<![CDATA[Planets in the Fourth Dimension]]>

<![CDATA[Planets in the Fourth Dimension]]>You probably that planets go around the sun in elliptical orbits. But do you know why?

In fact, they’re moving in circles in 4 dimensions. But when these circles are projected down to 3-dimensional space, they become ellipses!

This animation by Greg Egan shows the idea:

The plane here represents 2 of the 3 space dimensions we live in. The vertical direction is the mysterious fourth dimension. The planet goes around in a circle in 4-dimensional space. But down here in 3 dimensions, its ‘shadow’ moves in an ellipse!

What’s this fourth dimension I’m talking about here? It’s a lot like time. But it’s not exactly time. It’s the difference between ordinary time and another sort of time, which flows at a rate inversely proportional to the distance between the planet and the sun.

The movie uses this other sort of time. Relative to this other time, the planet is moving at constant speed around a circle in 4 dimensions. But in ordinary time, its shadow in 3 dimensions moves faster when it’s closer to the sun.

All this sounds crazy, but it’s not some new physics theory. It’s just a different way of thinking about Newtonian physics!

Physicists have known about this viewpoint at least since 1980, thanks to a paper by the mathematical physicist Jürgen Moser. Some parts of the story are much older. A lot of papers have been written about it.

But I only realized how simple it is when I got this paper in my email, from someone I’d never heard of before:

• Jesper Göransson, Symmetries of the Kepler problem, 8 March 2015.

I get a lot of papers by crackpots in my email, but the occasional gem from someone I don’t know makes up for all those.

The best thing about Göransson’s 4-dimensional description of planetary motion is that it gives a clean explanation of an amazing fact. You can take any elliptical orbit, apply a rotation of 4-dimensional space, and get another valid orbit!

Of course we can rotate an elliptical orbit about the sun in the usual 3-dimensional way and get another elliptical orbit. The interesting part is that we can also do 4-dimensional rotations. This can make a round ellipse look skinny: when we tilt a circle into the fourth dimension, its ‘shadow’ in 3-dimensional space becomes thinner!

In fact, you can turn any elliptical orbit into any other elliptical orbit with the same energy by a 4-dimensional rotation of this sort. All elliptical orbits with the same energy are really just circular orbits on the same sphere in 4 dimensions!

Jesper Göransson explains how this works in a terse and elegant way. But I can’t resist summarizing the key results.

The Kepler problem

Suppose we have a particle moving in an inverse square force law. Its equation of motion is

$\displaystyle{ m \ddot{\mathbf{r}} = - \frac{k \mathbf{r}}{r^3} }$

where $\mathbf{r}$ is its position as a function of time, $r$ is its distance from the origin, $m$ is its mass, and $k$ says how strong the force is. From this we can derive the law of conservation of energy, which says

$\displaystyle{ \frac{m \dot{\mathbf{r}} \cdot \dot{\mathbf{r}}}{2} - \frac{k}{r} = E }$

for some constant $E$ that depends on the particle’s orbit, but doesn’t change with time.

Let’s consider an attractive force, so $k > 0,$ and elliptical orbits, so $E < 0.$ Let's call the particle a 'planet'. It's a planet moving around the sun, where we treat the sun as so heavy that it remains perfectly fixed at the origin.

I only want to study orbits of a single fixed energy $E.$ This frees us to choose units of mass, length and time in which

$m = 1, \;\; k = 1, \;\; E = -\frac{1}{2}$

This will reduce the clutter of letters and let us focus on the key ideas. If you prefer an approach that keeps in the units, see Göransson’s paper.

Now the equation of motion is

$\displaystyle{\ddot{\mathbf{r}} = - \frac{\mathbf{r}}{r^3} }$

and conservation of energy says

$\displaystyle{ \frac{\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}}{2} - \frac{1}{r} = -\frac{1}{2} }$

The big idea, apparently due to Moser, is to switch from our ordinary notion of time to a new notion of time! We’ll call this new time $s,$ and demand that

$\displaystyle{ \frac{d s}{d t} = \frac{1}{r} }$

This new kind of time ticks more slowly as you get farther from the sun. So, using this new time speeds up the planet’s motion when it’s far from the sun. If that seems backwards, just think about it. For a planet very far from the sun, one day of this new time could equal a week of ordinary time. So, measured using this new time, a planet far from the sun might travel in one day what would normally take a week.

This compensates for the planet’s ordinary tendency to move slower when it’s far from the sun. In fact, with this new kind of time, a planet moves just as fast when it’s farthest from the sun as when it’s closest.

Amazing stuff happens with this new notion of time!

