# Wave and Wave Speeds (aka, the Boring Backround to Interesting Stuff)

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 LÃ©on 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.)

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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 be absorbed and re-emitted by each successive atom.

I've never encountered that putative explanation before, most likely because it is so obviously bogus. From basic spectroscopy we know that atoms and molecules only absorb and emit in certain frequency lines or bands, and at least for gases the absorption/emission probabilities do not depend on direction.

By Eric Lund (not verified) on 13 Apr 2011 #permalink

I've heard those words used as explanation before on several occasions, but now that I think about it, in print I'll usually see something like a general "interaction time" referenced rather than emission and re-absorption. It's a closer to correct classically, since it takes some time for the electron cloud of an atom to start wiggling appreciably in an applied field. Maybe I should re-word that sentence in the post for fairness.

I am not sure the hand-waving explanation is as bogus as you seem to think it is.

You are essentially saying that wave dispersion is responsible for variation of the group velocity, but it is the phase velocity that defines refractive index, not group velocity, so you still need to explain why the phase velocity depends on frequency. The answer, as you say, is due to the phase lag introduced by response time of electron charge clouds around the atoms, which effectively changes the wavelength of re-emitted wave.

Eric's objection from spectroscopy is irrelevant because it is not necessary that the electons make any discrete transitions in order to produce this phase lag.

[i](As always, I regret the slowness of posting. As much as I like doing this, paying work's gotta take the front seat.)[/i]

Well, at least your blog is still kickin' in the back seat.

"I've never encountered that putative explanation before, most likely because it is so obviously bogus."

Only if you're an idiot.

When a photon passes a charged particle the charged particle will react.

The reaction is delayed by a phenomenon called "inertia" and therefore, even though "absorbed" is not an accurate term for what has happened (what would be?), it still means that the charged particle will take up the energy and its movement in response will cause the EM field to propagate.

Reading comprehension fail there, Wow.

Matt edited the post, and the revised version is less wrong than the original (it does indeed moot the spectroscopy objection), but it is still obviously wrong to anybody who has ever worked with light filters or wave guides. The simple interaction model would predict that the response as a function of frequency would be monotonic; to explain any non-monotonic response would require handwaving on top of handwaving. I'm trying not to steal Matt's thunder here, but it turns out that there are features in the wave propagation known as "resonances" and "cutoffs". This would not be a fatal flaw if there were one of each, but I work with media which have several. And in my world things are not isotropic: it matters whether your wave is moving along or across any magnetic field which may be present.

In addition, as the Michaelson-Morley experiment demonstrate, EM fields *don't* require a medium.

By Eric Lund (not verified) on 14 Apr 2011 #permalink

No, the result of an inability to read backward in time, Eric Loon.

"but it is still obviously wrong to anybody who has ever worked with light filters or wave guides"

Doesn't mean it's wrong, it just means the model used for such people doesn't work with it. You see, people using waveguides and filters center on the time-independent solution. Not the transient.

Bugger, you missed that, didn't you?

"to explain any non-monotonic response would require handwaving on top of handwaving."

Odd that your proof of this is handwaving. Care to elucidate?

As far as I can see, there's no such problem: reaction and momentum transfer can easily cause frequency shifts, so still not seeing where your "obviously bogus" gets its obviosity. Absorption is RETENTION of the wave in the system to set a new (quasi-)static state. This has to be (near) a set quantised frequency because there is no other static state to exist in.

Therefore your insistence that quantised levels makes the description "obviously bogus" is merely a by[product of your muddled thinking.

"In addition, as the Michaelson-Morley experiment demonstrate, EM fields *don't* require a medium."

Well done for you. Never said it needed one.

However, without a medium, the light travels in a vacuum and the speed of light in vacuuo is constant. This thread is talking about light in a medium and explains why the light speed in a medium is less than that in a vacuum.

Making that little factette rather pointless except as further demonstration that your thoughts are muddled.

are we going to see some kind of new result pertaining to
this ancient topic ?

By IN Hell's Kitc… (not verified) on 14 Apr 2011 #permalink

Wow - could you make your points in a calm and objective manner without indulging in distracting personal insults?

By annonymous (not verified) on 15 Apr 2011 #permalink

I dunno. Could you?

I will sincerely try.

By annonymous (not verified) on 15 Apr 2011 #permalink

OK, me too.

#4: Zing!

#8: Eventually, but first we have to develop the conceptual tools to handle it.

@Wow: I appreciate a good argument, but at least on this site I'm holding the fort on keeping them collegial. The general rule here is debate on the merits or not at all.

(As always, I regret the slowness of posting. As much as I like doing this, paying work's gotta take the front seat.)

I'm glad you keep the blog going, I find these discussions fascinating. I don't know a lot of science, your posts inspire me to read more science.

OK.

But, and I realise it's childish, he started it.

I give back a little more than I'm given.