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Transformation of k_y in the wave 4-vector

  1. Feb 13, 2010 #1
    I'm checking how k_y in the wave 4-vector transforms, but not getting what I expect:

    The wave 4-vector is defined as [tex](\omega/c,\ \textbf{k} )[/tex] where [tex]\textbf{k} = 2\pi/ \boldsymbol{\lambda},\ \textbf{u}[/tex] is the velocity of propagation of the plane wave

    Let s' travel, as usual, along the x axis of s with velocity v, and k make an angle theta wrt x axis.

    [tex]\omega'\ =\ \gamma\omega(1-v/u\ \cos\theta),\ u'_{y'} = u_{y}/\gamma (1-vu_x/c^2)[/tex] are standard results and substituting into

    [tex]k'_{y'} = 2 \pi/\lambda'_{ x'}[/tex]

    [tex]= \omega'/ u'_{y'}[/tex]

    [tex]= \gamma\omega(1-v/u\ \cos\theta)\gamma (1-vu_x/c^2)/u_{y}[/tex]

    [tex] = \omega/u_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)[/tex]

    [tex] = k_{y}\gamma^2(1-vu_{x}/u^2)(1-vu_x/c^2)[/tex]

    Since [tex] k_{y}=k'_{y'}[/tex] then

    [tex] (1-vu_{x}/u^2)(1-vu_x/c^2) = 1 - v^2/c^2[/tex]

    which isn't generally true.

    Where have I gone wrong in my working?

    Thanks in advance.
  2. jcsd
  3. Feb 24, 2010 #2
    I've found where I've gone wrong. I assumed that the wave vector and phase velocity are pointing in the same direction in all frames, when in fact they don't, and the effect is called "relativistically induced optical anisotropy". It means that the wave 4 vector in its usual form is a 4 vector and the phase is a Lorentz scalar only for a phase velocity equal to c, as outlined in this paper:

    Is the phase of plane waves a frame-independent quantity?

    The invariance of the phase of plane waves among inertial frames is investigated in some details. The reason that eventually led the author of a recent EPL letter [EPL \textbf{79}, 1006 (2007)] to a spurious conclusion of the non-invariance of the phase of waves has been identified -- it is the ignorance of the effect of relativistically-induced optical anisotropy in the analysis of the problem. It is argued that the Lorentz-invariant expression for the phase of waves should be taken in the form $\Phi=\mathbf{k\cdot r}-\mathbf{k\cdot u}/c$, instead of the widely-used expression $\Phi=\mathbf{k\cdot r}-\omega t$ which has a limited validity.


    Had anyone come across this effect before?

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