Built on Facts

Last time, we did some slightly boring groundwork. This time, we’re going to look at something more interesting: the way a pulse of light propagates in something (like a piece of glass) with a frequency-dependent refractive index. As we discussed, the refractive index is just a way to express the phase velocity of a monochromatic continuous wave. If a medium has a refractive index of 1.5, then the speed at which the crests and troughs of a continuous wave travel is c/1.5, where c is the speed of light.

But here’s the interesting part – there’s really no such thing as a purely continuous monochromatic light wave. Inherent in the mathematics of Fourier analysis is the brute fact that any pulse of light that lasts a finite amount of time will actually contain a range of frequencies. For example, take this pulse:

This pulse, plotted as a function of time, wiggles about with a frequency of about 4.7 fs^-1 (ie, 4.7 complete wiggles per femtosecond). We can do some Fourier analysis on this pulse and plot as a function of frequency:

This pulse is peaked at 4.7 fs^-1 just like we said, but we see the pulse is actually made by the addition of a broad range of continuous waves with frequencies ranging from about 2.5 to 6.5 fs^-1. We know that something like glass can have a different refractive index for each frequency – hence the rainbow colors from a prism – and we might wonder what this pulse will look like after passing through a piece of glass that (say) slows down the blue light more than it slows down the red parts of the wave. For our made-up case, we’ll pretend that the refractive index looks something like this as a function of frequency:

“Woah, that can’t be right.”, you might say, “That refractive index drops below 1 for certain values. Would light with a frequency of 7 really be moving at 2c?” And this ought to bother you, it’s by no means an easy question to resolve. You might suspect that my made-up refractive index curve is simply wrong and that a refractive index of less than 1 is impossible forever and ever amen. But in fact it is possible, and it’s not even that uncommon. The resolution of this problem is the subject of the next post, for now we don’t have to worry about it because our pulse doesn’t contain frequencies in that range. After propagating 100 nanometers into the substance with that made-up refractive index curve, the pulse looks like this:

We notice several things. First, it’s shifted to the right. Remember the pulse is not moving to the right in terms of distance, because this graph has time as the x-axis. Rather, because we’re now looking at the pulse from an observation point 100nm from where it started, it takes light a few extra femtoseconds to get there from where it started. Thus, the shift toward the right. Then we notice that the pulse is wider. The slower frequencies lag back, and so the pulse is spread over a longer time interval. Then, we notice that the pulse is lower in amplitude. Mostly this is just due to the pulse being spread out in time. Finally, we notice that the pulse is sort of lopsided. It trails out to the right, but not so much to the left. This means that the pulse rises rapidly and trails off more gradually.

We notice that if any part of the pulse had shifted too far to the left, we’d be in trouble. Remember the x-axis is time. If the guy watching the pulse go by at the position origin x = 0 (the first graph) doesn’t yet see anything happening when his watch reads t = -1, the guy sitting at x = 100 had definitely better not be seeing anything at that same instant. The pulse is even farther from him, after all. But if the refractive index is less than 1, what’s to stop something so crazily impossible from happening?

Feel free to speculate in the comments – I know at least some of y’all know the answer. Also, I’m sorry for being so non-mathematical in this post but in this case it’s hard to avoid. This is the sort of thing where either you already know the math or you don’t, which means either way you probably won’t gain much by seeing a bunch of integrals.

Comments

#1 Eric Lund
April 20, 2011

There are actually two velocities that come up: the phase velocity ω/k, and the group velocity dω/dk. Whatever energy, momentum, or other information the pulse contains moves at the group velocity, so that must be bounded from above by c. No such constraint exists for the phase velocity, so it can get arbitrarily large (and does, in certain plasma physics applications).

