Shankar exercise-2.1.3

Q.

A particle of mass $m$ moves in three dimensions under a potential $V(r, \theta, \phi)=V(r)$. Write its $L$ and find the equation of motion.

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Trial.

Lagrangian

$L = T-V={1 \over 2}mv^2-V(r, \theta, \phi)$.

But, potentail $V$ depends only on $r$, $V(r)$,

$L = T-V={1 \over 2}mv^2-V(r)\tag{1}$

Line element is

$dS=(dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2)^{1 \over 2}$

$v={dS \over dt}=(\dot{r}^2+r^2\dot{\theta^2}+r^2\sin^2\theta \dot{\phi}^2)^{1 \over 2}\tag{2}$

From eq.(1) & eq.(2),

$L={1 \over 2}m(\dot{r}^2+r^2\dot{\theta^2}+r^2\sin^2\theta \dot{\phi}^2) -V(r)\tag{3}$

Lagrange’s equations for $r, \theta, \phi$

For $r$

${d \over dt}\left( {\partial L \over \partial \dot{r}}\right) - {\partial L \over \partial r} = 0\tag{4}$

From eq.(3),

${\partial L \over \partial \dot{r}} = m\dot{r}$
${\partial L \over \partial r} = m(r\dot{\theta}^2+r\sin^2\theta \dot{\phi}^2)-{\partial V(r) \over \partial r}$

Putting the above result to Eq.(4) gives

${d \over dt}\left( m\dot{r}\right)=m(r\dot{\theta}^2+r\sin^2\theta \dot{\phi}^2)-{\partial V(r) \over \partial r}\tag{5}$

For $\theta$

${d \over dt}\left( {\partial L \over \partial \dot{\theta}}\right) - {\partial L \over \partial \theta} = 0\tag{6}$

From eq.(3),

${\partial L \over \partial \dot{\theta}} = mr^2\dot{\theta}$
${\partial L \over \partial \theta} = m(r^2\sin\theta\cos\theta \dot{\phi}^2)$

${d \over dt}\left( mr^2\dot{\theta}\right)=mr^2\sin\theta\cos\theta \dot{\phi}^2\tag{7}$

${d \over dt}\left( r^2\dot{\theta}\right)=r^2\sin\theta\cos\theta \dot{\phi}^2\tag{7.1}$

For $\phi$

${d \over dt}\left( {\partial L \over \partial \dot{\theta}}\right) - {\partial L \over \partial \theta} = 0\tag{8}$

From eq.(3),

${\partial L \over \partial \dot{\phi}} = mr^2\sin^2\theta\dot{\phi}$
${\partial L \over \partial \phi} = 0$

${d \over dt}\left( mr^2\sin^2\theta\dot{\phi}\right)=0\tag{9}$

${d \over dt}\left( r^2\sin^2\theta\dot{\phi}\right)=0\tag{9.1}$

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Equations of motion

Classical mechanics $m = const.$

For $r$

From eq.(5)

$m\ddot{r} = m(r\dot{\theta}^2+r\sin^2\theta \dot{\phi}^2)-{\partial V(r) \over \partial r}\tag{10}$

For $\theta$

${d \over dt}\left( r^2\dot{\theta}\right)=2r\dot{r}\dot{\theta}+r^2\ddot{\theta}$

From eq.(7.1)

$2\dot{r}\dot{\theta}+r\ddot{\theta}=r\sin\theta\cos\theta \dot{\phi}^2\tag{11}$

For $\phi$

${d \over dt}\left( r^2\sin^2\theta\dot{\phi}\right)=2r\dot{r}\sin^2\theta\dot{\phi}+2r^2\sin\theta\cos\theta\dot{\theta}\dot{\phi}+r^2\sin^2\theta\ddot{\phi}$

From eq.(9.1)

$2\dot{r}\sin\theta\dot{\phi}+2r\cos\theta\dot{\theta}\dot{\phi}+r\sin\theta\ddot{\phi}=0\tag{12}$