Gravitational Potential of a Ring Mass in the Plane of the Ring

[utesfan100]

Upon reflection, it occurs to me that in GR a rotating ring would be a Kerr Metric. The radius, however, would make this an over-extreme solution. These solutions have a naked singularity at the ring, confirming that the infinity is the correct solution.

korjik: The dimensions I was referring to above was the topological dimension of the mass. At any point on a circle the topology is 1D.

onomatomanic: The angle of the boundary between a surface of the green equipotential surfaces being a torus shape and a red blood cell shape at the origin should be a function of R/M. The bigger this value the larger the angle should be, increasing the eggyness.

I remember once wanting to make a torus world, with a moon that orbited in a figure eight orbit through the center. All these years latter I finally have an outline to the solution to this problem.

ETA: It is possible that the equipotential curves in the classical theory are Cassini Ovals rotated about the Y axis. If this is true, it would be interesting to see if an orbit near the Lemniscate of Bernoulli is stable.

[onomatomanic]

^ Changing M while holding R doesn't seem to have any effect other than scaling the force-field homogeneously; the contours appear unaffected. This makes sense to me, as the force normal to the XY plane depends on the total mass of the ring while the force in the XY place depends on the portion of ring in the relevant direction. Changing M affects both of those equally, thus, the ratio remains fixed, thus, the curvature of the contours remains fixed.

Cassini ovals, courtesy of the 'pedia:



That looks very much like it, agreed. Also, the article mentions that "[c]urves orthogonal to the Cassini ovals [are] equilateral hyperbolas [...] pass[ing] through the foci", and it's fairly apparent that the corresponding field lines have to be hyperbolic as the ring has to be approximatable as a point mass for d >> R. Nice call. As to orbital stability, I do seem to remember proving in some dynamical systems class or other that figure-of-eight orbits are unstable a priori, regardless of the specifics of the underlying geometry. The approach escapes me for the moment, though.

ETA: Ah, yes, I think this was it, in outline: For a figure-of-eight symmetry, the orbit has to be asymptotically straight at (or near) the point at which it self-intersects. This requires the potential field at (or near) the point of self-intersection to be locally flat. Therefore, examine the nature of the three elementary cases that yield this: If the point is a local minimum, it means the orbit passes through or very near an attractor, which is unstable (small perturbation -> large effect). If the point is a local maximum, it means the orbit will tend to "fall off" it, which makes it unstable (ditto). If the point is a saddle, it is neutrally stable itself, but the relevant approaches to the point are necessarily "across slopes" which the orbit will want to "slide down" in the long run.

It is possible that there are non-elementary cases which are stable, though - like a saddle point with more than two axes. Not quite sure if that means something like a quadrupole field or something which isn't actually analytical at all. Bah.

[grapes]

In cartesian coords, F=-Gm_rm_p/r²x

Where
m_r is the mass of the ring
m_p is the mass of a little tiny bit of the ring
r is the radius of the ring
x is the x unit vector

with the ring centered on the origin and the point being measured at (1,0) on the xy plane

OK

I think with the OP numbers it is F=-x, but I am too lazy to actually look

In the OP, G and r are equal to 1, but the mass of the ring m_r is 2π, so your formula gives F=-2πm_px. The formula in the OP does not have the mass m_p of the test particle.

That's the value you get if you ignore about a one degree (.856, about twice times 0.00747 radians) section of the ring centered on the test particle. In a tenuous ring, the effect of most of its mass is going to be as if it were centered at the center of the circle--a long ways away. It is the nearby particles that produce the infinity.

Wolframalpha input: Integrate[2/(4*sin(\theta/2)),{\theta,0.00747,\pi}]

[utesfan100]

^ Changing M while holding R doesn't seem to have any effect other than scaling the force-field homogeneously; the contours appear unaffected.

Yeah. I should have saw this, at least in the classical limit.

ETA: Ah, yes, I think this was it, in outline: For a figure-of-eight symmetry, the orbit has to be asymptotically straight at (or near) the point at which it self-intersects. This requires the potential field at (or near) the point of self-intersection to be locally flat. Therefore, examine the nature of the three elementary cases that yield this: If the point is a local minimum, it means the orbit passes through or very near an attractor, which is unstable (small perturbation -> large effect). If the point is a local maximum, it means the orbit will tend to "fall off" it, which makes it unstable (ditto). If the point is a saddle, it is neutrally stable itself, but the relevant approaches to the point are necessarily "across slopes" which the orbit will want to "slide down" in the long run.

The potential field at the center is clearly a saddle point. I am interested in what you mean by slide down the across slopes.

Clearly the orbits will be nearly straight at the crossing point. Intuitively, a Lemniscate of Bernoulli orbit itself should have the same stability as a circular orbit of Newtonian physics.

