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Proper distance integral limits seem wrong

  1. Dec 24, 2015 #1
    I've seen in some lecture notes that the proper distance dp(t) can be written as
    ##\int_{t_e}^{t_0} c dt/a = \int_0^z c dz /H(z)##

    I can perform this integral ok using
    ##H =\dot a/a## and the fact that ##1 + z = 1/a(t_e)## but it requires associating the limits of the integration as te transforming to z and t0 to z= 0 - there is a minus sign which creeps in when you find dz/dt because da = - a2dz so the limits have to be switched. Thus they don't match as it appears when you read the integral.

    I don't see how to interpret this because I feel it should be the other way round. That is t 0 should be associated with red shift z (that's what we measure today) and at time t = te the red shift should be 0 . Why am I wrong here?
     
  2. jcsd
  3. Dec 24, 2015 #2

    Chalnoth

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    If the photon is emitted at t_0, then its observed redshift will be zero. If it's emitted at t_e, then its observed redshift will be z.
     
  4. Dec 25, 2015 #3
    Thank you Chalnoth that makes so much sense. I just couldn't get it right in my mind but this has settled the matter.

    Kindest regards
     
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