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?

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    $\begingroup$ 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. $\endgroup$ Mar 8, 2016 at 17:46

5 Answers 5


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).

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    $\begingroup$ You don't need general relativity, you just need Gauss's Law and stoke's theorem to hold to derive the d-1 rule. $\endgroup$ Jan 13, 2013 at 23:26
  • $\begingroup$ @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? $\endgroup$ Jan 14, 2013 at 0:00
  • $\begingroup$ +1. Also, so this is actually a question about $2+1$ right, not $1+1$? $\endgroup$
    – kηives
    Jan 14, 2013 at 4:55
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    $\begingroup$ @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. $\endgroup$
    – user10851
    Jan 14, 2013 at 18:34
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    $\begingroup$ This answer is mathematically correct, but I think the mathematics-to-english translation is wrong and it doesn't answer the original question. We're talking about stability, not closure. Others have said you can have stable orbits in 2-D space. I'll add that even in our 3-D space, elliptical orbits aren't closed, they precess, as per general relativity. (And when it comes down to it, they're not perfectly stable either.) $\endgroup$ Jan 14, 2013 at 21:42

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.


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    $\begingroup$ 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. $\endgroup$
    – user10851
    Jan 14, 2013 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.


Possibly also will be interesting to try Wolfram interactive plot with numerical solution for adjusted gravity law. It's possible to get most of the edge cases (~real precession, circular, instable closed, d < 3 edge case).
It's not dimension simulation, but gives stable orbits for "4D" law as well.



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.

  • $\begingroup$ 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. $\endgroup$ Mar 8, 2016 at 17:44

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