Does Coulomb's Law, with Gauss's Law, imply the existence of only three spatial dimensions?

Question

Coulomb's Law states that the fall-off of the strength of the electrostatic force is inversely proportional to the distance squared of the charges.

Gauss's law implies that the total flux through a surface completely enclosing a charge is proportional to the total amount of charge.

If we imagine a two-dimensional world of people who knew Gauss's law, they would imagine a surface completely enclosing a charge as a flat circle around the charge. Integrating the flux, they would find that the electrostatic force should be inversely proportional to the distance of the charges, if Gauss's law were true in a two-dimensional world.

However, if they observed a $\frac{1}{r^2}$ fall-off, this implies a two-dimensional world is not all there is.

Is this argument correct? Does the $\frac{1}{r^2}$ fall-off imply that there are only three spatial dimensions we live in?

I want to make sure this is right before I tell this to my friends and they laugh at me.

Answer 1

Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law).

You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or electric field) is proportional to 1/r, not 1/r^2, so Gauss's law is still perfectly valid.

I believe the same conclusion can be made from experiments performed in graphene sheets and the like, which are even better simulations of a true 2D universe, but I don't know of a specific reference to cite for that.

Answer 2

I would say yes !

Actually some theories explaining quantum gravity use also this reasoning: gravity is a very weak interaction at a quantum level because it "leaks" into other dimensions, not observable at our scale, but that are present at this scale.

The mathematical tools are different, but if you just think about gauss's law you can imagine one explanation why additional dimensions are present in these theories.

Answer 3

It's more the other way around, I would say. Gauss's law, together with the fact that we live in a world with 3 spatial dimensions, requires that the force between charges falls off as 1/r^2. But there are perfectly consistent analogues of electrostatics in worlds with 2 or more spatial dimensions, which each have their own ``Coulomb's law" -- with a different falloff of force with distance.

More to the point, it's a lot more obvious that we live in a world with 3 spatial dimensions (look around!) than it is that the force between charges has an inverse-square law. So empirically, as well as theoretically, the number of spatial dimensions is more fundamental than the force law.

Answer 4

Loosely speaking, (super)string theory considers additional spatial dimensions that are "wrapped up" (have unusual topologies of high curvature, I believe). Now it is of course complete speculation, but if these dimensions do exist, electromagnetism would not spread out much into those dimensions, hence it would appear as if there are only three dimensions still (to a very good approximation).

Saying that, your argument is more or less sound (though far from bulletproof). It certainly suggests we don't live in a 2D world, and that any possible extra dimensions are comparatively very small!

Answer 5

Yes. In fact the flux will not equal the charge density in other dimensionalities without similar modification of Coulomb's law according to the dimensionality.

In particular, in $d$ dimensions, we have that

$${\underbrace{\int \int \cdots \int}_{d}}_{\partial R} \mathbf{E} \cdot d\mathbf{A} = K \rho$$

over the boundary of some spatial region $R$ only if the field of a point charge (which integrates to give the field of a charge distribution) has the form

$$F_E \propto \frac{q_1 q_2}{r^{d-1}}$$

where $\rho$ is measured as charge per unit $d$-volume.

So Gauss' law constrains - but technically does not determine - Coulomb's law. To determine it, you need some additional assumptions about symmetry and homogeneity of space and/or electromagnetic laws on top of Gauss's law, or some of the remaining Maxwell equations (and this generalization to higher dimensionalities is more difficult because they involve curls. In higher dimensionalities these become polyvectors, I believe, as in exterior algebra), in particular at least $\nabla \times \mathbf{E} = \mathbf{0}$ for electrostatics (in 3D).

Also, interestingly, I believe there are no stable bound atoms in dimensions $d > 3$ with the usual electromagnetic laws. In particular, the $r^{-3}$ Coulomb's law for 4D, and its higher analogues, lead to the Schrodinger equation having energy levels unbounded below, meaning even the quantum mechanics we know will not save an atom from collapse. Effectively, the walls of the potential well are too steep. A universe in 4D will have to operate in some other ways on rather different kinds of physical laws for them to admit stable matter, much less life, even if it has some forces that are similar to our electromagnetism.

Answer 6

Not necessarily, because extra-dimensional theories can be equipped with the process of dimensional reduction, which justify the disappearance of the extra dimensions at our (actual) energy scale.. "Compactification is the necessity for the supplemented dimensions to be made small and finite or `curled up'". The Coulomb's Law will be the special case of an ultimate theory at the classical level, like the Newtonian gravity for general relativity.

Answer 7

Though others have already answered it, I would like to add my few thoughts on this.

Any force (like "coulomb's electrostatic force", "newton's force of gravity") that:

1. Originates from a point.
2. Spreads out uniformly in all directions.
3. Declines in intensity over distance.

Will have a flux = 0 (surface integral of the field) at all points in space except the point of origin.

$${\displaystyle \oint _{S}\mathbf {E} \cdot d\mathbf {A} = 0}$$

This implies that divergence must be 0 at all points in space.

$$\nabla \cdot \vec F(\vec r)=\begin{cases}0&,r\ne 0\\\\\text{non-zero}&,r=0\end{cases}$$

Applying the divergence formula on unit vectors we get the following:

1. For 3D Space

$ \nabla \frac{1}{r} \hat{r} = \frac{1}{r} $

$ \nabla \frac{1}{r^2} \hat{r} = 0 $

$ \nabla \frac{1}{r^3} \hat{r} = -\frac{1}{r^3} $

$ \nabla \frac{1}{r^4} \hat{r} = -\frac{2}{r^5} $

In 3-D space, the divergence is 0 only when the denominator is square of r. Hence forces are proportional to inverse square of distance from source.

2. For 2D Space

$ \nabla \frac{1}{r} \hat{r} = 0 $

$ \nabla \frac{1}{r^2} \hat{r} = -\frac{1}{r^3} $

In 2-D space, the divergence is 0 only when the denominator is just r. Hence forces are proportional to inverse from source.

3. For 4D Space

$ \nabla \frac{1}{r^3} \hat{r} = 0 $

In a 4-D space, the divergence is 0 only when the denominator is cube of r. Hence forces will be proportional to inverse cube of distance from source.

References:

https://math.stackexchange.com/questions/2136837/divergence-of-vecf-frac1r2-hatr/2136857#2136857