weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: B.4 If $\Delta E_{r m s}=5.54 \times 10^{-17} \mathrm{~J}$, calculate the rms speed of the Be nuclei, $\mathrm{V}_{\mathrm{Be}}$, and hence estimate $T_{\mathrm{c}}$. (Hint: $\Delta E_{r m s}$ depends on the rms value of the component of velocity along the line of sight).

Solution: Moving ${ }^{7} \mathrm{Be}$ nuclei give rise to Doppler effect for neutrinos. Since the fractional change in energy ( $\Delta E_{\mathrm{rms}} / E_{\nu} \sim 10^{-4}$ ) is small, the Doppler shift may be considered in the nonrelativistic limit (a relativistic treatment gives almost same answer). Taking the line of sight along the $z$-direction,

$$
\begin{aligned}
\frac{\Delta E_{\mathrm{rms}}}{E_{\nu}} & =\frac{v_{z, r m s}}{c} \\
& =3.85 \times 10^{-4} \\
& =\frac{1}{\sqrt{3}} \frac{V_{\mathrm{Be}}}{c}
\end{aligned}
$$

$\Rightarrow V_{\mathrm{Be}}=\sqrt{3} \times 3.85 \times 10^{-4} \times 3.00 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}=2.01 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}$.

The average temperature is obtained by equating the average kinetic energy to the thermal energy.

$$
\begin{gathered}
\frac{1}{2} m_{\mathrm{Be}} V_{\mathrm{Be}}^{2}=\frac{3}{2} k_{\mathrm{B}} T_{\mathrm{c}} \\
\Rightarrow \quad T_{\mathrm{c}}=1.13 \times 10^{7} \mathrm{~K}
\end{gathered}
$$



Answer: $V_{\mathrm{Be}}=2.01 \times 10^{5}$ , $T_{\mathrm{c}}=1.13 \times 10^{7}$
Problem: 1 Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 

Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.

Solution: 1) According to the first law of thermodynamics, the amount of heat transmitted $\delta Q$ to the gas in the bubble is found as

$$
\delta Q=v C_{V} d T+p d V
\tag{B.1}
$$

where the molar heat capacity at arbitrary process is as follows

$$
C=\frac{1}{v} \frac{\delta Q}{d T}=C_{V}+\frac{p}{v} \frac{d V}{d T}
\tag{B.2}
$$

Here $C_{V}$ stands for the molar heat capacity of the gas at constant volume, $p$ designates its pressure, $v$ is the total amount of moles of gas in the bubble, $V$ and $T$ denote the volume and temperature of the gas, respectively.

Evaluate the derivative standing on the right hand side of (B.2). According to the Laplace formula, the gas pressure inside the bubble is defined by

$$
p=\frac{4 \sigma}{r}
\tag{B.3}
$$

thus, the equation of any equilibrium process with the gas in the bubble is a polytrope of the form

$$
p^{3} V=\text { const. }
\tag{B.4}
$$

The equation of state of an ideal gas has the form

$$
p V=v R T
\tag{B.5}
$$

and hence equation (B.4) can be rewritten as

$$
T^{3} V^{-2}=\text { const. }
\tag{B.6}
$$

Differentiating (B.6) the derivative with respect to temperature sought is found as

$$
\frac{d V}{d T}=\frac{3 V}{2 T}
\tag{B.7}
$$

Taking into account that the molar heat capacity of a diatomic gas at constant volume is

$$
C_{V}=\frac{5}{2} R
\tag{B.8}
$$

and using (B.5) it is finally obtained that

$$
C=C_{V}+\frac{3}{2} R=4 R=33.2 \frac{\mathrm{J}}{\mathrm{mole} \cdot \mathrm{K}}
\tag{B.9}
$$

Answer: $33.2 $
Problem: 2 Find formula for the frequency $\omega$ of the small radial oscillations of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations.

Hint: Laplace showed that there is pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that $\Delta p=\frac{2 \sigma}{r}$.

