I’ve been working on a math project involving the periodic table of elements and the Kepler problem—that is, the problem of a particle moving in an inverse square force law. I started in 2021, the year I retired from U. C. Riverside. I had driven my wife across the country to DC, and spent the fall with her at the Center for Hellenic Studies, where she was working. They own a compound of dinky little houses right across the street from Hilary Clinton’s much grander home. The brilliant fall leaves inspired thoughts of chemistry: electrons and photons in resonance.

I felt very free, with nothing particular to do, so I did something just for fun: I explored a curious relation between the Kepler problem, the periodic table, and quantum field theory.
When I drove back to Riverside at the end of 2021 I got distracted by other things. But this year I’ve been in the mood for wrapping up half-done projects. I finished this project today after a week of hard work. It should appear on the arXiv soon.
I want to explain it! But I need some things to be fresh in your mind, so I’m going to republish some blog articles I wrote in 2021 to explain this work, starting with this one. (For Part 1 go here, but you won’t need that for what’s to come.)
Let’s start with the ‘eccentricity vector’. This is a conserved quantity for the Kepler problem. It always points in the direction of the orbit’s perihelion, and its magnitude equals the eccentricity of the orbit. There’s a simple formula for it in terms of the position, momentum and angular momentum of the particle:
But it never changes.
The eccentricity vector was named the ‘Runge–Lenz vector’ after Lenz used it in 1924 to study the hydrogen atom in the framework of the ‘old quantum mechanics’ of Bohr and Sommerfeld: Lenz cited Runge’s popular German textbook on vector analysis from 1919, which explains this vector. But Runge never claimed any originality: he attributed this vector to Gibbs, who wrote about it in his book on vector analysis in 1901!
Nowadays many people call it the ‘Laplace–Runge–Lenz vector’, honoring Laplace’s discussion of it in his famous treatise on celestial mechanics in 1799. But in fact this vector goes back at least to Jakob Hermann, who wrote about it in 1710, triggering further work on this topic by Johann Bernoulli in the same year.
Nobody has seen signs of this vector in work before Hermann. So, we might call it the Hermann–Bernoulli–Laplace–Gibbs–Runge–Lenz vector, or just the Hermann vector. But I prefer to call it the eccentricity vector, because for a particle in an inverse square law its magnitude is the eccentricity of that orbit!
Let’s suppose we have a particle whose position obeys this version of the inverse square force law:
where I remove the arrow from a vector when I want to talk about its magnitude. So, I’m setting the mass of this particle equal to 1, along with the constant saying the strength of the force. That’s because I want to keep the formulas clean! With these conventions, the momentum of the particle is
For this system it’s well-known that the following energy is conserved:
as well as the angular momentum vector:
But the interesting thing for me today is the eccentricity vector:
Let’s check that it’s conserved! Taking its time derivative,
But angular momentum is conserved so the second term vanishes, and
so we get
But the inverse square force law says
so
How can we see that this vanishes? Mind you, there are various geometrical ways to think about this, but today I’m in the mood for checking that my skills in vector algebra are sufficient for a brute-force proof—and I want to record this proof so I can see it later!
To get anywhere we need to deal with the cross product in the above formula:
There’s a nice identity for the vector triple product:
I could have fun talking about why this is true, but I won’t now! I’ll just use it:
and plug this into our formula
getting
But look—everything cancels! So
and the eccentricity vector is conserved!
So, it seems that the inverse square force law has 7 conserved quantities: the energy the 3 components of the angular momentum
and the 3 components of the eccentricity vector
. But they can’t all be independent, since the particle only has 6 degrees of freedom: 3 for position and 3 for momentum. There can be at most 5 independent conserved quantities, since something has to change. So there have to be at least two relations betwen the conserved quantities we’ve found.
The first of these relations is pretty obvious: and
are at right angles, so
But wait, why are they at right angles? Because
The first term is orthogonal to because it’s a cross product of
and
the second is orthogonal to
because
is a cross product of
and
.
The second relation is a lot less obvious, but also more interesting. Let’s take the dot product of with itself:
or in other words,
But remember this nice cross product identity:
Since and
are at right angles this gives
so
Then we can use the cyclic identity for the scalar triple product:
to rewrite this as
or simply
or even better,
But this means that
which is our second relation between conserved quantities for the Kepler problem!
This relation makes a lot of sense if you already know some facts about the eccentricity vector: it points in the direction of the orbit’s perihelion, and its magnitude equals the eccentricity of the orbit. From the relation above you can then see:
• if then
and the orbit is a hyperbola.
• if then
and the orbit is a parabola.
• if then
and the orbit is an ellipse (or circle).
But why is the eccentricity? And why does the particle move in a hyperbola, parabola or ellipse in the first place? We can show both of these things by taking the dot product of
and
Using the cyclic property of the scalar triple product we can rewrite this as
Now, we know that moves in the plane orthogonal to
. In this plane, which contains the vector
the equation
defines a conic of eccentricity
. I won’t show this from scratch, but it may seem more familiar if we rotate the whole situation so this plane is the
plane and
points in the
direction. Then in polar coordinates this equation says
or
This is well-known, at least among students of physics who have solved the Kepler problem, to be the equation of a conic of eccentricity .
Another thing that’s good to do is define a rescaled eccentricity vector. In the case of elliptical orbits, where we define this by
Then we can take our relation
and rewrite it as
and then divide by getting
or
This suggests an interesting similarity between and
which turns out to be very important in a deeper understanding of the Kepler problem. And with more work, you can use this idea to show that
is the Hamiltonian for a free particle on the 3-sphere. But more about that some other time, I hope!
For now, you might try this:
• Wikipedia, Laplace–Runge–Lenz vector.
and of course this: