Topology of equipotential surfaces

Let's think of planar equipotential surfaces, say they are parallel to the x-y plane, then apparently $p_x$ and $p_y$ are conserved quantities.

Next, let's move on to cylindrical equi-potentials. Then $p_z$ and $p_\phi$ are conserved from symmetries. The other way we can think about it is that a cylinder can be obtained from identifying two opposite borders of a rectangle. So we can map $p_x \rightarrow p_z, p_y\rightarrow p_\phi$.

Let's do the folding once more, so that the cylinder now becomes a torus. If you believe topology, then we should expect two conserved momenta, associated with orbiting the red circle in the image and the magenta circle. But careful calculation invoking Noether's theorem doesn't seem to support this belief because transformation along the red circle doesn't seem to preserved the Lagrangian and hence doesn't provide a conserved momentum.

To expand a little more, I use the following coordinates on the torus, where $\theta$ is the angle on minor circles, and $\phi$ is on the major circle. Then the Lagrangian expressed in terms of $\theta$ and $\phi$ is dependent on $\theta$, so $p_\theta$ is not conserved.

So, if you believe in topology, where is the other symmetry?

\begin{align}x(\theta,\varphi) &= (R+ r\cos\theta)\cos\varphi\\ y(\theta,\varphi) &= (R+ r\cos\theta)\sin\varphi\\ z(\theta,\varphi)&= r\sin\theta\end{align} • Are you talking about the linear momentum of a particle experiencing no change in potential energy moving on the different surfaces? – freecharly Sep 22 '16 at 17:22
• Cylindrical equipotentials lead to a Lagrangian that doesn't depend explicit on $z$ and $\phi$, so the corresponding CONJUGATE MOMENTUM are conserved by Noether's theorem. Then you fold the cylinder into a torus, the orbit would become the red line in picture, but it's does't give rise to a conserved conjugate momentum anymore. – JamieBondi Sep 22 '16 at 17:28
• I don't know about topology, but It seems to me from a physical point of view that the orbits you draw on the torus dont conserve momentum as any particle moving througth the purple particle will be drag outsie by intertia. Maybe you need to set a different system of coordinates. – Victor Sep 22 '16 at 17:31
• @Victor Purple line certainly gives a conserved momentum, because the Lagrangian in $\theta,\phi$ coordinates doesn't depend on $\phi$. – JamieBondi Sep 22 '16 at 17:35
• So what is the coordinate for the red line, whose conjugate momentum you are looking for? – freecharly Sep 22 '16 at 17:42

Your represented (in the nice figure!) torus is non-flat and is immersed in $\mathbb R^3$, the one obtained by the identifications of the opposite edges of a rectangle is instead flat and is not metrically immersed in $\mathbb R^3$.

These two kinds of torus are topologically identical (homeomorphic and also diffeomorphic actually), but they are metrically distinct (they are not isometric).

Here metrical notions matter. This is the reason why the immersed torus has one symmetry less than the flat torus.

The orbits of the angle $\theta$ are symmetries provided the metric on the torus is flat as it arises putting the metric of the plane on the torus with the standard identifications just to produce a flat torus. However, this metric is not the one the torus receives from the metric of $\mathbb R^3$ viewing it as an immersed surface: the curvature shows up here and $\theta$ is not a metrically invariant direction. In the flat torus, for instance, all the violet circles have the same length, in the immersed torus, their length is variable depending on $\theta$ as it is evident from the figure...

The Lagrangian possesses the corresponding symmetries depending on which notion of torus you consider.

In the limit case of a torus with an infinite radius $R$, that is a cylinder, the two metrics coincide. This is the reason why you cannot see the problem just looking at the cylinder.

This is an interesting example where topology is not enough to fix physics.

• I see. What's more interesting is that the two limit cases $R\rightarrow\infty$ (to become a cylinder) and $r\rightarrow\infty$ (to become a sphere) would give two conserved momenta, but in between the torus only has one. So there's sort of "discontinuity" here. – JamieBondi Sep 22 '16 at 19:28
• Yes, in both cases you remove an asymmetry, nice! – Valter Moretti Sep 22 '16 at 19:33
• I have a new question though. In the planar equipotential case, in addition to $p_x$ and $p_y$, $L_z$ is also conserved obviously, but is $L_z$ independent from $p_x$ and $p_y$? Note $L_z$ can be written as $yp_x-xp_y$. – JamieBondi Sep 23 '16 at 1:30
• What is the question? I do not understand well. $L_z$ is conserved because the potential depends on $z$ only, but not on $x$ and $y$. So that it does not depend on the angular coordinate around $z$ when passing to work in cylindrical coordinates. The conjugated momentum of that angle is in fact $L_z$. – Valter Moretti Sep 23 '16 at 4:47
• Never mind then. I figured out px, py and Lz are independent Killing vectors on R2. – JamieBondi Sep 23 '16 at 16:57