Why three dimensions?

This question is suggested by Stephen Hawking in A Brief History of Time, and it may be answered with classical mechanics plus the weak anthropic principle:

Weak Anthropic Principle (WAP): the observed values of all physical and cosmological quantities are not equally probable but they take on the values restricted by the requirement that there exist sites where carbon-based life can evolve and by the requirement that the Universe be old enough for it to have already done so. John Barrow and Frank Tipler, The Anthropic Cosmological Principle, p. 16

So, what is the connection between the WAP and the fact that we don't observe more than three spatial dimensions? We can take inspiration from general relativity and justify that it would be reasonable that, if we had a universe with \(d\) spatial dimensions, then as we move away from our sun a distance \(r\), the gravitational force would decay as

\[ \vec F_g = -\frac{k}{r^{d-1}}\,\hat r \]

But let's see what happens if we consider a particle of mass \(m\) (planet Earth) subject to a force proportional to the above gravitational force:

\[ \begin{aligned} m\vec a &= m\big[(\ddot r - r\dot\theta^2)\,\hat r + (2\dot r\dot\theta + r\ddot\theta)\,\hat\theta\big] \\ &= -\frac{k}{r^{d-1}}\,\hat r \end{aligned} \]

First, let's project the above dynamic equation onto \(\hat\theta\):

\[ \begin{aligned} 2\dot r\dot\theta + r\ddot\theta &= 0 \\ \Rightarrow\quad \frac{1}{r}\frac{d}{dt}\big(mr^2\dot\theta\big) &= 0 \\ \Rightarrow\quad mr^2\dot\theta &= \text{constant} \equiv L \end{aligned} \]

We have obtained our first constant of motion, the angular momentum \(L\), due to the fact that we are dealing with a central force. Now, let's project our dynamic equation onto \(\hat r\):

\[ m(\ddot r - r\dot\theta^2) = -\frac{k}{r^{d-1}} \]

Expressing \(\dot\theta\) in terms of \(L\) and \(r\), we get:

\[ \begin{aligned} m\Big(\ddot r - \frac{L^2}{m^2 r^3}\Big) + \frac{k}{r^{d-1}} &= 0 \\ \Rightarrow\quad \frac{d}{dt}\Big(\frac{m\dot r^2}{2} + \frac{L^2}{2mr^2} - \frac{k}{(d-2)\,r^{d-2}}\Big) &= 0 \\ \Rightarrow\quad \frac{m\dot r^2}{2} + \frac{L^2}{2mr^2} - \frac{k}{(d-2)\,r^{d-2}} &= E \end{aligned} \]

We have obtained the second constant of motion, the energy \(E\). The energy conservation equation written above can be seen as the first integral of motion of a 1-D problem, which we can express as

\[ E(r,\dot r) = T(\dot r) + V_{\text{eff}}(r) \qquad \text{(kinetic + effective potential energy)} \]

where

\[ V_{\text{eff}}(r) = \frac{L^2}{2mr^2} - \frac{k}{(d-2)\,r^{d-2}} \]

(For \(d = 2\) the second term becomes \(k\ln r\); nothing below changes.) If we need the existence of stable orbits, we have to be able to find stable equilibrium radii \(r_0\) for \(V_{\text{eff}}\). Let's search for equilibrium radii, which correspond to circular orbits:

\[ V'_{\text{eff}}(r) = -\frac{L^2}{mr^3} + \frac{k}{r^{d-1}} \] \[ V'_{\text{eff}}(r_0) = 0 \quad\Rightarrow\quad r_0 = \Big(\frac{mk}{L^2}\Big)^{\frac{1}{d-4}} \]

So there is one possible radius of equilibrium. Let's see if that radius corresponds to a stable equilibrium. If it does, then \(V''_{\text{eff}}(r_0)\) should be positive:

\[ V''_{\text{eff}}(r) = \frac{3L^2}{mr^4} - \frac{(d-1)\,k}{r^d} \] \[ \Rightarrow\quad V''_{\text{eff}}(r_0) = \frac{L^2}{mr_0^4}\,(4-d) > 0 \quad\Rightarrow\quad \boxed{d < 4} \]

So the only way to have planets orbiting around some star is to have a universe with at most 3 spatial dimensions! In a universe with more spatial dimensions, with a gravitational force decaying faster, any perturbation to a circular orbit would throw the planet out to the confines of the universe or into its star (we wouldn't have elliptic orbits), and we wouldn't be here asking this stupid damn question! (WA principle.)