## Posts Tagged ‘optics’

### Reflections on the Moon

September 26, 2009

Q: Why don’t elephants play tennis?
A: They prefer squash.

Two players compete in a fast-paced game at the gym. I exercise across from them, watching as they smash a blue rubber ball in turns. The game is in a small indoor court, and the ball moves very quickly. Its motion, followed from a distance, appears nearly rectilinear. All of the walls, the floor, and the ceiling are in play, and the ball’s wild meanderings, jots of wild color, mesmerize me as I stretch.

The game is squash (Wikipedia), which I had heard of but not seen played before. Watching the squash players, I was at first surprised by their ability to predict the seemingly-chaotic motion of the ball. However, a geometric property of the court aids them.

The squash ball moves very quickly (faster than a major-leaguer’s fastball), so over the short distances of the court, we can approximate its motion as roughly a straight line. The court is a rectangular prism, and this shape has the property that if a player smashes the ball at a corner, the ball will pop back out right at them, parallel to the way it came in. In two dimensions this is shown in the following diagram:

The squash ball follows the dark blue path, bouncing off the corners at points A and B. The light blue line shows the continuation of the paths, with a hypothetical meeting point D if the incoming and outgoing lines are not parallel.

The ball comes into the corner bouncing at point $A$, making an angle $\theta$ with the wall. By hypothesis, it bounces off with the same angle, then comes into the next wall with an angle $\phi$ at point $B$. It bounces off with this angle as well and returns to the court. We’re trying to show that the incoming and outgoing paths are parallel.

To do this, I’ve drawn in light blue the continuation of the incoming and outgoing paths. If they’re parallel, they never meet, and the angle drawn as $\delta$ should be zero. Notice that $\theta$ and $\phi$ are the small angles of a right triangle $ABC$, so they add to a right angle. Angle $CBD$ is opposite $\phi$, and so equal to it. That means angle $ABD$ is $2\phi$ and similarly angle $BAD$ is $2\theta$. Those two angles, along with $\delta$ form the triangle $ABD$. However, they add to a straight angle by themselves, and so we must have $\delta = 0$, showing that the ball pops out parallel to the way it came in, allowing the players to predict its motion easily.

In three dimensions, this is just the same, except that you have to work through the argument over again. A student of mine pointed out a different argument to come to the same result. Set up and $x-y-z$ coordinate system at the corner along the intersections of the planes. Then one wall works by flipping the $x$-coordinate of the incoming ball’s velocity vector, and the other two flip the ball’s $y$ and $z$-coordinates, so that after bouncing off all three, the velocity vector is reversed.

In squash, this result is far from perfect because gravity affects the ball’s motion, and its rotation, along with friction from the walls, may affect its angle of reflection. Energy is lost in each reflection as well, and the ball will slide some against the wall, so all in all it’s a rough approximation.

For light the approximation is much better as long as the wavelength of light is much smaller than the size of the mirrors. The setup with three orthogonal plane mirrors like this is a called a retroreflector because it reflects light back the way it came. If you look into one from any angle, you will see your own pupil at the corner, because at the corner the incoming and outgoing rays are not only parallel, but on top of one another, so light must start and end at the same place after reflecting there. All the light you see ends at your pupil, so that’s what you see in the corner. If you have three hand mirrors, it’s an easy experiment to try.

One interesting application of this idea is shown here:

A retroreflector on the moon's surface. The Apollo missions left this array of hundreds of reflectors intended for extremely accurate measurements of the Eart-Moon distance.

This is a retroreflector array on the moon. When an Earth-based laser sends a pulse of light at the moon, the retroreflectors send the pulse back to Earth. If you could measure the trip time very precisely, you can multiply by the speed of light to find the distance to the moon. The APOLLO project (not the lunar orbiters, but the ground-based Apache Point Observatory Lunar Laser-ranging Operation) is trying to do this to an accuracy of one millimeter.

