In three spatial dimensions, would it be possible to have a force that decreases with the inverse of the distance?

Question

I know that in our universe, electromagnetism and gravity decrease with the inverse of the distance squared. Would it be possible though for there to be a force in another universe with three spatial dimensions that decreases with the inverse of the distance instead of with the distance? If so, could such a force have an infinite range?

Answer 1

In our universe, electric and gravitational force decrease with $r^{-2}$ because of the inverse-square law.

Imagine a spherical shell around some point source. The area of a shell at some radius $R$ is $4\pi R^2$; thus, the flux through that shell is reduced by a factor of $r^{-2}$. This is true in three dimensions, and can be proved mathematically. In $N$ dimensions, a force will always decrease by $r^{N-1}$. This can be proven in a similar fashion.

So, unfortunately, your scenario is impossible in three dimensions.

If so, could such a force have an infinite range?

One of the great things about having a force that decreases by $r^{N-1}$ is that no matter how far away an object is, there will always be some non-zero force between two particles. Always. It might be very small, but it will be non-zero. The force will always have an infinite range.

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Proof of the coefficient of surface area (see these notes):

Technically, the term "sphere" refers only to the $N-1$-dimensional surface; "ball" refers to the $N$-dimensional region. I will use those terms in the following proof.

The volume of an $n$-ball is proportional to some power of its radius: $$V_N=C_Nr^N\tag{1}$$ Let us imagine an $n-1$-dimensional spherical shell. The volume-area relation is well known: $$\mathrm{d}V_n=A_n\mathrm{d}r\to A_n=\frac{\mathrm{d}V_N}{\mathrm{d}r}\tag{2}$$ We can differentiate $(1)$ by the power rule to obtain $$\frac{\mathrm{d}V_N}{\mathrm{d}r}=C_NNr^{N-1}\tag{3}$$ Take, for the sake of conjecture, the integral $$\int_{-\infty}^{\infty}e^{-r^2}\mathrm{d}V_N\tag{4}$$ We can re-write this as $$\int_{-\infty}^{\infty}\mathrm{d}x_1\cdots\int_{-\infty}^{\infty}\mathrm{d}x_Ne^{-x_1^2-x_2^2-\cdots-x_N^2}\tag{5}$$ This last step is done because $$V_N=x_1x_2\cdots x_N$$ We then have $$\left(\int_{-\infty}^{\infty}\mathrm{d}xe^{-x^2}\right)^N=\pi^{N/2}\tag{6}$$ Again, this is because of the equality of all $x_k$s. Substituting in $\mathrm{d}V_N=A_N\mathrm{d}r$, we have $$\int_0^{\infty}e^{-r^2}A_N\mathrm{d}r=C_NN\int_0^{\infty}e^{-r^2}r^{N-1}\mathrm{d}r=\frac{C_NN}{2}\int_0^{\infty}e^{-s}s^{N/2-1}\tag{7}$$ Solving this integral using the gamma function, we get $$\frac{C_NN}{2}\Gamma(N/2)=C_N(N/2)!\to C_N=\frac{\pi^{N/2}}{(N/2)!}\tag{8}$$ Finally, this gives us $$A_N=N\frac{\pi^{N/2}}{(N/2)!}r^{N-1}\to A_N\propto r^{N-1}\tag{9}$$ Why did we use $$\int_{\infty}^{\infty}e^{-r^2}\mathrm{d}V\text{ ?}$$ Well, it turns out that this becomes $\sqrt{\pi}^N$, a factor that we know will arise when finding the surface area or volume of any sphere. We're simply using the integrals to calculate that factor and the rest of the coefficient $C_N$.

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A much simpler proof (that gets to the point!)

This is actually a really simple method, but you might not think of it right away. An answer by Colin Pratt on Mathematics Stack Exchange gives a good 30-second introduction on the $N=3$ case.

First, let's calculate the volume of an $N$-sphere. We can reference what we did above and note that the volume of region in $N$-dimensional space can be found by $$\int\cdots\int_V\mathrm{d}x_1\cdots\mathrm{d}x_N$$ However, we can then translate this into hyperspherical coordinates. We find that the volume element is $$\mathrm{d}V=r^{N-1}\left(\prod_{i=2}^{N-1}\sin^{N-i}(\phi_i)\right)\mathrm{d}r\mathrm{d}\phi_1\cdots\mathrm{d}\phi_{n-1}$$ We then integrate this to find the volume of the sphere. Be careful about choosing your bounds, though.

Next, we can find surface area through volume. Interestingly enough, $$\frac{\mathrm{d}V}{\mathrm{d}r}=A$$ for all dimensions of $N$. It's not a coincidence, either. Therefore, if we know the volume, we can find the surface area.

Answer 2

If you assume that the natural state of the universe is for everything to be clumped together in one cosmic singularity then you can redefine gravity as 'the force that keeps that from happening', which increases with distance from an object.

Take the following statements:

1: The natural state of the universe if for everything to be in one place, and the universe exerts a force to cause this. This is the statement that must be assumed.

2: Not everything is in one place, and things are at rest.

3: Therefore there is a force keeping everything from being in one place that is balanced against the universal collapse. We know by experiment that we feel this force less strongly at the surface of the planet, and more strongly as we move away from it.

4: This force therefore increases as the inverse square of the distance from an object.

It's all a matter of perspective. Sadly that particular perspective is rather contrived, counterintuitive and really not that useful, as it's simpler to assume that the natural state of the universe is to be at rest and the force is pulling things in, which decreases as the square of distance (aka gravity). At this point HDE 226868's excellent answer on the nature of spheres and flux applies.

On the other hand if you are building a whole other universe the residents of that universe don't have to have the same idea of what defines simple. Maybe it's easier for them to assume that the universe is constantly trying to collapse. Who knows?