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Transformation of solid angle

  1. Feb 2, 2015 #1
    When considering a small beam of null-geodesics in spacetime it is possible to define the solid angle spanned by two of the rays at the observer.

    At page 111 in "Gravitational Lenses" by P.Schneider et. al. they state with reference to Figure (b) that:

    "The dependence of this distance on the 4-velocity of the observer, given
    the events 8 and 0, is due to the phenomenon of aberration. In fact, one
    infers from [Figure (b)] that, if ##k^\alpha## is any vector tangent to the ray ##SO## at ##O## and ##U^\rho##, ##\tilde{U}^\rho## are two 4-velocities at ##O## with corresponding solid angles ##d\Omega## and ##d\tilde{\Omega}##, then
    $$ \frac{d \tilde{ \Omega} }{d \Omega} = \frac{(k_\rho U^\rho )^2}{(k_\rho \tilde{U}^\rho)^2}."$$

    No further explanations are given, so I guess the relation above must somehow be easily inferred from Figure (b), however I can not see how.

    How is the relation above inferred from the figure?
     
  2. jcsd
  3. Feb 7, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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