Subfield: Modern Physics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Zed is trying to model the repulsive interaction between 2 objects, $A$ and $B$ (with masses $m_{A}$ and $m_{B}$, respectively), in a relativistic setting. He knows that in relativity, forces cannot act at a distance, so he models the repulsive force with a small particle of mass $m$ that bounces elastically between $A$ and $B$. Throughout this problem, assume everything moves on the x-axis. Suppose that initially, $A$ and $B$ have positions and velocities $x_{A}, v_{A}$ and $x_{B}, v_{B}$, respectively, where $x_{A}<x_{B}$ and $v_{A}>v_{B}$. The particle has an initial (relativistic) speed $v$.
For simplicity, assume that the system has no total momentum. You may also assume that $v_{A}, v_{B} \ll v$, and that $p_{m} \ll p_{A}, p_{B}$, where $p_{m}, p_{A}, p_{B}$ are the momenta of the particle, $A$, and $B$, respectively. Do NOT assume $v \ll c$, where $c$ is the speed of light.

Find the position (in $\mathrm{m}$ ) of $A$ when its velocity is 0 , given that $m_{A}=1 \mathrm{~kg}, m_{B}=2 \mathrm{~kg}, v_{A}=0.001 c$, $m=1 \times 10^{-6} \mathrm{~kg}, v=0.6 c, x_{A}=0 \mathrm{~m}, x_{B}=1000 \mathrm{~m}$.

Note: Answers will be tolerated within $0.5 \%$, unlike other problems.

Solution:
Since total momentum is 0 , we have $m_{A} v_{A}+m_{B} v_{B}=0$, so $v_{B}=-0.0005 c$ By conservation of energy:

$$
\frac{1}{2} m_{A} v_{A}^{2}+\frac{1}{2} m_{B} v_{B}^{2}+\gamma_{0} m c^{2}=\gamma m c^{2}
$$

where we define $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ to correspond to the final state of the particle and $\gamma_{0}$ to the initial state. This equation allows us to solve for $\gamma$. Since the speed of the particle is much larger than the speed of the masses $A$ and $B$, we can imagine the particle moving in a infinite well potential where the walls are slowly moving. Applying the adiabatic theorem, we get that the adiabatic invariant $p x$ is conserved, where $p$ is the particle's momentum, and $x$ is the distance between $A$ and $B$. Thus, $\gamma v x$ is conserved, so

$$
\gamma v x=\gamma_{0} v_{0}\left(x_{B}-x_{A}\right)
$$

We can solve for $x$, since we know $\gamma$ and $v$. Finally, we realize that the center of mass stays at $x_{c} m=\frac{m_{A} x_{A}+m_{B} x_{B}}{m_{A}+m_{B}}$, so the final position of $A$ is simply

$$
x_{c} m-\frac{m_{B}}{m_{A}+m_{B}} x=378 \mathrm{~m}
$$

Final answer: 378 m