dimensional relativistic Doppler shift, and the secret point of this problem was to derive it in a simple way.

Final answer: $f^{\prime}=\frac{\sqrt{1-\beta^{2}}}{1+\beta \cos \theta} f$
Subfield: Electromagnetism

## Electroneering



An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.



Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.

Question: 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.
Note that for a uniformly charged ring of mass $d m$, charge $d q$, and radius $r$, rotating with angular velocity $\omega$, the ratio of the magnetic moment and the







angular momentum is



$$

\frac{\mu}{L}=\frac{\pi r^{2}(\omega d q / 2 \pi)}{\left(r^{2} d m\right) \omega}=\frac{1}{2} \frac{d q}{d m}

$$



The ratio is independent of $r$ and $\omega$. Since the sphere can be decomposed into such rings, the total magnetic moment and total angular momentum must have the same ratio,



$$

\frac{\mu}{L}=\frac{1}{2} \frac{q}{m}

$$



Finally, we know that $L=(2 / 3) m R^{2} \omega$ for a spherical shell. Plugging this in and using $\omega=c / R$ gives $\mu=q c R / 3$ as before, but with no integration required.

Final answer: $\mu=q c R / 3$
Subfield: Electromagnetism

## Electroneering



An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.



Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.
Context question:
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}

$$
Context answer:
\boxed{$\mu=q c R / 3$}


Question: 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.

Final answer: $R=\frac{3}{2} \frac{\hbar}{m c}$
Subfield: Electromagnetism

## Electroneering



An electron is a particle with charge $-q$, mass $m$, and magnetic moment $\mu$. In this problem we will explore whether a classical model consistent with these properties can also explain the rest energy $E_{0}=m c^{2}$ of the electron.



Let us describe the electron as a thin spherical shell with uniformly distributed charge and radius $R$. Recall that the magnetic moment of a closed, planar loop of current is always equal to the product of the current and the area of the loop. For the electron, a magnetic moment can be created by making the sphere rotate around an axis passing through its center.
Context question:
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}

$$
Context answer:
\boxed{$\mu=q c R / 3$}


Context question:
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.
Context answer:
\boxed{$R=\frac{3}{2} \frac{\hbar}{m c}$}


Question: 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.)

Final answer: $U_{E}=\frac{1}{3} \alpha E_{0}$
Subfield: Electromagnetism


Question: e. How does your estimate for the total energy in the electric and magnetic fields compare to $E_{0}$ ?

Solution:
Both $U_{E}$ and $U_{B}$ are much smaller than $E_{0}$, by a factor of $\alpha \ll 1$. Thus, this classical model cannot explain the origin of the electron's rest energy.



There were many attempts to make classical models of the electron in the early $20^{\text {th }}$ century, but they all ran into difficulties like this one. For more on this subject, see chapter II-28 of the Feynman lectures.

Final answer: Both $U_{E}$ and $U_{B}$ are much smaller than $E_{0}$, by a factor of $\alpha \ll 1$. Thus, this classical model cannot explain the origin of the electron's rest energy.
Subfield: Mechanics


Question: a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$.



![](https://cdn.mathpix.com/cropped/2023_12_21_d0fccf1ebe41035d3cadg-1.jpg?height=474&width=491&top_left_y=907&top_left_x=833)



Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ?

Solution:
During the first leg of the trip, the disk has angular velocity



$$

\omega=-\frac{2 m v}{M R}

$$



by conservation of angular momentum. Thus, the disk rotates through an angle



$$

\phi_{1}=-\frac{2 m v}{M R} \frac{\theta R}{v}=-\frac{2 m \theta}{M}

$$



to leading order in $m / M$. (Here we have neglected the fact that the disk rotates under the ant as it is walking, somewhat reducing the distance it has to walk; this changes the answer only to higher order in $m / M$. The exact answer is a more complicated function of $m / M$. By going to "leading order", we mean we are expanding that exact answer in a series in $m / M$, such as with the binomial theorem, and keeping only the first nonzero term.)



