Central Forces Notes Section 9

• The equation of motion for a reduced mass in a central potential is written in polar coordinates $$f(r){\bf\hat{r}} = \mu {\bf \ddot{r}} = \mu \left((\ddot{r}- r \dot{\phi}^2) {\bf\hat r} + ( r \ddot{\phi} + 2 \dot{r}\dot{\phi}) {\bf\hat \phi}\right) $$ and broken into two coupled differential equations:

$$\mathbf{\hat{r}}:\; f(r)=\mu(\ddot{r}-r\dot{\phi}^2)$$ $$\mathbf{\hat{\phi}}:\;0=\mu(r\ddot{\phi}+2\dot{r}\dot{\phi})$$

• This is a good place to emphasize that when one is trying to solve coupled differential equations, it is a good idea to look for conserved quantities $$\ell=\mu r^2\dot{\phi}=\text{constant}.$$ SWBQ: Calculate $\frac{d\ell}{dt}$ in polar coordinates $$\dfrac{d\ell}{dt}=\dfrac{d}{dt}(\mu r^2\dot{\phi})=\mu(r\ddot{\phi}+2\dot{r}\dot{\phi})=0.$$ Thus, the $\hat{\phi}$ equation of motion is simply equivalent to a statement of conservation of angular momentum.
• Emphasize that when trying to de-couple the equations, we want only one variable in each equation. So we want to get rid of all of the $\dot{\phi}$'s in our $\bf\hat{r}$ equation and we can once again use conservation of angular momentum.

$$\ell=\mu r^2\dot{\phi}\longrightarrow \dot{\phi}=\dfrac{\ell}{\mu r^2}$$ $$f(r)=\mu(\ddot{r}-r\dot{\phi}^2)\longrightarrow \ddot r = \frac{l^2}{\mu^2 r^3}+\frac{1}{\mu} f(r)$$