Solution: Since the heat capacity of the gas is much smaller than the heat capacity of the soap film, and there is heat exchange between them, the gas can be considered as isothermal since the soap film plays the role of thermostat. Consider the fragment of soap film, limited by the angle $\alpha$ as shown in the figure. It's area is found as

$$
S=\pi(\alpha r)^{2}
\tag{B.10}
$$

and the corresponding mass is obtained as

$$
m=\rho S h \text {. }
\tag{B.11}
$$

Let $x$ be an increase in the radius of the bubble, then the Newton second law for the fragment of the soap film mentioned above takes the form

$$
m \ddot{x}=p^{\prime} S^{\prime}-F_{\text {surf }},
\tag{B.12}
$$

where $F_{\text {surf }}$ denotes the projection of the resultant surface tension force acting in the radial direction, $p^{\prime}$ stands for the gas pressure beneath the surface of the soap film and

$$
S^{\prime}=S\left(1+2 \frac{x}{r}\right)
$$

$F_{\text {surf }}$ is easily found as

$$
F_{\text {surf }}=F_{S T} \alpha=\sigma \cdot 2 \cdot 2 \pi[(r+x) \alpha] \cdot \alpha .
\tag{B.13}
$$

Since the gaseous process can be considered isothermal, it is

<img_4410>
written that

$$
p^{\prime} V^{\prime}=p V \text {. }
\tag{B.14}
$$

Assuming that the volume increase is quite small, (B.14) yields

$$
p^{\prime}=p \frac{1}{\left(1+\frac{x}{r}\right)^{3}} \approx p \frac{1}{\left(1+\frac{3 x}{r}\right)} \approx p\left(1-\frac{3 x}{r}\right)
\tag{B.15}
$$

Thus, from (B.10) - (B.16) and (B.3) the equation of small oscillations of the soap film is derived as

$$
\rho h \ddot{x}=-\frac{8 \sigma}{r^{2}} x
\tag{B.16}
$$



with the frequency

$$
\omega=\sqrt{\frac{8 \sigma}{\rho h r^{2}}}=108 \mathrm{~s}^{-1}
\tag{B.17}
$$

Answer: $108 $
Problem: A1 Estimate $b$ and express it in terms of the diameter of the molecules $d$.

Solution: If $V=b$ is substituted into the equation of state, then the gas pressure turns infinite. It is obvious that this is the moment when all the molecules are tightly packed. Therefore, the parameter $b$ is approximately equal to the volume of all molecules, i.e.

$$
b=N_{A} d^{3}
\tag{A1.1}
$$

Answer: $b=N_{A} d^{3}$
Problem: A1 Find $_{0}, a, b$ and express them in terms of $Z_{\text {ext }}$ and $r$.

Solution: Let us derive an equation describing the change of the electron number density with time. It is determined by the two processes; the generation of ion pairs by external ionizer and the recombination of electrons with ions. At ionization process electrons and ions are generated in pairs, and at recombination processthey disappear in pairs as well.Thus, their concentrations are alwaysequal at any given time, i.e.

$$
n(t)=n_{e}(t)=n_{i}(t)
\tag{A1.1}
$$

Then the equation describing the numberdensityevolution of electrons and ions in time can be written as

$$
\frac{d n(t)}{d t}=Z_{e x t}-r n(t)^{2}
\tag{A1.2}
$$

It is easy to show that at $t \rightarrow 0$ the function $\tanh b t \rightarrow 0$, therefore, by virtue of the initial condition $n(0)=0$,one finds

$$
n_{0}=0
\tag{A1.3}
$$

Substituting $n_{e}(t)=a \tanh b t$ in (A1.2) and separating it in the independent functions (hyperbolic, or 1 and $\left.e^{x}\right)$, one gets

$$
a=\sqrt{\frac{Z_{e x t}}{r}}
\tag{A1.4}
$$
$$
b=\sqrt{r Z_{e x t}}
\tag{A1.5}
$$

Answer: $n_{0}=0$ , $a=\sqrt{\frac{Z_{e x t}}{r}}$ , $b=\sqrt{r Z_{e x t}}$
Problem: A2 Find the electron number density $n_{e}$ at equilibrium when both external ionizers are switched on simultaneously.

