Sunday, February 26, 2012

symmetry is awesome

I heard this cute math puzzle recently (thanks Piotr :)

Three ants start at the vertices of a unit equilateral triangle. Ant 1 always faces ant 2, 2 always faces 3, and 3 always faces 1. They all start walking at the same time towards their target neighbor, at one unit per second. What happens?

Now I'll wait while you figure that one out.

In the meantime, here's another nice symmetry-related thing that Joanna showed me: The surface area on a sphere in between any two parallel slices through the sphere is only dependent on the vertical distance between the slices. So if you slice off the top mile of the Earth, and take out a one mile slice at the equator, those sections have the same surface area.

That's a little surprising, until you realize that since sphere are perfectly symmetric (they have constant curvature in any direction), the circumference of the slice you're taking is shrinking as you move away from the equator at exactly the same rate that it's flattening out, so to speak. That is, three inches "above" (3 inches along the axis of rotation) the equator translates to three inches on the ground, but three inches "below" the north pole translates to a looong way on the ground. And this exactly offsets the fact that the Earth is much bigger around at the equator than the pole.

...(ant spoiler alert)...

Ok back to the ants. By symmetry, they will collide at the center of the triangle. The question is, how long does it take for them to get there? They're spiraling inwards as they turn to keep facing each other, so they don't take a straight route there. But, by symmetry, at every point in time they're still in an equilateral triangle formation, essentially starting from scratch on a slightly smaller and rotated triangle! That means that if you figure out what's going on at the first instant, the same thing applies at every later instant.

From there it's easy. The inward distance from each vertex to the center of the triangle is √3/3. And the component of each ant's velocity that is pointing inward is cos(30)=√3/2. So they will collide in the center in 2/3 of a second.

Update: arithmetic corrected... again.

Thursday, February 23, 2012

Mercury and a 28-hour moon

Yesterday there was an aesthetically nice astronomical coincidence. Even better, I got to observe it from the comfort and convenience of the economics grad student lounge, which has a beautiful view towards the west from the 6th floor of a building up a hill in Berkeley. This means you can see all the way to the western horizon, so you can observe objects that fall right on the cusp of possibility, setting just after the sun itself.

Today Mercury and a 28-hour old moon coincided in this brief window. Mercury by itself is very difficult to observe: since it orbits close to the sun, you can only see it just after sunset or just before sunrise, and even then only rarely, when it's as far away from the sun as it ever appears to us (with the Earth and Mercury forming a right angle with the sun.) The last time I saw it definitively was in 2001 (although I can't say I've often tried; I'm more of a deep sky fan than a solar system buff...)

The same applies to extremely thin crescent moons. A goal of any lunar observer is to pick out the hairline crescent as soon as possible either before or after the new moon, when it passes directly between us and the sun and is momentarily invisible. 28 hours old is definitely not the youngest moon you can observe, but I think it's my personal record, and is definitely sufficiently strikingly beautiful.

So, some pictures:

Mercury and crescent moon side by side over bay, shortly after sunset (about half an hour)

moon through binocular eyepiece

Mercury and moon a little later, as it's getting dark

Jupiter (upper left), Venus (left of middle) and moon sliver (on horizon). Unfortunately Mercury didn't quite get picked up in this exposure, but you could still see it naked eye at the time.