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

The schematic below shows the Hadley circulation in the Earth's tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm \varphi_{d}$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth's spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale.

<img_4380>

Question: (a)  Assume that there is no wind velocity in the east-west direction around the point $\mathrm{X}$. What is the expression for the east-west wind velocity $u_{Y}$ at the points Y? Convention: positive velocities point from west to east.

(The angular velocity of the Earth about its spin axis is $\Omega$, the radius of the Earth is $a$, and the thickness of the atmosphere is much smaller than $a$.)

Solution:
As the problem is symmetric about the equator, we need only consider the northern hemisphere as shown below.

<img_4488>

Conservation of angular momentum about the Earth's spin axis implies that:

$$
\begin{aligned}
& \Omega a^{2}=\left(\Omega a \cos \varphi_{d}+u_{Y}\right) a \cos \varphi_{d} \\
& u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)
\end{aligned}
$$

Final answer: $u_{Y}=\Omega a\left(\frac{1}{\cos \varphi_{d}}-\cos \varphi_{d}\right)$
Subfield: Mechanics


Question: (b)  Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation?

Tick the correct answer(s). There can be more than one correct answer.

(I) There is friction from the Earth's surface.

(II) There is turbulence in the lower atmosphere, where different layers of air are mixed

(III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth.

(IV) The air is moist at the lower levels causing retardation to the wind velocity.

Solution:
(I),(II)

Final answer: (I),(II)
Subfield: Thermodynamics


Question: (f)  Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_{A}, p_{B}, p_{C}, p_{D}, p_{E}$, respectively at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}$ by a series of inequalities.

(Given that $p_{A}=1000 \mathrm{hPa}$ and $p_{D}=225 \mathrm{hPa}$. Note that $1 \mathrm{hPa}$ is $100 \mathrm{~Pa}$.)

Solution:
Since there is less and less air above as one climbs upward in the atmosphere, atmospheric pressure must decrease upwards.

So,

$$
p_{A}>p_{B}>p_{C} \quad \text { and } \quad p_{E}>p_{D}>p_{C}
$$

The process EA represents an isothermal expansion as heat is gained from the surface. So,

$$
p_{E}>p_{A}
$$

Since the total heat gain must equal the total heat loss, more heat must be lost in the isothermal compression CD than in the isothermal expansion BC. So net heat loss occurs from $\mathrm{B}$ to $\mathrm{D}$ and hence

$$
p_{D}>p_{B}
$$

So with the values of the pressure at $A$ and $D$, we deduce that:

$$
p_{A}>p_{D}
$$

Collecting all inequalities together,

$$
p_{E}>p_{A}>p_{D}>p_{B}>p_{C}
$$

Final answer: $p_{E}>p_{A}>p_{D}>p_{B}>p_{C}$
Subfield: Thermodynamics


Question: (l)  Which of the following statements best explains why $\varepsilon$ is less than the ideal value? Tick the correct answer(s). There can be more than one correct answer.

(I) We have ignored work done against surface friction.

(II) Condensation occurs at a temperature lower than the temperature of the heat source.

(III) There is irreversible evaporation of water at the surface.

(IV) The ideal efficiency is applicable only when there is no phase change of water.

Solution:
(II) and (III)

Final answer: (II) and (III)
Subfield: Electromagnetism

The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron "biprism" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram).

<img_4450>

The electron "biprism" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the "biprism" is $\ell$, and the distance between the distance between the "biprism" and the screen is $L$.

Question: (a)  Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.)

Solution:
Writing out $|\mathrm{E}|=\frac{\lambda}{2 \pi \epsilon_{0} r}=-\frac{\partial}{\partial r} V(r)$

$$
=-\frac{\partial}{\partial r} \frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{b}{r}
$$

Note that

$$
V(r)=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{b}{r}(=0 \text { at the mesh })
$$

Also at the edge of the filament, $V_{a}=V(r=a)$, so

$$
V_{a}=\frac{\lambda}{2 \pi \epsilon_{0}} \ln \frac{b}{a}
$$

Giving together

$$
V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}
$$



Final answer: $V(r)=V_{a} \frac{\ln (b / \sqrt{x^{2}+z^{2}})}{\ln (b / a)}$
Subfield: Electromagnetism

The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron "biprism" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram).

<img_4450>

The electron "biprism" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the "biprism" is $\ell$, and the distance between the distance between the "biprism" and the screen is $L$.
Context question:
(a)  Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.)
Context answer:
$V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$


Question: (b)  An incoming electron plane wave with wave vector $k_{z}$ is deflected by the "biprism" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the "biprism" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant.

