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USAPhO 2023, Полуфинал — B3. Starry Messengers

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2023 USA Physics Olympiad Exam · 25 т.

Внимание: Бележка към темата

The archive contains the complete question packet and the official interleaved solution packet. The separate official answer sheets referred to in the instructions are absent from this archive folder. Original printed errors are retained: A1(c)'s solution uses $z'_x$ although the question defines $y(x)$ ; A2(b.iii)'s solution writes radius $r+\delta r$ although the question defines $r=r_0+\delta r$ ; A3's key refers to “part (b)” in its part a.iii check, and B1's key calls the rod a “disc”. B2's printed energy expression uses $2.0\times10^7\ \mathrm{J}$ although the question gives energy density in J/kg. The “this paper” references in B2 and B3 have no recoverable URI in the supplied PDF.

AAPT UNITED STATES PHYSICS TEAM — AIP 2023; USA Physics Olympiad Exam

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Important Instructions for the Exam Supervisor

This examination has two parts. Each part has three questions and lasts for 90 minutes.
For each student, print out one copy of the exam and one copy of the answer sheets. Print everything single-sided, and do not staple anything. Divide the exam into the instructions (pages 2–3), Part A questions (pages 4–8), and Part B questions (pages 9–11).
Begin by giving students the instructions and all of the answer sheets. Let the students read the instructions and fill out their information on the answer sheets. They can keep the instructions for both parts of the exam. Also give students blank sheets of paper to use as scratch paper throughout the exam.
Students may bring calculators, but they may not use symbolic math, programming, or graphing features of these calculators. Calculators may not be shared, and their memory must be cleared of data and programs. Cell phones or other electronics may not be used during the exam or while the exam papers are present. Students may not use books or other references.
To start the exam, give students the Part A questions, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the Part A questions and answer sheets.
Then give students a 5 to 10 minute break. Then give them the Part B questions, and allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, collect everything, including the questions, the instructions, the answer sheets, and the scratch paper. Students may not keep the exam questions. Everything can be returned to the students after April 19th, 2023.
After the exam, sort each student’s answer sheets by page number. Scan every answer sheet, including blank ones.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

Tengiz Bibilashvili, Kellan Colburn, Samuel Gebretsadkan, Abi Krishnan, Natalie LeBaron, Kye Shi, Brian Skinner, and Kevin Zhou.

Instructions for the Student

You should receive these instructions, the reference table on the next page, answer sheets, and blank paper for scratch work. Read this page carefully before the exam begins.
You may use a calculator, but its memory must be cleared of data and programs, and you may not use symbolic math, programming, or graphing features. Calculators may not be shared. Cell phones or other electronics may not be used during the exam or while the exam papers are present. You may not use books or other outside references.
When the exam begins, your proctor will give you the questions for Part A. You will have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. If you finish all of the questions, you may check your work, but you may not look at Part B during this time.
After 90 minutes, your proctor will collect the questions and answer sheets for Part A. You may then take a short break.
Then you will work on Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time. When the exam ends, you must return all papers to the proctor, including the exam questions.
Do not discuss the questions of this exam, or their solutions, until after April 19th, 2023. Violations of this rule may result in disqualification.

Below are instructions for writing your solutions.

All of your solutions must be written on the official answer sheets. Nothing outside these answer sheets will be graded. Before the exam begins, write your name, student AAPT number, and proctor AAPT number as directed on the answer sheets.
There are several answer sheets per problem. If you run out of space for a problem, you may use the extra answer sheets, which are at the end of the answer sheet packet. To ensure this work is graded, you must indicate, at the bottom of your last answer sheet for that problem, that you are using these extra answer sheets.
Only write within the frame of each answer sheet. To simplify grading, we recommend drawing a box around your final answer for each subpart. You should organize your work linearly and briefly explain your reasoning, which will help you earn partial credit. You may use either pencil or pen, but sure to write clearly so your work will be legible after scanning.