I’ve sometimes heard silly things like “the Campbell’s soup cans thrown out by Americans in a single month could stretch to the Moon and back three times.” I say this is silly because

• Why would you want to do that?
• I totally just made the statistic up because it is meaningless and nonmemorable. Things like per-capita consumption, percentages of usable land being turned into dumps, and statistics about ecological impact actually mean something.
• No they can’t. They would fall down if they tried.

Regardless, if you know the distance to the moon with one millimeter accuracy, then your estimate of how many Campbell’s soup cans away it is is limited by how accurately you know the size of a Campbell’s soup can until you measure the can to an accuracy of single atom (and Campbell’s soup cans vary from one to another by a lot more than that, and don’t stack perfectly regularly, and shift around, and get hotter and colder, etc).

There are a few good reasons you’d want to know the position to the moon so precisely. Perhaps the most striking is as an extremely tight test of the equivalence principle of general relativity. The Earth and Moon have different densities, and so might conceivably fall towards the sun at different accelerations, even when they’re the same distance away. Modern cosmology and theoretical physics frequently explore theories of modified gravity in attempts to explain the acceleration of the universe’s expansion or create quantum theories of gravity. If the equivalence principle doesn’t hold, watching the acceleration of the moon very closely could be the first place we’d get a hint of it.

November 16, 2008

As always, check out the question.

A simple explanation for why surface brightness cannot increase is that it would violate the second law of thermodynamics. Instead of calculating, I’ll try to convince you of this with a gedankenexperiment.

Imagine a universe with one large ideal black body, one small “othello disk” (perfectly black on one side, perfectly white on the other), a parabolic reflecting mirror, and some dark matter to cause gravitational lensing. (In the picture below, the black body is orange, the mirror is gray, the disk is green, and optical paths are dotted black.)

The time scale on which the large black body comes to thermal equilibrium with its surroundings by radiation is much longer than the equivalent time scale for the disk, so throughout the course of the experiment we can consider the large black body to have a constant temperature. The disk is at the focus of the parabolic mirror. It is small enough and placed close enough to the large black body that the image of the large black body covers the surface of the disk. The mirror itself is smaller than the typical spatial scale of variations in the large black body’s radiation’s surface brightness (assuming it has some).

If there were no variations in surface brightness, the black side of the disk would be completely covered by the image of the large black body. Then the disk would come to equilibrium with the radiation coming in, and would reach the same temperature as the large black body.

Now imagine there are variations in surface brightness. Conservation of energy requires that if surface brightness decreases somewhere, it must increase somewhere else. So place the dark matter so that the mirror is at a patch of high surface brightness. Then the disk is still covered by the image of the black body, but that image is now brighter. When the disk comes to equilibrium, it must be hotter than the large black body, which was itself the source of the heat. So we have heat flowing from a cold body to a hot one, in violation of the second law. By R.A.A, the variations in surface brightness must not exist.

### New Problem: Surface Brightness

November 6, 2008

When I began studying gravitational lensing, I was told that a gravitational lens preserves a quantity called “surface brightness”. This is defined as the flux per unit area. In other words, if you look at the sun from out at Pluto, it’s very dim. But if you receive $\frac{1}{1000}$ as much light at Pluto, it’s because the size of the sun in the sky is $\frac{1}{1000}$ what it is here, not because it’s gotten intrinsically dimmer. The stars are just as bright here as they are right up close. They look really small, though (so small even a big space telescope like Hubble can’t see any details on a star (except the Sun, of course)).

Even though gravitational lensing can bend light and thereby make a star seem bigger in the sky, it cannot make it seem intrinsically brighter. The same is true for normal optics. A magnifying glass can make the words on a page larger, but if the lights in the room are dim the magnifying glass cannot make anything brighter.

Question: why not?

Hint: you don’t need to know anything about optics or gravity to answer this question, except that lenses and gravitational potential are completely passive. That is, they only bend light, not create it or change it.