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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Transformation of solid angle

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