When the ant is moving radially, $\omega=0$, so no rotation occurs. On the last leg of the trip, the disk has angular velocity



$$

\omega=\frac{2 m v(R-h)}{M R^{2}}

$$





which means the disk rotates through an angle



$$

\phi_{2}=\frac{2 m v(R-h)}{M R^{2}} \frac{\theta(R-h)}{v}=\frac{2 m \theta}{M}\left(1-\frac{h}{R}\right)^{2}

$$



The net rotation is



$$

\phi_{1}+\phi_{2}=\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right) \approx-\frac{4 m}{M} \frac{h \theta}{R}

$$



The sign is not important since it is convention-dependent. (Solutions that were not fully approximated were also accepted; however, not approximating early dramatically increases the amount of work you have to do.)



Incidentally, you might have thought the answer had to be zero, by angular momentum conservation. After all, when a system has zero total linear momentum, its center of mass can't move. But this problem shows that systems with zero total angular momentum can perform net rotations, which is the reason, e.g. that a falling cat can always land on its feet. In more advanced physics, this would be described by saying the constraint on the disc's motion coming from angular momentum conservation is not holonomic.

Final answer: $\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right)$
$-\frac{4 m}{M} \frac{h \theta}{R}$
Subfield: Mechanics

## Disk Jockey



A disk of uniform mass density, mass $M$, and radius $R$ sits at rest on a frictionless floor. The disk is attached to the floor by a frictionless pivot at its center, which keeps the center of the disk in place, but allows the disk to rotate freely. An ant of mass $m \ll M$ is initially standing on the edge of the disk; you may give your answers to leading order in $m / M$.

Question: a. The ant walks an angular displacement $\theta$ along the edge of the disk. Then it walks radially inward by a distance $h \ll R$, tangentially through an angular displacement $-\theta$, then back to its starting point on the disk. Assume the ant walks with constant speed $v$.



<img_4433>



Through what net angle does the disk rotate throughout this process, to leading order in $h / R$ ?

Solution:
During the first leg of the trip, the disk has angular velocity



$$

\omega=-\frac{2 m v}{M R}

$$



by conservation of angular momentum. Thus, the disk rotates through an angle



$$

\phi_{1}=-\frac{2 m v}{M R} \frac{\theta R}{v}=-\frac{2 m \theta}{M}

$$



to leading order in $m / M$. (Here we have neglected the fact that the disk rotates under the ant as it is walking, somewhat reducing the distance it has to walk; this changes the answer only to higher order in $m / M$. The exact answer is a more complicated function of $m / M$. By going to "leading order", we mean we are expanding that exact answer in a series in $m / M$, such as with the binomial theorem, and keeping only the first nonzero term.)



When the ant is moving radially, $\omega=0$, so no rotation occurs. On the last leg of the trip, the disk has angular velocity



$$

\omega=\frac{2 m v(R-h)}{M R^{2}}

$$





which means the disk rotates through an angle



$$

\phi_{2}=\frac{2 m v(R-h)}{M R^{2}} \frac{\theta(R-h)}{v}=\frac{2 m \theta}{M}\left(1-\frac{h}{R}\right)^{2}

$$



The net rotation is



$$

\phi_{1}+\phi_{2}=\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right) \approx-\frac{4 m}{M} \frac{h \theta}{R}

$$



The sign is not important since it is convention-dependent. (Solutions that were not fully approximated were also accepted; however, not approximating early dramatically increases the amount of work you have to do.)



Incidentally, you might have thought the answer had to be zero, by angular momentum conservation. After all, when a system has zero total linear momentum, its center of mass can't move. But this problem shows that systems with zero total angular momentum can perform net rotations, which is the reason, e.g. that a falling cat can always land on its feet. In more advanced physics, this would be described by saying the constraint on the disc's motion coming from angular momentum conservation is not holonomic.

Final answer: $\frac{2 m \theta}{M}\left(\left(1-\frac{h}{R}\right)^{2}-1\right)$
$-\frac{4 m}{M} \frac{h \theta}{R}$