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 $E$. In short, the plan is to establish the equation of movement "Newton's 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 $E$.
\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}
01 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 interacts with the 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}
Note: This equation is so useful because it links the angular acceleration $\ddot{\theta}$ with the angular velocity $\dot{\theta}$, in other words the angular momentum $L$.
Next, we calculate the angular momentum $L$ by evaluating the vector product between $\vec{r}$ and $\vec{v}$, which filters out the components that have the same direction; in other words, only $\hat{e}_\theta$ 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 plane of rotation, so basically we're using cylindrical coordinates.
02 The Equation of the Orbit
Using the conservation of angular momentum $\dot{\theta} = L/mr^2$, where $L$ 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 to reparameterize 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}
Plugging 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 solution into 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}
Rearranging this expression gives this useful form:
\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 above machinery 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 plugging everything in, we set $\theta=\pi/2$, since the constants $A$ 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:
\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}
03 Application for an electron orbiting a central potential
Knowing that the electron orbits "classically" 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}
04 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 hydrogen atom. This shows that the orbits are circular following the fact that $\varepsilon = 0$.
References
- Goldstein, H. (1980). Classical Mechanics. Addison-Wesley.
- Landau, L. D., & Lifshitz, E. M. (1976). Mechanics. Pergamon Press.
