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Question

I just became aware of Schwartz's paradox of surface area (explanation below for the unfamiliar). How does this effect mathematical modelling of real-life surfaces? For example, suppose I wanted to measure the surface area of a mountain and had the elevation data. I've found approaches that produce a polyhedral approximation (here), but how do we know this polyhedral approximation is actually approaching the surface area of the mountain? Thanks!

PS Maybe this is a better physics question? Also, Mandlebrot's first fractal paper comes to mind as a similar problem.

Schwartz's Paradox Explanation

If I understand correctly, Schwartz's Paradox shows that simply because a polyhedral approximation, $P_n$, of a curved surface $S$ approaches the curved surface as $n \to \infty$, the surface area of the polyhedral approximation, $A(P_n)$, does not approach the geometrically intuitive surface area of the surface, $A(S)$. In summary,

$$\lim_{n\to\infty} P_n = S \not\Rightarrow \lim_{n\to\infty} A(P_n) = A(S) $$

I surmised this from the following paper.

Nubok
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user126350
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  • You should just approximate the surface area directly, instead of finding an interpolating curve/surface and then doing calculus on that. Eliminate the intermediary steps. –  May 09 '14 at 02:18
  • By making that approximation to the surface area am I not implicitly doing the calculus? It seems to me the idea of such an approximation would be that it converges to the actual surface as I improve the resolution. (Consider the cylinder example in the latter paper; if I chose the wrong polyhedral approximation, I would conclude the wrong surface area for the cylinder.) – user126350 May 09 '14 at 02:22
  • @user126350 I believe he means do a surface integral to find the surface area as opposed to doing an approximation like this. It would appear in general you cannot be SURE that approximating in this fashion is correct (other methods have been introduced). As for saying to simply integrate, I'm skeptical of this method for some sets are so difficult to integrate over it would be more worth while to use a different method to find said surface area. – DanZimm May 09 '14 at 02:55
  • @user126350 yep thats exactly what I meant by other methods ;P. Also certain area-problems have had sort of dual problems elsewhere and have been solved that way. – DanZimm May 09 '14 at 03:09
  • Sorry, I accidentally posted that early and subsequently changed my response! I'm not quite sure which version you're responding to, but here is my response: This seems a bit circular, since outside of analytically solvable cases, I must then ask, what's a surface integral in this context? Schwarz's paradox basically shows that a naive extension of Riemman integration is insufficient. When you only have, say a finite set of elevation points, it basically becomes a numerical integration problem. – user126350 May 09 '14 at 03:11
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    A real mountain may have crevices and overhangs that are part of its surface but are not described by elevation data, at least not the way those elevation data are usually collected. And how do you deal with a tree growing on the mountain? All our data on mountains are in fact just numerical approximations after all. – David K May 09 '14 at 12:10
  • @DavidK: If we count a single tree, then I'm afraid that the surface area would be many orders of magnitude larger. Even the leaf is not as smooth as it looks. Same thing happens when we try to estimate the length of a shore-line... – user21820 Aug 04 '15 at 02:45
  • Schwartz (lantern) $\to$ Schwarz without "t" – Jean Marie May 21 '22 at 14:48

2 Answers2

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When approximating the cylinder, for example, such an approximation does work, so long as we also have convergence of the normals.

The approximation fails for Schwarz lanterns because the normals of the triangles in the Schwarz lanterns don't converge at all, much less to the normals of the cylinder.

See for example here of how we can use triangulations to approximate surface area: http://arxiv.org/pdf/1404.1823v1.pdf

Also look at the last paragraph of the paper linked to in the original question -- Lebesgue already solved this problem in 1902 by demanding uniform convergence instead of just pointwise convergence. Clearly Schwarz's lanterns converge pointwise but not uniformly -- the reason why the surface area blows up is because the triangular approximations can become arbitrarily jagged (which is also why the normals fail to converge). With uniform convergence, this doesn't happen.

See this Wolfram demo for how it works: http://demonstrations.wolfram.com/CylinderAreaParadox/

If it is reasonable to demand uniform convergence of functions instead of just pointwise convergence, then it should also be reasonable for surface areas.

Chill2Macht
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  • Are you using "uniform convergence" in a different sense from https://math.stackexchange.com/a/2219995/472818 ? That answer says the Schwarz lantern does converge uniformly. – mr_e_man Sep 12 '19 at 03:48
  • Could respond to mr_e_man's comment? I believe he/she is correct and Schwartz's lanterns actually do converge uniformly to the cylinder (although as you say, the normals do not converge). – tparker Apr 26 '20 at 22:14
  • I wrote this nearly four years ago, so I don't remember what I precisely meant, but the key point is whether the normals behave nicely. – Chill2Macht May 01 '20 at 14:30
  • @tparker - I think Lebesgue's solution was not about normals or the type of convergence. It was, instead of taking the limit of a single sequence of inscribed polyhedral areas, taking the infimum of all such limits. Though the Schwarz lantern's limiting area can be greater than the cylinder's area, it can never be less. – mr_e_man Dec 07 '23 at 01:51
  • In other words, we take the liminf of inscribed polyhedral areas, as the polygon diameters approach $0$. – mr_e_man Dec 07 '23 at 01:58
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In short, it doesn't.

I don't have a general proof, but let's take a look at the original paradox. For a cylinder, the area is given by:

$$A(n,m) = 2nr\sin (\pi/n)\sqrt{h^2+r^2m^2(1-\cos^2 (\pi/n))^2}$$

Now, it's true that if you let $m$ go to infinity independently of $n$, you'll end up with nonsense. But in all real life scenarios, you only have finite sampling, and therefore, the ratio $c=m/n$ is bound by some constant. You can then show that if $m=cn$, the limit is indeed $2\pi hr$ irrespective of $c$.

Nathaniel Bubis
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  • When approximating the cylinder, for example, such an approximation does work, so long as we also have convergence of the normals. The approximation fails for Schwarz lanterns because the normals of the triangles in the Schwarz lanterns don't converge at all, much less to the normals of the cylinder. – Chill2Macht Aug 01 '16 at 21:17