Classical Central Potential: Orbital Dynamics
We analyze the motion of a particle with mass $m$ in a central potential field $V(r) = -k/r$. We define the angular momentum $L$ and the Energy $E$:
\begin{equation} \vec{L} = m\vec{r}\times\vec{v} \label{eq:a2_1} \end{equation} \begin{equation} E = \frac{p^2}{2m} - \frac{k}{r} \label{eq:a2_2} \end{equation}In short, the plan is to establish the equation of movement "Newton second law" and solve it with the help of the conservation of the angular momentum $L$, which will give a solution that has a single free parameter $A$, this parameter will be determined with the help of the conservation of energy $E$, meaning that we will plug the solution in the expression of the energy, and solve it for a constant $E$ and $\theta$.
1. Kinematics in Polar Coordinates
Expressing position $\vec{r}$, velocity $\vec{v}$, and acceleration $\vec{a}$ in polar unit vectors:
\begin{equation} \vec{r} = r\hat{e}_r \label{eq:a2_3} \end{equation} \begin{equation} \vec{v} = \frac{dr}{dt}\hat{e}_r + r\dot{\theta}\hat{e}_\theta \label{eq:a2_4} \end{equation} \begin{equation} \vec{a} = \left( \frac{d^2r}{dt^2} - r\dot{\theta}^2 \right) \hat{e}_r + \left( 2\frac{dr}{dt}\dot{\theta} + r\ddot{\theta} \right) \hat{e}_\theta \label{eq:a2_5} \end{equation}Since only the normal component $\hat{e}_r$ of the acceleration is what interact with Coulomb force, we get:
\begin{equation} 2\frac{dr}{dt}\dot{\theta} + r\ddot{\theta}=0 \implies \ddot{\theta}=-\frac{2}{r}\frac{dr}{dt}\dot{\theta} \label{eq:a2_6} \end{equation}This equation is so useful becasue it links the angular acceleration $\ddot{\theta}$ with the angular velocity $\dot{\theta}$, in other word the angular momentum $L$.
Next, we calculate the angular momentum $L$ by evaluating the vector product between $\vec{r}$ and $\vec{v}$, which filter out the components that have the same direction, in other word, only the $\hat{e}_\theta$ remains.
\begin{equation} \vec{L}=mr^2\dot{\theta}\hat{e}_z \label{eq:a2_7} \end{equation}Where $\hat{e}_z$ is the unit vector out of the plan of rotation, so basically we're using cylindrical coordinates.
2. The Equation of the Orbit
Using the conservation of angular momentum $\dot{\theta} = L/mr^2$, $L$ is the norm of $\vec{L}$, the radial acceleration becomes:
\begin{equation} \frac{d^2r}{dt^2} = \frac{L^2}{m^2r^3} - \frac{k}{mr^2} \label{eq:a2_8} \end{equation}First we need reparamterize the differential equation from temporal to $\theta$-dependent, we can achieve this by the following:
\begin{equation} \frac{dr}{dt} = \frac{d\theta}{dt}\frac{dr}{d\theta}=\frac{L}{mr^2}\frac{dr}{d\theta} \label{eq:a2_9} \end{equation} \begin{equation} \frac{d^2r}{dt^2} = \frac{L^2}{m^2r^4}\frac{d^2r}{d\theta^2}-\frac{2L^2}{m^2r^5}\left( \frac{dr}{d\theta} \right)^2 \label{eq:a2_10} \end{equation}Pluging back this new expression into the differential equation gives the following:
\begin{equation} \frac{d^2r}{d\theta^2} - \frac{2}{r}\left( \frac{dr}{d\theta} \right)^2 = r - \frac{mk}{L^2} r^2 \label{eq:a2_11} \end{equation}Substituting $u = 1/r$, we get the following effects:
\begin{equation} \frac{dr}{d\theta} = -\frac{1}{u}\frac{du}{d\theta} \label{eq:a2_12} \end{equation} \begin{equation} \frac{d^2r}{d\theta^2} = -\frac{1}{u^2}\frac{d^2u}{d\theta^2} + \frac{2}{u^3} \left( \frac{du}{d\theta} \right)^2 \label{eq:a2_13} \end{equation}Substituting back to the differential equation gives:
\begin{equation} \frac{d^2u}{d\theta^2} + u = \frac{mk}{L^2} \label{eq:a2_14} \end{equation}This equation admits the following solution:
\begin{equation} u = A\cos(\theta)+\frac{mk}{L^2} \label{eq:a2_15} \end{equation}In order to find the constant $A$, we need to plug this soution to the energy $E$:
\begin{equation} E = \frac{1}{2}m \left( \frac{dr}{dt} \right)^2 + \frac{1}{2}m r^2 \dot{\theta}^2 - \frac{k}{r} \label{eq:a2_16} \end{equation}Re-arranging this expression gives this useful expression:
\begin{equation} \left( \frac{dr}{dt} \right)^2 = \frac{2E}{m} - \frac{L^2}{m^2r^2} + \frac{2k}{mr} \label{eq:a2_17} \end{equation}Applying the avove machinary we get something that we can use directly.
\begin{equation} \frac{L^2}{m^2}\left( \frac{du}{d\theta} \right)^2 = \frac{2E}{m} - \frac{L^2}{m^2}u^2 + \frac{2k}{m}u \label{eq:a2_18} \end{equation}We have all the ingredients to solve this equation, after pluging everything, we set $\theta=\pi/2$, since the constant $A$ and $E$ don't care about the position of the particle, they're always constant.
\begin{equation} \frac{L^2}{m^2} A^2 = \frac{2E}{m} - \frac{L^2}{m^2} \left( \frac{mk}{L^2} \right)^2 + \frac{2k}{m} \left( \frac{mk}{L^2} \right) \label{eq:a2_19} \end{equation}Finally, we get the solution:
\begin{equation} A = \frac{mk}{L^2} \sqrt{1+\frac{2EL^2}{mk^2}} \label{eq:a2_20} \end{equation}The general solution for the path $r(\theta)$ is[cite: 75, 81]:
\begin{equation} r(\theta) = \frac{L^2/mk}{1 + \sqrt{1+\frac{2EL^2}{mk^2}}\cos\theta} \label{eq:a2_21} \end{equation}Where the eccentricity $\varepsilon$ is related to the energy $E$.
\begin{equation} \varepsilon = \sqrt{1+\frac{2EL^2}{mk^2}} \label{eq:a2_22} \end{equation}3. Application for an electron orbiting a central potential created by a proton
Knowing that the electron orbits "classicaly" around the positive proton without losing energy, means that its angular momentum is conserved "constant". So as Bohr showed in his Bohr model, electrons orbit the nucleus only in stable, allowed orbits, defined as follows:
\begin{equation} L = n\hbar \label{eq:a2_23} \end{equation}When dealing with the Coulomb interaction, $k$ is written as follows:
\begin{equation} k = \frac{e^2}{4\pi\varepsilon_0} \label{eq:a2_24} \end{equation}4. Energy States
Finally, we relate the system energy to the fundamental constants:
\begin{equation} E(\varepsilon) = -\frac{1}{n^2} \frac{me^4}{32\pi^2\epsilon_0^2\hbar^2} (1 - \varepsilon^2) \label{eq:a2_25} \end{equation}The energy is minimized at $\varepsilon=0$, which gives the correct energy for the hydrgen atom. This shows that the orbits are circular following the fact that $\varepsilon = 0$.