## The Kepler Problem (Part 3)

The Kepler problem studies a particle moving in an inverse square force, like a planet orbiting the Sun. Last time I talked about an extra conserved quantity associated to this problem, which keeps elliptical orbits from precessing or changing shape. This extra conserved quantity is sometimes called the Laplace–Runge–Lenz vector, but since it was first discovered by none of these people, I prefer to call it the ‘eccentricity vector’

In 1847, Hamilton noticed a fascinating consequence of this extra conservation law. For a particle moving in an inverse square force, its momentum moves along a circle!

Greg Egan has given a beautiful geometrical argument for this fact:

• Greg Egan, The ellipse and the atom.

I will not try to outdo him; instead I’ll follow a more dry, calculational approach. One reason is that I’m trying to amass a little arsenal of formulas connected to the Kepler problem.

Let’s dive in. Remember from last time: we’re studying a particle whose position $\vec q$ obeys $\ddot{\vec q} = - \frac{\vec q}{q^3}$

Its momentum is $\vec p = m \dot{\vec q}$

Its momentum is not conserved. Its conserved quantities are energy: $H = \frac{1}{2} p^2 - \frac{1}{q}$

the angular momentum vector: $\vec L = \vec q \times \vec p$

and the eccentricity vector: $\vec e = \vec p \times \vec L - \frac{\vec q}{q}$

Now for the cool part: we can show that $\displaystyle{ \left( \vec p - \frac{\vec L \times \vec e}{L^2} \right)^2 = \frac{1}{L^2} }$

Thus, the momentum $\vec p$ stays on a circle of radius $1/L$ centered at the point $(\vec L \times \vec e)/L^2.$ And since $\vec L$ and $\vec e$ are conserved, this circle doesn’t change! Let’s call it Hamilton’s circle.

Now let’s actually do the calculations needed to show that the momentum stays on Hamilton’s circle. Since $\vec e = \vec p \times \vec L - \frac{\vec q}{q}$

we have $\frac{\vec q}{q} = \vec p \times \vec L - \vec e$

Taking the dot product of this vector with itself, which is 1, we get $\begin{array}{ccl} 1 &=& \frac{\vec q}{q} \cdot \frac{\vec q}{q} \\ \\ &=& (\vec p \times \vec L - \vec e) \cdot (\vec p \times \vec L - \vec e) \\ \\ &=& (\vec p \times \vec L)^2 - 2 \vec e \cdot (\vec p \times \vec L) + e^2 \end{array}$

Now, notice that $\vec p$ and $\vec L$ are orthogonal since $\vec L = \vec q \times \vec p.$ Thus $(\vec p \times \vec L)^2 = p^2 L^2$

I actually used this fact and explained it in more detail last time. Substituting this in, we get $1 = p^2 L^2 - 2 \vec e \cdot (\vec p \times \vec L) + e^2$

Similarly, $\vec e$ and $\vec L$ are orthogonal! After all, $\vec e = \vec p \times \vec L - \frac{\vec q}{q}$

The first term is orthogonal to $\vec L$ since it’s the cross product of $\vec L$ and some other vector. And the second term is orthogonal to $\vec L$ since $\vec L$ is the cross product of $\vec q$ and some other vector! So, we have $(\vec L \times \vec e)^2 = L^2 e^2$

and thus $\displaystyle { e^2 = \frac{(\vec L \times \vec e)^2}{L^2} }$

Substituting this in, we get $\displaystyle { 1 = p^2 L^2 - 2 \vec e \cdot (\vec p \times \vec L) + \frac{(\vec L \times \vec e)^2}{L^2} }$

Using the cyclic property of the scalar triple product, we can rewrite this as $\displaystyle { 1 = p^2 L^2 - 2 \vec p \cdot (\vec L \times \vec e) + \frac{(\vec L \times \vec e)^2}{L^2} }$

This is nicer because it involves $\vec L \times \vec e$ in two places. If we divide both sides by $L^2$ we get $\displaystyle { \frac{1}{L^2} = p^2 - \frac{2}{L^2} \; \vec p \cdot (\vec L \times \vec e) + \frac{(\vec L \times \vec e)^2}{L^4} }$

And now for the final flourish! The right hand is the dot product of a vector with itself: $\displaystyle { \frac{1}{L^2} = \left(\vec p - \frac{\vec L \times \vec e}{L^2}\right)^2 }$

This is the equation for Hamilton’s circle!