Solution: According to equation (A1.4) the number density of electronsat steady-state is expressed in terms of the external ionizer activity as

$$
n_{e 1}=\sqrt{\frac{Z_{e x t 1}}{r}}
\tag{A2.1}
$$
$$
n_{e 2}=\sqrt{\frac{Z_{e x t 2}}{r}}
\tag{A2.2}
$$
$$
n_{e}=\sqrt{\frac{Z_{e x t 1}+Z_{e x t 2}}{r}}
\tag{A2.3}
$$

Thus,the following analogue of the Pythagorean theorem is obtained as

$$
n_{e}=\sqrt{n_{e 1}^{2}+n_{e 2}^{2}}=20.0 \cdot 10^{10} \mathrm{~cm}^{-3}
\tag{A2.4}
$$

Answer: $20.0 \cdot 10^{10}$
Problem: A3 Express the electric current $I$ in the tube in terms of $U, \beta, L, S, Z_{\mathrm{ext}}, r$ and $e$ which is the elementary charge.

Solution: In the steady state, the balance equations of electrons and ions in the tube volume take the form

$$
Z_{\text {ext }} S L=r n_{e} n_{i} S L+\frac{I_{e}}{e}
\tag{A3.1}
$$
$$
Z_{\text {ext }} S L=r n_{e} n_{i} S L+\frac{I_{i}}{e}
\tag{A3.2}
$$

It follows from equations (A3.1) and (A3.2) that the ion and electron currents are equal, i.e.

$$
I_{e}=I_{i}
\tag{A3.3}
$$

At the same time the total current in each tube section is the sum of the electron and ion currents

$$
I=I_{e}+I_{i}
\tag{A3.4}
$$

By definition ofthe current density the following relations hold

$$
I_{e}=\frac{I}{2}=e n_{e} v S=e \beta n_{e} E S
\tag{A3.5}
$$
$$
I_{i}=\frac{I}{2}=e n_{i} v S=e \beta n_{i} E S
\tag{A3.6}
$$

Substituting (A3.5) and (A3.6) into (A3.1) and (A3.2), the following quadratic equation for the current is derived

$$
Z_{e x t} S L=r S L\left(\frac{I}{2 e \beta E S}\right)^{2}+\frac{I}{2 e}
\tag{A3.7}
$$

The electric field strength in the gas is equal to

$$
E=\frac{U}{L}
\tag{A3.8}
$$

and solution to the quadratic equation (A3.7) takes the form

$$
I=\frac{e \beta^{2} U^{2} S}{r L^{3}}\left(-1 \pm \sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}\right)
\tag{A3.9}
$$



It is obvious that only positive root does make sense, i.e.

$$
I=\frac{e \beta^{2} U^{2} S}{r L^{3}}\left(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1\right)
\tag{A3.10}
$$

Answer: $I=\frac{e \beta^{2} U^{2} S}{r L^{3}}(\sqrt{1+\frac{4 r Z_{e x t} L^{4}}{\beta^{2} U^{2}}}-1)$
Problem: A4 Find the resistivity $\rho_{gas}$ of the gas at sufficiently small values of the voltage applied and express it in terms of $\beta,L,Z_{ext},r$ and $e$

Solution: At low voltages (A3.10) simplifies and gives the following expression

$$
I=2 U e \beta \sqrt{\frac{Z_{e x t}}{r}} \frac{S}{L}
\tag{A4.1}
$$

which is actually the Ohm law.

Using the well-known relation

$$
R=\frac{U}{I}
\tag{A4.2}
$$

together with

$$
R=\rho \frac{L}{S}
\tag{A4.3}
$$

one gets

$$
\rho=\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}
\tag{A4.4}
$$

Answer: $\frac{1}{2 e \beta} \sqrt{\frac{r}{Z_{e x t}}}$
Problem: i. By adjusting the launching angle for a ball thrown with a fixed initial speed $v_{0}$ from the origin, targets can be hit within the region given by

$$
z \leq z_{0}-k x^{2}
$$

You can use this fact without proving it. Find the constants $z_{0}$ and $k$.