Solution:
There are several ways to work out the solution:

A charge in an electric field will experience a force and hence a change in momentum. Note that potential energy of the electron (charge $=-e_{0}$ ) is $-e_{0} V(r)$. Using impulse acting on the electron due to the electric field, 

$$
\begin{aligned}
\text { Impulse } & =\left.\frac{1}{v_{z}} \int_{-\infty}^{\infty}\left(-e_{0}\right)\left(-\frac{\partial V\left(x, z^{\prime}\right)}{\partial x}\right) d z^{\prime}\right|_{x=a} \\
& =-\left.\frac{1}{v_{z}} \int_{-\infty}^{\infty} \frac{-e_{0} V_{a} x}{\left(x^{2}+z^{\prime 2}\right) \ln \frac{b}{a}} d z^{\prime}\right|_{x=a} \\
& =\frac{e_{0} V_{a} \pi}{v_{z} \ln \frac{b}{a}} \\
\Rightarrow k_{x} & =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}
\end{aligned}
$$

Final answer: $k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$
Subfield: Electromagnetism

The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron "biprism" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram).

<img_4450>

The electron "biprism" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the "biprism" is $\ell$, and the distance between the distance between the "biprism" and the screen is $L$.
Context question:
(a)  Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.)
Context answer:
$V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$


Context question:
(b)  An incoming electron plane wave with wave vector $k_{z}$ is deflected by the "biprism" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the "biprism" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant.
Context answer:
\boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$}


Extra Supplementary Reading Materials:

(c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$,

Question: (i)  assuming relativistic effects can be ignored

Solution:
Equating the kinetic energy to $e V_{0}$ 

$$
\begin{aligned}
& \frac{h}{\lambda}=\sqrt{2 m\left|-e_{0}\right| V_{0}} \\
& \lambda=\frac{h}{\sqrt{2 m e_{0} V_{0}}}
\end{aligned}
$$

Final answer: $\frac{h}{\sqrt{2 m e_{0} V_{0}}}$
Subfield: Electromagnetism

The two-slit electron interference experiment was first performed by Möllenstedt et al, MerliMissiroli and Pozzi in 1974 and Tonomura et al in 1989. In the two-slit electron interference experiment, a monochromatic electron point source emits particles at $S$ that first passes through an electron "biprism" before impinging on an observational plane; $S_{1}$ and $S_{2}$ are virtual sources at distance $d$. In the diagram, the filament is pointing into the page. Note that it is a very thin filament (not drawn to scale in the diagram).

<img_4450>

The electron "biprism" consists of a grounded cylindrical wire mesh with a fine filament $F$ at the center. The distance between the source and the "biprism" is $\ell$, and the distance between the distance between the "biprism" and the screen is $L$.
Context question:
(a)  Taking the center of the circular cross section of the filament as the origin $O$, find the electric potential at any point $(x, z)$ very near the filament in terms of $V_{a}, a$ and $b$ where $V_{a}$ is the electric potential of the surface of the filament, $a$ is the radius of the filament and $b$ is the distance between the center of the filament and the cylindrical wire mesh. (Ignore mirror charges.)
Context answer:
$V(r)=V_{a} \frac{\ln (b / r)}{\ln (b / a)} \text { where } r=\sqrt{x^{2}+z^{2}}$


Context question:
(b)  An incoming electron plane wave with wave vector $k_{z}$ is deflected by the "biprism" due to the $x$-component of the force exerted on the electron. Determine $k_{x}$ the $x$-component of the wave vector due to the "biprism" in terms of the electron charge, $e, v_{\mathrm{z}}, V_{a}, k_{\mathrm{z}}, a$ and $b$, where $e$ and $v_{z}$ are the charge and the $z$-component of the velocity of the electrons $\left(k_{x} \ll k_{z}\right)$. Note that $\vec{k}=\frac{2 \pi \vec{p}}{h}$ where $h$ is the Planck constant.
Context answer:
\boxed{$k_{x} =\frac{e_{0} V_{a} \pi}{\hbar v_{z} \ln \frac{b}{a}}$}


Extra Supplementary Reading Materials:

(c) Before the point $S$, electrons are emitted from a field emission tip and accelerated through a potential $V_{0}$. Determine the wavelength of the electron in terms of the (rest) mass $m$, charge $e$ and $V_{0}$,
Context question:
(i)  assuming relativistic effects can be ignored
Context answer:
\boxed{$\frac{h}{\sqrt{2 m e_{0} V_{0}}}$}


Question: (ii)  taking relativistic effects into consideration.

