# Gravity in other than 3 spatial dimensions and stable orbits

I have heard from here that stable orbits (ones that require a large amount of force to push it significantly out of it's elliptical path) can only exist in a three spatial dimensions because gravity would operate differently in a two or four dimensional space. Why is this?

• As an aside: if you're talking about generalizing General Relativity (instead of Newtonian gravity) to different dimensions, then 2-D is very different indeed from 3-D. In particular, the spacetime curvature is exactly zero in vacuum. – Michael Seifert Mar 8 '16 at 17:46

Specifically what that is referring to is the 'inverse-square law', nature of the gravitational force, i.e. the force of gravity is inversely proportional to the square of the distance:

$F_g \propto \frac{1}{d^2}$.

If you expand this concept to that of general power-law forces (e.g. when you're thinking about the virial theorem), you can write:

$F \propto d^a$,

Stable orbits are only possible for a few, special values of the exponent '$a$'---in particular, and more specifically 'closed1', stable orbits only occur for $a = -2$ (the inverse-square law) and $a = 1$ (Hooke's law). This is called 'Bertrand's Theorem'.

Now, what does that have to do with spatial dimensions? Well, it turns out that in a more accurate description of gravity (in particular, general relativity) the exponent of the power-law ends up being one-less than the dimension of the space. For example, if space were 2-dimensional, then the force would look like $F \propto \frac{1}{d}$, and there would be no closed orbits.

Note also that $a<-3$ (and thus 4 or more spatial dimensions) is unconditionally unstable, as per @nervxxx's answer below.

1: A 'closed' orbit is one in which the particle returns to its previous position in phase space (i.e. its orbit repeats itself).

• You don't need general relativity, you just need Gauss's Law and stoke's theorem to hold to derive the d-1 rule. – Jerry Schirmer Jan 13 '13 at 23:26
• @JerrySchirmer thanks, good point --- but isn't Stoke's theorem requisite on the force being expressed as the divergence of a field---which itself is unique to a = -2, and 1? – DilithiumMatrix Jan 14 '13 at 0:00
• +1. Also, so this is actually a question about $2+1$ right, not $1+1$? – kηives Jan 14 '13 at 4:55
• The problem is, there could still be stable orbits with this. The initial velocity would just have to me equal to rad(Gm) instead of rad(Gm/r). The required orbital velocity would just be independent of distance. I suppose the only thing is that elliptical orbits would be impossible, which is probably what the video I linked was talking about. – PixelArtDragon Jan 14 '13 at 5:15
• @zhermes While everything you say about closed orbits is correct, there is indeed a true sense of instability that sets in in 4+ dimensions: any perturbation to any orbit will send the separation either to infinity or to 0. See nervxxx's answer, which basically follows the method in, e.g., Goldstein. – user10851 Jan 14 '13 at 18:34

I'll try to answer it by considering radial deviations from a circular orbit. First we have to assume two things about our n-dimensional universe: Newton's second law still holds, that is,

for a particle's position vector in n-dimensions $\vec{x} = (x_1, x_2, \cdots x_n)$, \begin{align} m \ddot{\vec{x}} = \vec{F}, \end{align} where $\vec{F}$ is some n-dimensional force,

and also that the law of gravity is given by Gauss' law: \begin{align} \nabla \cdot \vec{g} = -4\pi G\rho, \end{align} where $\vec{g}$ is the gravitational force field. (See wikipedia for more information).

The solution to that pde is \begin{align} \vec{g} \sim = - r^{1-n} \hat{e_r}, \end{align} for $n \geq 2$. (For $n = 1$ the motion is on a line and because it's always attractive the 'orbit' will still remain an 'orbit')

Since the motion will always be constrained to move in the 2-plane spanned by the initial radial vector $\vec{r}_0$ and the initial velocity vector $\vec{v}_0$, it is easiest to analyze the motion in cylindrical coordinates. That is, Newton's second law becomes \begin{align} m(\ddot{r} - \dot{\theta}^2r)&=F_r \\ m(r\ddot{\theta}+2\dot{r}\dot{\theta}) &= F_\theta \\ m \ddot{x_3} &= F_{x_3} \\ m \ddot{x_4} &= F_{x_4} \\ &\cdots \\ m \ddot{x_n} &= F_{x_n}, \end{align} where $x_1$ and $x_2$ are coordinates of the plane spanned by $\vec{v}_0$ and $\vec{r}_0$. Here $r$ really means $\sqrt{x_1^2 + x_2^2}$, but it turns out that because the motion is just 2-D i.e. $x_3 = x_4 = \cdots x_n = 0$, we can say $r = \sqrt{x_1^2 + \cdots + x_n^2}$.