Solution: When the stone is thrown vertically upwards, it can reach the point $x=0, z=v_{0}^{2} / 2 g$ (as it follows from the energy conservation law). Comparing this with the inequality $z \leq z_{0}-k x^{2}$ we conclude that

$$
z_{0}=v_{0}^{2} / 2 g
$$

Let us consider the asymptotics $z \rightarrow-\infty$; the trajectory of the stone is a parabola, and at this limit, the horizontal displacement (for the given $z$ ) is very sensitive with respect to the curvature of the parabola: the flatter the parabola, the larger the displacement. The parabola has the flattest shape when the stone is thrown horizontally, $x=v_{0} t$ and $z=-g t^{2} / 2$, i.e. its trajectory is given by $z=-g x^{2} / 2 v_{0}^{2}$. Now, let us recall that $z \leq z_{0}-k x^{2}$, i.e. $-g x^{2} / 2 v_{0}^{2} \leq z_{0}-k x^{2} \Rightarrow k \leq g / 2 v_{0}^{2}$. Note that $k<g / 2 v_{0}^{2}$ would imply that there is a gap between the parabolic region $z \leq z_{0}-k x^{2}$ and the given trajectory $z=-g x^{2} / 2 v_{0}^{2}$. This trajectory is supposed to be optimal for hitting targets far below $(z \rightarrow-\infty)$, so there should be no such a gap, and hence, we can exclude the option $k<g / 2 v_{0}^{2}$. This leaves us with

$$
k=g / 2 v_{0}^{2}
$$

Answer: $\frac{v_{0}^{2}}{ 2 g}$ , $\frac{g}{2 v_{0}^{2}}$
Problem: i. During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, i.e. the ball stays in thermodynamic equilibrium with its surroundings. What is the number of times, $n$, by which the pressure increases when the radius is halved to $r_{1}=0.5 r_{0}$ ? Assume that the gas density remains uniform.

Solution: $$
\begin{aligned}
& T=\text { const } \Longrightarrow p V=\text { const } \\
& V \propto r^{3} \\
& \therefore p \propto r^{-3} \Longrightarrow \frac{p\left(r_{1}\right)}{p\left(r_{0}\right)}=2^{3}=8 .
\end{aligned}
$$

Answer: $n=8$
Problem: ii. Estimate the time $t_{2}$ needed for the radius to shrink from $r_{0}$ to $r_{2}=0.95 r_{0}$. Neglect the change of the gravity field at the position of a falling gas particle.

Solution: During the period considered the pressure is negligible. Therefore the gas is in free fall. By Gauss' theorem and symmetry, the gravitational field at any point in the ball is equivalent to the one generated when all the mass closer to the center is compressed into the center. Moreover, while the ball has not yet shrunk much, the field strength on its surface does not change much either. The acceleration of the outermost layer stays approximately constant. Thus,

$$
t \approx \sqrt{\frac{2\left(r_{0}-r_{2}\right)}{g}}
$$

where

$$
\begin{aligned}
g & \approx \frac{G m}{r_{0}^{2}} \\
\therefore t & \approx \sqrt{\frac{2 r_{0}^{2}\left(r_{0}-r_{2}\right)}{G m}}=\sqrt{\frac{0.1 r_{0}^{3}}{G m}} .
\end{aligned}
$$

Answer: $\sqrt{\frac{0.1 r_{0}^{3}}{G m}}$
Problem: iii. Assuming that the pressure remains negligible, find the time $t_{r \rightarrow 0}$ needed for the ball to collapse from $r_{0}$ down to a much smaller radius, using Kepler's Laws.

Solution: Gravitationally the outer layer of the ball is influenced by the rest just as the rest were compressed into a point mass. Therefore we have Keplerian motion: the fall of any part of the outer layer consists in a halfperiod of an ultraelliptical orbit. The ellipse is degenerate into a line; its foci are at the ends of the line; one focus is at the center of the ball (by Kepler's $1^{\text {st }}$ law) and the other one is at $r_{0}$, see figure (instead of a degenerate ellipse, a strongly elliptical ellipse is depicted). The period of the orbit is determined by the longer semiaxis of the ellipse (by Kepler's $3^{\text {rd }}$ law). The longer semiaxis is $r_{0} / 2$ and we are interested in half a period. Thus, the answer is equal to the halfperiod of a circular orbit of radius $r_{0} / 2$ :

$$
\left(\frac{2 \pi}{2 t_{r \rightarrow 0}}\right)^{2} \frac{r_{0}}{2}=\frac{G m}{\left(r_{0} / 2\right)^{2}} \Longrightarrow t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}
$$

centre of the cloud initial position of a certain parcel

strongly elliptical orbit of the gas parcel of gas area covered by the radius vector

