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Is the rotation angle in Minkowski's diagram real or imaginary?

  1. May 13, 2009 #1
    In Figure_1(b) I have depicted a simplified version of Minkowski's diagram, where

    [tex]\beta = \frac{v}{c}= \tanh \psi= - i \tan i \psi= \frac{( e^{\psi} - e^{-\psi})}{( e^{\psi} + e^{-\psi})} [/tex],

    the rotation of [tex](x',t')[/tex]-axes being defined as imaginary (considering x and t real, and c=1).

    However, I have found books where this rotation is considered as real, while the rotation in Figure_1(a) is shown as imaginary (considering x real, t imaginary, and c=1).

    Could you explain which rotation is real and which imaginary?
     
  2. jcsd
  3. May 13, 2009 #2
    Figure_1 is here!
     

    Attached Files:

  4. May 13, 2009 #3
    Doing a Lorentz transformation (boost) by a velocity v to some 4-vector is equivalent to rotate it an imaginary angle [tex]\chi[/tex] such that:

    [tex]\chi = arctan( v )[/tex]
     
  5. May 13, 2009 #4
  6. May 13, 2009 #5
    Thanks Kuon!

    If I understand correctly, you come with a third possibility, right? :smile:
     
  7. May 14, 2009 #6

    Fredrik

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    That depends on how you define "rotation". I wouldn't call a Lorentz boost a "rotation". I don't mind calling it a "hyperbolic rotation" or a "Minkowski orthogonal transformation" but calling it a "rotation" is a bit weird in my opinion.

    There's no need to bring imaginary numbers into this.

    The angle of a rotation is always real and can be defined by writing the matrix as

    [tex]\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}[/tex]

    A 2×2 matrix represents a proper rotation if and only if it's orthogonal and has determinant 1. Alll 2×2 matrices with those properties can be put in the form above, with different values of [itex]\theta[/itex].

    The rapidity of a Lorentz boost in 1+1 dimensions is always real and can be defined by writing the matrix as

    [tex]\begin{pmatrix}\cosh\theta & -\sinh\theta\\ -\sinh\theta & \cosh\theta\end{pmatrix}[/tex]

    A 2×2 matrix [itex]\Lambda[/itex] represents a proper orthochronous Lorentz boost if and only if it satisfies [itex]\Lambda^T\eta\Lambda=\eta[/itex], [itex]\det\Lambda=1[/itex], and [itex]\Lambda_{00}\geq 1[/itex]. (That's just its upper left component). All 2×2 matrices with those properties can be put in the form above, with different values of [itex]\theta[/itex]. This [itex]\theta[/itex] is usually referred to as the "rapidity" of the Lorentz transformation, not the "angle", but I wouldn't mind calling it the "hyperbolic angle" or even the "angle of a hyperbolic rotation". I just don't want to call it a "rotation angle", as you did in the thread title.

    Note that you can't turn one of these matrices into the other just by substituting [itex]\theta\rightarrow i\theta[/itex]. That's why it doesn't quite make sense to describe a Lorentz transformation as a rotation by an imaginary angle.
     
    Last edited: May 14, 2009
  8. May 14, 2009 #7
    Scott Walter (http://www.univ-nancy2.fr/DepPhilo/walter/), who published several papers on Minkowski and his diagram, wrote on page 9 of this article http://www.univ-nancy2.fr/DepPhilo/walter/papers/nes.pdf

    We can see now not only that it does make sense to describe a Lorentz transformation as a rotation by an imaginary angle, but to also inquire why would Minkowski really want to use the imaginary rotation. (I have to say that I buy only partially into Walter's explanation!)

    Trying to understand why wanted Minkowski to use the imaginary angle, I found in different sources three ways (including Koun's version!) to explain the rotation angle in Minkowski's diagram:

    1. x real, t real and the rotation angle real;
    2. x real, t real and the rotation angle imaginary;
    3. x real, t imaginary and the rotation angle imaginary.

    So, given that Minkowski didn't want to use version 1, which between version 2 and version 3 is the correct one?
     
    Last edited by a moderator: Apr 24, 2017
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