I threw up a quick sketch of the tidal forces the other day to show how the oceans really follow the quadrupole field. The sketch looked like this:
But actually, the tides are more interesting than that. They don't track the moon directly, as seen in this picture...they are actually displaced. I'd like to say they lag behind the moon, trying to keep up, but that may or may not be right. It's also quite possible that they run ahead of the moon. It's not a question you can settle without some careful thought.
There are three frequencies you have to account for in this calculation. There is of course the 24-hour rotation of the earth, and the 29-day period of revolution of the moon. But there is also the very important natural frequency of the oceans. If you look at the bulging oceans in the sketch, you can imagine that if the moon suddenly disappeared, the oceans wouldn't just instantly return to their spherical equilibrium. They would slosh back and forth for awhile, and they would do so at a certain characteristic frequency. Just what is this frequency?
That promises to be not so easy to calculate. But anybody who does any work with waves knows that a standing wave can be analyzed as the superposition of two travelling waves. The present example is a case in point. We can analyze the sloshing of the oceans as the simultaneous rushing, east-to-west and west-to-east, of two massive travelling waves. We can then ask: how fast does the wave travel?
It's not that we're about to do the actual calculation, it's just that you can actually visualize a massive travelling mile-high monster wave on the oceans, and the point is it's got to be moving really fast...say 100, maybe 200 miles per hour. The significant thing to notice is that it's travelling slower than the equatorial velocity of the earths rotation, namely 1000 mph. So the sloshing of the oceans takes place on a time scale of something like five days; in other words, somewhere in between the rotational period of the earth and the orbital period of the moon. This ordering of frequencies will be critical in evaluating the true physics of the tides, including the lead/lag phase relationship with the moon, and the fascinating question of whether the effect of tidal friction might eventually slow down the moon and bring it crashing down to earth?
But I started off today's post by noting that I was getting way behind, and before we get around to doing the tides, we still have to tidy up some work on that other quadrupole field problem, the Stern Gerlach effect. Now, where were we?
As I recall, I had just come up with the solution for the problem of the polarized beam in the quadrupole field. When I say "came up with", I really mean guessed, because that's what I actually did. Or more accurately, what I did was to keep guessing and reguessing until I came up with something that made sense. This was the distribution I came up with:
Having "guessed" this solution, the question becomes: does it make sense? And the first thing to verify is that if we have a random beam, composed of equal parts spin-up and spin-down, we get a uniform donut pattern. And since the spin-down pattern has to look just the same except upside-down, the uniformity of the resulting pattern is easy to verify by summing the squares of the amplitude. (For random beams, square first, then sum!).
But the real test will be the case of the coherent beams, where you have up and down spins in coherent combinations. This time, of course, it's just the opposite: sum first, and then square! It is well known that in spinor algebra, the coherent combination gives you sideways polarization. And because of the four-fold symmetry of the quadrupole field, we know that we ought to then get a resultant distribution which is just the picture we already have, rotated by 90 degrees. (Here is the quadrupole field in case you've forgotten it:)
So the question is: what happens when we analyze a sideways polarized beam, using our up and down basis states? Will we get the same pattern we've already derived, except this time rotated by 90 degrees?
Stay tuned.
No comments:
Post a Comment