To see this, first rewrite conservation of energy using this new notion of time. I’ve been using a dot for the ordinary time derivative, following Newton. Let’s use a prime for the derivative with respect to $s.$ So, for example, we have

$\displaystyle{ t' = \frac{dt}{ds} = r }$

and

$\displaystyle{ \mathbf{r}' = \frac{dr}{ds} = \frac{dt}{ds}\frac{dr}{dt} = r \dot{\mathbf{r}} }$

Using this new kind of time derivative, Göransson shows that conservation of energy can be written as

$\displaystyle{ (t' - 1)^2 + \mathbf{r}' \cdot \mathbf{r}' = 1 }$

This is the equation of a sphere in 4-dimensional space!

I’ll prove that conservation of energy can be written this way later. First let’s talk about what it means. To understand it, we should treat the ordinary time coordinate $t$ and the space coordinates $(x,y,z)$ on an equal footing. The point

$(t,x,y,z)$

moves around in 4-dimensional space as the parameter $s$ changes. What we’re seeing is that the velocity of this point, namely

$\mathbf{v} = (t',x',y',z')$

moves around on a sphere in 4-dimensional space! It’s a sphere of radius one centered at the point

$(1,0,0,0)$

With some further calculation we can show some other wonderful facts:

$\mathbf{r}''' = -\mathbf{r}'$

and

$t''' = -(t' - 1)$

These are the usual equations for a harmonic oscillator, but with an extra derivative!

I’ll prove these wonderful facts later. For now let’s just think about what they mean. We can state both of them in words as follows: the 4-dimensional velocity $\mathbf{v}$ carries out simple harmonic motion about the point $(1,0,0,0).$

That’s nice. But since $\mathbf{v}$ also stays on the unit sphere centered at this point, we can conclude something even better: $v$ must move along a great circle on this sphere, at constant speed!

This implies that the spatial components of the 4-dimensional velocity have mean $0,$ while the $t$ component has mean $1.$

The first part here makes a lot of sense: our planet doesn’t drift ever farther from the Sun, so its mean velocity must be zero. The second part is a bit subtler, but it also makes sense: the ordinary time $t$ moves forward at speed 1 on average with respect to the new time parameter $s$ , but its rate of change oscillates in a sinusoidal way.

If we integrate both sides of

$\mathbf{r}''' = -\mathbf{r}'$

we get

$\mathbf{r}'' = -\mathbf{r} + \mathbf{a}$

for some constant vector $\mathbf{a}.$ This says that the position $\mathbf{r}$ oscillates harmonically about a point $\mathbf{a}.$ Since $\mathbf{a}$ doesn’t change with time, it’s a conserved quantity: it’s called the Runge–Lenz vector.

Often people start with the inverse square force law, show that angular momentum and the Runge–Lenz vector are conserved, and use these 6 conserved quantities and Noether’s theorem to show there’s a 6-dimensional group of symmetries. For solutions with negative energy, this turns out to be the group of rotations in 4 dimensions, $mathrm{SO}(4).$ With more work, we can see how the Kepler problem is related to a harmonic oscillator in 4 dimensions. Doing this involves reparametrizing time.

I like Göransson’s approach better in many ways, because it starts by biting the bullet and reparametrizing time. This lets him rather efficiently show that the planet’s elliptical orbit is a projection to 3-dimensional space of a circular orbit in 4d space. The 4d rotational symmetry is then evident!

Göransson actually carries out his argument for an inverse square law in n-dimensional space; it’s no harder. The elliptical orbits in n dimensions are projections of circular orbits in n+1 dimensions. Angular momentum is a bivector in n dimensions; together with the Runge–Lenz vector it forms a bivector in n+1 dimensions. This is the conserved quantity associated to the (n+1) dimensional rotational symmetry of the problem.

He also carries out the analogous argument for positive-energy orbits, which are hyperbolas, and zero-energy orbits, which are parabolas. The hyperbolic case has the Lorentz group symmetry and the zero-energy case has Euclidean group symmetry! This was already known, but it’s nice to see how easily Göransson’s calculations handle all three cases.

Mathematical details

Checking all this is a straightforward exercise in vector calculus, but it takes a bit of work, so let me do some here. There will still be details left to fill in, and I urge that you give it a try, because this is the sort of thing that’s more interesting to do than to watch.

There are a lot of equations coming up, so I’ll put boxes around the important ones. The basic ones are the force law, conservation of energy, and the change of variables that gives

$\boxed{ t' = r , qquad \mathbf{r}' = r \dot{\mathbf{r}} }$

We start with conservation of energy:

$\boxed{ \displaystyle{ \frac{dot{\mathbf{r}} \cdot \dot{\mathbf{r}}}{2} - \frac{1}{r} = -\frac{1}{2} } }$

and then use

$\displaystyle{ \dot{\mathbf{r}} = \frac{d\mathbf{r}/dt}{dt/ds} = \frac{\mathbf{r}'}{t'} }$

to obtain

$\displaystyle{ \frac{\mathbf{r}' \cdot \mathbf{r}'}{2 t'^2} - \frac{1}{t'} = -\frac{1}{2} }$

