< Classical Mechanics

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Consider a central potential V(r). A central potential is where the potential is dependent only on the field point's distance from the origin; in other words, the potential is isotropic.

The Lagrangian of the system can be written as

${\mathcal {L}}={\frac {1}{2}}m{\dot {\vec {x}}}^{2}-V(r)$

Since the potential is spherically symmetry, it makes sense to write the Lagrangian in spherical coordinates.

${\dot {\vec {x}}}^{2}=\left({\frac {d}{dt}}\left(r\sin \phi \sin \theta ,r\cos \phi \sin \theta ,r\cos \theta \right)\right)^{2}$

It can then be worked out that:

${\dot {\vec {x}}}^{2}={\dot {r}}^{2}+r^{2}{\dot {\theta }}^{2}+r^{2}{\dot {\phi }}^{2}\sin ^{2}\theta$

Hence the equation for the Lagrangian is:

${\mathcal {L}}={\frac {1}{2}}m\left({\dot {r}}^{2}+r^{2}{\dot {\theta }}^{2}+r^{2}{\dot {\phi }}^{2}\sin ^{2}\theta \right)-V(r)$

One can then extract three laws of motion from the Lagrangian using the Euler-Lagrange formula:

${\frac {d}{dt}}\left({\frac {\partial {\mathcal {L}}}{\partial {\dot {r}}}}\right)={\frac {\partial {\mathcal {L}}}{\partial r}}$
Substituting in ${\textstyle {\mathcal {L}}}$:
${\frac {d}{dt}}\left(m{\dot {r}}\right)=\left(mr{\dot {\theta }}^{2}+mr{\dot {\phi }}^{2}\sin ^{2}\theta -{\frac {\partial V}{\partial r}}\right)$
Calculating the derivatives:
$m{\frac {d^{2}r}{dt^{2}}}=mr{\dot {\theta }}^{2}+mr{\dot {\phi }}^{2}\sin ^{2}\theta -{\frac {\partial V}{\partial r}}$

This looks messy, but when we look at the Euler-Lagrange relation for $\phi$, we have

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

Hence $mr^{2}{\dot {\phi }}\sin ^{2}\theta$ is a constant throughout the motion.