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Fourier series

  1. Jul 23, 2006 #1
    Not sure if this should go into a math section, but I am trying to understand it in order to understand the Hesenberg uncertainy principle. I can't find a simple introduction to fourier series to answer this question.

    In my modern physics book, it does a quick introduction to fourier series without proving any of it (fine with me), but I get a little confused. Any periodic waveform can be the sum of an infinite number of sines and cosines. Now, my understanding of infinite here needs to be defined, as off into infinity on the frequency axises. So we are summing sizes and cosines with larger and larger frequencies.

    f = 0, 1/T, 2/T, 3/T, ..., presumably off into infinity. If we are apprioximating, the more terms we include the closer we get to the true answer.

    It then says we can begin to let T grow to make are signal's period larger and larger (being influenced to model a single wave packet), and in this case, the fundlemental frequency gets smaller and all the harmonics move closer together. WE then determine we can move all the way to one single pulse by making T infinity large, and therefor using a continuous spread of frequencies.

    The next section discusses that if we make our single pulse half as wide, it increases (perhaps doubles) the range of continues frequencies we need. Here is my confusion. Didn't we assume at the start the the frequencies needed to be exact went off into infinity. It seems now though, we have infinity many frequencies (since we are assuming a continious band), but the band itself is now finite. In other words, we may have a delta(f) centered around f at 100Mhz or so, with a delta of 10Mhz, (90-110Mhz), with infinitly many frequencies in between (to make the period T as large as we need). BUT(!), why don't we need to sum out the frequencies up to inifinity like we did at the start, 1000/T, 1000000/T, 50000343234/T, etc. What happen to all of these terms which would have frequencies greater then 110Mhz. My book makes no effort to explain this. I can't understand Heisenberg uncertainty principle until I understand why delta(f) is only a delta and does not have terms reaching out into infinite f.

    I hope my question makes sense.
     
  2. jcsd
  3. Jul 23, 2006 #2
    Excellent question. Look at a Gaussian waveform - that's a stereotypical waveform in frequency space. "Most" momenta are within one or two sigma of the packet center, those are the frequencies that matter. By "delta", you can freely substitute a "sigma" in a Gaussian waveform, that's what it means, a statistical standard deviation. In principle you worry about all those frequencies up to infinity - and to be rigorous, note that the integral does say to integrate [tex]\int_{-\infty}^{\infty} d \omega \quad[/tex] - but in terms of physical reasoning, those extreme terms don't really matter. Their contribution is truly neglible for all practical purposes.*

    *But for the math to work out analytically, you really do need all of them in the integral. The difference between a physicist and a mathematician is just this - a physicist throws out tiny numbers that don't matter.
     
    Last edited: Jul 24, 2006
  4. Jul 24, 2006 #3
    By the way - the Gaussian waveform is ubiquitous for many reasons - among which, it keeps the same shape under a Fourier transform (the parameters change).
     
  5. Jul 24, 2006 #4
    thank you for the reply. So if I can rephase just to see if I understand, in any given wave packet of a particle, we do actually have frequencies that increase all the way to infinity, but after a certain point they become so neglagibly small (the amplitute is neglagibly small) that they contribute nothing to any meaning measurment of momentum or energy of an actual particle.
     
  6. Jul 24, 2006 #5
    Exactly. In some treatments you will come accross "wavepackets" that do extend to infinity, things like plane waves - but these are only a mathematical convenience, and do not represent anything physical. Physical wavepackets vanish at infinity. (that's so they can be properly normalized)
     
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