Monday, March 06, 2006

I went through the geometry of the whole Weyl Tile thingy and I've come up with some interesting results regarding the percent deviation between Euclidean (Pythagorean) right triangles (specifically, 45-45-90 triangles) and the triangles produced in the Weyl tiling (those weird stair case psuedo triangles we discussed in class.)

Suppose that we have a 45-45-90 right triangle whose legs both measure n units in length. Then, by the Pythagorean Theorem, n^2+n^2=r^2, where r=the length of the hypotenuse. So:
n^2+n^2=r^2
2n^2=r^2
Thus, r=sqrt(2n^2).
Therefore, r=n*sqrt(2).

So, for any 45-45-90 right triangle on the Euclidean plane, the Pythagorean Theorem tells us that the length of the hypotenuse is given by the length of one of the legs multiplied by the square root of 2. For the Weyl Tiling, the psuedo triangle's hypotenuse has a length of n. Therefore, the perecent deviation between the length of a psuedo triangle's hypotenuse and a Euclidean right triangle's hypotenuse is found through the following:

%Deviation = ([(n-n*sqrt(2)])/(n*sqrt(2)))*100 where we take the value given by the Pythagorean theorem as the reference value. Note that the "[" and "]" are absolute value symbols. For those unfamiliar with %dev calculations, the equation is simply given by the absolute value of the difference of the value you found to the reference value divided by the reference value and then multiplied by 100 (to turn it into a percentage.) It gives the deviation between any two values as a percentage of the reference value.

Now here comes the important part, which most people have already noticed already - the n's cancel out. So, we get:

%Deviation = ([(n-n*sqrt(2)])/(n*sqrt(2)))*100 = ([1-sqrt(2)]/sqrt(2))*100 for any 45-45-90 triangle whatsoever, no matter its size! So what is that percent deviation? Well, using my trusty TI-83, I get a value of approx 29.29%. Once again, that's true for any 45-45-90 triangle (on the Euclidean plane, of course!)

Now let's calculate how certain physical measurements would change if we lived in a universe in which space was quantized in a manner identical to the Weyl tiling. According to http://www.newscientist.com/channel/fundamentals/quantum-world/mg18424752.300, the smallest distance that any one could ever measure even in principle is about 10^-33 meters (the Planck length), other wise the measuring device would collapse into a black hole of its own making. So, let's suppose that space quanta are as small as they possibly could be, while still being measurable (if it was any smaller, it wouldn't even matter because no one could ever even hope to measure it.) Therefore, based on that assumption, space quanta must be on the scale of 10^-33 meters. Now, let's also suppose that we have a 45-45-90 triangle whose base is one meter. Then, it would measure 10^33 space quanta units on its two bases. Therefore, we have (10^33)^2+(10^33)^2=2*10^66 units^2 - giving a hypotenuse of sqrt(2)*10^33 units (or the sqrt(2) meters.) But the hypotenuse of the corresponding psuedo triangle measures 10^33 units, or 1 meter. That's a difference of approx 0.414 meters, or approx 1.36 feet. Since it's entirely reasonable to measure 1.36 feet, I conclude that if our universe does have a quantized space-time, then it is not quantized in the form of a Weyl tiling (assuming that you actually believe our experimental results from over 4000 years of measuring things - the ancient Babylonians and Chinese had figured out some of the useful properties of 3-4-5 right triangles and so forth - which are derived from the Pythagorean theorem - more than 4000 years ago.)

Next, I will show that there is a very large set of plane tilings which have this property. I will make use of the fact that for any plane tiling there is a corresponding planar network. Suppose you have a line of dots running horizontally across a page. Then, vertically down the page, you have an equal number of dots. For each dot running horizontally, there is a corresponding dot in the vertical direction. Draw a line down from each horizontal dot, and another outwards from each vertical dot. End the line where the two intersect. You should now have two lines, at 90 degrees to each other, one coming down from a horizontal dot and the other outwards from a vertical dot. Repeat this procedure for each pair of vertical/horizontal dots. The nodes formed by all those intersections now becomes a diagonal running at 45 degrees between the horizontal and the vertical. This pattern is a network which can be mapped onto any uniform tiling to form the right triangle given by that particular tiling. For instance, if we take this network and map it onto the Weyl tiling, we produce that same old stair case pattern. If we take a tiling of circles on the plane, a similiar thing happens. In fact, for any uniform periodic tiling of the Euclidean plane, this turns out to be true. Notice that each node corresponds to a tile on the tiling, and that there is a one to one correspondence between the set of nodes in one of the legs of the triangle and the number of nodes in the hypotenuse. Therefore, for any uniform periodic tiling of the Euclidean plane the number of polygons in the legs correspond to the number of polygons in the hypotenuse, meaning that the deviations from the Pythagorean theorem I derived above hold in general for the entire set.

