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Question on hyperbolic rotation

  1. Feb 27, 2013 #1
    Hello,

    I see that hyperbolic rotation of frame F' about the (x2,x3)-plane of frame F is identical to a Lorentz transformation, corresponding to a linear motion along x1 of the frame F' with respect to F.
    Then hyperbolic rotation about (x1,x2) means motion along x3 and
    hyperbolic rotation about (x1,x3) means motion along x2

    Is it possible to hyperb. rotate about say (<any axis>,x4)? Does this even make sense?
    The x4 axis is the temporal axis
     
  2. jcsd
  3. Feb 27, 2013 #2

    Bill_K

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    Using your notation, a rotation about the (x2, x3) plane is a rotation in the (x1, x4) plane, and so is a boost along the x1 axis.

    If you rotated about the (x1, x4) plane, say, that would be a (normal) rotation in the (x2, x3) plane.
     
  4. Feb 27, 2013 #3
    My confusion came from this thread

    thread

    They mention 3 + 1 translations as well.

    Then there should be 3 hyperb. rotations, 3 normal (spatial) rotations, 3 spatial + 1 temporal translation
    I somehow understood this to mean that the translations are result of some kind of hyperbolic rotations
     
  5. Feb 27, 2013 #4
    So, rotation in 4D is hyperbolic expansion in 3d?
     
  6. Feb 28, 2013 #5

    Fredrik

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    Translations don't have anything to do with rotations. Rotations (even the hyperbolic ones) take the origin to itself. The only translation that does that is the identity map, i.e. the translation by 0.
     
  7. Feb 28, 2013 #6
    Fredrik, they probably do not ... It could be misleading though

    It seems that since hyperbolic rotation about (x2,x3) is identical to a spatial translation along x1 --> Then, one could be led to conclude that hyperbolic rotation about (<something>,x4) might lead to a temporal translation, e.g. translation along x4.
     
  8. Feb 28, 2013 #7
    Bill K,

    this means you are saying that a rotation about the plane (x1,x4) is the same thing as rotation about the x1 axis. I haven't checked this, it could be true, I don't think its obvious though :)
     
  9. Feb 28, 2013 #8

    Fredrik

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    I think you have misunderstood what translations and boosts are. A boost is a change of velocity. A translation is a change of origin. A boost in the 1-direction is a hyperbolic rotation in the 0-1 plane, i.e. a map of the form
    $$\begin{pmatrix}x^0\\ x^1\\ x^2\\ x^3\end{pmatrix} \mapsto\begin{pmatrix}\cosh\theta & \sinh\theta & 0 & 0\\ \sinh\theta & \cosh\theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x^0\\ x^1\\ x^2\\ x^3\end{pmatrix}.$$ A translation in the 1-direction is a map of the form
    $$\begin{pmatrix}x^0\\ x^1\\ x^2\\ x^3\end{pmatrix} \mapsto\begin{pmatrix}x^0\\ x^1\\ x^2\\ x^3\end{pmatrix} +\begin{pmatrix}0\\ s\\ 0\\ 0\end{pmatrix}.$$ Edit: I should have mentioned that I like to number the components from 0 to 3. The 0 component is the one that corresponds to time.
     
    Last edited: Feb 28, 2013
  10. Feb 28, 2013 #9

    Fredrik

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    That's not what he's saying. What he means by a rotation in the (x2,x3) plane is exactly what you would call a rotation about the (x1,x4) plane.
     
    Last edited: Feb 28, 2013
  11. Mar 3, 2013 #10
    I did some reading... so boosts or Lorentz transformations are the same thing as hyperbolic rotations about (x_i,x_j) - e.g. they correspond to spatial motion along one of the spatial coordinate axes.

    Could there be a hyperbolic rotation such, that it would correspond to motion along the temporal axis - I haven't read anything yet to suggest that this is possible. In this article (http://en.wikipedia.org/wiki/Lorentz_transformation) there is a section (very hard to read) about non-orthocronous Lorentz transformations, where a time reversal is mentioned. It says that these are compound transformations which seem to involve more than just hyperbolic rotation. In reading the article though, one is unsure how is time reversal related to the non-orthocronous transformations.
     
  12. Mar 4, 2013 #11

    Fredrik

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    Let's consider the 1+1-dimensional case for simplicity. In units such that c=1, a Lorentz transformation is of the form
    $$\frac{\sigma}{\sqrt{1-v^2}}\begin{pmatrix}1 & -v\\ -\rho v & \rho\end{pmatrix},$$ where ##\sigma,\rho\in\{-1,1\}## and ##|v|<1##. A proper Lorentz transformation is one with ρ=1. An orthochronous Lorentz transformation is one with σ=1. Boosts are both proper and orthochronous.

    A good way to understand the significance of the values of σ and ρ is to look at the zero-velocity transformations. If we set v=0, the above turns into
    $$\sigma\begin{pmatrix}1 & 0\\ 0 & \rho\end{pmatrix}.$$ Time reversal is the transformation
    $$\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}.$$ This is a Lorentz transformation with v=0 and σ=ρ=-1. It's called time reversal because for all real numbers t,x, we have
    $$\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix} \begin{pmatrix}t\\ x\end{pmatrix} =\begin{pmatrix}-t\\ x\end{pmatrix}.$$
    You also asked about "motion along the temporal axis". It's not clear to me what that would mean. Note however that a boost in the 1 direction affects the 0 and 1 coordinates (or the 0 and 4 coordinates if you prefer the label 4 for the temporal axis) in the same way. It tilts both axes of a spacetime diagram by the same angle towards the t=x line. It's called a boost in the 1 direction, because the velocity vector v is in the 1 direction. The velocity of an arbitrary Lorentz transformation ##\Lambda## in 3+1 dimensions is defined by
    $$v=\frac{1}{(\Lambda^{-1})_{00}} \begin{pmatrix}(\Lambda^{-1})_{10}\\ (\Lambda^{-1})_{20}\\ (\Lambda^{-1})_{30}\end{pmatrix}.$$ This vector doesn't even have a fourth component, so it can't be in the 0 (or 4) direction.
     
    Last edited: Mar 4, 2013
  13. Mar 11, 2013 #12
    Fredrik,

    thank you for your detailed reply and for clarifying the orthocronous and arbitrary transformations. I will surely refer to this thread in the future :)
     
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