Angular Momentum in Spherical Coordinates

[PeroK]

I've started on "Noether's Theorem" by Neuenschwander. This is page 35 of the 2011 edition.

We have the Lagrangian for a central force:

$L = \frac12 m(\dot{r}^2 + r^2 \dot{\theta}^2 + r \dot{\phi}^2 \sin^2 \theta) - U(r)$

Which gives the canonical momenta:

$p_{\theta} = mr^2 \dot{\theta}$

And

$p_{\phi} = mr^2\dot{\phi} \sin^2 \theta$

Which then should be constants of the motion. I recognise $p_{\phi} = l_z$, but I'm puzzled by what $p_{\theta}$ is.

An an example, I took a circular path (in the x-y plane) and titled it about the x-axis by an angle $\alpha$ to get:

$\vec{r} = r(\cos \omega t, r\cos \alpha \sin \omega t, r\sin \alpha \sin \omega t)$

But, I can't see how $mr^2 \dot{\theta}$ is constant for this motion. As $r$ is constant, we must have a constant $\dot{\theta}$. We have:

$\cos{\theta} = \sin \alpha \sin \omega t$

Which doesn't lead to constant $\dot{\theta}$ by my calculations:

$\dot{\theta}^2 = \frac{w^2 \sin^2 \alpha \cos^2 \omega t}{1- \sin^2 \alpha \sin^2 \omega t}$

By constrast, I tried examples of titled circular motion about all three axes and $p_{\phi}$ always came out constant.

But, $p_{\theta}$ is only constant in the one case where $\theta$ is constant.

Can anyone shed any light on this?

A second question is that the book says:

When we calculate the particle's angular momentum $\vec{l} = \vec{r} \times (m\vec{v})$ about the origin we obtain:

$\vec{l} = p_{\theta} \hat{\theta} - p_{\phi} \hat{\phi}$

I'm not sure how he gets these from a vector product. In general, I'm not sure of the validity of expressing angular momentum in spherical coordinates, as it's the vector product of a displacement vector from the original with a velocity vector at the particle's location.

Most references I've found online take the operator definition of angular momentum, which I'm familiar with. But, Neuenschwander seems to do it all with vectors. Can anyone explain what he is doing?

Thanks

 

[kuruman]

Which then should be constants of the motion.

Not both of them. If you derive the equations of motion from the Lagrangian, you get for $\phi$ $$\frac{d}{dt} (mr^2 \dot \phi \sin^2 \theta)=0$$ which certainly implies that $mr^2 \dot \phi \sin^2 \theta=const.$ Do the same for $\theta$ and you get $$\frac{d}{dt} (mr^2 \dot \theta)=2r \dot \phi^2 \sin \theta \cos \theta$$

 

[PeroK]

Not both of them. If you derive the equations of motion from the Lagrangian, you get for $\phi$ $$\frac{d}{dt} (mr^2 \dot \phi \sin^2 \theta)=0$$ which certainly implies that $mr^2 \dot \phi \sin^2 \theta=const.$ Do the same for $\theta$ and you get $$\frac{d}{dt} (mr^2 \dot \theta)=2r \dot \phi^2 \sin \theta \cos \theta$$

Of course! Thanks.