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Rotation of coordinate system in minkowsky spacetime

  1. Jun 26, 2011 #1
    Does performing a rotation of the usual coordinate system [itex]ct,x[/itex] in the minkowsky spacetime makes sense?

    I guess it doesn't, but more than this i think that there is something that forbids it, since i could make coincident the 'lenght' axis of the non rotated coordinate system (observer A) with the 'time' axis of the rotated coordinate system (observer B), and that seems ridiculous to me (but you never know..)!

    From this I suppose that the Lorentz trasformations has to have some particular costraint, so I checked the propriety of the Lorentz group (to which they all belong) but i couldn't see it!

    Can you help me??
  2. jcsd
  3. Jun 26, 2011 #2


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    In space, you can rotate the axes using a transformation
    [tex]x' = x \, \cos \theta - y \, \sin \theta[/tex][tex]y' = x \, \sin \theta + y \, \cos \theta[/tex]
    and this preserves the metric
    [tex]ds^2 = dx^2 + dy^2[/tex]
    You can't apply the above rotation to (ct,x) coordinates to get an inertial frame, but you can apply the transformation
    [tex]ct' = ct \, \cosh \phi - x \, \sinh \phi [/tex][tex]x' = -ct \, \sinh \phi + x \, \cosh \phi [/tex]
    We can call this a "hyperbolic rotation". Note this is actually nothing more or less than a Lorentz transformation with [itex] v = c \, \tanh \phi [/itex] and [itex] \gamma = \cosh \phi [/itex], and which preserves the metric
    [tex]ds^2 = c^2 \, dt^2 - dx^2[/tex]
  4. Jun 26, 2011 #3


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    The Lorentz Transformations are in fact rotations in 4-D Minkowsky space-time. The Lorentz transforms are constrained by the fact that v<c. However, because of the weird metric (+,-,-,-) or (-,+,+,+), the rotations are described by hyperbolic functions rather than regular trigonometric functions, and the constraint v<c does not impose a restriction on the "angle of rotation". But, also because of the weird metric, one cannot make the analogy of this "rotation" by simply expanding the 3-D rotation that we are used to. From the form of the Lorentz transforms themselves, it is obvious that no matter what value for v we plug in, we can never get t'=x or x'=t. In that sense, we can't actually rotate the t-axis onto the x-axis or vice-versa, even though we can rotate through an "infinite" angle.

    For example, we have for some rotation angle (in the (ct,x) plane):

    [tex] x=x'\cosh(\psi)+ct'\sinh(\psi)[/tex]
    [tex] ct=x'\sinh(\psi)+ct'\cosh(\psi)[/tex]

    Nowhere does cosh(x) approach 0, so we can never get rid of the x' term in x, or the ct' term in ct (obviously we can get rid of the ct' term in x and vice versa for a velocity of 0).

    All this weirdness occurs due to the weird - sign that is present in the metric.
  5. Jun 26, 2011 #4
    Thanks a lot pals!!!

    So it's all becaouse it has always to be[tex]\eta_{\nu\mu}=\Lambda_{\nu}^{a}\Lambda_{\mu}^{b} \eta_{ab} [/tex] to preserve lenght and stuff!!! and my silly example obviously won't satisfy this!!!

    But new question...all of the Lorents transformation are reconducible to hyperbolic rotations?
    If so this means that both the lenght and time axis 'squeeze' by the same angle [itex]\phi[/itex] towards the world line of a particle which moves whit the speed of light! isn'it?
    Last edited: Jun 26, 2011
  6. Jun 26, 2011 #5


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    The Lorentz transforms are the 6 rotations in the 4-D space time (3 rotations in space, 3 boosts) yes. However, there are 4 additional symmetries of Minkowski spacetime (3 translations, and 1 identity transform), and together these 10 symmetries make up the Poincare group.

    In all, there are 3 boosts, 3 rotations in space, 3 translations, and 1 identity. I don't know where are you are getting this "squeezing" from.
  7. Jun 26, 2011 #6


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    There are four translations (three in space and one in time) and usually the identity isn't counted.
  8. Jun 26, 2011 #7

    George Jones

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    The (homogeneous restricted) Lorentz group: 1) does not include translations; 2) has elements that are generated by boosts and spatial rotations. These elements are not necessarily rotations or boosts.
  9. Jun 26, 2011 #8


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    Ah right, I miscounted. The identity is obtained by setting v=0 in a boost, or by rotating through angle 0, or by translating 0! 4 translations, 6 "rotations".

    George, what do you mean specifically that these elements are not necessarily rotations or boosts? You mean that the elements of the Lorentz group may not be pure rotations or boosts, and could be a combination? This property is guaranteed by the closure property of groups.
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