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In cartesian coordinates, a particle under an arbitrary potential $U(x,y,z)$ will have a Lagrangian $$L=\frac{m}{2}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)-U(x,y,z)$$ Consequently, the canonical conjugate momenta are $p_{x} =\frac{\partial L}{\partial \dot{x}}=m \dot{x}$, $p_{y} =\frac{\partial L}{\partial \dot{y}}=m \dot{y}$, and $p_{z} =\frac{\partial L}{\partial \dot{z}}=m \dot{z}$. They represent the linear momentum of the particle over each coordinate $x$, $y$ and $z$.

Similarly, in cylindrical coordinates the lagrangian will be $$L=\frac{m}{2}\left(\dot{\rho}^{2}+\rho^{2} \dot{\phi}^{2}+\dot{z}^{2}\right)-U(\rho, \phi,z)$$

With the associated conjugated canonical momenta $p_{\rho} =\frac{\partial L}{\partial \dot{\rho}}=m \dot{\rho}$, $p_{\phi} =\frac{\partial L}{\partial \dot{\phi}}=m \rho^{2} \dot{\phi}$, and $p_{z} =\frac{\partial L}{\partial \dot{z}}=m \dot{z}$. As I understand it, in this case, $p_\rho$ and $p_z$ would represent the linear momentums over the radial and vertical directions given by $\hat{u}_\rho$ and $\hat{u}_z$, whereas $p_\phi$ would correspond to the angular momentum of the particle rotating around the $Z$ axis.

Finally, in spherical coordinates, $L=T-U=\frac{m}{2}\left(\dot{r}^{2}+r^{2} \dot{\theta}^{2}+r^{2} \sin ^{2} \theta \dot{\phi}^{2}\right)-U(r, \theta, \phi) $, and

$$ \begin{aligned} p_{r} &=\frac{\partial L}{\partial \dot{r}}=m \dot{r} \\ p_{\theta} &=\frac{\partial L}{\partial \dot{\theta}}=m r^{2} \dot{\theta} \\ p_{\phi} &=\frac{\partial L}{\partial \dot{\phi}}=m r^{2} \sin ^{2} \theta \dot{\phi} \end{aligned} $$

While I see that $p_r$ would have a similar meaning to $p_\rho$ in cylindrical coordinates, what would be the meaning of $p_\theta$ and $p_\phi$ in this case?

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    $\begingroup$ $p_\phi$ is the same as it is in the cylindrical case. Did you mean to ask about $p_\theta$ instead? $\endgroup$ Aug 15, 2021 at 15:34
  • $\begingroup$ I meant both $p_\phi$ and $p_\theta$, sorry. I have edited the question. $\endgroup$
    – Asd
    Aug 15, 2021 at 15:42

4 Answers 4

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As in the cylindrical case, $p_\phi$ is the angular momentum about the $z$-axis, $L_z$. This makes sense because in both cases $\phi$ is defined in the same way: as the angle of rotation about the $z$-axis.

The conjugate momentum $p_\theta$ is harder to interpret. The best I've been able to come up with is to note that for an arbitrary particle, we can show that $$ T = \frac{|\vec{p}|^2}{2m} = \frac{1}{2m r^2} \left[ (\vec{r} \cdot \vec{p})^2 + (\vec{r} \times \vec{p})^2\right] = \frac{p_r^2}{2m} + \frac{|\vec{L}|^2}{2mr^2} $$ but also $$ T = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2 m r^2} + \frac{p_\phi^2}{2 m r^2 \sin^2 \theta} $$ from which we can conclude that $$ p_\theta^2 + \frac{L_z^2}{\sin^2 \theta} = |\vec{L}|^2. $$ where $L^2 = \vec{L} \cdot \vec{L}$ is the norm squared of the angular momentum vector. This can be simplified a little to yield $$ p_\theta^2 = L_x^2 + L_y^2 - (\cot^2 \theta) L_z^2. $$ but this is not terribly illustrative.

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The figure at the end of this answer illustrates the spherical coordinates $r, \theta,\phi$ and the associated unit vectors. $\theta$ is the polar angle and $\phi$ is the azimuthal angle.

$p_{\phi} = m r^2 \sin^2\theta \dot \phi$. This is the angular momentum in the z direction, shown as follows. The projection of $\vec r$ in the xy plane is $ \vec \rho = r \sin\theta \hat h$ where $\hat h$ is a unit vector in the direction of increasing $\vec \rho$. For constant r and constant $\theta$, $\vec v = r \dot \phi \sin\theta \hat m$ where $\hat m$ is a unit vector in the increasing $\phi$ direction. The angular momentum in the z direction $m(\vec \rho \times \vec v)$ for this situation is $m r^2 \sin^2\theta \dot \phi \hat h \times\hat m = m r^2 \sin^2\theta \dot \phi \hat k$ where $\hat k$ is a unit vector in the $z$ direction.

