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Minkowski spacetime geodesics

  1. Jun 30, 2010 #1
    I have some difficulties understanding how Minkowski spacetime is flat and therefore its geodesics should remain parallel, but at the same time I see it described in other sites as hyperbolic and then geodesics should diverge. Any comment on my confusion about this will be welcome.

    Thanks
     
  2. jcsd
  3. Jun 30, 2010 #2

    Dale

    Staff: Mentor

    You could make this same statement about a Euclidean flat space. The metric there has the form of a sphere. This does not imply that the space is curved nor that parallel geodesics will converge.

    What is spherical/hyperbolic is a surface of constant distance. Such surfaces are curved, but the space is not.
     
  4. Jun 30, 2010 #3
    After reading some more I think there is not really contradiction, the hyperbolicity of Minkowski spacetime refers to the space component or the space-like 3D slice you get if you freeze time at one point.



    I was talking about flat spacetime , not flat space, and of course spce can be curved.

    Regards
     
  5. Jun 30, 2010 #4

    bapowell

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    Minkowski is flat. The hyperboloids that arise are a result of the action of the Lorentz group. The Lorentz transformations leave invariant the interval:

    [tex]s^2 = c\Delta t^2 - \Delta x^2[/tex].

    Consider in 2D a frame K and a frame K' that we will take to be moving relative to K with some velocity v. If we take a specific point (x',t') in frame K' and boost frame K' across the full range of velocities from -c to c, this point traces out a hyperbola in the frame K. So, for example, if we choose the point (0,1) in K', then it traces out the hyperbola

    [tex]x^2 - t^2 = 1[/tex]

    in K. I like to think of the Lorentz transformations in Minkowski space as playing an analogous role to ordinary rotations in Euclidean space: rotations in Euclidean space leave invariant the circle ([tex]x^2 + y^2 = 1[/tex]), whereas Lorentz transformations leave invariant hyperbolas. This becomes more transparent when you write the Lorentz transformations in terms of hyperbolic sines and cosines, a form in which the action of the Lorentz transformations as hyperbolic rotations is manifest.

    So, Minkowski space is flat, but because of the signature (the minus sign in there), the locus of points traced by its symmetry group (Lorentz) is a hyperbola.
     
  6. Jun 30, 2010 #5

    Yes, I understand that is what they call the "velocity space" of Special Relativity.

    But does what I said in my previous post about the space-like slice make sense or not really?
     
  7. Jun 30, 2010 #6

    bapowell

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    I don't think so. The space-like hypersurfaces of Minkowski space (as you say, freezing time -- looking at the 3-space that is orthogonal to the time direction) are flat. After all, the Minkowski metric is Euclidean on the subspace of spacelike hypersurfaces. Perhaps I'm misunderstanding you.
     
  8. Jun 30, 2010 #7
    No, I think you are right, maybe changing the coordinates signature to (+,-,-,-) ?
     
  9. Jun 30, 2010 #8

    bapowell

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    No, because that just changes the sign of the inner products (spatial distances become negative quantities as a result). The curvature still vanishes for this metric.
     
  10. Jun 30, 2010 #9

    dx

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    I think its better to say spatial intervals become negative. Changing sign convention of the interval doesnt make spatial distances negative.
     
  11. Jun 30, 2010 #10

    bapowell

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    Well, negative in the sense that, eg:

    [tex]\Delta s^2 = -x^2 -y^2[/tex].
     
  12. Jun 30, 2010 #11
    That is correct and yet I found this thread about SR :
    https://www.physicsforums.com/showthread.php?t=317880 ,posts 7 and 27 that I think reflect what I meant about some change in coordinates that can shift the sign of the spatial curvature from k=0 to k=-1.

    Regards
     
  13. Jun 30, 2010 #12

    bapowell

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    Thanks for posting. Yes, you are correct that you can choose a coordinatization such that the spatial slices have non-vanishing curvature -- my mistake. However, the intrinsic curvature of the full 4D Minkowski space vanishes, and will remain zero under any global diffeomorphism. Your question regarding the 'hyperbolic nature' of Minkowski space is, however, not related to the Milne coordinatization, but rather corresponds to the orbits of the Lorentz group in Minkowski space.
     
  14. Jun 30, 2010 #13
    Yes, I got that.

    Ok, thanks.
     
  15. Jul 1, 2010 #14
    I gave it some more thought. Would it be right to say that in order to make this change of coordinates we can imagine that in hyperbolic coordinate space the two sheets hyperboloid you obtain in Minkowski flat coordinate space gets transformed into a 3D hypersphere and geodesics follow the line element : ds^2=dt^2-dr^2-sinh^2r(dθ2 + sin2 θdφ2) ?
     
  16. Jul 1, 2010 #15
    And just by adding a scale factor a(t)=t one gets the Milne model line element in FRW metric.
     
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