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Geometry of a conformal disk

  1. Nov 23, 2009 #1
    For better or worse, I've begun to work Penrose's book Road to Reality. In the second chapter, one of his exercises (supposedly simple) has entirely eluded me. Penrose presents the conformal Poincaré disk and gives the equation for hyperbolic distance between two points [tex]A[/tex] and [tex]B[/tex] inside the circle: [tex]\log\frac{QA\cdot PB}{QB\cdot PA},[/tex] where [tex]P[/tex] and [tex]Q[/tex] are the points where the Euclidean circle through [tex]A[/tex] and [tex]B[/tex] orthogonal to the bounding circle meets the bounding circle and where [tex]QA, QB, PA, PB,[/tex] are Euclidean distances. See the figure below.


    Now, Penrose simply says that you can include the [tex]C[/tex] of Lambert's area formula, which is given by the equation [tex]\pi-(\alpha+\beta+\gamma)=C\Delta[/tex] (where [tex]\Delta[/tex] is the area of the triangle and [tex]C[/tex] is some constant), by multiplying the distance formula above by [tex]C^{-1/2}[/tex], where [tex]C\neq 1[/tex]. He asks if you can see a simple reason why.

    My feeling is that it's because [tex]AB[/tex] is just one side of a triangle, and [tex]C[/tex] must be adjusted thereby. But this feels incomplete at best, and entirely off-track at worst. Any ideas? A hint in the right direction would be better than the outright correct answer. Thanks in advance!
  2. jcsd
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