Brad DeLong is compiling a hundred simple but interesting calculations and puzzles from mathematics (and physics). Here’s my contribution:
One of the more important ideas in physics is that of curvature. The basic idea is that if you carry an object around a loop while trying to keep it unchanged, you sometimes find that when you get back to where you started the object has, despite your best efforts, changed. For example, suppose you’re standing on the equator of the Earth facing northward and holding a spear pointing directly in front of you. You then walk to the North pole, keeping the spear pointing parallel to the ground in the same direction as best you can. When you get to the pole, you start walking sideways until you reach the equator again. You then walk backwards till you arrive back where you started. You’ll then find that despite your best efforts the spear has rotated through a right angle on its journey. This shows that the Earth’s surface is curved.
The concept of carrying things without changing them (“parallel transport”) and curvature are central to our theories of gravitation and the other forces of nature. Unfortunately, the mathematics is in general quite hard, but it’s possible to get a feeling for curvature using polyhedra, and along the way to get a taste of an important result called the Gauss-Bonnet theorem that links geometry and topology!
Let’s start with something easy to imagine. Suppose you’re standing in a cubic surface in the vicinity of one of the cube’s vertices (corners). Once again, you’re holding a spear. Now, walk around a little loop in the face of the cube and the spear will come back facing just the same direction. Therefore, it looks at first glance as if the surface has no curvature - it’s “flat”. However, this turns out to be a misleading first impression. Let’s go on another journey. Start by facing one of the edges of the cube. Walk perpendicularly towards that edge, and over it onto another face. Now walk sideways towards a second edge, still holding the spear in the same direction. Pass over that one, and start walking backwards until you pass over the edge only the face where you started. Now return to your starting point by walking just in that face. You’ll find that the spear has rotated through 90 degrees again. (It’s purely a coincidence that this is the same angle as that in our trip around part of the Earth!)
Now, you should soon be able to convince yourself that it’s not the faces or the edges causing this effect. In fact, all of the curvature resides in the vertices of the cube, and the rest of the cube is flat. We’ve more or less tamed the curvature by pushing it all to a discrete set of points. Let’s take a closer look at one of the vertices. Three squares meet at the vertex, which means that the sum of the angles at the vertex is 270 degrees. The amount that this is short of 360 degrees is the “deficit angle” at the vertex - it’s a measure of how far away the vertex is from being flat. If you think a little bit, you’ll realise that the deficit angle is the same as the amount that the spear rotates on parallel-transporting it around the vertex. A bit more thought, and you’ll realise that this is true of paths around any vertex in a polyhedron. The deficit angle is a measure of the curvature at the vertex.
Now, this in itself is very interesting, but the next part is even more interesting. The cube has eight vertices, so if we add up all the deficit angles of the vertices, we get a grand total of 720 degrees. That in itself isn’t so remarkable. But there’s more! Let’s try the same calculation with a tetrahedron. Here three equilateral triangles meet at a vertex and each angle of an equilateral triangle is 60 degrees, so each vertex has a deficit angle of 180 degrees. There are four such vertices, so the total deficit is again 720 degrees. This is also true for octohedra: six vertices with a deficit of 120 degrees each makes 720 degrees. And for dodecohedra: three pentagons at each vertex for a deficit angle of 36 degrees per vertex, so all twenty vertices add up to 720 degrees. And for icosohedra: five equilateral triangles per vertex gives a deficit of 60 degrees per vertex, and there are twelve vertices so the total deficit is once more 720 degrees.
But perhaps there’s something special about regular polyhedra. Let’s try something a bit trickier: an Egyptian-style pyramid. Here there’s a square base and four equilateral triangular faces stuck on top. So at the top where the four triangles meet is a vertex with a deficit angle of 120 degrees. At the bottom two triangles and a square meet at each vertex, for a deficit of 150 degrees. There are four vertices at the base, so once more the total deficit is the same 720 degrees! And if you try other polyhedra, you’ll get just the same result!
(Actually, that’s not quite true. It’s true for every polyhedron that’s topologically equivalent to a sphere. If you make polyhedra with “holes” through them, such as polyhedral approximations to a torus or whatever, then you’ll get different totals. But almost magically the total depends just on the number of holes and not on any details of how you make the polyhedron! This is the version of the Gauss-Bonnet theorem for polyhedra. The full theorem applies to more general manifolds.)
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Another complex thing explained simply (rather like the... don't know the name for it... the thing in the top bit of my IE window... the bit that says "Sharp Blue: making the complex simple"... sort of the opposite of what I've just done with my aside... or rather tangent... well NOW it's a tangent... where was I?) well done Rich! Just wondered whether you could insert "(corner)" at the end of the second sentence in the fourth paragraph. I was struggling for a bit because I mistook vertex for (whatever the mathmatical term for edge is), my maths vocabulary has been in decline for the last 9 years. Barney |
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The thing at the top is the "title" in HTML speak, or the "tagline". I've also made the change you suggest. |
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DF said:
I'm not sure I've ever seen a proof of the polyhedral version. I've skimmed Chern's proof of the theorem for closed, orientable Riemannian surfaces, but haven't worked through it in detail. Physicists have a shocking tendency to just accept theorems! And I, at least, tend to get intimidated by mathematicians when I learn things like the fact that the Gauss-Bonnet theorem is a special case of the Atiyah-Singer index theorem!
Undoubtedly. Although Gauss tried measuring the curvature of space long, long before Einstein. It's also interesting that Einstein resisted Minkowski's attempt to geometrise special relativity, and didn't know anything about the work of Riemann et al when he formulated general relativity. Similarly, physicists were working with theories of the other forces based on gauge symmetries, gauge-covariant derivatives and potentials long before they realised that such theories could be formulated as theories about connections on fibre bundles. |
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your stuff doesn't help! |
Which is to say that it applies to (smooth) manifolds (possibly with boundary); a polyhedron is not a smooth manifold! There are combinatorial proofs of the Gauss-Bonnet and umlaufsatz for polygons/polyhedra, but the smooth proofs I know use a smooth approximation to the pw-smooth polyhedra near the singularities.
And though it's certainly useful for physics, Riemann and Levi-Civita were certainly defining and working with curvature well before Einstein and his followers!