Reference table of possibly useful information

$g=9.8\ \mathrm{N/kg}$	$G=6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k=1/4\pi\epsilon_0=8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$	$k_m=\mu_0/4\pi=10^{-7}\ \mathrm{T\cdot m/A}$
$c=3.00\times10^8\ \mathrm{m/s}$	$k_B=1.38\times10^{-23}\ \mathrm{J/K}$
$N_A=6.02\times10^{23}\ (\mathrm{mol})^{-1}$	$R=N_Ak_B=8.31\ \mathrm{J/(mol\cdot K)}$
$\sigma=5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$	$e=1.602\times10^{-19}\ \mathrm{C}$
$1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$	$h=6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$m_e=9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$	$(1+x)^n\approx1+nx$ for $\vert x\vert \ll1$
$\sin\theta\approx\theta-\theta^3/6$ for $\vert \theta\vert \ll1$	$\cos\theta\approx1-\theta^2/2$ for $\vert \theta\vert \ll1$

Possibly useful integrals

\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}(x)+C

\int\frac{dx}{1-x^2}=\tanh^{-1}(x)+C

\int\frac{dx}{(1-x^2)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{\sqrt{1+x^2}}=\sinh^{-1}(x)+C

\int\frac{dx}{1+x^2}=\tan^{-1}(x)+C

\int\frac{dx}{(1+x^2)^{3/2}}=\frac{x}{\sqrt{1+x^2}}+C

\int\sqrt{\frac{1+x}{1-x}}\,dx=-\sqrt{1-x^2}-2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{\sqrt{1+x}}{(1-x)^{3/2}}\,dx=2\sqrt{\frac{1+x}{1-x}}+2\sin^{-1}\!\left(\sqrt{\frac{1-x}{2}}\right)+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{1/2}}=\sqrt{\frac{1+x}{1-x}}+C

\int\frac{dx}{(1-x)^{3/2}(1+x)^{3/2}}=\frac{x}{\sqrt{1-x^2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{1/2}}=\frac{(2-x)\sqrt{1+x}}{3(1-x)^{3/2}}+C

\int\frac{dx}{(1-x)^{5/2}(1+x)^{3/2}}=\frac{1+2x-2x^2}{3(1-x)^{3/2}(1+x)^{1/2}}+C

You may use this sheet for both parts of the exam.

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Условие

In 1987, light from supernova SN1987A was detected by telescopes on Earth. The supernova occurred in the Large Magellanic Cloud, a distance $d = 1.5 \times 10^{21}\ \mathrm{m}$ away, making it the closest in centuries. Observations of this event tell us a remarkable amount about elementary particles.

a.

Both light and neutrinos were produced in the core of the supernova. Neutrinos are elementary particles which interact extremely weakly with ordinary matter. Detectors on Earth saw a few dozen of these neutrinos, in a burst which occurred about $T = 3\ \mathrm{hours}$ before the light arrived.

i. One explanation of these observations is that the neutrinos' speed $v$ was faster than the speed of light $c$ , violating special relativity. If this is the case, find $v - c$ in m/s.

ii. Another explanation is that the light was slowed down by the gas in the solar system, while the neutrinos always moved at speed $c$ . Suppose the solar system has a uniform index of refraction $n$ within a radius $D = 10^{13}\ \mathrm{m}$ . What would $n$ have to be to explain the time delay?

Neither of these explanations seem plausible; the modern accepted explanation is that the light was trapped for some time inside the supernova, while the neutrinos were able to leave immediately. Therefore, for the rest of this problem you should assume special relativity holds. The results listed on the previous page may be helpful.

The neutrinos did not all arrive at once. The first arrived with an energy of about $E_1 = 40\ \mathrm{MeV}$ , and the last arrived about $t = 10\ \mathrm{s}$ later with an energy of about $E_2 = 20\ \mathrm{MeV}$ .

b.

One explanation of these observations is that neutrinos have a small mass $m$ , so that when they have energy $E \gg mc^2$ , their speed $v$ is slightly slower than the speed of light.

i. Find an approximate expression for $c - v$ , to leading nontrivial order in $mc^2/E$ .

ii. Using the information above, numerically estimate the neutrino mass $m$ , in units of $\mathrm{eV}/c^2$ .

c.

Another explanation is that the neutrinos did not travel in straight lines, but rather were deflected by the intergalactic magnetic field. Suppose this field is uniform, $B = 10^{-13}\ \mathrm{T}$ , and directed perpendicular to the line joining Earth and the supernova, and that neutrinos have charge $q = \epsilon e$ .

i. If a neutrino has momentum $p$ , then in the presence of the magnetic field, it travels in a circle of radius $r = p/(qB) \gg d$ , and its path to the Earth has a total length $\ell$ . Find an approximate expression for $\ell - d$ , to leading nontrivial order in $d/r$ .

ii. Using the information above, and assuming the neutrino mass is very small so that the effect in part b is negligible, numerically estimate $\epsilon$ .

