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Dispersion relation

  1. Oct 24, 2007 #1
    The group velocity of traveling wave is defined as [tex] v_g =\partial \omega/\partial k[/tex]. I am confused about how to actually calculate this. For instance, in the Schrodinger equation, we find that plane waves solve the equation provided that
    [tex] \omega = k^2 \hbar/2m[/tex]
    Does this mean that the group velocity of "Schrodinger waves" is [tex] k\hbar/m[/tex]? Won't this in general depend on the amplitude of the frequency components of a given wave?
    Given a specific solution to the wave equation how does one answer the question, what is the group velocity of this wave?

    Edit: related question....
    In elementary texts, it is shown how the superposition of two sine waves of equal amplitude and phase but slightly different frequency and speed gives rise to a "traveling envelope", the speed of which we associate with the group velocity. How do we know in general that that superposition of waves gives rise to a well defined envelope?
     
    Last edited: Oct 24, 2007
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  3. Oct 24, 2007 #2

    Gokul43201

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    Yes.

    Not sure what you are asking, but in general, the group velocity of a wave traveling through a dispersive medium is a function of frequency.

    Just as above, take the dispersion relation in its standard form, and find its derivative.
     
    Last edited: Oct 24, 2007
  4. Oct 24, 2007 #3
    Ah now I've confused myself. I guess I mean to say that a given wave may be composed of many wavelength components. So for what k do I evaluate the group velocity equation [tex] v_g = k\hbar/m[/tex]?
     
  5. Oct 24, 2007 #4
    Ok here's an example to illustrate my confusion. Take a plane wave:
    [tex] \Psi(x,t) = A e^{i(kx-\omega t)}[/tex]
    It's phase velocity is [tex] \omega/k=\hbar k/2m[/tex]. But its "group" velocity should be the same thing, no?
     
  6. Oct 24, 2007 #5

    Gokul43201

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    No, it's group velocity will be twice that number (and equal to the classical speed of the "free particle" described by the plane wave).
     
  7. Oct 24, 2007 #6
    But that doesn't make any sense to me. What envelope is involved here?

    In particular, I am really interested in finding out the details behind what's hinted at here:
    https://www.physicsforums.com/showthread.php?t=173138
    Where does the Fourier transform come into this?
     
  8. Oct 25, 2007 #7

    Claude Bile

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    Typically the centre value or the mean value for k is used, keeping in mind that the expression for the group velocity is only valid where the spread of values for k is small compared to the central/mean value of k.

    Regarding the Fourier variables, I'm not exactly sure what Meir Achuz was alluding to but I suspect that it might be linked to the fact that w and t are Fourier conjugate variables, as are k and x. The Fourier relationship between these variables is the key between obtaining an expression for velocity (i.e. x/t) in terms of w and k.

    Claude.
     
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