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 obeys

Its **momentum** is

Its momentum is not conserved. Its conserved quantities are **energy**:

the **angular momentum** vector:

and the **eccentricity** vector:

Now for the cool part: we can show that

Thus, the momentum stays on a circle of radius centered at the point And since and 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

we have

Taking the dot product of this vector with itself, which is 1, we get

Now, notice that and are orthogonal since Thus

I actually used this fact and explained it in more detail last time. Substituting this in, we get

Similarly, and are orthogonal! After all,

The first term is orthogonal to since it’s the cross product of and some other vector. And the second term is orthogonal to since is the cross product of and some other vector! So, we have

and thus

Substituting this in, we get

Using the cyclic property of the scalar triple product, we can rewrite this as

This is nicer because it involves in two places. If we divide both sides by we get

And now for the final flourish! The right hand is the dot product of a vector with itself:

This is the equation for Hamilton’s circle!

Now, beware: the momentum 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 and

so the distance of this center from the origin is

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

Summarizing:

• If 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 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 so its momentum goes around Hamilton’s circle exactly once as time passes.

• If 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 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.