Alternatively, one may write the energy conservation law $\frac{\dot{r}^{2}}{2}-\frac{G m}{r}=E$ (that in turn is obtainable from Newton's II law $\left.\ddot{r}=-\frac{G m}{r^{2}}\right)$ with $E=-\frac{G m}{r_{0}}$, separate the variables $\left(\frac{d r}{d t}=-\sqrt{2 E+\frac{2 G m}{r}}\right)$ and write the integral $t=-\int \frac{d r}{\sqrt{2 E+\frac{2 G m}{r}}}$ This integral is probably not calculable during the limitted time given during the Olympiad, but a possible approach can be sketched as follows. Substituting $\sqrt{2 E+\frac{2 G m}{r}}=\xi$ and $\sqrt{2 E}=v$, one gets

$$
\begin{aligned}
\frac{t_{\infty}}{4 G m} & =\int_{0}^{\infty} \frac{d \xi}{\left(v^{2}-\xi^{2}\right)^{2}} \\
& =\frac{1}{4 v^{3}} \int_{0}^{\infty}\left[\frac{v}{(v-\xi)^{2}}+\frac{v}{(v+\xi)^{2}}+\frac{1}{v-\xi}+\frac{1}{v+\xi}\right] d \xi
\end{aligned}
$$

Here (after shifting the variable) one can use $\int \frac{d \xi}{\xi}=\ln \xi$ and $\int \frac{d \xi}{\xi^{2}}=-\frac{1}{\xi}$, finally getting the same answer as by Kepler's laws.

Answer: $t_{r \rightarrow 0}=\pi \sqrt{\frac{r_{0}^{3}}{8 G m}}$
Problem: iv. At some radius $r_{3} \ll r_{0}$, the gas becomes dense enough to be opaque to the heat radiation. Calculate the amount of heat $Q$ radiated away during the collapse from the radius $r_{0}$ down to $r_{3}$.

Solution: By Clapeyron-Mendeleyev law,

$$
p=\frac{m R T_{0}}{\mu V}
$$

Work done by gravity to compress the ball is

$$
W=-\int p d V=-\frac{m R T_{0}}{\mu} \int_{\frac{4}{3} \pi r_{0}^{3}}^{\frac{4}{3} \pi r_{3}^{3}} \frac{d V}{V}=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}
$$

The temperature stays constant, so the internal energy does not change; hence, according to the $1^{\text {st }}$ law of thermodynamics, the compression work $W$ is the heat radiated.

Answer: $Q=\frac{3 m R T_{0}}{\mu} \ln \frac{r_{0}}{r_{3}}$
Problem: v. For radii smaller than $r_{3}$ you may neglect heat loss due to radiation. Determine how the temperature $T$ of the ball depends on its radius for $r<r_{3}$.

Solution: The collapse continues adiabatically.

$$
\begin{aligned}
& p V^{\gamma}=\text { const } \Longrightarrow T V^{\gamma-1}=\text { const. } \\
& \therefore T \propto V^{1-\gamma} \propto r^{3-3 \gamma} \\
& \therefore T=T_{0}\left(\frac{r_{3}}{r}\right)^{3 \gamma-3} .
\end{aligned}
$$

Answer: $T(r)=T_{0}\left(\frac{r_{3}}{r}\right)^{3 \gamma-3}$
Problem: vi. Eventually we cannot neglect the effect of the pressure on the dynamics of the gas and the collapse stops at $r=r_{4}$ (with $r_{4} \ll r_{3}$ ). However, the radiation loss can still be neglected and the temperature is not yet high enough to ignite nuclear fusion. The pressure of such a protostar is not uniform anymore, but rough estimates with inaccurate numerical prefactors can still be done. Estimate the final radius $r_{4}$ and the respective temperature $T_{4}$.

