## The Kepler Problem (Part 4)

The Kepler problem is the study of a particle moving in an attractive inverse square force. In classical mechanics, this problem shows up when you study the motion of a planet around the Sun in the Solar System. In quantum mechanics, it shows up when you study the motion of an electron around a proton in a hydrogen atom.

In Part 2 we saw that the classical Kepler problem has, besides energy and the three components of angular momentum, three more conserved quantities: the components of the eccentricity vector!

This was discovered long ago, in 1710, by the physicist Jakob Hermann. But thanks to Noether, we now know that in classical mechanics, conserved quantities come from symmetries. In the Kepler problem, conservation of energy comes from time translation symmetry, while conservation of the angular momentum comes from rotation symmetry. Which symmetries give conservation of the eccentricity vector?

As we shall see, these symmetries are rotations in 4-dimensional space. These include the obvious rotations in 3-dimensional space which give angular momentum. The other 4-dimensional rotations act in a much less obvious way, and give the eccentricity vector.

In fact, we’ll see that the Kepler problem can be rephrased in terms of a free particle moving around on a sphere in 4-dimensional space. This is a nice explanation of the 4-dimensional rotation symmetry.

After that we’ll see a second way to rephrase the Kepler problem: in terms of a massless, relativistic free particle moving at the speed of light on a sphere in 4-dimensional space. Our first formulation will not involve relativity. This second will.

All this is very nice. You can read some fun explanations of the first formulation here:

• Greg Egan, The ellipse and the atom.

• John Baez, Planets in the fourth dimension.

But how could you guess this 4-dimensional rotation symmetry if you didn’t know about it already? One systematic approach uses Poisson brackets. I won’t explain these, just dive in and use them!

Remember, the particle in the Kepler problem has various observables, which are all ultimately functions of its position and momentum:

• position: $\vec q$

• momentum: $\vec p$

• energy: $H = \tfrac{1}{2} p^2 - \tfrac{1}{q}$

• angular momentum: $\vec L = \vec q \times \vec p$

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

I’ll use conventions where the Poisson brackets of the components of position $q_k$ and momentum $p_\ell$ are taken to be

$\{q_k,p_\ell\} = \delta_{jk}$

From this, using the rules for Poisson brackets, we can calculate the Poisson brackets of everything else. For starters:

$\{H, L_k\} = \{H,e_h\} = 0$

These equations are utterly unsurprising, since they are equivalent to saying that angular momentum $\vec L$ and the eccentricity vector $\vec e$ are conserved. More interestingly, we have

$\begin{array}{ccl} \{L_k, L_\ell\} &=& \epsilon_{jk\ell} L_\ell \\ \{e_k, L_\ell\} &=& \epsilon_{jk\ell} e_\ell \\ \{e_k, e_\ell \} &=& -2H \epsilon_{jk\ell} L_\ell \end{array}$

where all the indices go from 1 to 3, I’m summing over repeated indices even if they’re both subscripts, and $\epsilon_{jk\ell}$ are the Levi–Civita symbols.

Now, the factor of $-2H$ above is annoying. But on the region of phase space where $H < 0$—that is, the space of bound states, where the particle carries out an elliptical orbit—we can define a new vector to deal with this annoyance:

$\displaystyle{ \vec M = \frac{\vec e}{\sqrt{-2H}} }$

Now we easily get

$\begin{array}{ccl} \{L_k, L_\ell\} &=& \epsilon_{jk\ell} L_\ell \\ \{L_j, M_k\} &=& \epsilon_{jk\ell} M_\ell \\ \{M_j, M_k \} &=& \epsilon_{jk\ell} M_\ell \end{array}$

This is nicer, but we can simplify it even more if we introduce some new vectors that are linear combinations of $\vec L$ and $\vec M,$ namely half their sum and half their difference:

$\vec A = \tfrac{1}{2} (\vec L + \vec M), \qquad \vec B = \tfrac{1}{2}(\vec L - \vec M)$

Then we get

$\begin{array}{ccl} \{ A_j, A_k\} &=& \epsilon_{jk\ell} A_\ell \\ \{ B_j, B_k\} &=& \epsilon_{jk\ell} B_\ell \\ \{ A_j, B_k\} &=& 0 \end{array}$

So, the observables $A_j$ and $B_k$ contain the same information as the angular momentum and eccentricity vectors, but now they commute with each other!

