Newton's Gravity in N Dimensions - Life and Stability of Planetary Orbits

“What if” not 3 space dimensions?

The physics describing the universe have a range of fixed numbers (“free parameters”), which just are what they are, and it is quite well known that they happen to have precise values needed for a “fun to be in” universe and “life as we know it” existence. What are we to make out of it will not be a question for us today beyond feeling very comfortable and very nice in the “life can exist” universe. What we will note however is that we happen to live also in “space” which has three directions (“dimensions”) - back and forth / left and right / up and down. It is so natural that it becomes invisible, but here is a question - if we consider that there could be any number of space dimensions, are 3 dimensions in any way special? A question of this sort was asked by P. Ehrenfest “In what way does it become manifest in the fundamental laws of physics that space has three dimensions?” (1917), and we will be looking into how having different numbers of space directions beyond three could impact gravity and planetary motion. Why gravity and planetary motion? For life to exist planets need to be at a steady distance of a warm fireplace - the sun. If they fall in, they burn, or if they become lonely wanderers they freeze. Neither works for having life on planets. Now, P. Ehrenfrest’s observation is - that the planetary orbits around the sun are not stable in 4 and above space dimensions. Thus life would not be able to exist. Physics deals with what is, so generalizing Newton’s Laws of Gravity will be just a mathematical game of “what if”. But it is fun to ask “what if”!

In what follows, familiarity with vector calculus and Newton’s Law of Gravity is assumed.

Gravity “source”

The first question is, what would be a “natural” way to generalize Newton’s Law of Gravitation [1] to having \(N\) space dimensions as a parameter. A common form for the law is \(F\sim mM/r^2\). There is a rather suspicious 2 in exponent for \(r\), which is the only variable involving “space”, but it is not quite clear how this would generalize. For that, we will consider Gauss’s Law formulation of gravity.

Let us consider fluids first. If there is a fluid “source” that fills up the volume with the fluid, then the amount of new fluid from the source is equal to fluid flowing out crossing the surface enclosing the source. This is just Gauss’s Law for incompressible fluids. Let’s consider a spherical surface enclosing the source. Now, the volume is N-dimensional, and velocity is a 1D vector then the area of the enclosing surface is N-1, as N-1 dimensional area multiplied by perpendicular 1D vector forms N-dimensional volume. As example if we are in 3D, then the area is 2D. So the velocity of water multiplied by surface area gives the volume of water crossing the surface per unit of time.

What is the area of a sphere in N dimensions? One could do some calculus, but we will make the following observation on units. The area for the sphere in SI units would be given in units of meters\(^{N-1}\). For a sphere, the only parameter involving units of meters is radius \(r\), thus the only formula for the area for the sphere in N-dimensions giving the right units is \(A \sim r^{N-1}\) which will suffice for our purposes.

Given fluid flux though surface with velocity \(v\) and fluid point source \(\rho\), then Gauss’s Law for surface at distance \(r\) is $$ vr^{N-1} \sim \rho $$ How does this apply to gravity? We in effect take the aforementioned discussion off the shelf, and state that there is a “gravity source” with strength \(M\), and “gravitational flux” \(g\) $$ gr^{N-1} \sim M $$ The force by the “flux” upon particle with mass \(m\) is to be \(F = mg\), so combining with above $$ F \sim mM/r^{N-1} $$ We can see that this reduces to the familiar “inverse square law” when N is 3.

