# 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?

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

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.

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 $$\vec \rho$$.

• 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 $\phi$? Aug 18, 2021 at 17:04
• Yes, I was wrong about the direction pθ. See my updated answer. I also added more explanation for pϕ. Thanks. Aug 18, 2021 at 21:23

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.

\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 $$\mathbf{e}_\theta$$

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