# Spherical co-ordinates are not canonical?

One of the conditions for canonical transformations is that all momentum variables should commute. But $$(L_x ,L_y)=L_z \neq 0$$. Does that mean these are not canonical co ordinates? But aren't point transformations a special case of canonical transformations? How can these transformations be point but not canonical?

EDIT- Spherical co-ordinates are $$(r, \theta, \phi)$$. The momenta corresponding to $$\theta$$ and $$\phi$$ are $$L_x$$ and $$L_z$$. They don't commute as I already wrote.

By point transformations, I mean the co ordinate transformations in the Lagrangian formalism. But even disregarding that, we use spherical co-ordinates all the time in Hamiltonian and Quantum mechanics. They can't be non-canonical, right?

• How did you find the momentum corresponding to $\theta$? Your conclusion that it's $L_x$ looks wrong. Feb 3 at 10:13
• @Ruslan I wrote them backwards sorry. $\theta$ has $L_z$ as it rotates about the $z$ axis. Feb 3 at 11:12
• Regardless of naming, only one angle has a well-defined axis: the one that rotates around $\hat e_z$. The other angle is that of a standing spherical wave along (and into) the $z$-axis, and if you find corresponding momentum, it won't coincide with any of Cartesian components of angular momentum. Feb 3 at 11:17
• @Ruslan Do you know of the other momentum? Why doesn't it have a popular name? Feb 3 at 11:21
• I don't know what it's called. I suppose it doesn't have much use because the interpretation of the variable is not so enlightening, nor are such states with $|m|\ll \ell$ very relevant to the classical limit. Feb 3 at 11:31

Spherical co-ordinates are $$(r,\theta,\phi)$$. The momenta corresponding to $$\theta$$ and $$\phi$$ are $$L_x$$ and $$L_z$$.

No they aren't. It's true that $$p_\phi=L_z$$, but $$p_\theta= \cos(\phi) L_y - \sin(\phi) L_x$$.

The free-particle Lagrangian in spherical coordinates is $$L = \frac{1}{2} m\big(\dot r^2 + r^2\dot \theta^2 + r^2\sin^2(\theta) \dot \phi^2\big)$$ $$\implies \pmatrix{p_\theta \equiv \frac{\partial L}{\partial \dot\theta} = mr^2\dot \theta\\p_\phi \equiv \frac{\partial L}{\partial \dot \phi} = mr^2\sin^2(\theta) \dot \phi}$$

Recall that $$L_x = m(y \dot z - z \dot y) = m r^2\big(-\dot \theta \sin^2(\theta)\sin(\phi)- \dot \phi\sin(\theta)\cos(\phi)\cos(\theta) - \dot \theta \cos^2(\theta)\sin(\phi)\big)$$ $$= -mr^2\big(\dot \theta \sin(\phi) + \dot \phi \sin(\theta)\cos(\theta)\cos(\phi)\big)$$ $$L_y= m(z\dot x - x\dot z)= mr^2(\dot \theta \cos^2(\theta) \cos(\phi) - \dot \phi \sin(\theta)\cos(\theta)\sin(\phi) + \dot \theta\sin^2(\theta)\cos(\phi)\big)$$ $$= mr^2\big(\dot \theta \cos(\phi) -\dot \phi \sin(\theta)\cos(\theta) \sin(\phi)\big)$$ $$\implies -\sin(\phi)L_x + \cos(\phi)L_y = mr^2\dot\theta = p_\theta$$ as promised.

The Poisson bracket of the two angular momentum coordinates is then

$$\{p_\theta,p_\phi\} = \{\cos(\phi)L_y,L_z\} - \{\sin(\phi) L_x,L_z\}$$ Noting that $$\{AB,C\}=A\{B,C\}+\{A,C\}B$$ and that $$\{f(\phi),L_z\} = \{f(\phi),p_\phi\} = f'(\phi)$$, we quickly find that $$\{p_\theta,p_\phi\}=0$$ as expected.