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I have always found it clear that since a spherical potential has all components of angular momentum conserved since the entire system is symmetric under any rotation. However, I was trying to prove this to myself using Noether's theorem. The Lagrangian in spherical coordinates for such a system:

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

Since the lagrangian does not have an explicity $\phi$ dependence, the conservation holds. However, when I try this on the $\theta$ axis, I am running into some troubles. Let $\theta \rightarrow \theta + \lambda$. Then by Noether's theorem, I need to show that the lagrangian does not change in the first order of $\lambda$:

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

The problem is with the sine term. My first intuition is to taylor expand:

$$\sin(x) = \sin(\theta) + \cos(\theta)(x - \theta) - \frac{1}{2} \sin(\theta) (x - \theta)^2+...$$

$$\sin(\theta + \lambda) = \sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...$$

$$\sin^2(\theta + \lambda) = (\sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...)(\sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...)$$ $$ = \sin^2(\theta) + 2\sin(\theta)\cos(\theta) \lambda + ....$$

Which is not a first order change. In other words:

$$\frac{\partial L}{\partial \lambda} = \frac{1}{2}mr^22\sin(\theta)\cos(\theta)\dot{\phi^2} \neq0.$$

However, I know intuitively that around $\theta$, there is symmetry. What is exactly wrong in my line of logic?

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The issue is that the transformation $\theta\to\theta+\delta\theta$ is not a rotation. This explains why the Lagrangian is not conserved ad does not give you an integral of motion. Remember the definition of $\theta$ is spherical coordinates, it is not periodic like $\phi$ so your transformation does not even make much sense.

Viewing other rotations in spherical coordinates is complicated. It’s best to revert back to cartesian ones.

Polar coordinates are more useful in the Hamiltonian formalism. Using separation of variables, you can easily get the conservation of $L^2$ without noticing the conservation of the other components of angular momentum. The corresponding Hamiltonian is: $$ H=\frac{1}{2m}\left(p_r^2+\frac{p_\theta^2}{r^2}+\frac{p_\phi^2}{r^2\sin^2\theta}\right)+\frac{K}{r^n} $$ The component $L_z=p_\phi$ is conserved corresponding to rotations about $z$ and you also get the conservation of $L^2=p_\theta^2+\frac{p_\phi^2}{\sin^2\theta}$ as announced.

Hope this helps.

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