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Free fall and blue shift question.

  1. Apr 19, 2012 #1
    Lets say I am in free-fall towards some massive body and I have 2 mirrors and photons bouncing in between them. Now as I fall closer to the massive body the light should get blue-shifted right? And lets say that the frequency of light that I started with is x. As I fall closer I put a jar of neon atoms in between the mirrors. And lets say that is takes x+d frequency to excite these neon atoms. By now the light has been shifted to x+d. But in my frame wouldn't I still see frequency x. Or does the energy levels of the atom change as we fall down?
     
  2. jcsd
  3. Apr 19, 2012 #2

    Matterwave

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    In your freely falling frame, you would not see any blue or red shift of the light. This is easily derivable from the strong equivalence principle.

    The blue/red shifting of light occurs for observers stationary with respect to the gravitating body.
     
  4. Apr 19, 2012 #3
    ok so what if someone is watching me fall. But I guess when the light reaches them it will get shifted back.
     
  5. Apr 19, 2012 #4

    mfb

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    If someone is watching you fall, he will observe different frequencies for the upwards and the downwards photons, as the mirrors are moving and therefore inducing a doppler effect. At the same time, the neon atoms would move, too. If you calculate these numbers, you get the same physics (e.g. the frequency, observed by the neon atoms) in both frames.
     
  6. Apr 22, 2012 #5
    Not necessarily. In Schwarzschild coordinates, an observer that is free falling from infinity sees light from a source at infinity as red shifted by a factor of 1/(1+√(GM/r)) while an observer that is stationary will see the light as blue shifted by a factor of √(1-GM/r) using units of c=1. (That is without the mirrors.)

    Now if we have two observers free falling radially together and one is continuously measuring the light from the source at infinity and the other is bouncing light from the same source between mirrors, then I assume they would see the same redshift when the light reaches the lower mirror. If my assumption is correct then the free falling observer would see an increasing red shift of his photon over time, each time it reaches the lower mirror.
     
  7. Apr 22, 2012 #6
    It kind of hinges on whether we consider the distance between the two mirrors as significant. If we don't there is no shifting whatsoever, if we do it is more complicated.
     
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