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Eddington Finkelstein Coordinates + Black Holes

  1. Jun 4, 2006 #1
    When deriving the Eddinton-Finkelstein Coordinates from the Schwarzschild metric, we start to examine light rays. However, in my relativity book, it states that ds^2=0: why do we assume that? :bugeye:
  2. jcsd
  3. Jun 4, 2006 #2


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    Hi, and welcome to these Forums discjockey!

    Not knowing exactly what text you are using makes a response a little risky, however the most obvious answer is that light rays travel on null-geodesics with ds2 = 0.

    In flat space-time:
    ds2 = dx2 + dy2 + dx2 - c2dt2.

    As a light ray travels at the speed of light c then
    dx2 + dy2 + dx2 = c2dt2.

    Hence ds2 = 0.

    The result also holds in curved space-time, it is a basic property of light to travel always at c as measured by local rulers and clocks.

    Last edited: Jun 4, 2006
  4. Jun 4, 2006 #3
    Thanks a lot!
  5. Aug 4, 2010 #4
    My question about Eddington-Finkelstein coordinates:
    if we use ingoing null coordinate (v) and the radius (r) for basis
    then we have a null-coordinate, and a space-coordinate above the horizon,
    and a null-coordinate, and a time-coordinate below the horizon.
    Usually we use time ad space coordinates together,
    and we evolve a system in the direction increasing time.
    In the case of EF-coordinates we can evolve the system
    using the null-coordinate (v)?
    But under the horizon r is a time coordinate,
    we can evolve the system in r also?
  6. Aug 4, 2010 #5

    George Jones

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  7. Aug 28, 2010 #6
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