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Group velocity is equal to the particle velocity?

  1. Feb 3, 2010 #1
    How can we show that the group velocity is equal to the particle velocity?
  2. jcsd
  3. Feb 4, 2010 #2

    Meir Achuz

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    The wave function of a moving particle is a wave packet.
    The group velocity is velocity of the envelope of the wave packet, while the phase velocity is the velocity of the waves within the packet.
    The derivation that the group velocity, so defined, is given by v_g=dw/dk is in many advanced textbooks.
  4. Feb 4, 2010 #3
    For the non relativistic case I did like this,but don't know how to do it relativistically
    Energy of a photon E = hν or ν = E/h ------ (1.17)
    We know ω = 2Πν or ω = (2ΠE)/h
    dω = (2Π/h)dE ------(1.18)
    further, k = 2Π/ λ = (2Πp)/h
    dk= (2Π/h)dp ------(1.19)
    dividing (1.18) by (1.19)
    dω/dk = dE/dp ------(1.20)
    by definition group velocity vg = dω/dk
    vg = dω/dk = dE/dp ------ (1.21)
    If a particle of mass m is moving with a velocity vparticle
    Then the Non relativistic energy
    E = (1/2)mv2particle = p2/2m -------(1.22)
    Differentiate with respect to p
    We get dE = (2p/2m)dp = (p/m)dp
    dE/dp = p/m = (mvparticle)/m = vparticle
    Hence vg = vparticle ------ (1.23)
  5. Feb 4, 2010 #4


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    In special relativity, E and p are related by E^2=p^2c^2 + m^2c^4.
    Starting with your Eq. (1.21):
    vg=dE/dp = pc^2/E=v, the velocity of the particle in SR.
  6. Feb 5, 2010 #5

    Claude Bile

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    Indeed, though the group velocity is actually a Taylor series of which dw/dk is the first term. Higher order terms govern the dispersion of the wave.

    It is worth keeping in mind (perhaps not specifically for this thread, but in general) that v_g=dw/dk is a first order approximation.

    /end nitpick.

  7. Feb 5, 2010 #6


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    Continue nitpick: Good point, but a common interpretation is that the 'group velocity' vg is defined as dw/dk, but if higher terms are important vg is not the 'signal velocity'.
  8. Feb 5, 2010 #7


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    I neglected to mention that the next term in the expansion, d^2/dw^2, produces a spread of the wave packet with vg still being the velocity of the peak of the wave packet.
    Since the front of the packet would go faster than vg, this could be interpreted as a faster signal velocity.
    The third term in the expansion would change everything.
  9. Feb 6, 2010 #8
    Interesting point. But isn't the essential usefulness of QM in specifying that the energy of non-radiating particles isn't dispersed and therefore the higher order terms will normally be null?

    Can you recommend a book for a thorough and understandable treatment of the subject? Would that be the Brillouin book?
    Last edited: Feb 6, 2010
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