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Hyperbolic geometry problem

  1. Aug 16, 2008 #1

    Mentz114

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    This is not homework. I've been analysing some space-time diagrams and found I need to know the biggest angle in the illustrated triangle in terms of the lengths and the angle A which are known.

    I found the equivalent of the cosine rule of Euclidean space for the hyperbolic plane but I'm confused about the definition of 'Cos' on this plane. Is it still adjacent/hypoteneuse ?

    Any hints much appreciated.
     

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  3. Aug 16, 2008 #2

    tiny-tim

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    Hi Mentz114! :smile:

    No, it's only adjacent/hypoteneuse for infinitesimal triangles.

    For a real triangle, adjacent/hypoteneuse = sine of the opposite angle. :rolleyes:
     
  4. Aug 16, 2008 #3

    HallsofIvy

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    The trig functions can be defined, in the Euclidean plane, in terms of a circle, as the (cos(t),sin(t)) coordinates of a point a distance t around the unit circle, [itex]x^2+ y^2= 1[/itex], counterclockwise from (1, 0)

    The equivalent in the hyperbolic plane are (x, y) coordinates of a point a distance t along the unit hyperbola, [itex]x^2- y^2= 1[/itex], upward from (1, 0) and are given by cosh(t), sinh(t).
     
  5. Aug 16, 2008 #4

    Hurkyl

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    Just to clarify -- you mention space-time diagrams, so I assume you meant to referring to the Minkowski plane (which is the geometry of 1+1-dimensional space-time in special relativity) and not the hyperbolic plane (which satisfies all of the usual axioms of Euclidean plane geometry except for the parallel postulate).
     
  6. Aug 16, 2008 #5

    tiny-tim

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    … oops!

    ooh, I didn't spot that! :redface:

    Ignore my previous post … it only applies to hyperbolic space …

    thanks, Hurkyl! :smile:
     
  7. Aug 16, 2008 #6

    Mentz114

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    Thanks for the replies. Obviously my confusion is deeper than I thought. I am in the Minkowski plane. Do the ratios of the lengths of right angle triangles define sinh, cosh etc instead of sin, cos etc because of the different length definition ?

    One of the sides of my triangle has zero length !
     
  8. Aug 16, 2008 #7

    tiny-tim

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    Hi Mentz114! :smile:

    Minkowski space is non-isotropic, and so far as I know there isn't any general equation for triangles.
     
  9. Aug 16, 2008 #8

    Hurkyl

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    However, all spacelike directions are the same, as are all timelike directions and all lightlike directions. You even have a sort of antisymmetry between space and time in that swapping them just contributes some negative signs here and there. Also, the lightlike directions are expressable as limits of spacelike or timelike directions.


    Incidentally, the angle measure in a mixed spacelike-timelike plane is given by rapidity (and relate to hyperbolic trig functions) -- attempting to use 'Euclidean' angles will give nonsensical results.
     
  10. Aug 16, 2008 #9

    Mentz114

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    Right, I'm getting there, thank you.
    I've finished this ( see pic ) and I actually don't need that angle. I thought it might have physical significance but it isn't interesting.
     

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