.

where

. ]]>

Thanks for working this out, guys! Very nice.

It’s interesting how the two branches of the same hyperbola correspond to motions in equal but opposite attractive and repulsive forces. I’m not sure what that means mathematically, since from some mathematical perspectives (like algebraic geometry) the two branches of a hyperbola are part of the same curve.

]]>No, all that is correct. The important point is that the momentum position angle in its circle is the planet position angle i.e. true anomaly, advanced by 90 degrees. With the hyperbolic orbit the asymptotes are parallel to the momentum vector itself. Such a momentum vector, drawn from the origin, makes a 90 degree angle with the radius; hence, tangents from the origin mark the limits of the momentum on each branch. With the parabolic orbit the circle goes through the origin so there is just one tangent, parallel to the axis of the parabola.

The momentum circle is offset +he/L in the 2-direction and has radius h/L. So as e tends to +infinity the offset becomes a larger and larger multiple of the radius. We can then visualise the circle as a point, compared to the actual momentum at any time. Basically if the “planet” goes fast enough the Sun might as well not be there.

A sketch makes it clear that the “other” hyperbola branch, i.e. curving away from the Sun, is for a repelling 1/r^2 force.

]]>OK, thinking about it physically again, there is no “symmetry argument” for the two branches. After all, one branch has the sun in the interior of the curve, the other has the sun in the exterior. It seems perfectly reasonable that the two branches have different lengths of the circle arc, as long as together they add up to the full circle.

Now I am trying to imagine (again, physically, without doing the maths) what would cause the two circle arcs (from the two branches) to converge to exactly a semicircle each.

From your model, it is clearly the transition through e=1. In physical terms it is easier for me to visualise as we increase e through 1… e<1 is the ellipse, and all momenta on the circle occur at some point around the ellipse. At e=1 we have the parabola – all momenta occur except one particular value, although it can be reached arbitrarily closely from either side in the far reaches of the parabola.

As e becomes just >1, that one missing momentum value splits into two missing values, with a large arc and a small arc, each of which corresponds to the two branches of the hyperbola. So, depending which branch you are on you can only experience one of the two momenta arcs. And as e tends to infinity, those two arcs become closer and closer to semicircles.

Now, as e tends to infinity, the two branches become more and more like two straight lines, and closer and closer together (is that right? my mental model of conic sections seems to tell me that but I am not certain it is right). But if that is right, surely the momentum vector never changes – and just becomes a point.

Have I pushed a mental model too far?

]]>I investigated using my parametric model

x1 = cL/(1+ec)

x2 = sL/(1+ec)

p1 = -(h/L)s

p2 = (h/L)(e+c)

where L=semi-latus rectum, e=ellipticity, h=angular momentum, s=sin(theta), c=cos(theta), theta=true anomaly i.e. polar angle at the origin i.e. a focus.

With e>1, the momentum describes the circle arc from theta=-arccos(-1/e) to arccos(-1/e). The rest of the circle is for the other hyperbola branch which is never visited. The result seems asymmetrical but there is an arbitrary choice of focus involved.

]]>Thanks! I was worrying about how the heck one could get just part of the circle: it violated my intuitions about how this sort of math should work. But I foolishly forgot about the other branch of the hyperbola.

I’m not sure this solves the problem, though. The momentum along one branch of the hyperbola will be exactly the negative of the momentum of the other branch of the hyperbola. To divide a circle into two parts, one part consisting of points that are the negative of those in the other part, the circle needs to be centered at the origin. Right?

]]>Thanks very much! These are really interesting. I’m eventually going to talk about the quantum version myself here. But there are many approaches to this business, and I think what I say will look very different. It’s a very rich subject.

]]>Yes, and this is what we need to do when studying the Kepler problem in higher dimensions. (This is the particle moving in a central force in )

]]>Let c=cos(theta), s=sin(theta) with theta the true anomaly; that is, the polar angle of the planet from perihelion measured at the Sun. The radial distance is r=L/(1+e*c) where e is eccentricity and L is the semi-latus rectum (sorry symbol overlap). The 2D position vector (x1,x2) is thus given parametrically by

x1 = c*L/(1+e*c)

x2 = s*L/(1+e*c)

Claim: the 2D momentum vector (p1,p2) is given by

p1 = -(h/L)*s
p2 = (h/L)*(e+c)

where h is angular momentum.

Proof:

(1) The cross product (x1,x2)^(p1,p2) evaluates to h.

(2) Employing partial derivatives wrt c and s, (p1,p2) is found to be parallel to the time derivative of (x1,x2)

(3) Thus (p1,p2) is correctly determined by conditions (1) and (2)

The Hamilton result follows from the fact that as (c,s) describes a circle so does (p1,p2).

Exploring this scenario in the projective plane: the momentum circle and the orbit ellipse are conics so they can be projectively related. In fact one can find a projectivity M between the instantaneous position and momentum, thus:

Introduce homogeneous coordinates such that (c,s) is the point (C1,C2,C3); (x1,x2) is (X1,X2,X3) and (p1,p2) is (P1,P2,P3) with c1=C1/C3, s=C2/C3 etc. Let (C1,C2,C3) be the entries in a column 3-vector C, etc. Then (denoting equality up to ratio by ~)

X ~ M1*C
P ~ M2*C

with

M1 =

[ L, 0, 0]

[ 0, L, 0]

[ e, 0, 1]

M2 =

[ 0, -h, 0]

[ h, 0, e*h]

[ 0, 0, L]

From these we can solve

P~M*X, M~M2/M1 such that

M =

[ 0, -h, 0]

[ h*(1-e^2), 0, L*e*h]
[ -L*e, 0, L^2]

I’ve no idea what this means!

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