So we left off with the most basic mathematical description of a wave. It’s a function of the form f(x – vt), or in words a disturbance that moves from one place to another at a constant speed without changing shape.
This is a nice start, but it’s both too general and not general enough to be useful. To general because f can be any function at all, and few mathematical techniques can conveniently handle anything you throw at it. Not general enough because waves can and do change shape, move at varying velocities, and so forth. We need to do better.
“Better”, to a physicist looking at waves, means solving the wave equation in terms of orthonormal eigenfunctions. The details of that are not important at the moment. What is important is that these orthonormal eigenfunctions are just the sine and cosine functions. For the moment, we’ll just pretend we’re only using sine functions. In this representation, a basic wave looks like this:
This doesn’t look too much like f(x – vt) at first, but if you identify v = ω/k the overall form is preserved. Of course φ is just an overall phase constant that sets the position of the wave at x = t = 0. Some names for these symbols: ω is the angular frequency, which is just the number of cycles per second, times 2π. k is the wavenumber, which is just the number of waves per meter, times 2π. (The pi factors come from the periodicity of sine and cosine.) Dimensionally, this makes our identification of v a little more intuitive, since (1/time)/(1/length) is length/time, which is velocity.
Now one important property of waves is that they’re linear. If one wave crosses another, the linearly add together and you get interference. So let me plot, as a function of t with x = 0, two waves with equal amplitudes but different angular frequency and wavenumber. We’ll arbitrarily pick units such that v = 1. The first wave has ω = 2π and k = 2π The second has twice the frequency and wavenumber: ω = π and k = π.
Now plot the same graph with (say) x = 1/4. In this time-domain view, we’ve just scooted our chair over 1/4 of a unit and are watching the wave oscillate up and down in time:
It looks like the wave has traveled over a bit to the right, but since we’re looking at the wave as a function of time (ie, a cork at a fixed location bobbing up and down) it’s better to think of the wave as being delayed. In essence, it takes a given feature of the wave a little longer to get to where you are. But what if the waves travel at different speeds? Let’s say the wave with ω = π is traveling at v = 2/3. This will mean k = 3ω/2. When we look at it at x = 1/4, it looks like this:
The long-wavelength wave looks exactly the same as above, shifted over by exactly as much as before. But the short-wavelength wave was traveling at a different speed and has thus been delayed by a different amount. The shape of the overall wave has been changed as this waveform propagates that 1/4 unit distance through space. This is dispersion.
“But wait,” you might say, “this is well and good for sound and surf and stuff like that. But light waves always travel at the speed of light, according to Einstein. What relevance can these different-velocity waves have for optics?” Well, the brief handwaving answer you might see in an intro textbook might be that light travels at c ~ 300,000,000 m/s between the atoms in a material, but it takes some time to interact with each successive atom. The real answer is that the handwaving answer is pretty close to bogus and this is in fact a pretty darn nontrivial question. It wasn’t honestly answered until some work in 1914 by Arnold Sommerfeld and Léon Brillouin. This post and the last post laid some groundwork, and next time we’ll be able to talk about pulses of light which consist of a continuous band of different frequencies, and what happens when those frequencies each propagate with a different speed. And from there, we’ll get to the relationship between dispersion and absorption and look at how a realistic pulse propagates in a so-called causal medium. And from there, a discussion of optical precursors. That’ll answer the nontrivial question… I hope!
(As always, I regret the slowness of posting. As much as I like doing this, paying work’s gotta take the front seat.)