Stable orbits in higher dimensions

In what way does it become manifest in the fundamental laws of physics that space has three dimensions? (Ehrenfest to Lorentz, private conversation)

Solving the Poisson equation $\nabla^2 V(r)=\rho$ , we find that the potential goes as $V(r)=-\alpha r^{2-n}$ in an arbitrary number of dimensions $n$ . This is intuitive as the force field has more dimensions to spread into in higher $n$ and as a result it dies faster with increasing distance.

As a concrete example, let's study the motion of two bodies of mass $m$ moving under the influence of gravity. The motion takes place in a plane. We introduce the polar coordinates in the plane $(r,\theta)$ . The two integrals of motion are the total energy of the system $E$ and the angular momentum $L$ :

$\frac{m}{2} (\dot{r}^2+r^2\dot{\theta}^2) + V(r) = E$

$m r^2 \dot{\theta}=L$

By elimination of $\dot{\theta}$ , we find for $\dot{r}$

$\dot{r}=\sqrt{\frac{2E}{m}-\frac{2 V}{m} - \frac{L^2}{m^2 r^2} }$

which we can rearrange as

$\dot{r}=\frac{1}{r} \sqrt{\frac{2E}{m} r^2+\frac{2 \alpha}{m} r^{4-n} - \frac{L^2}{m^2} }$

In a stable bound system the radial velocity $\dot{r}$ has to oscillate between positive and negative values, which means that it has to vanish in at least two points (it can also be identically zero for example). In fact, the quantity under the square root must always be positive between the two values of r for which it is zero. In other words the function $\frac{2E}{m} r^2+\frac{2 \alpha}{m} r^{4-n} - \frac{L^2}{m^2}$ must vanish in (at least) two points and be positive in between in order for a stable bound orbit to exist. You can easily convince yourself that this is impossible for $n>3$ .

The reason is simple. For $n=4$ the function has at most one real root in $r>0$ and therefore evidently does not satisfy the requirement. For $n>4$ and $E<0$ , the function has again only one real root in $r>0$ . For $n>4$ and $E>0$ , the function has two real roots in $r>0$ and it diverges to $+ \infty$ for both $r \rightarrow 0$ and $r \rightarrow \infty$ which means that it is negative in between. On the other hand, for the critical value of $n=3$ and for $E<0$ , the function has two real roots and it is positive in between. Therefore no stable bound states exist for $n>3$ .

We conclude that in a space with more than three dimensions, there can be no traditional atoms and perhaps no stable structures. In particular, higher dimensional universes cannot contain self-aware observers such as us. This is a hint that there is something special about 3 dimensions. For an interesting discussion along these lines, see this paper .

I don't know if its valid but i thought of the answer on a different way,

In a 4 dimensional space our infinite 3-dimensional space is bound as a singular point relatively ,

considering this bounded orbit is present inside 3 dimensions,

evolving these orbits into 4 dimensions means we have to extend orbits beyond infinity , thus breaking stability(relatively)

Since this is true for 4-dimensional space it will also be true for higher dimensions.

In other words for an observer in 4-D this orbit doesn't exist unless the system extends beyond infinity relative to 3-dimensional observer.

See also Bertrand's Theorem .