# A Technical Post – On more than 3 dimensions

I’ve made the occasional passing comment about how fortunate we are to live in a 3-dimensional universe. In an old ““Ask a Physicist” column (in a precursor to a fairly important discussion in my new book) I dropped a little anthropic truth on the world:

You may remember, dimly, something about gravity being an inverse square law. The idea is that if you double the distance between two objects, their force of gravity drops by a factor of four. The same rule holds for electromagnetism.

The inverse square law isn’t an accident. It turns out that it’s entirely a function of the fact that we live in a three-dimensional universe. If we lived a four-dimensional one then we’d have an inverse cube law.

I then concluded with:

It turns out, though, that an inverse square law is very special. Higher dimensional universes (with their inverse cube or inverse-fourth gravity laws or whatever) don’t have any stable orbits. In other words, in a 4-dimensional universe, the earth would either spiral in toward the sun or fly away. We wouldn’t get to enjoy the five billion or so years of nearly constant sunlight that we do in our universe.

That statement of fact wasn’t enough for one of our Drexel physics grad students who asked why, exactly, $D > 3$ matters so much. To answer that, I’m going to give a fairly technical answer (though there are some nice figures to help illustrate the point). If you’re scared off by equations, and haven’t seen any intermediate-level undergraduate mechanics, this may not be the post for you.

Have the faint of heart left? Good. Let’s get started.

Consider a particle of mass $m$ in an equatorial orbit at radius, $r$, in a central potential $U(r)$. The choice of an equatorial orbit is arbitrary, since in a central potential, we can always rotate the coordinates until the particle is moving around in a fixed plane. For an attractive force, there’s always a perfectly balanced circular orbit:

(though we’ll derive the properties of that orbit in a moment).

The Lagrangian

Generalizing the orbit requires a little bit of work, and introduction of a few physical principles that you may or may not remember. Unfortunately, this isn’t the time for a full course in variational mechanics, but a good book like Marion and Thornton or even the wikipedia page on Lagrangian Mechanics is a good place to start.

But here’s the 10 second version.

In ordinary particle mechanics, the trajectory of a particle can be computed based on the kinetic and potential energies via a quantity called the Lagrangian:

$L(q_i,\dot{q}_i,t)=K-U$

That’s the difference (rather than the sum) of the kinetic and potential energy. According to Hamilton’s Principle the Lagrangian is important because particles will travel along paths that minimize a quantity known as the “Action” (which is traditionally given by an S):

$S=\int_{t_1}^{t_2} L dt$

How do we figure out what path works? We use the Euler-Lagrange Equations

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)=\frac{\partial L}{\partial q_i}$

where $q_i$ is some coordinate of the system (in our case, $r$ or $\phi$) and $\dot{q}_i$ is the time derivative of that coordinate.

For a particle in orbital motion around a central potential, the Lagrangian can be written as:

$L=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\phi}^2-U(r)$

The Euler-Lagrange equation for $\phi$ produces:

$\frac{d}{dt}\left( mr^2\dot{\phi}\right)=0$

which is just the conservation of angular momentum. Thus, we could define:

$l\equiv mr^2\dot{\phi}$

as a constant of motion.

Our second E-L equation yields:

$m\ddot{r}=mr\dot{\phi}^2-\frac{\partial U}{\partial r}$

You may recognize that last term on the right as the radial component of the gravitational force, which would (in a 3-d universe) be:

$F_r=-\frac{\partial U}{\partial r}=-\frac{GMm}{r^2}$

but in a D-dimensional universe, a D-Sphere has a surface area proportional to $r^{D-1}$ so we’ll get a result like:

$F_r=-\frac{\partial U}{\partial r}=-\frac{C}{r^{D-1}}$

where $C$ contains the equivalent of the gravitational constant and the mass of the central body.

We’ll define an “effective” radial force by combining both terms on the RHS of the numbered equation above:

$F_{eff}\equiv mr\dot{\phi}^2-\frac{\partial U}{\partial r}$

or, writing it in terms of the fixed angular momentum:

$F_{eff}\equiv \frac{l^2}{mr^3}-\frac{\partial U}{\partial r}$

The circular and nearly circular orbit

At any given radius, there is clearly a solution that yields a circular orbit, supposing that $U(r)$ is a monotonically increasing function. That is, the circular orbit solution is given by:

$l_{\circ}=\sqrt{mr^3 \frac{\partial U}{\partial r}}$

and thus the radial force is identically zero.

But is that orbit stable?

Consider what happens if the particle gets a small, positive radial kick. In a 3D universe, this is not a problem. As $r$ increases, the restoring force is negative. If $r$ decreases, the restoring force is positive.

This is a force corresponding to a Simple Harmonic Oscillator. (You would also find, if you choose to plug in the numbers that the frequency of radial oscillation is the same as the frequency of the orbit. Thus, the orbit is closed and produces an ellipse, a result first observed by Kepler, and shown mathematically by Newton.)

In general, the effective force is not quite so simple:

$F_{eff}\equiv \frac{l^2}{mr^3}-\frac{1}{C r^{D-1}}$

or, pulling out factors of $r$:

$F_{eff}=\frac{1}{r^3}\left( \frac{l^2}{m}-\frac{1}{C} r^{4-D}\right)$

It’s the last term that you should pay attention to. If $Dle 4$, then the “restoring force” gets larger for perturbations away from the circular orbit. In other words, for a 3.1 dimensional universe (which is essentially the sort of thing you get near strong gravitational fields) you might get:

High Dimensions

For $D ge 4$, the problem is even worse. By inspection, the effective restoring force gets weaker and weaker for larger $r$. As a result, an angular momentum even a small amount greater than $l_{circ}$ causes a particle to fly off to infinity:

Because it matters, here I’m using $l=1.01 l_{circ}$ in a 4-D universe. The green dot indicates the starting point of the orbit. As you can see, the planet quickly spirals outward.

For a planet orbiting ever so slightly less than the circular velocity, the fate is even worse:

Although a sun can’t even exist in a universe with no stable orbits.

If you think this is just about macroscopic scales, you’re wrong. As I noted in the original article:

Because electromagnetism also obeys an inverse square law, it turns out that atoms wouldn’t be stable. They’d all spontaneously collapse. It’s really hard to imagine complex life without atoms, and even tougher to imagine having this conversation without the existence of life.

A note to the experts. Somebody is likely to point out in the comments section that electrons don’t “orbit” atoms in the same way that planets do the sun. True enough, but if you grind through the equations in quantum mechanics and do the problem correctly, you hit the same problem. No stable atoms. Sorry.

Finally, if you’d like to play around with non inverse-square laws on your own (and you have the vpython library installed on your machine), feel free to download my source code.

Thanks for indulging a bit of mathematical excess.

-Dave

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