What's the curse of the Schwarz lantern?

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Welcome to another Mathologer video. Have you  heard of the mysterious Schwarz lanterns and   the existential crisis in maths that was triggered  by them? Well that’s Schwarz over there with one   of his lanterns. Looks like something from IKEA,  doesn’t it? There is also this Japanese coffee can   version of the lantern. And what about this?  A hedgehog version of the lantern? Hmm :) Now   before I inflict some serious existential crisis  level Schwarz lanterns on you, let’s warm up with   a superviral maths meme :) The pi = 4 paradox.  What’s that about again? Well, let’s quickly go   over the whole thing to refresh our memory.  First, we are supposed to draw a circle of   diameter 1. Not a big deal. For later, what’s the  circumference of this circle? Well, the formula   for the circumference is pi times the diameter  1 and so the circumference is pi. Next, draw a   square around the circle. It says here that the  perimeter of this square is 4. Why is that? Well,   because we need 4 of those diameters to build the  square. Now, remove the four corners like this.   It says that the perimeter of the new shape is  still 4. Really, why is that? Why is it still the   same? Well let’s have a close look. Cutting off  this corner simply amounts to taking the dotted   part and moving it here. This means that when we  cut the corners the perimeter does not change,   the perimeter is still 4. Now, in the same way,  cut off all the corners of the new shape. As   before the perimeter stays unchanged. It’s still  4. Cool :) If we keep doing this, the perimeter   will always be 4. At the same time, our zigzag  curves will turn into the circle. We conclude   that the perimeter of the circle is equal to 4. In  other words pi = 4. Very weird and wonderful. Have   you seen this one before? Versions of this little  paradox have been around for a couple hundred   years. A real classic. Now at the end of this meme  it also asks “Problem Archimedes?” Well, clearly   we have a problem here, but why Archimedes? Well,  because Archimedes did something very similar to   estimate the value of pi. Start again with the  square around the circle. Okay. But this time   we are chopping the corners off like this. So the  new shape is a regular octagon. And since we are   literally cutting corners here, it’s clear that  the perimeter of this octagon will be less than   4. In fact, it’s 3.31 and a bit. Cutting all  corners of the octagon gives a regular 16-gon   with perimeter 3.18 and a bit, etc. Keep cutting  corners and the curves we come across turn into   the circle again. At the same time the perimeters  of these chopped curves really hone in on the true   value of pi. Anyway, as we’ve seen, approximating  the circle in different ways can give different   values. How can we be sure which is the right  one? And when it comes to a curvy curve like   a circle what is its length anyway? Right, not  obvious? Of course, when you are confronted with   something straight, measuring its length is not a  problem with a straight ruler :) But what about a   nice smooth, curvy curve? How do you measure its  length? Well, we could roll the ruler along the   curve like this. Some of Andrew’s animation magic  at work here :) Remember Andrew Kepert my new best   friend from the last video? He also suggested the  topic for this video and contributed a few of the   animations that you’ll see today. Well, rolling  a ruler, easier said than done but at least in   theory that’s a perfectly fine way to define  and measure the length of a smooth curve like   this :) How about in practice? How would you go  about getting a good estimate for the length of   this curve? Well, something like what we just did  seems like a good idea: just approximate the curve   by another curve consisting of straight-line  segments and measure its total length,   segment by segment, to get an approximation of the  length of the curvy curve. And how would you go   about finding a good straight-line approximation  of the curve? Well, the first thing that comes   to mind here doesn’t involve chopping off corners,  right? Instead just pick a couple of points on the   curve and connect them by straight lines. And now  measure like this. 14 and a bit, somewhat short   of the actual length of about 15, but not bad. If  we are after a better approximation of the length,   we just go for a finer approximation by  straight-line segments. And then choosing   finer and finer straight-line approximations with  the individual segments shrinking to zero will   give the exact length of the curve. In fact, if we  are dealing with a nice smooth curvy curve, it’s   possible to prove mathematically that no matter  how you refine your straight-line approximations,   as long as the lengths of the individual segments  shrink to zero, this will always give you the   exact length of the curvy curve. No ambiguity  there as we experienced with the two different   types of chopping corners. Of course, most of you  will know that apart from chopping off corners,   Archimedes also honed in on the circle exactly  like this to approximate pi. And using his two   ways of