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Doppler effect in matter waves

  1. Oct 17, 2012 #1
    If I have a matter wave (for example, electron waves in a electronic microscope) at a given wavelength λ, and I move with respect (towards) them at speed [itex]v[/itex], I will measure a Doppler shift in the wave given by:

    [itex]\frac{1}{{\lambda '}} = \frac{1}{{\lambda '}}\left( {1 + \frac{v}{{v_e }}} \right)[/itex]

    where [itex]v_e[/itex] is the velocity of the electron wave. But the phase velocity? or the group velocity?

    I know that [itex]v_{ph} v_{gr} = c^2[/itex], so the result can be very different.
     
  2. jcsd
  3. Oct 17, 2012 #2

    mfb

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    I think you can calculate the relative velocity of the particle (with the group velocity and your own velocity), and work back to the wavelength and phase velocity afterwards.
    The formula might look different - it is not a classical wave.
     
  4. Oct 17, 2012 #3
    But for [itex]v<<<v_e[/itex] it cannot be much different...
     
  5. Oct 18, 2012 #4
    Ok, let us suppose that, instead of matter waves, I speak of sound in a medium where phase and group velocities are different. Which one will go in the Doppler formula?
     
  6. Oct 18, 2012 #5

    mfb

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    For sound, I would measure the wavelength relative to the medium, independent of the observer. To calculate the frequency, I would use that invariant wavelength and velocity addition with the phase velocity.
     
  7. Oct 18, 2012 #6
    The product equals c*c for electromagnetic waves.
     
  8. Oct 19, 2012 #7

    mfb

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    In vacuum. In matter, both phase and group velocity can be below c, for example, therefore the product has to be smaller, too.
     
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