# Light waves, clarified

Transverse or longitudinal waves, purely as a matter of aesthetic preference? Transverse all the way, of course.

Not that there's a huge difference mathematically speaking. It's more of a species-type categorization than a rigorously formal one. Transverse waves oscillate perpendicular to their direction of motion, longitudinal waves oscillate parallel. This stuff does however give me a chance to spread the light (ahem) on a common misconception my 208 students often seem to have. This is the type of picture usually presented in physics textbooks of a transverse wave (a light wave in this case), with its twin perpendicular electric and magnetic oscillations:

So if you picture a water wave propagating outward you notice that the wave has a certain size. The amplitude is just the height of each wave. Light ain't like that. There's no physical "size" of the up and down oscillations in terms of spatial extent, certainly not in this purely classical plane-wave sense. When students perfectly reasonably look at pictures of waves, they built up a misconception in their heads that light in some sense takes up space and needs room to wiggle. But what's oscillating is not a physical substance moving up and down. What's oscillating is the strength of the electric and magnetic fields. The up-and-down displacement of the printed wave in the diagram represents the strength of the field at that point. Nothing's actually moving up and down, and thus the wave isn't taking up space - not by virtue of its amplitude, anyway. A wave with higher amplitude will not get "stuck" in its passage through a small opening any more than a wave with small amplitude will.

Thus light as a wave is a little more abstract than waves of spatial displacement. A subtle point, but an important one.

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I understand what you're getting at (that the model causes people to develop wrong impressions)...but I also don't see a problem with people imagining light as propagating "photon soup". That has some useful points as well.

"Well, actually wave-particle duality is perfectly normal."

Okay, pedant, it's counter-intuitive when you are first exposed to it.

A wave with higher amplitude will not get "stuck" in its passage through a small opening any more than a wave with small amplitude will.But a wave with a higher frequency will get stuck (i.e. the perforated metal shield in a microwave door.) This stuff can be baffling to the layman.

"Amplitude" comes from degenerate multi-photon summation (e.g., ridiculously large laser electric fields) not frequency. Magnetic fields about a factor of 10 above those of magnetars cause the vacuum to go dichroic. Nuclear electric fields beyond about Z = 1/(fine structure constant) spark the vacuum. Can a realy nasty coherent in time and space laser amplitude "not fit" into the vacuum?

(Higher frequencies fit through smaller holes, x-ray crystallography to ultrasonic microscopes.)

I'm not sure where you guys are coming from. Matt didn't mention photons at all. The students' misconception is purely a misunderstanding of classical electromagnetism.

Hmm. this was somewhat of a basis for my understand of polarization but now i am somewhat convinced i need to rethink all of that.

Oh the wonders of being an undergrad.

By Paul Johnson (not verified) on 21 May 2009 #permalink

OK, but doesn't the size of a slit through which light is passing interact with the wavelength, as though size mattered somehow? Kinda like what Paul says...

@paul and @greg: I always found this puzzling. I think I got it now, but anybody please correct me if not:
The wavelength (in contrast to the amplitude) is a real length in usual space, so that the wavelength determines things like diffraction.
Also, the electric and magnetic fields have a orientation in usual space, nevertheless their amplitudes don't form a vector (and thus don't have a length) in usual space. Still the orientation is responsible for polarization effects.

By Anonymous (not verified) on 22 May 2009 #permalink

@Anonymous #11: I think you muddied the waters further with that explanation.

Wavelength is a physical length in space, but it is not a vector. There is a vector called the wave vector; its magnitude is called the wave number and is inversely proportional to the wavelength. Yes, the wavelength does matter in things like slit diffraction: waves much longer than the slit dimensions don't get through, waves much shorter pass without noticing the slit. Things get interesting when you are dealing with wavelengths comparable to the slit size.

Electric and magnetic field are vectors in Euclidean space (in special relativity they are elements of an antisymmetric tensor), but they have different units than physical length. They are also time varying. The amplitude is important, because it tells you about things like whether the wave is traveling, standing, or a combination due to partial reflection from a boundary. For a traveling electromagnetic wave the ratio E/B (if both are measured in SI units) gives the propagation speed of the wave--this will be c in vacuum, but other interesting possibilities arise if there is a medium (solid, liquid, gas, or plasma) present.

I think the problem with this visualization of a wave is that it is restricted to a special case: surface waves on a body of water. That's the kind of wave most people are familiar with because it's easy to actually see the wave, and its amplitude is the displacement height of the water's surface, which is a physical length. But for most waves (not just electromagnetic waves), amplitude is not a length, and thinking of amplitude as a length can lead you astray.

By Eric Lund (not verified) on 22 May 2009 #permalink

This misunderstanding is in fact a well-known fact, described, for example, in the paper "The influence of student understanding of classical physics when learning quantum mechanics" by Steinberg, Wittman, Bao and Redish (Should be available on the internet).

Hmmm. Where to begin.

I'll begin by thanking Matt for his insight into the struggles his students have, and share mine.

@7: "The students' misconception is purely a misunderstanding of classical electromagnetism." I'd go further and say it starts with a misunderstanding of functions that might go back to excessive use of graphing calculators that only plot y vs x, one that is barely confronted in first semester physics and even calculus.