Solution:
Consider

$$
\begin{aligned}
E^{2} & =(p c)^{2}+\left(m c^{2}\right)^{2} \\
& =\left(\frac{h}{\lambda} c\right)^{2}+\left(m c^{2}\right)^{2} \\
\frac{h^{2} c^{2}}{\lambda^{2}} & =\left(m c^{2}+\left|-e_{0}\right| V_{0}\right)^{2}-\left(m c^{2}\right)^{2} \\
& =2 m c^{2} e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right) \\
\lambda & =\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}
\end{aligned}
$$

Final answer: $\frac{h}{\sqrt{2 m e_{0} V_{0}\left(1+\frac{e_{0} V_{0}}{2 m_{0} c^{2}}\right)}}$
Subfield: Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.

Question: 1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.

Solution:
Using Gauss law

$$
\oint \mathbf{E} \cdot \mathbf{d} \mathbf{A}=\frac{q}{\epsilon_{0}} .
\tag{1}
$$

From symetry we know that the electric field only has radial component. Choose a cylinder (with a line charge as the axis) as the Gaussian surface, we obtain

$$
E .2 \pi r l=\frac{\lambda l}{\epsilon_{0}}
$$

Simplify to obtain

$$
\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}
\tag{2}
$$



Final answer: $\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$
Subfield: Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.
Context question:
1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
Context answer:
$\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$


Question: 2.  The potential due to the line charge could be written as

$$
V(r)=f(r)+K,
$$

where $K$ is a constant. Determine $f(r)$.

Solution:
The potential is given by

$$
\begin{aligned}
V & =-\int_{\mathrm{ref}}^{r} \mathbf{E} \cdot \mathbf{d} \mathbf{l} \\
& =-\int_{\mathrm{ref}}^{r} E \cdot d r \\
V & =-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r+K,
\end{aligned}
\tag{3}
$$

so $f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$. where $K$ is a constant.

Final answer: $f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$
Subfield: Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.
Context question:
1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
Context answer:
$\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$


Context question:
2.  The potential due to the line charge could be written as

$$
V(r)=f(r)+K,
$$

where $K$ is a constant. Determine $f(r)$.
Context answer:
\boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$}


Context question:
3.  Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces.
Context answer:
<img_4501>


Extra Supplementary Reading Materials:

For the following questions, ignore any edge effects.
Context question:
4.  Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin.
Context answer:
the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$

The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$.

The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$.


Question: 5.  Calculate the capacitance $C$ of the system.

Solution:
From eq.(18), we have

$$
V_{0}=\frac{q}{l \pi \epsilon_{0}} \ln 3
\tag{20}
$$

so we get

$$
C=\frac{q}{V_{0}}=\frac{l \pi \epsilon_{0}}{\ln 3}
\tag{21}
$$

Final answer: $\frac{l \pi \epsilon_{0}}{\ln 3}$
Subfield: Electromagnetism

A system consisted of two conductor bodies is immersed in a uniform dielectric and weakly conducting liquid. When a constant voltage difference is applied between both conductors, the system has both electric and magnetic fields. In this problem we will investigate this system.
Context question:
1.  First consider an infinitely long line with charge per unit length $\lambda$ in vacuum. Calculate the electric field $\mathbf{E}(\mathbf{r})$ due to the line.
Context answer:
$\mathbf{E}=\hat{r} \frac{\lambda}{2 \pi \epsilon_{0} r}$


Context question:
2.  The potential due to the line charge could be written as

$$
V(r)=f(r)+K,
$$

where $K$ is a constant. Determine $f(r)$.
Context answer:
\boxed{$f(r)=-\frac{\lambda}{2 \pi \epsilon_{0}} \ln r$}


Context question:
3.  Calculate the potential in all space $V(x, y, z)$ due to an infinitely long line with charge per unit length $\lambda$ at $x=-b, y=0$ and another infinitely long line with charge per unit length $-\lambda$ at $x=b, y=0$. Both lines are parallel to the $z$-axis. Take $V=0$ at the origin. Sketch the equipotential surfaces.
Context answer:
<img_4501>


Extra Supplementary Reading Materials:

For the following questions, ignore any edge effects.
Context question:
4.  Now consider two identical conducting cylinders, both with radius $R=3 a$ in vacuum. The length of each cylinders are the same and much larger than its radius $(l \gg R)$. The axis of both cylinders are on the $x z$-plane and parallel to the $z$-axis, one at $x=-5 a, y=0$ and the other at $x=5 a, y=0$. An electrical potential difference of $V_{0}$ is applied between the two cylinders (the cylinder at $x=-5 a$ has the higher potential) by connecting them to a battery. Calculate the potential in all regions. Take $V=0$ at the origin.
Context answer:
the potential outside the two cylinders are $V=\frac{V_{0}}{4 \ln 3} \ln \frac{(4 a-x)^{2}+y^{2}}{(4 a+x)^{2}+y^{2}}$

The potential inside the cylinder centered at $(x=5 a, y=0)$ is $V=-V_{0} / 2$.

