Subfield: Mechanics


Question: The logo of OPhO describes two objects travelling around their center-of-mass, following the same oval-shape trajectory. For simplicity, we assume these objects are point-like, have identical mass, and interacts via an interacting potential $U(d)$ depends on the distance $d$ between them. Choose the polar coordinates $(r, \theta)$ as shown in the figure, where the origin is located at the center of the logo, then the shared trajectory obeys the equation:
$$
r(\theta)=\frac{L}{2}[1-\epsilon \cos (2 \theta)]^{-(1+\gamma)},
$$

in which we consider $\epsilon=0.12$ and $\gamma=0.05$. Here the smallest and largest separation between the objects are $d_{\min }$ and $d_{\max }$. Since the interacting potential $U(d)$ is defined up to a constant, let us pick $U(L)=0$. Find the ratio $U\left(d_{\min }\right) / U\left(d_{\max }\right)$.
![](https://cdn.mathpix.com/cropped/2023_12_21_7cb4a6b8f7c2770434e1g-1.jpg?height=364&width=1332&top_left_y=392&top_left_x=381)

Solution:
Define the dimensionless inverse-radius $u=\frac{L}{d}$ and the angular-derivative ${ }^{\prime}=\frac{d}{d \theta}$, then for any central interacting potential we have Binet equation (can be derived directly from the equations of motion in polar coordinates):
$$
\frac{d}{d u} U(u) \propto-\left(u^{\prime \prime}+u\right)
\tag{26}
$$
For the given oval-shape trajectory:
$$
d=L[1-\epsilon \cos (2 \theta)]^{-(1+\gamma)} \Longrightarrow u=(1-\epsilon \cos \phi)^{(1+\gamma)}, \cos \phi=\epsilon^{-1}\left[1-u^{\left(\frac{1}{1+\gamma}\right)}\right]
\tag{27}
$$
where $\phi=2 \theta$. The angular 2nd-derivative of the dimensionless inverse-radius $u$ can be calculated:
$$
\begin{aligned}
u^{\prime \prime} & =4 \frac{d^{2}}{d \phi^{2}} u=4 \epsilon(1+\gamma)\left[\frac{\gamma \epsilon\left(1-\cos ^{2} \phi\right)}{1-\epsilon \cos \phi}+\cos \phi\right](1-\epsilon \cos \phi)^{\gamma} \\
& =4 \epsilon(1+\gamma)\left\{\gamma \epsilon\left(1-\epsilon^{-2}\left[1-u^{\left(\frac{1}{1+\gamma}\right)}\right]^{2}\right) u^{-\left(\frac{1}{1+\gamma}\right)}+\epsilon^{-1}\left[1-u^{\left(\frac{1}{1+\gamma}\right)}\right]\right\} u^{\left(\frac{\gamma}{1+\gamma}\right)} \\
& =4(1+\gamma)\left\{\gamma \epsilon^{2} u^{-\left(\frac{1-\gamma}{1+\gamma}\right)}-\gamma u^{-\left(\frac{1-\gamma}{1+\gamma}\right)}\left[1-2 u^{\left(\frac{1}{1+\gamma}\right)}+u^{\left(\frac{2}{1+\gamma}\right)}\right]+\left[u^{\left(\frac{\gamma}{1+\gamma}\right)}-u\right]\right\} \\
& =-4 \gamma(1+\gamma)\left(1-\epsilon^{2}\right) u^{-\left(\frac{1-\gamma}{1+\gamma}\right)}+4(1+\gamma)(1+2 \gamma) u^{\left(\frac{\gamma}{1+\gamma}\right)}-4(1+\gamma)^{2} u .
\end{aligned}
$$
Plug this back into Eq. (26), we can arrive at:
$$
\begin{aligned}
\frac{d}{d u} U(u) & \propto\left[4(1+\gamma)^{2}-1\right] u+4 \gamma(1+\gamma)\left(1-\epsilon^{2}\right) u^{-\left(\frac{1-\gamma}{1+\gamma}\right)}-4(1+\gamma)(1+2 \gamma) u^{\left(\frac{\gamma}{1+\gamma}\right)} \\
\Longrightarrow U(u) & \propto\left(2-\frac{1}{2(1+\gamma)^{2}}\right) u^{2}+2\left(1-\epsilon^{2}\right) u^{\left(\frac{2 \gamma}{1+\gamma}\right)}-4 u^{\left(\frac{1+2 \gamma}{1+\gamma}\right)}+W
\end{aligned}
\tag{28}
$$
where $W$ is a constant so that we can assign $U(u=1)=0$. For $\epsilon=0.12$ and $\gamma=0.05$, we get $W=0.48231$. Following that, we use Eq. (27), the maximum distance corresponds to $\theta=0$ and thus $u_{\triangleright}=0.87439$, the minimum distance corresponds to $\theta=\pi / 2$ and thus $u_{\triangleleft}=1.1264$. The ratio of interests, therefore, can be found directly:
$$
\frac{U\left(u=u_{\triangleleft}\right)}{U\left(u=u_{\triangleright}\right)} \approx-0.6864
$$
We plot the interacting potential, up to a pre-factor as given in Eq. (28), in the figure below.



![](https://cdn.mathpix.com/cropped/2023_12_21_1b9e28ea40fa6fb3d058g-1.jpg?height=828&width=1568&top_left_y=236&top_left_x=274)

Final answer: -0.6864