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Please excuse if the question is too easy; I'm just not familiar enough with PDEs.

I'd like to understand a little bit classical implications of "adding compact dimensions" in Physics, that is, what would change if one considers classical Physics in $M\times \mathbb{R}^3$ instead of $\mathbb{R}^3$, where $M$ is compact and "small".

Here's a handwaving example of what I'm looking for. It's well known that a classical wave "detects" the parity of the dimension of $\mathbb{R}^n$: only for odd $n$ the wave originated at a point source would reach the observer once. Now, what would happen in $M\times \mathbb{R}^1$ when $M=S^1\times S^1$ of diameter $\epsilon$ when the distance between the source and the observer $l$ is much larger than $\epsilon$? One could replace a single point source in $M\times \mathbb{R}^1$ with a 2-d square grid of sources at distances $\epsilon$ from each other in $\mathbb{R}^3$, with the observer at distance $l >> \epsilon$ from the plane containing the grid. Then the waves from different sources in the grid would reach the observer at different times. This makes the point source wave in $M\times \mathbb{R}^1$ act qualitatively different from what one expect from a point source in $\mathbb{R}^1$: one would, er, "feel the disturbance in the force" long after one should, even if $M$ is "small".

My question is something like this: assuming that $\mathbb{R}^3$ is a reasonable model of a classical space, what makes it possible to consider $M\times \mathbb{R}^3$ and still arrive to reasonable conclusions?

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  • $\begingroup$ Why a 2-d square grid of sources rather than an infinite array of sources? $\endgroup$
    – André Henriques
    Commented Jan 7, 2016 at 22:49
  • $\begingroup$ @AndréHenriques: unrolling $S^1\times S^1$ with a single point source into $\mathbb{R}^2$ with a 2-d grid of sources, with adjacent ones at the distance equal to $S^1$ length $\epsilon$. $\endgroup$
    – Michael
    Commented Jan 8, 2016 at 0:02
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    $\begingroup$ The standard way to compare the $M \times \mathbb{R}^3$ case to the $\mathbb{R}^3$ one is to expand the wave field in modes of the compact factor (simple Fourier modes, in your example). Each mode coefficient will satisfy a PDE only on $\mathbb{R}^3$, possibly with extra terms that are proportional to $\epsilon$. All quantitative or qualitative differences between $\mathbb{R}^3$ and $M\times \mathbb{R}^3$ are then expected to be small of order $\epsilon$, or some higher power of $\epsilon$. $\endgroup$
    – Igor Khavkine
    Commented Jan 8, 2016 at 0:55

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The compactification introduces periodic boundary conditions, and therefore the momentum in the extra dimension can take only discrete values. A particle that propagates in the extra compactified dimension then is associated with a ladder of momenta in our three-dimensional world. In a scattering experiment, one would observe this ladder as copies of the particle with different mass (because of the energy-momentum relation), socalled Kaluza-Klein copies. See On Kaluza-Klein States from Large Extra Dimensions.

For wave propagation in extra dimensions without compactification (where still we would only be able to observe three dimensions), see Pulse propagation in a hyper-lattice.

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  • $\begingroup$ The question is about classical physics. When you refer to momentum, is that really interchangeable with wavenumber (inverse wavelength)? $\endgroup$
    – user21349
    Commented Jan 8, 2016 at 16:24
  • $\begingroup$ @BenCrowell --- in the case of a classical wave, for momentum read wavenumber (discretized at $k_n$), and for mass read cutoff frequency $\omega_n=ck_n$, so that the dispersion relation in three-dimensional space is $\omega^2=(ck)^2 +\omega_n^2$. $\endgroup$
    – Carlo Beenakker
    Commented Jan 8, 2016 at 18:18

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