Physical meaning of the canonical conjugate momenta in spherical coordinates 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?
 A: 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.
A: 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 $\vec \rho$.

A: 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.
A: \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 $
