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Wave group question

  1. Aug 17, 2005 #1
    Why is the narrorer the wave group, the greater the range of wavelengths involved?
     
  2. jcsd
  3. Aug 17, 2005 #2

    Galileo

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    Have you tried to look at the expression for a wave-packet and considered its Fourier transform?
     
  4. Aug 17, 2005 #3

    jtbell

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    I think it's easier to answer this question qualitatively if you turn it around: why does a small range of wavelengths produce a broad wave packet, and vice versa?

    Set up the waves so that they're all in phase at one point (say, the origin). If their wavelengths are nearly the same, they are still nearly in phase and interfere mostly constructively at a large distance from the origin. On the other hand, if they have a wide range of wavelengths, the destructive interference is significant at a short distance from the origin.
     
  5. Aug 18, 2005 #4
    so you mean that the smaller the amount of waves, the less they interfere with each other, so the bigger the range of wavelengths?
    @@a
    still a little confused~
     
  6. Aug 18, 2005 #5
    @@a
    i haven't studied the fourier transformation yet, so i don't know how to use it~
    but i believe that most complicated things can be explained in simple ways~
     
  7. Aug 18, 2005 #6

    jtbell

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    No, I'm talking about the range of wavelengths that the waves have. You're integrating an infinite number of them together via a Fourier integral:

    [tex]\psi(x,t) = \int_{\infty}^{-\infty} {A(k) e^{i(kx - \omega t)} dk} [/tex]

    [itex]A(k)[/itex] gives the amplitude of the wave with wavelength [itex]2 \pi / k[/itex]. Although we normally write the integral using in infinite range (limits) in [itex]k[/itex], what counts is the range where [itex]k[/itex] is significantly different from zero.

    You might say that a large range in [itex]k[/itex] has a larger number of waves in it (i.e. a larger number of values of [itex]k[/itex]), but that's not really correct because any continuous range has an infinite number of values in it!
     
  8. Aug 18, 2005 #7
    @@a
    sorry, but I'm still a little confused...
     
  9. Aug 18, 2005 #8

    jtbell

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    Well, I guess I'm still a little confused about what you're confused about... :confused:
     
  10. Aug 18, 2005 #9

    reilly

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    In reality, most physicists would say that fourier transforms are simple.

    However try building a wave packet graphically -- take several waves with different frequencies, and add them, by hand or whatever, and graph the result. Add some more -- big freqs and small ones. As you do this, you will see the interference patterns that ultimately produce a delta function -- make sure that all waves are of the form exp(kx), where k is the wave number, so that the exp functions are all = one at x=0.

    Regards,
    Reilly Atkinson
     
  11. Aug 18, 2005 #10

    quasar987

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  12. Aug 19, 2005 #11
    thanks! :)
     
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