The easiest wave to confront is a transverse wave on a string where you actually see y as a function of x at various times. Thus a snapshot of y(x,t) looks like y(x) and y actually is a distance in m that you can measure during a demonstration. Problems with all other waves are little different from the problem of graphing v(t) - or df/dx - and asking the student to describe what how the particle is moving - or how the graph of f looks - when v>0 but dv/dt<0. They will usually say it is "moving down". The abstraction of what is on the "vertical" axis is a huge problem. If you doubt me, ask a chemist trying to get their students to interpret a graph of log(this) vs sqrt(that) in the lab!

There is little difference between struggling with a longitudinal wave where the pressure P(x,t) or displacement deltaX(x,t) is a function of x, or a transverse EM wave where the E(x,t) vector field is a function of x. The longitudinal wave might be easier at this point! And, now that I think about it, we probably make things worse when we walk across the room using our arms to represent the mutually perpendicular fields of a traveling EM wave. We are giving it an actual "physical" size while trying to show them the unit vector of the field.

@12: I don't think @11 muddled anything. E and B are vectors in a space that has dimensions of V/m and T, respectively, not those of the usual space (m) -- but the component directions are tied to those of the actual space. Hence polarization properties are tied to E-hat (unit vector), but only the actual spatial dimension (wavelength) has anything to do with "fitting" through something like a slit.

One can go further. The Poynting vector and the wave vector both point in the same direction, but only the size of the latter has anything to do with "fitting" somewhere in space ... and then certainly NOT in the sense most students think it means (see below).

PS @13-
Pasting that title into google produces a direct link to the pdf for that paper. The newish Cummings, etc, Redish textbook grew out of that sort of research, by the way.

Sorry. Part of one paragraph got eaten by the HTML monster because of how I used the greater and less than signs:

... Problems with all other waves are little different from the problem of graphing v(t) - or df/dx - and asking the student to describe how the particle is moving - or how the graph of f looks - when v > 0 but dv/dt < 0. They will often say it is "moving down". The abstraction of what is on the "vertical" axis is a huge problem. If you doubt me, ask a chemist trying to get their students to interpret a graph of log(this) vs sqrt(that) in the lab! Or ask other physics students.

Again!

... when dv/dt IS LESS THAN 0. They will often say it is "moving down". The abstraction of what is on the "vertical" axis is a huge problem. If you doubt me, ask a chemist trying to get their students to interpret a graph of log(this) vs sqrt(that) in the lab! Or ask other physics students.

By CCPhysicist (not verified) on 22 May 2009 #permalink

Nothing's actually moving up and down, and thus the wave isn't taking up space - not by virtue of its amplitude, anyway.

But if you place a free charged particle on the path of this wave, its spatial "up and down" movement is governed by the amplitude of the electric field.

I think the important matter to consider here is from where these abstractions of electric and magnetic fields stem: from the actual forces acting on charged particles and currents.

The easiest wave to confront is a transverse wave on a string where you actually see y as a function of x at various times. Thus a snapshot of y(x,t) looks like y(x) and y actually is a distance in m that you can measure during a demonstration.
thank you

My understanding is that the vectors for E and B represent actual forces (in principle) on little test charges that were exposed to the EM wave. Their directions are indeed "real". This is very true for radio waves where photon issues are minimal. But what about a single photon? In what way does the EM character apply? We talk of amplitude for the photon (as for constructing a superposed state with combination of RH and LH components) but how literally is that per an EM field? Photons do have actual "length" (coherence length) too, based on how many "wavings" they underwent during emission. IS the amplitude of the longer photons less than that of the shorter ones, is the E field somehow effectively higher (one photon can indeed kick an electron), does the chance of absorption really oscillate with square of amplitude? Interesting questions.

BTW, I have presented a curious paradox of relativistic forces at my blog "Tyrannogenius" (OK, trying for the hippest tough-guy name), check out the latest post there and feel free to comment.

sorry for the following dumb question:

How do polarized lenses work then? I was always under the impression that they worked on the fact that light waves had a physical 'up and down' movement.

Thanks!

I hate to throw a wrench in this, but it appears that we're dealing with various models to represent the behavior and properties of electromagnetic radiation. We start with the wave on a string, then proceed to the more complex E and B field combination.

But isn't the photon the primary unit?

It is my understanding that these particles (photons) have a probability of being detected at some time in the future based on a probability distribution, which has a time-based oscillation, so it is this that is demonstrating the wave-like properties that we observe. However, we must sum up the probability amplitudes over ALL possible paths between point A and B to determine the likelihood of detecting the photon. If we try to find out something about the intermediate path by placing another detector, we've changed the problem and have to recalculate.

At least that's my paraphrasing of what Feynman taught.

So all this discussion about the EM field is arguing about an earlier model of light. No?

By Chris Becke (not verified) on 31 May 2009 #permalink

What I was trying to say:

The photon is a particle.

Where you find it is based on its wave function.

So why are we talking about an EM wave?

By Chris Becke (not verified) on 31 May 2009 #permalink

We are talking about a classical (and thus approximate) description, but for many macroscopic purposes the classical Maxwell description is effectively perfect. This is one of those times. It's like neglecting relativity when calculating travel times between your house and the store - sure it's not technically correct, but it's so close as to make no difference.

In any event, photons can't really be treated entirely as particles either. They don't even have wavefunctions in any normal sense. But none of that's necessary for this kind of macroscopic situation where the classical field theory of light is more than adequate.