$p_{\theta} = mr^2 \dot \theta$. This is the magnitude of the angular momentum for a mass moving at constant r, constant $\phi$, in a circle with changing angle $\theta$. As a vector the angular momentum for this motion is $m(\vec r \times \vec v)$ where $\vec r = r \hat n$ and $\vec v$ is $r \dot \theta \hat l$, $\hat r$ and $\hat l$ being unit vectors in the increasing r and $\theta$ directions, respectively. $\hat n \times \hat l = \hat m$ where $\hat m$ is a unit vector in the increasing $\phi$ direction. So the angular momentum vector is $mr^2 \dot \theta \hat m$; magnitude $mr^2 \dot \theta$ in the $\hat m$ direction.

An earlier answer by @Michael Seifert expresses $p_{\theta}$ in terms of $L_x, L_y$, and $L_z$.

The total angular momentum in spherical coordinates can be expressed as $\vec L = m(\vec r \times \vec v)$ where $\vec r = r \hat n$ and $\vec v = \dot r \hat n + r \dot \theta \hat l + r \dot \phi \sin\theta \hat m$. The result is $\vec L = m(r^2 \dot \theta \hat n \times \hat l + r^2 \dot \phi \sin \theta \hat n \times \hat m) = mr^2(\dot \theta \hat m - \dot \phi \sin \theta \hat L)$. The $mr^2\dot \theta \hat m$ term is the component of $\vec L$ in the $\hat m$ direction. Since $\hat l = -\hat k \sin \theta + \hat h \cos \theta$, the component of $\vec L$ in the $\hat k$ ($z$) direction is $+mr^2 \dot \phi \sin^2 \theta$.

The figure below illustrates the spherical coordinates. Relating my notation above to the figure: $\hat n = \hat e_r$, $\hat m = \hat e_{\phi}$, $\hat l = \hat e_{\theta}$, and $\hat k$ is a unit vector in the z direction. $\vec \rho = r\sin\theta \hat h$ is the vector for the component of $\vec r$ in the xy plane, and $\hat h$ is a unit vector in the direction of increasing $\vec \rho$.

enter image description here

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    $\begingroup$ I don't think your second statement is correct. If $p_\theta$ is the angular momentum in the direction of increasing $\theta$, why isn't $p_\phi$ the angular momentum in the direction of increasing $\phi$? $\endgroup$ Aug 18, 2021 at 17:04
  • $\begingroup$ Yes, I was wrong about the direction pθ. See my updated answer. I also added more explanation for pϕ. Thanks. $\endgroup$
    – John Darby
    Aug 18, 2021 at 21:23
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The interpretations in general are coordinate-dependent. For some coordinate $q^i$ take $\frac{\partial L}{\partial \dot{q}^i} = \frac{\partial T}{\partial \dot{q}^i} = m {\bf v} \cdot \frac{\partial {\bf v}}{\partial \dot{q}^i} = m {\bf v} \cdot \frac{\partial {\bf r}}{\partial {q}^i}$.

Therefore, you see that these conjugate momenta are nothing more than the linear momentum $m {\bf v}$ projected along covariant basis vectors $\frac{\partial {\bf r}}{\partial {q}^i}$. For some coordinate choices like cylindrical and spherical, $\frac{\partial {\bf r}}{\partial {q}^i}$ contains a length and will turn $m {\bf v} \cdot \frac{\partial {\bf r}}{\partial {q}^i}$ into a component of the angular momentum.

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\begin{align*} & \text{ the position vector}\\\\ &\mathbf R= \left[ \begin {array}{c} r\cos \left( \theta \right) \sin \left( \phi \right) \\ r\sin \left( \theta \right) \sin \left( \phi \right) \\ r\cos \left( \phi \right) \end {array} \right]\\\\ &\text{from here you obtain the velocity}\\\\ &\mathbf v=\frac{\partial\mathbf{R} }{\partial r}\,\dot{r}+ \frac{\partial\mathbf{R} }{\partial \phi}\,\dot{\phi}+ \frac{\partial\mathbf{R} }{\partial \theta}\,\dot{\theta}\\ &\mathbf v=\mathbf e_r\,\dot{r}+\mathbf e_\phi\,r\,\dot{\phi}+\mathbf e_\theta\,r\,\sin(\phi)\,\dot{\theta}\\\\ &\text{where $~\mathbf{e}~$ are unit vectors} \end{align*} those

$p_r$ momenta towards $\mathbf{e}_r$

$p_\phi$ momenta towards $\mathbf{e}_\phi$

$p_\theta$ momenta towards $\mathbf{e}_\theta $

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  • $\begingroup$ This is not exactly right. $p_\phi \neq m{\bf v} \cdot {\bf e}_\phi$ and $p_\theta \neq m {\bf v} \cdot {\bf e}_\theta$ $\endgroup$
    – Evan
    Aug 17, 2021 at 3:20
  • $\begingroup$ @even “ momenta towards” not equal $\endgroup$
    – Eli
    Aug 20, 2021 at 6:54

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