approximating the circle from the outside  and the inside Archimedes came up with his   legendary boxing in of pi. There Archimedes famous  22/7 on the right and, on the left, 223/71 which,   when you have a closer look, is really just  a nifty refinement of 22/7. Cute :) Anyway,   ready for some Schwarz Lantern magic? Great :)  Again, for a nice smooth curvy curve the length is   just the limit of the lengths of finer and finer  straight-line approximations. Oookay, and now,   obviously, something similar should also be true  for nice smooth curvy surfaces, right? Take,   for example, a sphere, and highlight a couple of  points on the sphere. Connect those points up into   triangles. Wait can’t see anything so let’s make  the sphere transparent:) Use more points on the   sphere to get a better approximation. More points.  And more. That’s a pretty good approximation of   the sphere, right? And the easily measured  surface area of this triangle approximation   must definitely be very close to the surface area  of the sphere. Okay, keep refining such that all   those triangle edges shrink to zero aaand, in this  way 1) our triangle approximations should become   indistinguishable from the sphere and 2) the areas  of our triangle approximations should converge to   the area of the sphere. Obvious, right? :) And  the same should be the case for any reasonable   surface. Right? Well, yes and no, not at all  :( Now what’s true is that as long as the edges   shrink to 0 the triangle approximations become  indistinguishable from the surface. However,   their areas do not necessarily approach  that of the surface. Bummer :) Around 1880,   the mathematician Hermann Schwarz published his  notorious counterexample. In his counterexample,   a nice cylinder surface gets honed in on by some  of its nicest imaginable triangle approximations,   the Schwarz lanterns. Surprisingly, if you choose  these special triangle approximations just right,   as you refine them, their areas do not approach  the true surface area of the cylinder. In fact,   you can make the areas of these lanterns  approach any value greater or equal to   the cylinder’s surface area, you can even  make them approach infinity. That’s wild,   isn’t it? And of course, that’s a big problem. If  our straightforward method of calculating surface   area by refining triangle approximations keels  over even for a simple surface like a cylinder,   how can we calculate surface area reliably at  all? Sorting out this conundrum turned out to   be a big task for mathematicians. In the rest  of the video, I’ll show you how exactly the   lanterns foils our method of calculating area and  how our method can be evolved to give the correct   surface area. I’ll also show you how the pi=4  paradox is resolved. And to finish off, we’ll   also have some origami and can-crushing fun with  Schwarz lanterns. Lots to look forward to :) Okay,   so how exactly do the Schwarz lanterns mess up  our method of calculating the surface area of   a cylinder? Let me show you. In keeping with our  pi = 4 puzzle, let’s approximate a very special   cylinder, one with surface area equal to pi.  There, both the diameter of the circle and the   height of the cylinder is 1. This means that that  there is not just one but two obvious pi’s baked   into this cylinder. First the circumference of  the circle is diameter times pi, so that’s pi.   And then there is the area of the blue cylinder,  which is simply this circumference times the   height which is also pi. To construct one of  the lantern approximations of this cylinder,   split the cylinder into four equal horizontal  bands. Put seven equally spaced red points on each   of the circles like this. Put in the triangles,  et voila, there is our lantern. This particular   lantern has four horizontal bands of triangles and  has 7 equally spaced red points at every level.   Here are a couple of other lanterns with different  numbers of points and bands. Now, to get finer and   finer lantern approximations of the cylinder, we  simply crank up the numbers of bands and points.   So, let’s do some cranking. Okay, start with this  simple lantern with just one band, stick with this   one band throughout but crank up the number of  points. There. The lanterns look more and more   like a cylinder and their areas approaches pi.  So far so good :) Okay, now something different:   let’s crank up the number of bands but let’s  not meddle with the number of points. So the   number of points stays 5 throughout. The lantern  area is 2.95 to start with. That’s less than   pi but now watch that area as we crank up the  number of bands. Area is going up as expected,   but look at that, the area is now greater than pi  and keeps rising. In fact, the surface areas of   these lanterns explode to infinity. Really? Yes!  To see why this is the case without torturing any   formulas, just have a look inside the same setup  as we add bands. Go. There, more and more of these   bands. Alright. So from this viewpoint it becomes  clear that, as we crank up the number of bands,   basically the same area is added to the lantern  over and over. Right? As the bands go thinner and   thinner the triangles will go flatter and flatter  and every band essentially adds this grey area   here. And this same area added infinitely often  gives infinite area, the areas of the lanterns   explode to infinity! Easy :) Well, at least that’s  the gist of the argument. Okay, getting there.   