Next, I will show that this mapping is even true for a large number of non-Euclidean geometries. For this I will use the topological property of a surface known as homeomorphism. Two geometries are said to be homeomorphic (i.e. topologically identical) if their topological properties remain invariant as one is continuously transformed into the other. In other words, two shapes are homeomorphic if a continous transformation can change one into the other without tearing or connecting the surface in a way that it wasn't connected before. In this way, a coffee cup is homeomorphic to a donut. Since the surfaces would remain internally connected in the same way as it was before under a topological transformation (that is, topological invariants would be, well, invariant) the mapping I outlined above wouldn't really change - it would just get stretched or bent or what not. Imagine that you have a picture laid down onto a piece of silly putty. When you stretch or bend the silly putty, you are performing a topological transformation. Under that transformation, the network I constructed above would also remain topologically invariant. In other words, a given network when laid down upon a surface remains topologically invariant under a given transformation of the surface so long as the resultant surface is homeomorphic to the original surface. So, the resultant network/tiling would be homeomorphic to the original network/tiling. Of course, under such transformation, the relationship between the lengths of the legs of a right triangle and its height is not invariant - that is, the Pythagorean Theorem does not apply to surfaces other than the Euclidean plane. So, if the Weyl tiling's psuedo triangle was consistant with the non-Euclidean surface’s analogue of the Pythagorean Theorem, then the whole thing would be consistent and space could be quantized (because there would be either little or no deviation between the result given by the Pythagorean theorem and the result obtained by constructing the surface’s tiling.)

So, now we must show that the Euclidean version of the Pythagorean Theorem truly does hold in our universe – in other words, that the Pythagorean Theorem isn’t simply just an approximation. We could suppose that space-time was bent in such a way that both the pseudo triangle’s hypotenuse and the Euclidean triangle had corresponding hypotenuses – as outlined above. Therefore, we could ask what is it that is known to bend space-time. From General Relativity, it is known that the only thing that affects the curvature of space-time is the mass density of whatever occupies a given spatial region. So, for instance, space-time is curved far more around the Sun than it is around the Earth. However, space-time curvature around the Earth, for instance, is small enough that we can assume it to be negligible locally – that is, at sufficiently small scales, it appears as though space-time isn’t curved at all, just as a circle’s edge appears flat upon being sufficiently magnified. Therefore, I will conclude that the Pythagorean Theorem does hold in our universe, if only on the small scale. Since space-time quanta are presumably extraordinarily small (beyond our current ability to measure) I do not believe that they are affected in a noticeable way by typical space-time curvature (except, perhaps, at singularities or discontinuities in space-time – as in a black hole.)

From all of this, I conclude that our observations of the universe in which we live is inconsistent with the idea that space-time is quantized. If our observations guarantee us a correct answer, then space-time is not quantized.

One possible response to my argument is that perhaps the quanta of space-time are not uniform or aperiodic. There are two ways in which this could occur – there are special space-time quanta that appear any time a right triangle is formed or measured. These quanta lie along the hypotenuse and trick us into measuring a different value than we would have measured otherwise. But I think the idea that space-time would “know” that we’re measuring it is ridiculous.

Another way in which space-time could be non-uniform or aperiodic is if it were tiled in an aperiodic way. For example, space-time quanta could take on the form of the shapes in the Penrose tiling, which is aperioidic, but has no special points whatsoever. This is plausible, although I see no reason for space-time to be aperiodic – why should one region of space differ from any other region of space in any appreciable way? Invoking Occam’s Razor, I deduce that this is an unreasonable assumption (i.e. space-time cannot be tiled aperiodically without a motivation for doing so. There is no motivation for doing so. Therefore, space-time is not tiled aperiodically.)

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