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Once the light is emitted as a pulse, and once that pulse has passed, what's behind it? Is light a continuous wave, or only a wavefront? What is left in the wake?

Qmechanic
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Gabri Botha
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  • Not clear what you are asking. What is behind a car when it has passed? Are you asking if light is a wave or a particle? This has been asked many times already, eg Light, a wave or a particle or something else? – sammy gerbil May 06 '18 at 21:08
  • The top answer in the linked question does not explain it well enough. If a photon travels as a wave, say 400nm wavelength, and the electromagnetic wave propagates, what is left behind, like the last 100nm after the wave passed, or does it not work like that? Im really asking whats left in light's wake? – Gabri Botha May 06 '18 at 21:50

1 Answers1

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A "pulse" is a wave packet that starts and then ends. There is nothing in front of it, and nothing behind it. An example is what happens when you turn a flashlight on and then turned it off, leaving you in complete darkness.

Jerrold Franklin
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  • Don't tell that to someone who has taken the Fourier transform of a wave packet! Mathematically, a wave packet can be represented as a superposition of a continuum of infinite-length monochromatic components. Does that make physical sense? As it turns out, yes. An experiment that filters out a very narrow wavelength band yields a monochromatic pulse that is much longer than an original broadband pulse. The leading and trailing parts of the broadband pulse aren't detectable without filtering, but in a strange sense they are there. – S. McGrew May 07 '18 at 01:51
  • @S.McGrew No, they're not. If you add a sharp filter then you'll get a long quasi-monochromatic 'tail' behind the pulse, but that's not the pulse itself you're seeing, it's the inertia of the filter itself, which takes a long time (inversely proportional to its bandwidth) to wind down. And no matter what you do, all physical filters are causal, which means that they will never produce light in front of the pulse. – Emilio Pisanty May 07 '18 at 08:04
  • There's no argument about your causality point. My real point in the comment was that the Fourier transform representation of a pulse has some problems. Attributing the "tail" to inertia in the filter itself needs more explanation though. Two cases worth examining: the standard pulse-shaping apparatus that turns a Gaussian femtosecond pulse into an extended pulse of near-arbitrary shape using diffraction gratings, lenses and apertures, and a similar apparatus using a prism and lens. If I pose that as a question, will you provide a detailed answer? – S. McGrew May 07 '18 at 14:10
  • Please provide a link to your question about the filter and the tail it creates. – Gabri Botha May 08 '18 at 07:17
  • "Don't tell that to someone who has taken the Fourier transform of a wave packet!" I wouldn't bother to tell that to someone who has no idea about what a Fourier transform does. Fourier's theorem says that the inverse transform gives back the original pulse. Putting a white pulse through a pure green filter just gives pure green waves in the pulse without changing its shape. You can't produce light before or after a pulse by using a filter. – Jerrold Franklin May 08 '18 at 10:41
  • @Gabri_Botha, here is the question: [https://physics.stackexchange.com/questions/404785/unphysical-results-of-fourier-transforms] – S. McGrew May 08 '18 at 19:38
  • @Jerrold_Franklin, please see Emilio_Pisanty's comment above, agreeing that there is a "tail" produced by filtering a pulse, but pointing out that there is nothing that appears in front of the pulse. – S. McGrew May 08 '18 at 19:43
  • @JerroldFranklin That last assertion is incorrect. You can't produce light before a pulse by using a filter, but you most definitely can turn a short pulse into a long wind-down oscillation by using any filter whose bandwidth is smaller than the original pulse's; indeed under those bandwidth conditions, you must produce light after the pulse, solely from the presence of the filter. – Emilio Pisanty May 08 '18 at 21:57
  • $f(x,0)= e^{in\pi x/L},\quad 0< x < L$ is the wave packet that results when a pulse of white light passes through a pure wave length filter. – Jerrold Franklin May 13 '18 at 18:16