Question 7.1: Obtain Hamilton’s equations for a particle in a central pote......

In spherical coordinates the Lagrangian is

L = T − V = \frac{m}{2}\left(\dot{r}^{2}+{r}^{2}\dot{\theta }^{2} +{r}^{2}\sin^{2}\theta \dot{\phi }^{2} \right) -V\left(r\right), (7.12)

whence

p_{r}=\frac{\partial L}{\partial \dot{r}}=m\dot{r}, p_{\theta}=\frac{\partial L}{\partial \dot{\theta}}=mr^{2} \dot{\theta} , p_{\phi }=\frac{\partial L}{\partial \dot{\phi }}=mr^{2} \sin^{2}\theta \dot{\phi }.   (7.13)

Solving these equations for the velocities we find

\dot{r}= \frac{p_{r}}{m}, \dot{\theta}= \frac{p_{\theta}}{mr^{2}}, \dot{\phi }=\frac{p_{\phi }}{mr^{2}\sin^{2}\theta},  (7.14)

so that

H =\dot{r}p_{r}+\dot{\theta}p_{\theta}+\dot{\phi }p_{\phi }-L= \frac{1}{2m} \left(p^{2}_{r}+\frac{p^{2}_{\theta}}{r^{2}}+\frac{p^{2}_{\phi }}{r^{2}\sin^{2}\theta}\right)+V\left(r\right).   (7.15)

This Hamiltonian equals the total energy and Hamilton’s equations are

\dot{r}=\frac{\partial H}{\partial p_{r}}=\frac{p_{r}}{m}, \dot{\theta}= \frac{\partial H}{\partial p_{\theta}}=\frac{p_{\theta}}{mr^{2}}, \dot{\phi }=\frac{\partial H}{\partial p_{\phi}}=\frac{p_{\phi}}{mr^{2}\sin^{2}\theta},   (7.16)

\dot{p}_{r}=-\frac{\partial H}{\partial r}= \frac{p^{2}_{\theta}}{mr^{3}}+\frac{p^{2}_{\phi}}{mr^{3}\sin^{2}\theta}-\frac{dV}{dr},

\dot{p}_{\theta}=-\frac{\partial H}{\partial \theta}= \frac{p^{2}_{\phi} \cot \theta}{mr^{2}\sin^{2}\theta},\dot{p}_{\phi}=-\frac{\partial H}{\partial \phi} =0.  (7.17)

According to a general remark previously made, Eqs. (7.16) are the inverses of (7.13). With their use Eqs. (7.17) become identical to the Lagrange equations generated by the Lagrangian (7.12).