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Little problem with group velocity

  1. Jul 2, 2008 #1
    I have a little trouble in demonstrating that the group velocity of an e.m. wave transform (in special relativity, by passing from an inertial frame to another) like the velocity of a particle with mass.

    So, [tex] \left( \frac{w}{c}, k_x, k_y, k_z\right) = \left( \frac{w}{c}, \vec k \ \right)[/tex] is a 4-vector, where [tex]w[/tex] is [tex]2\pi\nu[/tex] and [tex]\vec k[/tex] is the wave vector. So, if another inertial frame moves with velocity [tex]v[/tex] in the direction of the [tex]x[/tex] axis, we have

    [tex]\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}[/tex]
    [tex]dw' = \gamma\left(dw-v\cdot dk_x\right)[/tex]
    [tex]dk_x' = \gamma\left(dk_x-\frac{v}{c^2}dw\right)[/tex]
    [tex]dk_y' = dk_y[/tex]
    Let be [tex]u_x = \frac{dw}{dk_x}[/tex], [tex]u_y = \frac{dw}{dk_y}[/tex].
    So, I find [tex]u_x' = \frac{dw'}{dk_x'} = \frac{u_x-v}{1-\frac{v\cdot u_x}{c^2}}[/tex] (correct, because it's the same transformation for the velocity in [tex]x[/tex] of a particle with mass), and
    [tex]u_y' = \frac{dw'}{dk_y'} = \gamma\frac{dw-v\cdot dk_x}{dk_y} = \gamma\left(u_y-\frac{dk_x}{dk_y}\cdot v\right) = \gamma\left( u_y-v\cdot\frac{dw}{dk_y}\cdot\frac{1}{\frac{dw}{dk_x}}\right) = \gamma\left(u_y-v\cdot u_y\cdot\frac{1}{u_x}\right) = \gamma\cdot u_y\left(1-\frac{v}{u_x}\right)[/tex]
    But this is different from the relationship for particle with mass velocity, that is
    [tex]u_y' = \frac{u_y}{\gamma\cdot \left(1-v\cdot\frac{u_x}{c^2}\right)}[/tex]

    Where's the problem with this derivation? How can I find the right relationship?

    Maybe the problem is with the fact that I should do the partial derivation
    [tex]u = \frac{\partial w}{\partial \vec k}[/tex]

    (I know the transform must be the same for group velocity of an e.m. wave AND for velocity of particles with mass). Please, if you can, show the passages of the derivation.
  2. jcsd
  3. Jul 2, 2008 #2
    Firstly, to clarify, the vacuum group velocity of EM waves is c, which is *not* like the velocity of a particle with mass.

    You can see that the wave 4-vector is a tensor (in SR) just by recognising that gWx (the phase difference between two events on a wave) is a scalar (while the displacement x and inner product g are both also tensors); W must be contravariant (same as velocity dx/dl) since that equation is valid in all frames.

    Hence, gWW (=b) is a constant in all reference frames (although frequency and wave-vector components are not constant, in fact this equation gives you the dispersion relation for waves of this "type", that is, waves that are related by the symmetry of SR). From this you can write the relation between group and phase velocity (they will be inverse fractions of c). Then by simple geometry you can recognise that the group velocity four-vector is parallel to W. QED (and moreover, if you wish to identify massive particles with waves obeying SR then you have proven E=-ih(d/dt) and p=-ih(d/dx) for some constant h, which is of course the basis of QM).

    This argument is sketched at the beginning of some QFT texts.
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