Before we get to the heart of yesterday’s quiz, let me briefly define solid angle for those who may not be familiar with it. Regular angles are measured in degrees or radians, and solid angles are to angles as square meters are to meters. Solid angle is more or less angular area, and is measured in steradians (or sometimes square degrees). Here’s one steradian, from Wikipedia. There are 4π steradians in a sphere:
Now let’s take a look at the sun. Make yourself a square frame made with 1-meter sides. Fly down and stand on the surface of the sun (this is a thought experiment, after all), and hold the frame so that the sunlight passes through the frame. The total power passing through the frame is just the power output of 1 square meter of the sun’s surface. The solid angle covered by the sun from the perspective of the frame is more or less that entire half of of the sky, so if we define a quantity called radiance which is watts per steradian per square meter then we have a particular value for the radiance at the frame. Now bring the frame back to earth and point it at the sun. The total power passing through the frame is now lower since it’s spread out by the usual 1/r^2 factor. But the angular diameter of the sun is also reduced by a factor of r, so the solid angle covered by the sun is also reduced by that 1/r^2 factor. If the power and the angular size have decreased by the same factor, then the total radiance is the same.
This glosses over some important issues involving the angle at which the rays pass through the frame, but with the understanding that those issues won’t end up affecting the result, it turns out to be a general principle of optics that the radiance is conserved. You can’t increase it with a passive optical system. When you burn paper with sunlight and a magnifying glass, you’ve increased the solid angle worth of sun seen by the paper. This means more power per area and thus more temperature, but the best you can possibly do is to increase the apparent size of the sun to just what is is when you were standing on it in the thought experiment. This will make the spot on the paper plenty hot, but no hotter than the surface of the sun.
But that geometric argument which I haven’t proved (The proof is not difficult, but it’s more trouble than it’s worth here. It’s in any good optics text.) might not be entirely convincing. What about, for instance, the case where we put the entire solar system in a huge elliptical reflector with the sun at one focus? What if we put a basketball-sized blackbody at the other focus? Won’t it be blasted away instantly with the entire power output of the sun?
No, it turns out. The reason is that the sun is not at the other focus. The sun is an extended body with only its center at the focus. The light emitted at the solar north pole is nearly half a million miles away from the focus, and except for the microscopic fraction of light leaving perfectly along the outward normal, the light will correspondingly miss the basketball by a comparable amount. Only a tiny, microscopic fraction of the sun’s output will actually be aligned precisely enough to hit the basketball. It will end up being heated to the temperature of the sun.
I’ll grant that it’s not obvious. It’s not terribly common to invoke the rules of geometric optics to look at thermodynamics. But nature doesn’t really worry about the boundaries between our textbook names.