Now, beware: the momentum $\vec p$ doesn’t usually move at a constant rate along Hamilton’s circle, since that would force the particle’s orbit to itself be circular.

But on the bright side, the momentum moves along Hamilton’s circle regardless of whether the particle’s orbit is elliptical, parabolic or hyperbolic. And we can easily distinguish the three cases using Hamilton’s circle!

After all, the center of Hamilton’s circle is the point $(\vec L \times \vec e)/L^2,$ and $(\vec L \times \vec e)^2 = L^2 e^2$

so the distance of this center from the origin is $\displaystyle{ \sqrt{\frac{(\vec L \times \vec e)^2}{L^4}} = \sqrt{\frac{L^2 e^2}{L^4}} = \frac{e}{L} }$

On the other hand, the radius of Hamilton’s circle is $1/L.$ So his circle encloses the origin, goes through the origin or does not enclose the origin depending on whether $e < 1, e = 1$ or $e > 1.$ But we saw last time that these three cases correspond to elliptical, parabolic and hyperbolic orbits!

Summarizing:

• If $e < 1$ the particle’s orbit is an ellipse and the origin lies inside Hamilton’s circle. The momentum goes round and round Hamilton’s circle as time passes.

• If $e = 1$ the particle’s orbit is a parabola and the origin lies exactly on Hamilton’s circle. The particle’s momentum approaches zero as time approaches $\pm \infty,$ so its momentum goes around Hamilton’s circle exactly once as time passes.

• If $e > 1$ the particle’s orbit is a hyperbola and the origin lies outside Hamilton’s circle. The particle’s momentum approaches distinct nonzero values as time approaches $\pm \infty,$ so its momentum goes around just a portion of Hamilton’s circle.

By the way, in general the curve traced out by the momentum vector of a particle is called a hodograph. So you can learn more about Hamilton’s circle with the help of that buzzword.

### 13 Responses to The Kepler Problem (Part 3)

1. amarashiki says:

Just an stupid question: can the cross product be substituted by the wedge product in the LRL vector?

• John Baez says:

Yes, and this is what we need to do when studying the Kepler problem in higher dimensions. (This is the particle moving in a $-1/r^2$ central force in $\mathbb{R}^n.$)

2. Jesper says:

Here’s a recent blog post about the related quantum Kepler problem:

• John Baez says:

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.

3. Peter Haggstrom says:

John

It is really good to go over this stuff that people like Hamilton thought about deeply. Between you and Greg there is a great foundation for understanding celestial mechanics. From an historical perspective it is fascinating how Laplace and Lagrange (among many others) pounded away on these celestial problems without modern vector calculus. I have actually worked my way through slabs of Laplace’s magnum opus on celestial mechanics. He has my undying respect. I must get on to Lagrange. I loved Greg’s derivations. It was in the spirit of Principia! I’ve been working on a piece dealing with Bessel functions which at the purely mathematical level are fascinating in themselves. Few students know that the Bessel function they study arose from his analysis of the “mean anomaly” in the context of elliptical orbits.

Cheers
Peter

4. Graham says:

I haven’t done the maths to check, but in the e>1 case, would the other branch of the hyperbola cause the momentum to go around the rest of the circle? In which case, presumably, symmetry implies each branch leads to exactly a semicircle of the momentum circle?

• John Baez says:

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?

• pwmiles says:

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.

• Graham says:

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?

• pwmiles says:

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.

• John Baez says:

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.

5. pwmiles says:

Here is a streamlined derivation of the Hamilton momentum circle result:

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 = cL/(1+ec)
x2 = sL/(1+ec)

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 ~ M1C
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, Leh]
[ -L
e, 0, L^2]

I’ve no idea what this means!

6. AK says:

Not to forget Hamilton vector $\vec{h}=\vec{p} -\frac{1}{L}\vec{e}_\theta$.
where $\vec{e}=\vec{h}\times\vec{L}$.

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