The following results from relativity may be helpful:

The Lorentz factor is defined as $\gamma = 1/\sqrt{1 - v^2/c^2}$ .
An object of mass $m$ and velocity $\mathbf{v}$ has momentum $\mathbf{p} = \gamma m \mathbf{v}$ and energy $E = \gamma m c^2$ . The force is defined by $\mathbf{F} = d\mathbf{p}/dt$ .
The momentum and energy of light are related by $E = pc$ .
In a frame $S'$ with velocity $v\hat{\mathbf{x}}$ relative to a frame $S$ , the energy and momentum are

E' = \gamma(E - v p_x), \quad p_x' = \gamma(p_x - vE/c^2).

Решение

In 1987, light from supernova SN1987A was detected by telescopes on Earth. The supernova occurred in the Large Magellanic Cloud, a distance $d = 1.5\times10^{21}\ \mathrm{m}$ away, making it the closest in centuries. Observations of this event tell us a remarkable amount about elementary particles.

a. i. If the light took a time $t$ to arrive at the Earth, then $ct = v(t-T) = d$ . Approximately solving for $v-c$ , using the fact that $v$ is very close to $c$ , gives

v - c = \frac{c^2 T}{d} = 0.65\ \mathrm{m/s}.

ii. The time delay is $T = (n-1)D/c$ , and plugging in the numbers gives $n = 1.3$ . Given how thin the gas in the solar system is, such a large value is implausible.

b. i. We start with the equation for relativistic energy:

E = \gamma m c^2

where $\gamma$ is the Lorentz factor defined as $\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}$ where $\beta \equiv \frac{v}{c}$ .

Let $\beta = 1-\delta$ , where $\delta \ll 1$ . Then,

\gamma \approx \frac{1}{\sqrt{1-1+2\delta}} \implies \delta \approx \frac{1}{2\gamma^2}.

Note that $\delta = 1 - v/c$ , so

c - v = c\delta \approx \frac{c}{2\gamma^2} = \frac{m^2 c^5}{2E^2}.

ii. We compute a relationship between $v_1$ and $v_2$ , the velocities of the first and second set of neutrinos, using the time delay. From the same equation as 1(a),

v_1 - v_2 = \frac{10\mathrm{s}\times c^2}{d} = 6\times 10^{-4}\ \mathrm{m/s}.

We have

c - v_1 \approx \frac{m^2 c^5}{2E_1^2}, \quad c - v_2 \approx \frac{2 m^2 c^5}{E_1^2}.

Then,

v_1 - v_2 = \frac{3m^2 c^5}{2E_1^2}.

Solving for $m$ gives

m = \sqrt{\frac{2(v_1-v_2)}{3c}}\frac{E_1}{c^2} = 46\ \mathrm{eV}/c^2.

c. Another explanation is that the neutrinos did not travel in straight lines, but rather were deflected by the intergalactic magnetic field. Suppose this field is uniform, $B = 10^{-13}\ \mathrm{T}$ , and directed perpendicular to the line joining Earth and the supernova, and that neutrinos have charge $q = \epsilon e$ .

i. We are computing the difference between the arc length of a small arc and the distance connecting the endpoints. If the arc length is $\ell$ , the angle subtended is $\theta = \ell/r$ . The distance connecting the end points is

d = 2r\sin(\theta/2) = 2r\sin\left(\frac{\ell}{2r}\right).

Then, Taylor expanding the sine gives

\ell - d \approx \ell - \ell + \frac{\ell^3}{24r^2} = \frac{\ell^3}{24r^2} \approx \frac{d^3}{24r^2}.

ii. The radius of the path is

r \approx \frac{E}{qBc}.

Substituting gives

\ell_2 - \ell_1 = \frac{d^3 q^2 B^2 c^2}{8E_1^2}.

Then,

q = \frac{2\sqrt{2}E_1}{Bcd}\left(\frac{\ell_2-\ell_1}{d}\right)^{1/2} = \frac{2\sqrt{2}E_1}{Bcd}\left(\frac{c\Delta t}{d}\right)^{1/2} \approx 3.6\times 10^{-15}\, e.

Then,

\epsilon \approx 3.6\times 10^{-15}.

Since the effects of a neutrino mass and charge add, and we know neutrinos have mass, this result yields a (very strong) upper bound on the possible charge of a neutrino, which as far as we know could be exactly zero. For more about the physics of SN1987A, see this paper.

Оригинал в Архива: 2023-USAPhO-Exam.pdf · официални решения: 2023-USAPhO-solutions.pdf