It gets even weirder in anisotropic media. In the definition of group velocity the d/dk operator becomes the k-space gradient operator. In that case the phase and group velocities need not be aligned (just as an arbitrarily shaped object spinning about an arbitrary axis need not have its angular velocity and angular momentum in the same direction). In certain regimes of parameter space, you get self-focusing waves. See, e.g., A. V. Streltsov (1999), J. Geophys. Res. 104, 22657, doi:10.1029/1999JA900281.
#2 Matt Springer
April 20, 2011

In some circumstances you can even have a group velocity greater than c, if the refractive index is sharply-varying enough. In these cases though, higher-order terms become significant and dω/dk simply doesn’t correspond well to velocity any more. The pulse shape becomes highly distorted and “group” loses its meaning. It’ll turn out, as you’d expect, that even with phase and/or group velocity greater than c you still can’t propagate a signal faster than c.
#3 Remo
April 20, 2011

Interesting. I think you just want us to integrate the pulse over the refractive index. This gives you a little pulse leading the big pulse where the little pulse appears to be exceeding the speed of light. I’d draw an analogy to tunneling.

Very interested in where you are going with this. Nice challenge for my slowly aging neurons.
#4 Keçiören Nakliyat
April 21, 2011

This is kind a wierd and a little different then what we see at university.
Science changes soo fast nowadays
#5 davmab
April 21, 2011

PZ NEEDS HIS MEDS

MINDPHOQUE

clubconspiracy.com/forum/f29/my-special-poem-randis-head-13401.html
#6 In Hell's Kitchen (NYC)
April 21, 2011

If one wants to maintain contact with reality one should
think strictly in the time-domain where pulses turn on and
off, and nothing propagates with a speed greater than c.
#7 Anonymous Coward
April 21, 2011

Just to echo #2′s reply to #1, #1 is incorrect. Neither energy, nor momentum, nor information travels at the group velocity.

This is a little fib we tell students when they encounter a phase velocity > c. When, later, they encounter a group velocity > c, then we have to reveal the white lie.
#8 altın çilek
April 21, 2011

Very interested in where you are going with this. Nice challenge for my slowly aging neurons.
#9 Robert S.
April 21, 2011

One little thing that bothers me. Wouldn’t a single photon count as having a single frequency? Or is the uncertainty in its exact energy the same as frequency spreading.
#10 Matt Springer
April 21, 2011

Robert: for now we’re in the purely classical domain. It’s all Maxwell, without worrying about any particle-like aspects of light. But even if we bit the bullet and did this with a full quantum-mechanical treatment it wouldn’t change the conclusions.

The very, very approximate answer to your question though is that single photons have single frequencies, but a classical pulse of light consists of a lot of photons.
#11 Ben (NTF)
April 22, 2011

Very interesting and certainly has me thinking…..

Excellent post.
#12 Anonymous Coward
April 25, 2011

Re: Matt at #10

I see no reason why a single photon would have a single frequency any more than a classical pulse of light would have a single frequency.

Of course, if you MEASURED the frequency of a single photon you would only get one number, but that’s a different issue.

But a single photon with finite uncertainty in its time will most certainly have a nonzero uncertainty in its frequency.
#13 Wow
May 4, 2011

“But a single photon with finite uncertainty in its time will most certainly have a nonzero uncertainty in its frequency.”

To me that’s the clincher.

You can only tell what frequency your light is by counting how many up-down transitions it has and how long you counted for (which can only be less or equal to the time the photon exists for) then divvying up one by the other. But that means it could have been about to do another up but your photon ended “early”. Ergo, you can’t know precisely the frequency from the count of peaks.

Similarly for wavelength: count the peaks over the length of the photon at one snapshot of time and divvy up one to another.

But what if your photon measure ended part way through a transition?

So even if you counted a photon as having only one frequency, but as being a classical ball vibrating at the photon’s frequency, you still cannot measure the frequency/wavelength accurately and you therefore have an error.

Which “looks” like a spread of frequencies.

And to physics, if it “looks” like something, that’s as real as any other phenomena, unless you can find a way that it doesn’t look like it.

Which lead on to wave-particle duality.