What does this mean?

Well, when you’re first learning math the Levi–Civita symbols $\epsilon_{jk\ell}$ may seem like just a way to summarize the funny rules for cross products in 3-dimensional space. But as you proceed, you ultimately learn that $\mathbb{R}^3$ with its cross product is the Lie algebra of the Lie group $\mathrm{SO}(3)$ of rotations in 3-dimensional space. From this viewpoint, the Levi–Civita symbols are nothing but the structure constants for the Lie algebra $\mathfrak{so}(3):$ that is, a way of describing the bracket operation in this Lie algebra in terms of basis vectors.

So, what we’ve got here are two commuting copies of $\mathfrak{so}(3),$ one having the $A_j$ as a basis and the other having the $B_k$ as a basis, both with the Poisson bracket as their Lie bracket.

A better way to say the same thing is that we’ve got a single 6-dimensional Lie algebra

$\mathfrak{so}(3) \oplus \mathfrak{so}(3)$

having both the $A_j$ and $B_k$ as basis. But then comes the miracle:

$\mathfrak{so}(3) \oplus \mathfrak{so}(3) \cong \mathfrak{so}(4)$

The easiest way to see this is to realize that $S^3,$ the unit sphere in 4 dimensions, is itself a Lie group with Lie algebra isomorphic to $\mathfrak{so}(3).$ Namely, it’s the unit quaternions!—or if you prefer, the Lie group $\mathrm{SU}(2).$ Like any Lie group it acts on itself via left and right translations, which commute. But these are actually ways of rotating $S^3.$ So, you get a map of Lie algebras from $\mathfrak{so}(3) \oplus \mathfrak{so}(3)$ to $\mathfrak{so}(4),$ and you can check that this is an isomorphism.

So in this approach, the 4th dimension pops out of the fact that the Kepler problem has conserved quantities that give two commuting copies of $\mathfrak{so}(3).$ By Noether’s theorem, it follows that conservation of angular momentum and the eccentricity vector must come from a hidden symmetry: symmetry under some group whose Lie algebra is $\mathfrak{so}(4).$

And indeed, it turns out that the group $\mathrm{SO}(4)$ acts on the bound states of the Kepler problem in a way that commutes with time evolution!

But how can we understand this fact?

Historically, it seems that the first explanation was found in the quantum-mechanical context. In 1926, even before Schrödinger came up with his famous equation, Pauli used conservation of angular momentum and the eccentricity to determine the spectrum of hydrogen. But I believe he was using what we now call Lie algebra methods, not bringing in the group $\mathrm{SO}(4).$

In 1935, Vladimir Fock, famous for the ‘Fock space’ in quantum field theory, explained this 4-dimensional rotation symmetry by setting up an equivalence between hydrogen atom bound states and functions on the 3-sphere! In the following year, Valentine Bargmann, later famous for being Einstein’s assistant, connected Pauli and Fock’s work using group representation theory.

All this is quantum mechanics. It seems the first global discussion of this symmetry in the classical context was given by Bacry, Ruegg, and Souriau in 1966, leading to important work by Souriau and Moser in the early 1970s. Since then, much more has been done. You can learn about a lot of it from these two books, which are my constant companions these days:

• Victor Guillemin and Shlomo Sternberg, Variations on a Theme by Kepler, Providence, R.I., American Mathematical Society, 1990.

