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

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

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 $ \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.