Now we make use of the fact that gravity is always radial, so $F_\theta = 0$ and we can combine the first two equations to get \begin{align} \ddot{r} - \frac{L^2}{r^3} = F_r = f(r), \end{align} where $L$ is a constant of motion (in 3D this is the angular momentum).

For a circular orbit at $r = r_c$, $\ddot{r} = 0$, so we are left with \begin{align} -\frac{L^2}{r^3} = f(r). \end{align} Consider small deviations from $r_c$: $x = r-r_c$. Plugging this into newton's law and expanding to first order, one gets \begin{align} \ddot{x} + \left[-3f(r_c)/r_c-f'(r_c) \right]x = 0. \end{align} This is a simple harmonic equation if the stuff in the parenthesis is positive. So we obtain a stability condition \begin{align} \left[-3f(r_c)/r_c-f'(r_c) \right] > 0. \end{align}

Let's check this on a radial force $f(r) = -kr^d$. The stability condition gives \begin{align} -k r_c^d -\frac{kd}{3}r_c^d < 0, \end{align} which implies $d > -3$. So if the force law goes as $r^d$ where $d > -3$, then the orbit is not stable. One can, with a bit more work, show that $d = -3$ is also unstable.

So for dimensions $n \geq 4$, the orbit is unstable. It appears, however, that for $d = -1$ or $-2$, the orbit is stable, so this gives us the result that orbits in 3-dimensions (our world) and also that of 2-dimensions are stable, in disagreement with the video's statement. I might be wrong, though.

cheers.

• Agreed. (Except I think you meant to say $d < -3$ implies not stable.) $n \geq 4$ is unstable, while $n = 2$ doesn't have closed orbits but is stable. – user10851 Jan 14 '13 at 7:52

One brief point to add to the answers posted above, even though I can't pretend to understand all the math:

As far as I know, orbits in 2D are stable in the sense the orbiting body does not escape or collapse to the primary. See e.g.

In fact, as the 2D force decays to 1/r, its potential is logarithmic -- meaning that escape velocity is infinite. This is rather easy to show; even I can do it.

However, if I have understood correctly, it seems that the orbits are usually not closed, but shaped like flower petals. For near-circular orbits that wouldn't necessarily be too much of an issue.

I assume this is talking about Newtonian gravity (i.e., not relativity). Let's consider the effective potential:

$$V_\text{eff}(r) = \frac{L^2}{2mr^2} + V(r)$$

where $V$ is the ordinary potential energy, and $L$ is the angular momentum. First, you may ask why the effective potential has this form. Remember that for a single particle, $L = mr^2 \omega$, so this is equivalently,

$$V_\text{eff}(r) = \frac{\omega^2 r^2}{2m} + V(r)$$

This first term appears from the equations of motion for a free particle. Phrasing it in terms of angular momentum is convenient because under central forces, angular momentum is a conserved quantity.

Why do we use the effective potential? Because it helps us talk solely about the radial motions of a particle, lumping the angular motions in with the real potential. A local extremum of the effective potential tells us about an equilibrium distance.

Now, in 3d, the potential $V(r)$ for gravity is $-GMm/r$. What this means is that, as $r \to 0$, the effective potential will eventually blow up, thanks to the angular momentum part, overcoming the gravitational part and forcing the particle outward again unless it lies on a direct infall trajectory.

In 2d, the potential is different. Why is this? Newtonian gravity deals with differential equations of the form $\nabla^2 V \propto \rho$. The point-source solution to this equation (the Green's function) is proportional to $\ln r$--compare, for example, the electric potential of an infinite line charge. This is exactly the same geometry and differential equation, at least in structure.

Let's check for a second that this is the case. Let $V = C \ln r$ in 2d for some constant $C$. Then the gravitational force is

$$F = - \frac{\partial V}{\partial r} = -C/r$$

which is inward for all positive $C$. This is important. In 2d, then, our effective potential looks like,

$$V_\text{eff} = Kr^{-2} + C \ln r$$

for two constants $K, C$. The force is

$$F_{\text{eff}} = 2 K r^{-3} - C r^{-1} = -r^{-1} (-2K r^{-2} + C)$$

So $r_\text{eq} = \sqrt{2K/C}$. But is this equilibrium stable?

$$\frac{\partial F_\text{eff}}{\partial r} = -6 Kr^{-4} + C r^{-2}$$

At $r_\text{eq}$, this evaluates to $-6C^2/4K + C^2/2K = -C^2/K$.

Hm. That would suggest the equilibrium point is stable. So, perhaps someone has a reference to suggest this. I'm stuck.

• If the effective potential goes to $\infty$ as both $r \to 0$ and $r \to \infty$, then it must have at least one stable minimum, and therefore there is at least one stable circular orbit. A potential energy proportional to $\ln r$ would satisfy this criterion. – Michael Seifert Mar 8 '16 at 17:44