• Bruno Cordani, The Kepler Problem: Group Theoretical Aspects, Regularization and Quantization, with Application to the Study of Pertubation, Birkhäuser, Boston, 2002.

But let me try to summarize a bit of this material.

One way to understand the $\mathrm{SO}(4)$ symmetry for bound states of the Kepler problem is the result of Hamilton that I explained last time: for a particle moving around an elliptical orbit in the Kepler problem, its momentum moves round and round in a circle.

I’ll call these circles Hamilton’s circles. Hamilton’s circles are not arbitrary circles in $\mathbb{R}^3$. Using the inverse of stereographic projection, we can map $\mathbb{R}^3$ to the unit 3-sphere:

$\begin{array}{rccl} f \colon &\mathbb{R}^3 &\to & S^3 \subset \mathbb{R}^4 \\ \\ & \vec p &\mapsto & \displaystyle{\left(\frac{p^2 - 1}{p^2 +1}, \frac{2 \vec p}{p^2 + 1}\right).} \end{array}$

This map sends Hamilton’s circles in $\mathbb{R}^3$ to great circles in $S^3.$ Furthermore, this construction gives all the great circles in $S^3$ except those that go through the north and south poles, $(\pm 1, 0,0,0).$ These missing great circles correspond to periodic orbits in the Kepler problem where a particle starts with momentum zero, falls straight to the origin, and bounces back the way it came. If we include these degenerate orbits, every great circle on the unit 3-sphere is the path traced out by the momentum in some solution of the Kepler problem.

Let me reemphasize: in this picture, points of $S^3$ correspond not to positions but to momenta in the Kepler problem. As time passes, these points move along great circles in $S^3...$ but not at constant speed.

How is their dynamics related to geodesic motion on the 3-sphere?
We can understand this as follows. In Part 2 we saw that

$L^2 + M^2 = - \frac{1}{2H}$

and using the fact that $\vec L \cdot \vec M = 0,$ an easy calculation gives

$H \; = \; -\frac{1}{8A^2} \; = \; -\frac{1}{8B^2}$

In the 3-sphere picture, the observables $A_j$ become functions on the cotangent bundle $T^\ast S^3$. These functions are just the components of momentum for a particle on $S^3$, defined using a standard basis of right-invariant vector fields on $S^3 \cong \mathrm{SU}(2).$ Similarly, the observables $B_j$ are the components of momentum using a standard basis of left-invariant vector fields. It follows that

$K = 8A^2 = 8B^2$

is the Hamiltonian for a nonrelativistic free particle on $S^3$ with an appropriately chosen mass. Such a particle moves around a great circle on $S^3$ at constant speed. Since the Kepler Hamiltonian $H$ is a function of $K$, particles governed by this Hamiltonian move along the same trajectories—but typically not at constant speed!

Both $K$ and the Kepler Hamiltonian $H = -1/K$ are well-defined smooth functions on the symplectic manifold that Souriau dubbed the Kepler manifold:

$T^+ S^3 = \{ (x,p) : \; x \in S^3, \, p \in T_x S^3, \, p \ne 0 \}$

This is the cotangent bundle of the 3-sphere with the zero cotangent vectors removed, so that $H = -1/K$ is well-defined.

All this is great. But even better, there’s yet another picture of what’s going on, which brings relativity into the game!

We can also think of $T^+ S^3$ as a space of null geodesics in the Einstein universe: the manifold $\mathbb{R} \times S^3$ with the Lorentzian metric

$dt^2 - ds^2$

where $dt^2$ is the usual Riemannian metric on the real line (‘time’) and $ds^2$ is the usual metric on the unit sphere (‘space’). In this picture $x \in S^3$ describes the geodesic’s position at time zero, while the null cotangent vector $p + \|p\| dt$ describes its 4-momentum at time zero. Beware: in this picture two geodesics count as distinct if we rescale $p$ by any positive factor other than 1. But this is good: physically, it reflects the fact that in relativity, massless particles can have different 4-momentum even if they trace out the same path in spacetime.