Solution: During the collapse, the gravitational energy is converted into heat. Since $r_{3} \gg r_{4}$, The released gravitational energy can be estimated as $\Delta \Pi=-G m^{2}\left(r_{4}^{-1}-r_{3}^{-1}\right) \approx-G m^{2} / r_{4}$ (exact calculation by integration adds a prefactor $\frac{3}{5}$ ); the terminal heat energy is estimated as $\Delta Q=c_{V} \frac{m}{\mu}\left(T_{4}-T_{0}\right) \approx$ $c_{V} \frac{m}{\mu} T_{4}$ (the approximation $T_{4} \gg T_{0}$ follows from the result of the previous question, when combined with $r_{3} \gg r_{4}$ ). So, $\Delta Q=\frac{R}{\gamma-1} \frac{m}{\mu} T_{4} \approx \frac{m}{\mu} R T_{4}$. For the temperature $T_{4}$, we can use the result of the previous question, $T_{4}=T_{0}\left(\frac{r_{3}}{r_{4}}\right)^{3 \gamma-3}$. Since initial full energy was approximately zero, $\Delta Q+\Delta \Pi \approx 0$, we obtain

$$
\frac{G m^{2}}{r_{4}} \approx \frac{m}{\mu} R T_{0}\left(\frac{r_{3}}{r_{4}}\right)^{3 \gamma-3} \Longrightarrow r_{4} \approx r_{3}\left(\frac{R T_{0} r_{3}}{\mu m G}\right)^{\frac{1}{3 \gamma-4}}
$$

Therefore,

$$
T_{4} \approx T_{0}\left(\frac{R T_{0} r_{3}}{\mu m G}\right)^{\frac{3 \gamma-3}{4-3 \gamma}}
$$

Alternatively, one can obtain the result by approximately equating the hydrostatic pressure $\rho r_{4} \frac{G m}{r_{4}^{2}}$ to the gas pressure $p_{4}=\frac{\rho}{\mu} R T_{4}$; the result will be exactly the same as given above.

Answer: $r_{3}\left(\frac{R T_{0} r_{3}}{\mu m G}\right)^{\frac{1}{3 \gamma-4}}$,$T_{0}\left(\frac{R T_{0} r_{3}}{\mu m G}\right)^{\frac{3 \gamma-3}{4-3 \gamma}}$
Problem: A.1 Find the exact expression for the final velocity $v$ of the protons as a function of the accelerating voltage $V$, and physical constants.

Solution: Conservation of energy:

$$
m_{p} \cdot c^{2}+V \cdot e=m_{p} \cdot c^{2} \cdot \gamma=\frac{m_{p} \cdot c^{2}}{\sqrt{1-v^{2} / c^{2}}}
$$

Penalties

$$
\text { No or incorrect total energy }
$$

Solve for velocity:

$$
v=c \cdot \sqrt{1-\left(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e}\right)^{2}}
$$

Answer: $v=c \cdot \sqrt{1-(\frac{m_{p} \cdot c^{2}}{m_{p} \cdot c^{2}+V \cdot e})^{2}}$
Problem: A.2 For particles with high energy and low mass the relative deviation $\Delta=(c-v) / c$ of the final velocity $v$ from the speed of light is very small. Find a first order approximation for $\Delta$ and calculate $\Delta$ for electrons with an energy of $60.0 \mathrm{GeVusing}$ the accelerating voltage $V$ and physical constants.

Solution: velocity (from previous question):

$$
v=c \cdot \sqrt{1-\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}} \text { or } c \cdot \sqrt{1-\left(\frac{m_{e} \cdot c^{2}}{V \cdot e}\right)^{2}}
$$

relative difference:

$$
\begin{gathered}
\Delta=\frac{c-v}{c}=1-\frac{v}{c} \\
\rightarrow \Delta \simeq \frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2} \text { or } \frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{V \cdot e}\right)^{2}
\end{gathered}
$$

relative difference

$$
\Delta=3.63 \cdot 10^{-11}
$$

Answer: $\frac{1}{2}\left(\frac{m_{e} \cdot c^{2}}{m_{e} \cdot c^{2}+V \cdot e}\right)^{2}$, $\Delta=3.63 \cdot 10^{-11}$
Problem: A.3 Derive an expression for the uniform magnetic flux density $B$ necessary to keep the proton beam on a circular track. The expression should only contain the energy of the protons $E$, the circumference $L$, fundamental constants and numbers. You may use suitable approximations if their effect is smaller than precision given by the least number of significant digits.

