Motion in a Central Potential

Consider a particle of mass $m$ moving in two dimensions in the central potential $U(r)$. This is clearly a two degree of freedom dynamical system. As described in Sect. 6.5, the particle's instantaneous position is most conveniently specified in terms of the plane polar coordinates $r$ and $\theta$. These are our two generalized coordinates. According to Eq. (298), the square of the particle's velocity can be written

$$v^2 = \dot{r}^{\,2} + (r\,\dot{\theta})^2.$$ (684)

Hence, the Lagrangian of the system takes the form

$$L = \frac{1}{2}\,m\,(\dot{r}^{\,2} + r^2\,\dot{\theta}^{\,2}) - U(r).$$ (685)

Note that

$$\frac{\partial L}{\partial\dot{r}} = m\,\dot{r}, ~~~~~~ \frac{\partial L}{\partial r} = m\,r\,\dot{\theta}^{\,2} - \frac{dU}{dr},$$ (686)

$$\frac{\partial L}{\partial \dot{\theta}} = m\,r^2\,\dot{\theta}, \frac{\partial L}{\partial\theta} = 0.$$ (687)

Now, Lagrange's equation (683) yields the equations of motion,

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{r}}\right)- \frac{\partial L}{\partial r} = 0,$$ (688)

$$\frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot{\theta}}\right)- \frac{\partial L}{\partial \theta} = 0.$$ (689)

Hence, we obtain

$$\frac{d}{dt}\!\left(m\,\dot{r}\right) - m\,r\,\dot{\theta}^{\,2} + \frac{dU}{dr} = 0,$$ (690)

$$\frac{d}{dt}\!\left(m\,r^2\,\dot{\theta}\right) = 0,$$ (691)

or

$$\ddot{r} - r\,\dot{\theta}^{\,2} = - \frac{d V}{dr},$$ (692)

$$r^2\,\dot{\theta} = h,$$ (693)

where $V = U/m$, and $h$ is a constant. We can recognize Eqs. (692) and (693) as the equations we derived in Sect. 6 for motion in a central potential. The advantage of the Lagrangian method of deriving these equations is that we avoid having to express the acceleration in terms of the generalized coordinates $r$ and $\theta$.