It’s pretty easy to see that for the same reason,  fixing the number of points to be any other number   and then cranking up the number of bands will  lead to infinite area. Here is another example,   with the number of point 7 instead of 5. Again, we  are essentially adding the same stuff infinitely   often as we crank up the number of bands and  therefore the areas of the lanterns again   explode to infinity. Alright, are the lantern  refinements that we just looked at Schwarz’s   famous counterexamples? Well, we are refining  and the areas head off to infinity instead of pi.   Looking good :) However, these particular lantern  refinements are not quite what we want. Why? Well,   remember, the triangle refinements we did for the  sphere? There, all the triangle edges shrink to   zero. There. And, as the edges shrink to zero, the  triangle approximations become indistinguishable   from the sphere. On the other hand, the edges in  our lantern triangle refinements don’t shrink to   zero. At the same time, our lantern approximations  don’t turn into the cylinder. Right? From the top   the lanterns don’t look more and more like a  circle, the top view of a cylinder. From the   top our lanterns look more and more like this.  The reason for this is that in our lantern   refinements so far, we are only upping the number  bands but not the number of points. To make the   triangle edges shrink to zero, and in this way  guarantee that the lantern approximations become   indistinguishable from the cylinder, we have to  let both the number of bands and the number of   points go to infinity. At the same time, we want  to make sure that the surface areas explode to   infinity. Sounds tricky but here is an important  point to keep in mind: There are MANY different   ways to simultaneously let both numbers go to  infinity. Let’s make up one such edge shrinking   way that ticks all the boxes. This then amounts  to the counterexample we are after. Okay, we start   with the number of points being 4. For B we choose  the smallest number of bands such that the area of   the lantern is greater than 4. Why do we know that  there is such a special number of bands? Well, of   course, because if we leave the number of points  fixed at 4 and crank up the number of bands,   the area explodes to infinity and so at some point  this area has to be greater than 4. Anyway that   special b happens to be 7. So, the lantern with 7  bands and 4 points is our first lantern. The next   lantern has 5 points. What’s the b this time  around? We choose b to be the smallest number   of bands such that the lantern has area greater  than 5. This special number happens to be 15. And   so on. So, the next p is 6, we also want to have  the area greater than 6, and the special number   of bands that does this is 26. And then p=7. and  so on. Now this particular triangle refinement   ticks all our boxes. Right? Both the number of  points and bands goes to infinity. This implies   that the edges of the triangles shrink to zero.  There, shrink, shrink, shrink. And because the   edges shrink to zero, the lantern approximations  do become indistinguishable from the cylinder and   after a while you will only be able to see the  buckling under a microscope. Finally, the surface   areas explode to infinity because we’ve made sure  that the surface of every lantern is greater than   its number of points. Right the first lantern has  area greater than 4, the next greater than 5, the   next greater than 6, then 7, 8, 9 and so on.  Explodes to infinity :) Great :) Alright, So,   this particular proper triangle refinement of our  cylinder surface behaves, literally, infinitely   badly. Professor Schwarz wins. :) And he’s happy.  But there are other lantern triangle refinements   that behave nicely and give the correct area.  For example this one. There is a pattern here 4   points 4 bands, 5 points and 5 bands, 6 points and  6 bands, and so on. Now when you do the maths, it   turns out that the area of this lantern refinement  actually does converge to pi, the area of the   cylinder. The lantern on the left with 10 points  and 10 bands exemplifies the long term behaviour   of the lanterns. Unlike in our counterexample, the  buckling of the lantern surface completely dies   out meaning that basically every little triangle  fully contributes to making up the area of the   cylinder. With a more detailed analysis using the  area formula of our lanterns we can devise lantern   refinements whose areas approach any number  between pi and infinity. For example, we can make   the limit area 4. If we wanted our Schwarz lantern  meme over there to look more like the famous 2d   version we started with, we could replace the  cylinder with a sphere of area pi. Just like a   cylinder, the sphere and really any curved surface  admits one of our pathological buckling triangle   refinements whose limit areas can be anything  between the actual surface area and infinity.   