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.