With a little algebra this gives

$\boxed{ \displaystyle{ \mathbf{r}' \cdot \mathbf{r}' + (t' - 1)^2 = 1} }$

This shows that the ‘4-velocity’

$\mathbf{v} = (t',x',y',z')$

stays on the unit sphere centered at $(1,0,0,0).$

The next step is to take the equation of motion

$\boxed{ \displaystyle{\ddot{\mathbf{r}} = - \frac{\mathbf{r}}{r^3} } }$

and rewrite it using primes ( $s$ derivatives) instead of dots ( $t$ derivatives). We start with

$\displaystyle{ \dot{\mathbf{r}} = \frac{\mathbf{r}'}{r} }$

and differentiate again to get

$\ddot{\mathbf{r}} = \displaystyle{ \frac{1}{r} \left(\frac{\mathbf{r}'}{r}\right)' } = \displaystyle{ \frac{1}{r} \left( \frac{r \mathbf{r}'' - r' \mathbf{r}'}{r^2} \right) } = \displaystyle{ \frac{r \mathbf{r}'' - r' \mathbf{r}'}{r^3} }$

Now we use our other equation for $\ddot{\mathbf{r}}$ and get

$\displaystyle{ \frac{r \mathbf{r}'' - r' \mathbf{r}'}{r^3} = - \frac{\mathbf{r}}{r^3} }$

$r \mathbf{r}'' - r' \mathbf{r}' = -\mathbf{r}$

$\boxed{ \displaystyle{ \mathbf{r}'' = \frac{r' \mathbf{r}' - \mathbf{r}}{r} } }$

To go further, it’s good to get a formula for $r''$ as well. First we compute

$r' = \displaystyle{ \frac{d}{ds} (\mathbf{r} \cdot \mathbf{r})^{\frac{1}{2}} } = \displaystyle{ \frac{\mathbf{r}' \cdot \mathbf{r}}{r} }$

and then differentiating again,

$r'' = \displaystyle{\frac{d}{ds} \frac{\mathbf{r}' \cdot \mathbf{r}}{r} } = \displaystyle{ \frac{r \mathbf{r}'' \cdot \mathbf{r} + r \mathbf{r}' \cdot \mathbf{r}' - r' \mathbf{r}' \cdot \mathbf{r}}{r^2} }$

Plugging in our formula for $\mathbf{r}''$ , some wonderful cancellations occur and we get

$r'' = \displaystyle{ \frac{\mathbf{r}' \cdot \mathbf{r}'}{r} - 1 }$

But we can do better! Remember, conservation of energy says

$\displaystyle{ \mathbf{r}' \cdot \mathbf{r}' + (t' - 1)^2 = 1}$

and we know $t' = r.$ So,

$\mathbf{r}' \cdot \mathbf{r}' = 1 - (r - 1)^2 = 2r - r^2$

and

$r'' = \displaystyle{ \frac{\mathbf{r}' \cdot \mathbf{r}'}{r} - 1 } = 1 - r$

So, we see

$\boxed{ r'' = 1 - r }$

Can you get here more elegantly?

Since $t' = r$ this instantly gives

$\boxed{ t''' = 1 - t' }$

as desired.

Next let’s get a similar formula for $\mathbf{r}'''.$ We start with

$\displaystyle{ \mathbf{r}'' = \frac{r' \mathbf{r}' - \mathbf{r}}{r} }$

and differentiate both sides to get

$\displaystyle{ \mathbf{r}''' = \frac{r r'' \mathbf{r}' + r r' \mathbf{r}'' - r \mathbf{r}' - r'}{r^2} }$

Then plug in our formulas for $r''$ and $\mathbf{r}''.$ Some truly miraculous cancellations occur and we get

$\boxed{ \mathbf{r}''' = -\mathbf{r}' }$

I could show you how it works—but to really believe it you have to do it yourself. It’s just algebra. Again, I’d like a better way to see why this happens!

Integrating both sides—which is a bit weird, since we got this equation by differentiating both sides of another one—we get

$\boxed{ \mathbf{r}'' = -\mathbf{r} + \mathbf{a} }$

for some fixed vector $\mathbf{a},$ the Runge–Lenz vector. This says $\mathbf{r}$ undergoes harmonic motion about $\mathbf{a}.$ It’s quite remarkable that both $\mathbf{r}$ and its norm $r$ undergo harmonic motion! At first I thought this was impossible, but it’s just a very special circumstance.

The quantum version of a planetary orbit is a hydrogen atom. Everything we just did has a quantum version! For more on that, see

• Greg Egan, The ellipse and the atom.

For more of the history of this problem, see:

• John Baez, Mysteries of the gravitational 2-body problem.

This also treats quantum aspects, connections to supersymmetry and Jordan algebras, and more! Someday I’ll update it to include the material in this blog post.

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