As an example, the central force problem used all knowledge of the basic principles in classical mechanics, which is a good practice to review all knowledge that have learnt in the first two chapters of Classical Mechanics (Goldstein)

Reduce Central Force Problem to Equivilent One-body Problem

\(L=T(\dot{\mathbf{r}}_1,\dot{\mathbf{r}}_2)-U(\mathbf{r}_1-\mathbf{r}_2,\dot{\mathbf{r}}_1-\dot{\mathbf{r}}_2)\) Use the center of mass Coordinate $R=\sum_i m_ir_i$, we have:

\(\begin{align*} T&=\frac{1}{2}(m_1+m_2)\dot{\mathbf{R}}^2+\frac{1}{2}M(m_1\dot{\mathbf{r}}_1-m_2\dot{\mathbf{r}}_2)^2\\ &=\frac{1}{2}m_1\dot{\mathbf{r}}_1^2+\frac{1}{2}m_2\dot{\mathbf{r}}_2^2 \end{align*}\) We can fond that $M=\frac{(m_1+m_2)m_1m_2}{(m_1\dot{\mathbf{r}}_1-m_2\dot{\mathbf{r}}_2)^2}(\dot{\mathbf{r}}_1-\dot{\mathbf{r}}_2)^2$, let $\mathbf{r}=\mathbf{r}_1-\mathbf{r}_2$, so

\(L=\frac{1}{2}(m_1+m_2)\dot{\mathbf{R}}^2+\frac{1}{2}\frac{m_1m_2}{m_1+m_2}\dot{\mathbf{r}}^2\) Recall Lagrange’s Equation: \(\frac{d}{dt}\left[\frac{\partial L}{\partial \dot{q}}\right]-\frac{\partial L}{\partial q}=0\) We can found that the equation of $R$ and $r$ is independent, so we can divide the $L$ into two Lagrangian: \(L_1=\frac{1}{2}(m_1+m_2)\dot{\mathbf{R}}^2, \quad L_2=\frac{1}{2}\frac{m_1m_2}{m_1+m_2}\dot{\mathbf{r}}^2-U(\mathbf{r},\dot{\mathbf{r}})\) We can give a reduced mass $m=\frac{m_1m_2}{m_1+m_2}$, and solving the one-body problem: \(L=\frac{1}{2}m\dot{\mathbf{r}}^2-U(\mathbf{r},\dot{\mathbf{r}})\) We consider the simplified $U$ that depends only on the length $r$ of $\mathbf{r}$.

“Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, about any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic. These symmetry properties result in a considerable simplification in the problem.”

This means angular momentum is conserved, since $\frac{dL}{dt}=\mathbf{r}\times \nabla U=\mathbf{r}\times \frac{dU}{dr}\hat{\mathbf{r}}=0$. So each time when observed a sperical symmetry, try to give a image of a fixed pole that the angular momentum is conserved to.

Lagrangian in Sperical Coordinate

\(\begin{align*} x&=r\sin\phi\cos\theta\\ y&=r\sin\phi\sin\theta\\ z&=r\cos\phi \end{align*}\)

Since the angular momentum is conserved, which means there is always a pole direction that $\mathbf{r}$ is perpendicular to. So we fix the direction of $L$ to $\mathbf{z}$ axis, and find that $z=0, x=r\cos\theta, y=r\sin\theta$. So $\dot{\mathbf{r}}^2=\dot{r}^2+r^2\dot{\theta}^2$ and: \(L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)\)

Conservation of Angular Momentum

We can found the momentum corresponding to $\theta$ is $p_\theta=mr^2\dot{\theta}$ and it conserve since the equation of motion of $\theta$ requires \(\frac{d}{dt}\left(mr^2\dot{\theta}\right)=0\) we can denote $mr^2\dot{\theta}=l$.

Conservation of Energy

Next we derive the conservation of energy from equation of motion of coordinate $r$. First we noticed that: \(E=T+V=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)+V(r)\) is conserved in the system. We start from the Langrange’s Equation of $r$ that: \(\frac{d}{dt}(m\dot{r})-mr\dot{\theta}^2+\frac{\partial V(r)}{\partial r}=0\) $\dot{\theta}$ can be eliminated using $l$, that \(m\ddot{r}-\frac{l^2}{mr^3}+\frac{\partial V(r)}{\partial r}=0\) Rewrite as a derivitive of $t$, where \(m\ddot{r}\dot{r}=\frac{d}{dt}\left(\frac{1}{2}m\dot{r}^2\right)=\dot{r}\left(\frac{l^2}{mr^3}-\frac{\partial V(r)}{\partial r}\right)=-\frac{dr}{dt}\frac{d}{dr}\left(\frac{l^2}{2mr^2}+V(r)\right)\) So, we have both side of derivitive with respective to $t$ as: \(\frac{d}{dt}\left(\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}+V(r)\right)=0\) Where $E=\frac{1}{2}m\dot{r}^2+\frac{l^2}{2mr^2}+V(r)$, we have the conservation of energy.

Solving the Equation using Conservative Quantities

With $E$ and $l$, the problem can be reduced to integrations by two conservative equation, that: \(\dot{r}=\sqrt{\frac{2}{m}\left(E-V-\frac{l^2}{2mr^2}\right)}\) $r(t)$ can be attained given $r_0$ and $E,l,m$. Then $\dot{\theta}$ can be solved for: \(mr^2\dot{\theta}=l\) Given $\theta_0$. So, with four constant term $(E,l,\theta_0, r_0)$, the solution of $r(t),\theta(t)$ can be determined, and so as their derivitives. Also, we can use $(r_0,\dot{r}_0, \theta_0, \dot{\theta}_0)$,

“In quantum mechanics, such constants as the initial values of $r$ and $\theta$, or $\dot{r}$ and $\dot{\theta}$ become meaningless, but we can still talk in terms of the system energy or of the system angular momentum.”

Image of one dimensional orbital

The above orbital is equivilent to a atom moving in potential $V’=V+\frac{l^2}{2mr^2}$. Take $V=\frac{k}{r^2}$ , $V=\frac{1}{r^3}$ or others potential, we can find different orbital types. Like when $V=\frac{k}{r^2}$, the orbital can be Ellipse, hyperbola and parabola.