Subfield: Mechanics

- 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: A tesseract is a 4 dimensional example of cube. It can be drawn in 3 dimensions by drawing two cubes and connecting their vertices together as shown in the picture below:

<img_4271>

Now for the 3D equivalent. The lines connecting the vertices are replaced with ideal springs of constant $k=10 \mathrm{~N} / \mathrm{m}$ (in blue in the figure). Now, suppose the setup is placed in zero-gravity and the outer cube is fixed in place with a sidelength of $b=2 \mathrm{~m}$. The geometric center of the inner cube is placed in the geometric center of the outer cube, and the inner cube has a side-length $a=1 \mathrm{~m}$ and mass $m=1.5 \mathrm{~kg}$. The inner cube is slightly displaced from equilibrium. Consider the period of oscillations

- $T_{1}$ : when the springs have a relaxed length of 0 ;
- $T_{2}$ : when the springs are initially relaxed before the inner cube is displaced.

What is $T_{1}+T_{2}$ ?

Solution:
First let us prove that there is a net external torque of $\vec{\tau}=0$ on the cube for small displacements which means the inner cube behaves like a point mass. Consider a simple case when the cube is pushed to one side.
<img_4314>

If we label the vertices of the cube from 1 to 4 clockwise, where 1 is the top left side, it is apparent that sides 1 and 2 provide a positive torque while sides 3 and 4 provide a negative torque. As the displacement is small, the angles created are small enough such that $\sin \theta \approx \theta$. As force is



proportional to the extension of the spring as $F \propto x$, we can write that

$$
\tau \propto \theta\left(r_{+}+r_{-}-r_{+}-r_{-}\right) \propto 0
$$

If torque is zero when the cube is displaced in the $x$-direction, then by symmetry, the torque is zero when the cube is displaced in the $y$-direction. Superposing both solutions implies that torque as a function of displacements in the $x$ and $y$ directions $\alpha \hat{x}+\beta \hat{y}$ is

$$
\tau(\alpha x+\beta y)=\tau(\alpha x)+\tau(\beta y)=\alpha \tau(x)+\beta \tau(y)=0
$$

1.Label the vertices of the outer cube as $1,2, \ldots, 8$ and the vectors that point to these vertices from the inner cube as $\vec{r}_{1}, \vec{r}_{2}, \ldots, \vec{r}_{2}$. Consider when the inner cube deviates from equilibrium with a vector $\vec{r}$. The force as a function of $\vec{r}$ is
$$
\begin{aligned}
F(\vec{r}) & =k\left[\left(\vec{r}_{1}-\vec{r}\right)+\left(\vec{r}_{2}-\vec{r}\right)+\ldots\left(\vec{r}_{8}-\vec{r}\right)\right] \\
& =k\left(\sum_{i=1}^{8} \vec{r}_{i}-8 \vec{r}\right) \\
& =-8 k \vec{r}
\end{aligned}
$$

This implies the period of oscillations is

$$
T_{1}=2 \pi \sqrt{\frac{m}{8 k}}
$$

2.Let the center of the inner cube be $(0,0,0)$. Consider the coordinates $(a / 2, a / 2, a / 2)$ and $(b / 2, b / 2, b / 2)$ which correspond to the vertex of the inner and larger cube respectively. Consider moving the cube in the $x$-direction. From defining $y=b / 2-a / 2$, the compressional/extension of each spring $\pm \Delta \ell$ is then
$$
\begin{aligned}
\Delta \ell & = \pm \sqrt{(x+y)^{2}+2 y^{2}}-\sqrt{3} y \\
& = \pm \sqrt{3} y \sqrt{1+\frac{2 x}{3 y}+\mathcal{O}\left(x^{2}\right)}-\sqrt{3} y \\
& \approx \pm \sqrt{3} y \frac{x}{3 y} \\
& = \pm \frac{x}{\sqrt{3}}
\end{aligned}
$$

The total energy in all springs together are then

$$
E=8 \times \frac{1}{2} k\left(\frac{x}{\sqrt{3}}\right)^{2} \Longrightarrow F=-\frac{8 k}{3} x \Longrightarrow T_{2}=2 \pi \sqrt{\frac{3 m}{8 k}}
$$

Hence, our total answer is

$$
T_{1}+T_{2}=2 \pi(1+\sqrt{3}) \sqrt{\frac{m}{8 k}}
$$

Final answer: 2.35 s