Calculate the magnetic flux density $B$ for a proton energy of $E=7.00 \mathrm{TeV}$, neglecting interactions between the protons.

Solution: Balance of forces:

$$
\frac{\gamma \cdot m_{p} \cdot v^{2}}{r}=\frac{m_{p} \cdot v^{2}}{r \cdot \sqrt{1-\frac{v^{2}}{c^{2}}}}=e \cdot v \cdot B
$$

In case of a mistake, partial points can be given for intermediate steps (up to max 0.2). Examples:

$$
\begin{array}{ll}
\text { Example: } & \text { Lorentz force } \\
\text { Example: } & \frac{\gamma \cdot m_{p} \cdot v^{2}}{r}
\end{array}
$$

Energy:

$$
E=(\gamma-1) \cdot m_{p} \cdot c^{2} \simeq \gamma \cdot m_{p} \cdot c^{2} \rightarrow \gamma=\frac{E}{m_{p} c^{2}}
$$

Therefore:

$$
\frac{E \cdot v}{c^{2} \cdot r}=e \cdot B
$$

With

$$
v \simeq c \text { and } r=\frac{L}{2 \pi}
$$

follows:

$$
\rightarrow B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}
$$

Solution:

$$
B=5.50 \mathrm{~T}
$$

Answer: $B=\frac{2 \pi \cdot E}{e \cdot c \cdot L}$ , $B=5.50$
Problem: A.4 Use dimensional analysis to find an expression for the radiated power $P_{\text {rad }}$.

Solution: Ansatz:

$$
P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}
$$

Dimensions: $[\mathrm{a}]=\mathrm{ms}^{-2},[\mathrm{q}]=\mathrm{C}=\mathrm{As},[\mathrm{c}]=\mathrm{ms}^{-1},\left[\epsilon_{0}\right]=\mathrm{As}(\mathrm{Vm})^{-1}=\mathrm{A}^{2} \mathrm{~s}^{2}\left(\mathrm{Nm}^{2}\right)^{-1}=\mathrm{A}^{2} \mathrm{~s}^{4}\left(\mathrm{kgm}^{3}\right)^{-}$

Answer: $P_{\text {rad }}=a^{\alpha} \cdot q^{\beta} \cdot c^{\gamma} \cdot \epsilon_{0}^{\delta}$
Problem: A.5 Calculate $P_{\text {tot }}$, the total radiated power of the LHC, for a proton energy of E= 7.00 $\mathrm{TeV}$ (Note table 1). You may use suitable approximations.

Solution: Radiated Power:

$$
P_{r a d}=\frac{\gamma^{4} \cdot a^{2} \cdot e^{2}}{6 \pi \cdot c^{3} \cdot \epsilon_{0}}
$$

Energy:

$$
E=(\gamma-1) m_{p} \cdot c^{2} \text { or equally valid } E \simeq \gamma \cdot m_{p} \cdot c^{2}
$$

Acceleration:

$$
a \simeq \frac{c^{2}}{r} \text { with } \quad r=\frac{L}{2 \pi}
$$

Therefore:

$$
\begin{gathered}
P_{\text {rad }}=\left(\frac{E}{m_{p} c^{2}}+1\right)^{4} \cdot \frac{e^{2} \cdot c}{6 \pi \epsilon_{0} \cdot r^{2}} \text { or }\left(\frac{E}{m_{p} c^{2}}\right)^{4} \cdot \frac{e^{2} \cdot c}{6 \pi \epsilon_{0} \cdot r^{2}} \\
\left(\text { not required } P_{\text {rad }}=7.94 \cdot 10^{-12} \mathrm{~W}\right)
\end{gathered}
$$

Total radiated power:

$$
P_{t o t}=2 \cdot 2808 \cdot 1.15 \cdot 10^{11} \cdot P_{r a d}=5.13 \mathrm{~kW}
$$

Answer: 5.13