Okay some of our triangle refinements work and  some don’t. Now, figuring out which do and which   don’t is not only of theoretical importance  but also becomes important when we represent   curved surfaces by triangulated surfaces using  visualisation software. If the software blindly   refines the triangulation, buckling is likely to  occur and this means that because the triangles   end up pointing all over the place, virtual light  is not reflected in the right way and leads to   weird visual artefacts. So when constructing  our refinements, apart from ensuring that   all corners of the triangles are on the target  surface, and the triangle edges shrink to zero,   we also have to ensure that the triangles hug  the target surface better and better. Just   exactly how well a triangle hugs the target  surface is usually measured by focussing on   one of its surface normals, that is, a spike  at right angles to the triangle like in this   hedgehog lantern that I showed you before. If  the spikes are pointing all over the place,   then there is a lot of buckling. What you want  to aim for when you construct your refinements,   is for the spikes in the approximations to point  as much as possible perpendicular to the target   surface, like in this example. The same sort of  extra requirement also makes sure that linear   refinements of curves do the right thing. Take,  for example, the pi = 4 paradox refinement from   the beginning. Bad because the spikes don’t end  up aligning themselves with the circle and so   the buckling of the curve leads to excess area in  the limit. On the other hand, with straight-line   approximations in which all corners are on the  nice smooth curvy curve and the edges shrink to 0,   the spikes automatically align themselves in the  right way. Great. A very nice resolution of our   problems don’t you think? Archimedes would also  have been happy with all this :) Okay, but you   may be thinking: Didn’t he say something about an  existential crisis at the beginning? In the end,   the resolution of our problem wasn’t that hard to  guess. Yes, that’s true. But remember I’ve been   stressing throughout this video that all the curvy  curves and surfaces that we are approximating are   nice and smooth. If you try to measure the  length or area of curves and surfaces that   are themselves buckling like crazy, things like  fractal curves and surfaces our solution that   works for smooth curvy curves and surfaces does  not work anymore and things get really tricky.   Mathematical missions like the quest to properly  define length and area gave rise to a whole new   branch of mathematics called measure theory.  But that’s a topic for another video. Anyway,   that was all very serious. Now to finish, let’s  have some fun and let’s have another close look   at the lanterns themselves. You probably already  noticed that all the triangles in this lantern are   clones of this isosceles triangle. Isosceles, so  the two green angles are the same and therefore   2 green plus one red is equal to 180 degrees. Now  focus on the six triangles around a corner. What   can we say about the angles around the corner?  4 green + 2 red What’s that? Well, double the   angle sum that I highlighted up there of course.  So the angles around a corner add to 360 degrees.   This means that if we cut the lantern here the  whole thing unfolds into a flat rectangle. Andrew   went wild with this idea and put together a nice  Geogebra app that animates the unfolding process.   Here is a clip of one of those seriously buckled  lanterns unfolding into a very tall rectangle,   virtually superimposed onto Andrew’s backyard.  Very impressive :) You can download Andrew’s   app via the link in the description of this  video. Does anybody know what the noise in   the background is? Andrew’s also made a video  that has more details about all this. The fact   that our lanterns fold flat also means that  every Schwarz lantern can be origamied from   a flat piece of paper. I’ll include a link to a  video of an origami expert folding one of these   lanterns in the description of this video. Very  satisfying to watch and do yourself. Anyway,   the fact that our lanterns can be folded from flat  sheets also explains to some extents why under the   right conditions lantern patterns show up in the  buckling patterns of thin real-world cylinders.   Right, a cylinder is just a flat sheet rolled  up? Again, if you are interested in more details,   follow some of the links in the description. In  turn, this buckling business suggests some lantern   fun to be had with empty soda cans. For the rest  of your life you have to fold up empty cans like   this and before you discard them :) As usual  let me know what worked and what did not work   for you in this video in the comments. If there  is anything you don’t understand please ask. And   that’s it for today. Merry Christmas. To finish  off let me show you how you can transform an empty   soda can into a nice lantern   sculpture.
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Channel: Mathologer
Views: 166,102
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Length: 25min 10sec (1510 seconds)
Published: Sat Dec 23 2023
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