# Classical Mechanics/Central Field

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

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

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

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

It can then be worked out that:

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

Hence the equation for the Lagrangian is:

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

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

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

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

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

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