Classical Mechanics/Central Field

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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

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

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

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

It can then be worked out that:

$${\dot{\overrightarrow{x}}}^2={\dot{r}}^2+r^2{\dot{\theta }}^2+r^2{\dot{\varphi }}^2{\sin }^2\theta$$

Hence the equation for the Lagrangian is:

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

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

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

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

$$\frac{d}{dt}\left(m{r}^2\dot{\varphi }{\sin }^2\theta \right)=0$$

Hence $m{r}^2\dot{\varphi }{\sin }^2\theta$ is a constant throughout the motion.