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

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.

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 $C_N$.

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.

posted almost 9 years ago

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