In short, the Kepler manifold $T^+ S^3$ also serves as the classical phase space for a free massless spin-0 particle in the Einstein universe!

And here’s the cool part: the Hamiltonian for such a particle is

$\sqrt{K} = \sqrt{-1/H}$

So it’s a function of both the Hamiltonians we’ve seen before. Thus, time evolution given by this Hamiltonian carries particles around great circles on the 3-sphere… at constant speed, but at a different speed than the nonrelativistic free particle described by the Hamiltonian $K.$

In future episodes, I want to quantize this whole story. We’ll get some interesting outlooks on the quantum mechanics of the hydrogen atom.

### 11 Responses to The Kepler Problem (Part 4)

1. Pedro says:

Is a quantum particle in a Kepler potential with positive energy equivalent to a free quantum particle on some sort of hyperboloid?

2. allenknutson says:

Curiously, there’s a diffeomorphism of the Kepler manifold with the space of 2×2 complex matrices of rank exactly 1 (I’m pretty sure). That has not just the SO(4) action you mentioned but also an extra circle, multiplication by scalars. I wonder if that’s going to appear in the story.

• allenknutson says:

OK yeah that circle action is exactly the one whose Hamiltonian is that square root at the end of what you wrote.

Consider the map $\det: M_2(\mathbb C) \to \mathbb C$. The general fiber looks like $SL_2(\mathbb C) \equiv T^* S^3$. It has an $SU(2)$ action from each side, which put together, are your $SO(4)$ action. The special fiber, $\det=0$ matrices, also has the scaling action. By Harada-Kaveh there is a symplectic map from the general fiber to the special, and it exactly collapses the $SU(2)$ to the zero matrix. Geometrically quantizing, we get the vector space of holomorphic functions on the variety of determinant zero matrices. How’m I doing.

• John Baez says:

Sounds cool! I hadn’t thought about $\mathrm{SL}(2,\mathbb{C}) \cong T^\ast S^3$ in this context, though it’s in my realm of knowledge.

What’s the space of determinant zero 2 × 2 matrices like? What does it mean in terms of the Kepler problem?

I’ll talk a bit about geometric quantization later—a lot of people have tried geometrically quantizing the Kepler problem, with mixed success.

• allenknutson says:

“What’s it like…” Well, it’s the toric variety associated to the square-based pyramid with no base, if you like that sort of thing. If you leave out the zero matrix, it’s the total space of a $\mathbb C^\times$-bundle over $(\mathbb P^1)^2$.

3. AK says:

The following paper is relevant too.

“Existence of the Dynamic Symmetries $O_4$ and $SU_3$ for All Classical Central Potential Problems”

Progress of Theoretical Physics, Vol. 37, No. 5, May 1967

4. MH says:

I feel a bit cheeky asking, but, do you have any plans to write a post discussing the finer details of the zero energy case?

• John Baez says:

No, I don’t plan to write anything about the zero or positive energy cases, even though they’re interesting. I plan to write a lot more about the negative energy case. In fact I already have written a lot more: I just haven’t had time to turn it into blog articles yet.

• Richard says:

Surprisingly, this mystery “click to access” link isn’t blog comment spam or phishing!

Less cryptically:

arXiv:2104.14416 “Global Symmetries of the Kepler Problem”

Joanna Gonera, Piotr Kosi´nski, Patryk Michel

The global symmetry transformations generated by Runge-Lenz vector of twodimensional Kepler problem are explicitly described. They are given in terms of SU(2) left group multiplication with group elements being suitably parametrized by phase space points. The resulting non-linear action of SU(2) on the phase space is characterized in terms of the theory of nonlinear realizations developed in Phys. Rev. 177 (1969) 2239.

• John Baez says:

Thanks! I often expand cryptic links provided by readers of this blog, but I hadn’t gotten around to this one.

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