1
$\begingroup$

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?

$\endgroup$
12
  • $\begingroup$ How did you find the momentum corresponding to $\theta$? Your conclusion that it's $L_x$ looks wrong. $\endgroup$
    – Ruslan
    Feb 3, 2022 at 10:13
  • $\begingroup$ @Ruslan I wrote them backwards sorry. $\theta$ has $L_z$ as it rotates about the $z$ axis. $\endgroup$
    – Ryder Rude
    Feb 3, 2022 at 11:12
  • $\begingroup$ 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. $\endgroup$
    – Ruslan
    Feb 3, 2022 at 11:17
  • $\begingroup$ @Ruslan Do you know of the other momentum? Why doesn't it have a popular name? $\endgroup$
    – Ryder Rude
    Feb 3, 2022 at 11:21
  • $\begingroup$ 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. $\endgroup$
    – Ruslan
    Feb 3, 2022 at 11:31

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.