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Invariance of uxw (velocityxphase velocity)

  1. Mar 19, 2007 #1
    Please inform me if you know places where the invariance of the product
    velocityxphase velocity=cc is discussed (derived?). Thanks
  2. jcsd
  3. Mar 19, 2007 #2
    invariance to what?
  4. Mar 19, 2007 #3
    I don't understand the question. How could it possibly be invariant?

    By counter example, choose a frame where a particle is moving: velocity x phase velocity is non-zero. Now choose a frame where that same particle isn't moving: velocity is zero so the product is as well.
  5. Mar 19, 2007 #4
    electron velocity u and asociated wave w uxw=cc

    Thanks. Please have a look at K. Moller, "The Theory of relativity" Clarendon Press Oxford 1972 Chapter 2.9." Consider that u represents the velocity of a tardyon and w represents the phase velocity of the associated wave. I mean by invariace the fact that uw=u'w'. I am interested if you have found mentioned that fact elsewhere as well, in order to enlarge my reference. Excuse please the inexact formulation of my question.
  6. Mar 19, 2007 #5

    Meir Achuz

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    I haven't read Moller in years, but he must be using the classical expression
    v=p/E for the "velocity of a tardyon". This gives v=k/w k and w are divided by hbar. The phase velocity of a wave is v_p=w/k, and he gets his result.
    BUT, in wave mechanics, the particle's position is described by a
    wave packet, whose peak moves with a group velocity, v_g=dw/dk.
    This only equals k/w for a massless particle.
    Moller seems to be using classical mechanics for v, and wave mechanics
    for v_p.
  7. Mar 19, 2007 #6

    Thanks. For you and for others interested in my thread I recall what Moller does. He starts with the phase of a plane wave propagating with phase velocity w in I and w' in I'. Among others he derives the addtion law of phase velocities and the transformation equation of the angles along which the wave propagates when detected from I and I' respectively. He has derived previously the transformation equation for the angles along which a tardyon moves with velocity u (u') and the addtion law for u and u'. The conclusion is:"A comparison of the transformation equations for the addition law of u and w respectively and of the angles along which the particle moves and the wave propagates become equal to each other respectively when we put u=cc/w and u'=cc/w'. In other words the velocity of a particle u and its direction n are transformed in the same manner as the corresponding quantities for a wave with the phase velocity w=cc/u and direction n. In his wave theory of elementary particles de Broglie made use of the circumstances by attributing to a particle with the direction of propagation n and the phase velocity w=cc/u a procedure which thus is relativistically invariant."
    Please comment.
  8. Mar 19, 2007 #7
    Sorry for the question. So, in wave mechanics, it's wrong to write:
    p = mv*gamma and E = mc^2*gamma for a particle?
  9. Mar 19, 2007 #8

    Meir Achuz

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    Whoops, I made a silly mistake. Thank you light--> for questioning the result.
    The sentence starting with BUT (I only capitalize when I am wrong.)
    should have read:
    A particle's position is described by a
    wave packet, whose peak moves with a group velocity, v_g=dw/dk.
    This equals k/w for a particle of any mass.
    Then scrap the last sentence.
    The v in the expressions is v_g.

    The algebra is:(d/dk)\sqrt{k^2+m^2}=k/\sqrt{k^2+m^2}.
    (All hbare=1=c)
    Sometimes I click submit too before I think.
  10. Mar 20, 2007 #9

    Do you think that the following derivation holds
    Using results of quantum mechanics: wavelength L =h/mv, where v is the
    particle velocity.
    Using relativity: energy E = mc^2 = hf, where f is the frequency
    associated with E.

    Multiply: hfL = mc^2 h/mv
    fL = c^2/v
    u = c^2/v, where u is the phase velocity

    so uv=c^2
    Note that this uses for frequency the total energy,including the rest
    energy, not just the kinetic energy. And the phase velocity gets larger
    as the particle velocity gets smaller!
  11. Mar 20, 2007 #10
    I would instead write:
    wavelength L =h/mv*gamma;
    energy E = mc^2*gamma.
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