I do think that a general orbit would be offset from the center by some amount. These can be classified by the points where they cross the plane of the ring. This happens at the two outermost points and at a point near the center between each lobe.

Eccentricity: A parameter of the orbit similar to eccentricity (actually (1+e)/(1-e)) can be represented by the ratio of consecutive closest approaches to the ring, the left and right most points on the Lemniscate.

Precession: Considering that the force is confined to the plane defined by the radius and the axis of the ring, any momentum normal to this plane will be a constant of motion. Considering one lobe of the orbit, this will cause the two adjacent center points to be offset from the center normal to the plane defined by the axis of the ring and the line from the center of the ring to the outer intersection of the lobe. Considering one of the adjacent lobes shows that the next center point will have rotated relative to the previous orbit. This would also replace the point of self intersection with a circular ring of crossover points.

[onomatomanic]

I'm going on one third memory, one third naive visualization, and one third guesswork here, so this may turn out to be garbage - but I'll try to explain why I think your intuition is faulty regarding the stability of a near-lemniscate orbit. All references are to the diagram in my previous post.

Firstly, consider a particle orbiting about the left-hand focus, on the purple contour*, in a clockwise direction. Now, assume the particle is given some excess kinetic energy as it passes the leftmost point of the orbit. What happens? The excess speed results in its trajectory curving less than the purple contour (since the central force is now insufficient to supply the centripetal force required for the previous orbit); hence, the particle migrates gradually outside its previous orbit; hence, the particle climbs to a higher potential; hence, the particle loses speed. The only way for it to stop climbing to progressively higher potentials is for its trajectory to curve more than the local countour, which will only happen once its speed drops below the orbital speed appropriate for that local contour. Overall, the effect of excess speed is speed loss at least to the point at which there is no more excess speed with respect to a new orbit, and possibly beyond that. That is, we have unlimited negative feedback.

Similarly, assume another such particle is subjected to a loss of some kinetic energy as passes the leftmost point of the orbit. By the analogous line of reasoning, the particle migrates inwards to a lower potential and gains speed, and will continue to do so until it has attained or exceeded the orbital speed appropriate for the local contour. Overall, the effect of lack of speed is speed gain at least to the point of acquiring sufficient speed for a new orbit; we again have unlimited negative feedback. Unlimited negative feedback with respect to perturbations either way guarantees stability.

Secondly, consider a particle orbiting about both foci, on the blue contour, in a clockwise direction in its left-hand lobe and a counter-clockwise direction in its right-hand lobe. Again, assume the particle is given some excess kinetic energy as it passes the leftmost point of the orbit. Now what happens? Once more, the short-term effect will be for the particle to curve less than the contour and thus to climb to a higher potential and thus lose speed. However, as the particle proceeds along its new orbit, something happens which did not happen in the previous situation: The contours begin to straighten out and eventually to curve the other way. Consequently, our particle can now transition from a trajectory climbing in potential to a trajectory falling in potential without first having to increase its curvature beyond that of the contour itself. Meaning, the particle may find itself once more gaining speed while still having speed in excess of that appropriate for a local orbit. Concretely, this transition happens somewhere in the region between the locus of the inflexion points of the "red blood cell shaped" contours to the left of z-axis and the right-hand lobe of the Lemniscate. If one mentally draws such a trajectory between the blue contour and the next-larger grey contour, this is fairly apparent, IMO. Overall, the effect of excess speed here is some indeterminate amount of speed loss followed by some equally indeterminate amount of speed gain. That is, we have limited negative feedback and also limited positive feedback.

Vice versa, the same holds for subjecting the particle to a speed lose at it passes our designated point of interest. First, it migrates inwards and thus drops to lower potentials and thus regains some of the lost speed, but ultimately, it has to once again climb that same potential difference to be able to cross out of the left-hand lobe of the Lemniscate, let alone enter the right-hand half of the diagram. Speed loss leads to some amount of speed gain followed by a second instance of speed loss. We have feedback, but it is limited and can and does flip. Evidently, stability is far from guaranteed here.

That's the long version of what "slides down the slope" was the short version of. For orbits with no inflection, the slopes crossed by the orbital trajectory are always concave, which makes sliding impossible. For orbits with inflection, some of the slopes crossed by the orbital trajectory are bound to be convex, which makes sliding almost inevitable. Convinced?

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* Note: I'm unclear on whether on-contour orbits, i.e. orbits in which the distribution between potential and kinetic energy does not vary with the orbital phase, are actually possible within the lobes. However, it is clear that they are asymptotically possible for orbits arbitrarily close to a focus, which ought to be good enough for the argument I'm making.