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Curvature of reciprocal Euclidean space

  1. Jul 1, 2004 #1
    A triangle in Euclidean space can be described as having a hypotenuse of one, and legs of Lorentz parameters [tex]\beta[/tex] and [tex]\gamma[/tex]. What spatial curvature underlies a triangle with hypotenuse one, and legs [tex]1/ \beta[/tex] and [tex]1/ \gamma[/tex]?
     
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  3. Jul 2, 2004 #2

    robphy

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    If so, then [tex]1^2=\beta^2+\gamma^2[/tex].
    Since [tex]\gamma^2=\frac{1}{1-\beta^2}[/tex], you get [tex]1-\beta^2=\frac{1}{1-\beta^2}[/tex]. Then, [tex]\beta=0[/tex] and [tex]\gamma=1[/tex].
    Am I misunderstanding something?
     
  4. Jul 2, 2004 #3
    robphy,

    Mea culpa.

    Thank you for the reminder: [tex]\beta=v/c[/tex] and [tex]\gamma=1/ \sqrt{1-v^2/c^2}[/tex]

    Revised: A triangle in Euclidean space can be described as having a hypotenuse of one, and legs of Lorentz parameters beta and 1/gamma. What spatial curvature underlies a triangle with hypotenuse one, and legs of 1/beta and gamma?
     
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