The potential inside the cylinder centered at $(x=-5 a, y=0)$ is $V=V_{0} / 2$.


Context question:
5.  Calculate the capacitance $C$ of the system.
Context answer:
\boxed{$\frac{l \pi \epsilon_{0}}{\ln 3}$}


Context question:
6.  Now both cylinders are totally immersed in a weakly conducting liquid with conductivity $\sigma$. Calculate the total current that flows between both cylinders. Assume the permittivity of the liquid is equal to that of vacuum, $\epsilon=\epsilon_{0}$.
Context answer:
\boxed{$I=\frac{V_{0} \pi \sigma l}{\ln 3}$}


Question: 7.  Calculate the resistance $R$ of the system. Calculate $R C$ of the system.

Solution:
The resistance is given by

$$
R=\frac{V_{0}}{I}=\frac{\ln 3}{\pi \sigma l}
\tag{27}
$$

and therefore

$$
R C=\frac{\epsilon_{0}}{\sigma}
\tag{28}
$$

Final answer: $R C=\frac{\epsilon_{0}}{\sigma}$
Subfield: Modern Physics

Global Positioning System (GPS) is a navigation technology which uses signal from satellites to determine the position of an object (for example an airplane). However, due to the satellites high speed movement in orbit, there should be a special relativistic correction, and due to their high altitude, there should be a general relativistic correction. Both corrections seem to be small but are very important for precise measurement of position. We will explore both corrections in this problem.

First we will investigate the special relativistic effect on an accelerated particle. We consider two types of frame, the first one is the rest frame (called $S$ or Earth's frame), where the particle is at rest initially. The other is the proper frame (called $S^{\prime}$ ), a frame that instantaneously moves together with the accelerated particle. Note that this is not an accelerated frame, it is a constant velocity frame that at a particular moment has the same velocity with the accelerated particle. At that short moment, the time rate experienced by the particle is the same as the proper frame's time rate. Of course this proper frame is only good for an infinitesimally short time, and then we need to define a new proper frame afterward. At the beginning we synchronize the particle's clock with the clock in the rest frame by setting them to zero, $t=\tau=0$ ( $t$ is the time in the rest frame, and $\tau$ is the time shown by particle's clock).

By applying equivalence principle, we can obtain general relativistic effects from special relavistic results which does not involve complicated metric tensor calculations. By combining the special and general relativistic effects, we can calculate the corrections needed for a GPS (global positioning system) satellite to provide accurate positioning.

Some mathematics formulas that might be useful

- $\sinh x=\frac{e^{x}-e^{-x}}{2}$
- $\cosh x=\frac{e^{x}+e^{-x}}{2}$
- $\tanh x=\frac{\sinh x}{\cosh x}$
- $1+\sinh ^{2} x=\cosh ^{2} x$
- $\sinh (x-y)=\sinh x \cosh y-\cosh x \sinh y$



- $\int \frac{d x}{\left(1-x^{2}\right)^{\frac{3}{2}}}=\frac{x}{\sqrt{1-x^{2}}}+C$
- $\int \frac{d x}{1-x^{2}}=\ln \sqrt{\frac{1+x}{1-x}}+C$


Part A. Single Accelerated Particle 

Consider a particle with a rest mass $m$ under a constant and uniform force field $F$ (defined in the rest frame) pointing in the positive $x$ direction. Initially $(t=\tau=0)$ the particle is at rest at the origin $(x=0)$.

Question: 1.  When the velocity of the particle is $v$, calculate the acceleration of the particle, $a$ (with respect to the rest frame).

Solution:
The equation of motion is given by

$$
\begin{aligned}
F & =\frac{d}{d t}(\gamma m v) \\
& =\frac{m c \dot{\beta}}{\left(1-\beta^{2}\right)^{\frac{3}{2}}}
\end{aligned}
\tag{1}
$$

$$
F=\gamma^{3} m a
\tag{2}
$$

where $\gamma=\frac{1}{\sqrt{1-\beta^{2}}}$ and $\beta=\frac{v}{c}$. So the acceleration is given by

$$
a=\frac{F}{\gamma^{3} m}
\tag{3}
$$

Final answer: $a=\frac{F}{\gamma^{3} m}$