Sunday, September 21, 2014

V. I. Arnold's Topological Proof

I just came across the strangest thing. This Israeli guy put up a youtube video about how you can prove the fifth-degree equation is unsolvable, based on the ideas of a Russian mathematician named V. I. Arnold. I haven't worked it all out yet, but it starts off with the strangest idea. You take an equation, and then plot out the roots in the complex plane. Then you also plot out the coefficients  in the complex plane. Why the complex plane, you ask? The cooefficients are all integers. Why not just plot them on the number line? Answer: because we're about to mess with them.

This is the funny thing. Take the simplest possible equation, like:

x^2 - 2 = 0.

The coefficients are 1 and 2, and the roots are +/- sqrt(2). Here is what we are going to do. We are going to take the "2" (the constant coefficient) from the equation and move it slowly along a circle about the origin. And we're going to observe what happens to the solutions of the equation as it changes.

Try it yourself! You'll see that when you complete the circle, so that the equation returns to its original form, that the roots of the equation also go back to their original values. But they are reversed! You have to go around the circle twice to put the roots back where they started. It's the strangest thing.

In the video, the Israeli guy claims (without really explaining it all that clearly) that in general, you can construct loops in the map of the coefficient such that by dragging the coefficients around those loops, you can arbitrarily force every possible permutation of the roots. I've shown you what happens when you drag the "2" about a loop in the complex plane - in the map of the roots, the two square roots of two switch places. Arnold's idea is that in general, you can force every permutation of the roots by dragging the coefficients around the complex plane.

I haven't yet figured out why this must be so (EDIT: Okay, I've thought about it and it's true: Boaz explains it around 4 minutes into his video), but if  since it's true then it has consequences. The idea is that if there is a solution to the fifth degree, it has to be written in terms of the coefficients arranged somehow within a complicated nested system of radicals...but for any such representation, there are restrictions as to where the roots can go when you mess with the coefficients.

How does this help us? Well, for one thing, in the example I've just shown, it proves that the solutions of x^2 - 2 cannot be rational. (EDIT: No, that's not quite right: it only shows that you need to write them with a formula that includes a square root sign). Why? Because the loop we constructed flips them around. But it's not so hard to see that if the roots are given by rational expressions, then any loop of the coefficients in the complex plane has to bring each of those rational expressions (for the roots) right back to where it started. So rational expressions don't flip around with each other, the way we did with the square roots of two.

Anyhow, that's the basis of Arnold's proof, which I'm not able to go much farther into right now. But it's something to think about.

If you're a follower of my blog, you know I've written a lot about the fifth degree equation. I think I explain it pretty well here at Why You Can't Solve The Qunitic. But I've never seen anything like Arnold's method before.,

Thursday, September 4, 2014

How Light and Sound are Different

The other day I started doing the Doppler shift as a relativity problem. It starts off looking a lot like the Doppler shift for ordinary sound in air. I had Alice sending a series of sound/light pulses to a train travelling at 4/5 the speed of sound/light; and I had Bob on the back of the train with a mirror reflecting the sound/light back to Alice. It's not too hard to calculate that if Alices pulses are 1 microsecond apart, then when they come back to Alice they are 9 microseconds apart, for a Doppler shift of 900%. You can see it easily from the picture below. The calculation is correct for light, and it's correct for sound:

What gets funny if we asked how much Doppler shift Bob sees. It's not hard to do the calucation for sound in air. We just drop vertical lines from Bob's pulse detection points down to the time axis. It's not hard to see that Bob detects the pulses 5 microseconds apart, for a Doppler shift of 500%:

Here's the thing: Bob is sending out pulses 5 microseconds apart, and Alice is measuring them 9 microseconds apart. So Alice's Doppler Shift is 180%. That's not the same as Bob's. So by analyzing who has a greater doppler shift, they can figure out that Alice is stationary and Bob must be moving.

And that's not how relativity works. In relativity we're not allowed to distinguish the stationary from the moving observer, so both Alice and Bob have to measure the same doppler shift. It's almost impossible to see how they can do that...unless we realize that time is moving slower for Bob than for Alice.

The total doppler for the reflected pulses is 900%. And the only way to make it the same for both observers is for them both to see 300%. Bob sees the pulses 3 microseconds apart, and Alice sees them 9 microseconds apart.

We did the caluclation for the special case of the train going at 4/5 the speed of light. But if we let the speed vary from 0 to 100%, and trace the contour defined by the equal-interval ticks, we will find that they are the hyperbolas defined by the equation x^2 - t^2 = constant:

Tuesday, September 2, 2014

The Doppler Shift in Relativity

It's been quite a while since I've posted any new physics, but I had a visit on the weekend from an old physics buddy from univeristy days. Richard is teaching at Waterloo these days, and relativity is his thing. I've mentioned once or twice that it's not really my territory, but last year I thought I did a pretty nice series on how to do those first-year compound velocity problems with pictures. I didn't assume any formulas except for the fact that x^2 - t^2 is invariant (the relativistic Law of Pythagoras.) You can see what I wrote here. Near the bottom of the page you can see how hyperbolas on the x-t diagram are the same "distance" from the origin. So someone moving along the orange line counts of his seconds "one, two, three..." accorcing to when he intersects those hyperbolas:

Well, on the weekend Richard and I were talking physics, and he told me about a problem he gave his students to calculate the doppler shift in relativity. He wanted them to do it without using the formula, just deriving it from basic principles. Naturally I wanted to try it for myself - and without assuming anything about x^2-t^2. I think I got it right, and as a consequence, basically derived the fact that x^2-t^2 is invariant. Here is what I did.

Now, what you normally do is have an observer on the ground (Alice) and an observer in the train (Bob). Alice stands on the tracks after the train has passed and broadcasts a radio signal to Bob, say 900 kHz. Bob measures the frequency and finds it is slower than what Alice sent out. This is normal: it's just the same way sound works. If Alice blows a whistle at 900 hz, Bob hears it at a lower pitch. If Bob blows a whistle, Alice hears the pitch change from high to low as the train passes by her. That's how sound works, and that's how light works. Either way, the picture looks like this:

The orange line is the moving train (which I'm going to take as going 4/5 the speed of sound (or light) in this picture, and the blue lines represent the "pulses" passing between the two observers. You can think of them as pulses or you can think of them as wavefronts, where I've drawn the waves in purple. I've done something a little bit odd: instead of Alice blowing one whistle and Bob blowing another one, I have Bob holding up a mirror so that Alices waves get reflected right back to her. The sound is reflected back at a lower pitch, and so is the light.

It's not hard to do the calculation for Alice. It's not so hard to see (from the basic geometry) that if she blows a whistle at 900 Hz, the echo that relfects back reaches her ear at 100 Hz. And it's the same for light waves: if she radios Bob at 900 kHz,  it comes back at 100 kHz.

(It's actually a little more natural to work with periods instead of frequency. If alice sends out pulses (of sound or light) one microsecond apart, they come back to her nine microseconds apart.)

But where it gets interesting is when we consider what Bob hears. For Alice, it's all the same if she's sending out light pulses or sound pulses. But for Bob it makes a difference, and there are big imlications that flow from that difference. That's what we'll talk about when I return.