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What do the eigennumbers of the Lorentz transformation represent?

  1. Oct 31, 2012 #1
    I am currently looking a bit into special relativity. Consider the matrix

    [itex]\Lambda=\left( \begin{array}{cc}
    \gamma & -\gamma \beta c \\
    -\gamma \beta c & \gamma \end{array} \right)[/itex]
    where
    [itex]\beta=\frac{v}{c},\quad \gamma=\frac{1}{\sqrt{1-\beta^2}}[/itex]
    and c is the speed of light.

    Then, an observer moving in the positive x direction will observe

    [itex]x^{\mu '}=\Lambda^{\mu '}_\nu x^\nu[/itex]

    The eigenvectors of the transformation matrix describes light rays and are

    [itex]u^\mu_\pm=\left( \begin{array}{c}
    t \\
    \pm ct \end{array} \right)[/itex]

    That eigenvectors of the Lorentz transformation are light rays simply means that light rays stay light rays under any Lorentz transformation which is the starting point of special relativity.

    But the eigenvalue of the Lorentz transformation is
    [itex]e_\pm=\frac{1\mp \beta}{\sqrt{1-\beta^2}}[/itex]

    And I was wondering what the meaning of this eigenvalue is. Do anyone know?

    And, we have the interesting identity (which may be somewhat intuitive when you think about it)
    [itex]e_+e_-=1[/itex]
     
    Last edited: Oct 31, 2012
  2. jcsd
  3. Oct 31, 2012 #2

    Bill_K

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    Doppler shift.
     
  4. Oct 31, 2012 #3
    Care to elaborate?
     
  5. Oct 31, 2012 #4

    Bill_K

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    If you ask how a null vector kμ transforms under a Lorentz transformation, in general there are two effects: it gets stretched and its direction changes. If k stands for the frequency 4-vector of a light ray these are the Doppler shift and aberration respectively.

    For the particular case where the Lorentz transformation is in the same direction as k (i.e. k is an eigenvector) there is no aberration, and the expressions e± you've written are the Doppler shift factors.
     
  6. Oct 31, 2012 #5

    Mentz114

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    My take on this is that when a null vector representing a light ray is boosted, it remains null ( obviously) but the frequency is shifted by
    [tex]\frac{1\mp \beta}{\sqrt{1-\beta^2}}[/tex]

    [I didn't see the reply above]
     
  7. Oct 31, 2012 #6

    robphy

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    As others have said, these eigenvalues are the Doppler factor [itex]\sqrt{\frac{1+v}{1-v}}[/itex] and its reciprocal.
    Another name for this factor is the k (and 1/k) factors in Bondi's k-calculus.
    In terms of rapidities, it is [itex]k_{+}=e^\theta[/itex] and [itex]k_{-}=e^{-\theta}[/itex].
    With a little algebra, you can write v and [itex]\gamma[/itex] in terms of these eigenvalues.
     
  8. Nov 1, 2012 #7
    Thanks for the answers!
     
  9. Nov 3, 2012 #8
    Here is a different perspective. If the people in the S and S' frames of reference were unaware that spacetime is Lorentzian rather than Euclidean, then the people in the S frame of reference might mistakenly conclude that, according the Lorentz Transformation, the S' frame of reference has has suffered a homogeneous deformation relative to their own. In this case, they would determine that the principle directions of deformation are along the 45 degree lines x = [itex]\pm[/itex]ct, and the principle extension ratios are the eigenvalues of the Lorentz Transformation. This, if fact, is one way of geometrically constructing a Minkowski diagram.
     
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