Before moving forward, it is important to note, that we require that there exists a scalar function \(\phi\), such that \(\textbf{g} = \nabla\phi\) [2]

Stability of orbits

In the last section, we have generalized Newton’s Law of Gravity to allow for arbitrary dimensions \(N\). We will introduce proportionality constant \(G\) (different constant for different N, but not important for our discussion) $$ F = GmM/r^{N-1} $$

So, how will a planet with mass \(m\) move in this gravitational field? First, it should be noted, that the mass is only pulled radially inwards. Ie. there is no force towards “sides” to its motion of direction and radial force. This implies that the mass is moving in a fixed plane, so we will use polar coordinates for vector equation \(\textbf{F} = m\textbf{a}\) with noting that the magnitude of F only depends on distance \(r\) $$ m(\ddot{r} - r\dot{\theta}^2) = -F $$ $$ \partial_t(r^2\dot{\theta})/r = 0 $$ Let’s note, that the second equation implies \(r^2\dot{\theta} = C\), where \(C\) is some constant. This allows us to eliminate \(\dot{\theta}\) in the first equation, and then combining the above equations and with some shuffling $$ \ddot{r} = C^2/r^{3} - GM/r^{N-1} $$

There are two terms acting opposite to each other as \(r\) changes. Lets write out the right side for a few \(N\)

$$ N = 2: 1/r^3(C^2 - GMr^2) $$ $$ N = 3: 1/r^3(C^2 - GMr) $$ $$ N = 4: 1/r^3(C^2 - GM) $$ $$ N = 5: 1/r^4(C^2r - GM) $$ $$ N = 6: 1/r^5(C^2r^2 - GM) $$ $$ … $$

Let us in particular investigate how the sign of the above equations changes as \(r\) changes. The factor in front does not change the sign of the acceleration. Let’s consider case \(N=4\). Clearly, as all values are just constants, the \(C^2 - GM\) term is either negative or positive, so acceleration is constantly inwards or outwards, so the planet either falls into the sun or tumbles away - no stable orbit. Considering all other cases there is a magical distance \(r\) where the radial acceleration is zero. But what happens if the planet is disturbed a bit from said distance? For cases \(N>4\), when disturbance is towards smaller \(r\), the acceleration will become negative, so the planet will keep tumbling inwards, while if the disturbance is towards bigger \(r\) the acceleration will be outwards, so the planet will tumble away outwards. So, no stable orbits for \(N>4\) as disturbance gets reinforced. How about the \(N<4\) cases? Following the reasoning about the sign of acceleration, the situation is the opposite - disturbances get counteracted. So for 2D and 3D cases planets can have orbits with finite distance [3].

So, just on grounds of stability of orbits, we see that dimensions \(N>3\) are not viable for the existence of life [4].

[1] We will stick to Newton’s Gravity, however, one could play the “what if” game with General Relativity too

[2] In solving for force we assumed that the “gravity source” in our problem is a single point, so when finding “gravitational flux” the problem was spherically symmetric. But this is not true in general, as for one, there is the other particle with mass \(m\) which is also a source of “gravitational flux”. There needs to be another ingredient. We required that there exists a scalar function, but we could have stated a requirement which is more familiar from electrostatics - that “curl” of gravitational field is zero. If there is a scalar potential, then the “curl” of the gravitational field will be zero. But is the reverse true? The answer is that not necessarily. But it would take us further afield to deeper waters, so we go with stating that potential exists.

[3] While 2D and 3D both can have stable orbits, the physics of both situations is not the same in all ways. It is worth noting, that the potential energy of gravitational field in 2D takes form \(\log(r)\) (derivative of which is force), while for 3D it takes form \(1/r\). Ie. if we took mass in 3D and tried to carry it infinitely far, then \(1/r \rightarrow 0\), so the energy needed would be finite as the total energy needed is the difference between potential at \(r\) and infinity, while for 2D case \(\log(r) \rightarrow \infty\). Ie. in 3D it is possible to give enough kick to a particle for it to escape the mass \(M\), while in 2D it is not.

[4] We are not considering 1D as it is too simple for life, but what about 2D? On the ground of stability of orbits, it is not eliminated as a possibility. Does a 2D universe suffice for complex life? It is a quite restrictive universe. For example, 3D folded proteins form the basis for the machinery of life, but in a 2D world 3D folded protein shapes could not exist. But what 2D would look like and if it would be viable for life is a story for another time.