Why isn't the Lagrangian invariant under ##\theta## rotations?

[PhysicsRock]

TL;DR Summary

Lagrangian of a particle in radial potential isn't invariant under $\theta$ rotations.

I just calculated the Lagrangian of a particle of mass $m$ in a radially symmetric potential $V(r)$. I thought it would be a good idea to switch to spherical coordinates for that matter. What I get is

$$
L = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \dot{\varphi}^2 \sin^2(\theta) \right) - V(r)
$$

This is typically a well-known result and I am very confident that it's correct. We instantly see that this Lagrangian does not depend on $\varphi$, only on it's time derivative. Thus, we can introduce an infinitesimal rotational shift $\delta\varphi$, which will not change the Lagrangian. However, a shift $\delta\theta$ does not leave this expression invariant, since $\theta$ explicitly appears in the Lagrangian. With radial symmetry though, I would expect that both angles can be altered infinitesimally without changing the physics of the system. Since $\theta$ appears in the sine squared term, we could introduce a discrete shift $\delta\theta = n \pi$ with $n \in \mathbb{N}$. This, however, would not yield a conservation law, since Noether's theorem demands continuous symmetries. I would've expected that angular momentum is conserved here, but using said theorem, the only conserved quantity here would be

$$
\frac{d}{dt} \left( m r^2 \dot{\varphi} \sin^2(\theta) \right) = 0
$$

which looks like a component of angular momentum (EDIT: it is in fact the $L_\theta$ component), but not angular momentum completely.

Why is that?

 

[Ibix]

A movement in the $\theta$ direction isn't a rigid rotation. It maps lines of constant latitude onto each other, and they don't have equal radius. Contrast a rotation in the $\phi$ direction that maps lines of constant longitude onto each other - they are all great circles.

There are Killing fields corresponding to rotations around the $x$ and $y$ axes (assuming rotation around $z$ is the $\delta\phi$ in your post), but they are not so cleanly expressed on spherical polars (slightly surprisingly, I agree).

 

[PhysicsRock]

A movement in the $\theta$ direction isn't a rigid rotation. It maps lines of constant latitude onto each other, and they don't have equal radius. Contrast a rotation in the $\phi$ direction that maps lines of constant longitude onto each other - they are all great circles.

There are Killing fields corresponding to rotations around the $x$ and $y$ axes (assuming rotation around $z$ is the $\delta\phi$ in your post), but they are not so cleanly expressed on spherical polars (slightly surprisingly, I agree).

So it is correct that only $L_\theta \sin(\theta)$ is conserved in such a system?

 

[TSny]

So it is correct that only $L_\theta \sin(\theta)$ is conserved in such a system?

A particle moving in a central potential will have an angular momentum (about the origin), $\vec L$, that is conserved. The direction of $\vec L$ is perpendicular to the plane of motion of the particle.

Since $\vec L$ remains constant, the projection of $\vec L$ along any fixed axis in the inertial frame will be conserved. The quantity $mr^2\sin^2\theta \, \dot{\varphi}$ is the canonical momentum associated with the variable $\varphi$ and this quantity is the projection of $\vec L$ along the fixed z-axis. So this quantity is conserved.

The quantity $mr^2 \dot{\theta}$ is the canonical momentum associated with the variable $\theta$. At some instant of time, this quantity is the projection of $\vec L$ along an axis that lies in the x-y plane and makes an angle $\varphi$ to the y-axis. But this axis is not a fixed axis (in general) since $\varphi$ will vary with time (in general). Thus, $mr^2 \dot{\theta}$ is generally not conserved.

I don't know if this helps.

 

[PhysicsRock]

A particle moving in a central potential will have an angular momentum (about the origin), $\vec L$, that is conserved. The direction of $\vec L$ is perpendicular to the plane of motion of the particle.

Since $\vec L$ remains constant, the projection of $\vec L$ along any fixed axis in the inertial frame will be conserved. The quantity $mr^2\sin^2\theta \, \dot{\varphi}$ is the canonical momentum associated with the variable $\varphi$ and this quantity is the projection of $\vec L$ along the fixed z-axis. So this quantity is conserved.

The quantity $mr^2 \dot{\theta}$ is the canonical momentum associated with the variable $\theta$. At some instant of time, this quantity is the projection of $\vec L$ along an axis that lies in the x-y plane and makes an angle $\varphi$ to the y-axis. But this axis is not a fixed axis (in general) since $\varphi$ will vary with time (in general). Thus, $mr^2 \dot{\theta}$ is generally not conserved.

I don't know if this helps.

It does help a lot. Thank you.

 

[vanhees71]

TL;DR Summary: Lagrangian of a particle in radial potential isn't invariant under $\theta$ rotations.

I just calculated the Lagrangian of a particle of mass $m$ in a radially symmetric potential $V(r)$. I thought it would be a good idea to switch to spherical coordinates for that matter. What I get is

$$
L = \frac{1}{2} m \left( \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \dot{\varphi}^2 \sin^2(\theta) \right) - V(r)
$$

This is typically a well-known result and I am very confident that it's correct. We instantly see that this Lagrangian does not depend on $\varphi$, only on it's time derivative. Thus, we can introduce an infinitesimal rotational shift $\delta\varphi$, which will not change the Lagrangian. However, a shift $\delta\theta$ does not leave this expression invariant, since $\theta$ explicitly appears in the Lagrangian. With radial symmetry though, I would expect that both angles can be altered infinitesimally without changing the physics of the system. Since $\theta$ appears in the sine squared term, we could introduce a discrete shift $\delta\theta = n \pi$ with $n \in \mathbb{N}$. This, however, would not yield a conservation law, since Noether's theorem demands continuous symmetries. I would've expected that angular momentum is conserved here, but using said theorem, the only conserved quantity here would be

$$
\frac{d}{dt} \left( m r^2 \dot{\varphi} \sin^2(\theta) \right) = 0
$$

which looks like a component of angular momentum (EDIT: it is in fact the $L_\theta$ component), but not angular momentum completely.

Why is that?

From a symmetry point of view, it's because with spherical coordinates you introduce a perferred direction, i.e., the direction of the polar axis, along which the spherical coordinates are singular.

Rotational invariance is better expressed in terms of Cartesian coordinates. An infinitesimal rotation is defined by
$$\delta t=0, \quad \delta \vec{r} = \delta \vec{\varphi} \times \vec{r}.$$
Full rotation symmetry means you can take $\delta \vec{\varphi}$ in an arbitrary direction, and the Lagrangian invariant under this infinitesimal rotation.

By introducing spherical coordinates you distinguish one direction by introducing the polar axis, and the full rotation symmetry is thus broken to the symmetry under rotations only around the polar axis, and this residual symmetry is parametrized by $\varphi$, which is thus a cyclic variable for a particle in a central potential.