## Tuesday, February 7, 2012

### How Fast is a Tidal Wave?

When I worked out the effect of the tides on the earth-moon system, I said at one point that the natural frequency of the oceans was going to be a factor. Then I didn't really make much use of it in my analysis. At one point I guessed that a giant tidal wave might travel at something like 200 mph, but that was just a wild guess. Today I'm going to actually calculate it.

You ought to know that what I'm going to show you today are some of my very best tricks. I don't know how the rest of the world does this calculation, but I use tricks: good tricks that really work. One thing I really don't like is differential equations, so I've come up with ways of doing stuff with energy that lets me avoid a lot of hard math. The other thing I don't like is formulas: I like numbers. So instead of letting the height be h and the velocity v, I "special-case" the hell out of everything. Let's start by considering the case of a giant tidal wave one kilometer high.

We're going to assume the oceans are everywhere one kilometer in depth, and that there are no continents to get in the way of our nice calculations. You will recall that the moon creates a tidal bulge on both sides of the planet: so I've sketched a slice of the ocean taken half way around the equator, going from one bulge to the other. The full circumference of the earth is just twice as much as I've shown here:

Below my sketch I've drawn a graph of the potential energy of the system. You can see that I'm showing a maximum for the potential energy where the water is piled the highest, but I'm also showing the same maximum for the deepest troughs. I think that's the right way to do it. The potential energy is zero where the water is at its equilibrium level, shown by the thin red line.

We can consider this sketch as representing the traveling wave pulled along by the moon, racing around the planet; but that's actually not what I have in mind. I want to treat this as a standing wave, a stationary system sloshing up and down. So the peaks and troughs stay where they are, and the neutral points stay where they are. You ought to know that such a system is equivalent to two travel waves passing through each other in opposite directions. We'll come back to that perspective later.

As a standing wave, this system goes through a full cycle like so: first, as shown in the sketch, it is stationary at its maximum displacement. The potential energy is distributed as I've shown in my graph, and there is no kinetic energy because the water is at its moment of reversal. Next it starts to fall, and after a certain time, the bulges have reversed: what was high is now low, and what was then low is now high. Oddly enough, the graph of potential energy will look exactly the same at that moment, because I calculated it based on the displacement from equilibrium, regardless of whether it was positive or negative. But the more interesting question is what happens halfway between those extremes?

It should be clear that at some point the water is completely flat. So what happened to all the potential energy we had calculated? Does it just simply disappear? Obviously not. In fact, it is converted into kinetic energy. When the ocean is flat, the water is rushing to fill in the giant void, and at the moment of flatness the kinetic energy of rushing water is at its greatest. I've drawn a graph to show how the potential energy compares to the kinetic energy at those critical points of the cycle:

All of the energy that was potential is now converted into kinetic form. Most significantly, the concentration of energies at the respective peaks of the two graphs is obviously the same: a given quantity of rushing water has exactly the same kinetic energy as did the same quantity of elevated water, a quarter of the way around the world, in potential form.From this we can calculate how fast the water is rushing. Now, I haven't yet put any numbers on the graph yet, and the nice thing is that I don't have to. The potential energy of that mountain of water at longitude 0 is converted into kinetic energy of the same quantity of water at longitude 45, 5000 kilometers away: and we already know (or we ought to know) how to convert potential to kinetic: the speed is the same speed you get if you just let a stone fall through that distance under gravity. Since the change in elevation is one kilometer (that's an average change, based on the maximum difference of 2 kilometers from peak to trough) we just have to ask: how fast does a stone travel if you let it fall freely through a distance of one kilometer? I make it 140 meters per second, and you can check that if you like.

This is actually such a cool calculation that I don't want to spoil it for you at home. I hope you can see where I'm going with this: I need to figure out how long it will take to drain that giant bulge and fill in the void if the water is travelling with a maximum velocity of 140 meters per second. That will give me a pretty good ballpark of the natural frequency of oscillation. I have to tell you I've already looked up the "answer" on Wikipedia (go to the section on Amplitude and Cycle Time ) and it comes out pretty close if you do it right. So I'm going to leave you in suspense until tomorrow, in case you want to try the calculation for yourself.