Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I How does flux density scale with redshift?

  1. Sep 3, 2017 #1
    I'm not sure I understand how to correctly scale flux density with redshift. That is, if I observe say 10 Jy at my observing frequency coming from a source at z = 0.3, how can I estimate the flux density I would expect from the same source at z=2? From what I understand, the final scaling is given by

    $$S \propto \frac{1}{(1+z)}$$,

    but I'm not sure I understand how that comes about. I believe there are two factors of 1/(1+z) due to the photon energy and time dilation, but I'm not sure what other factors to take into account, such that it reduces to only one factor of (1+z) in the denominator.

    Furthermore, it seems strange to me that in the example I give, a 10 Jy source at z=0.3 is only dimmed to a flux density of 4.3 Jy at z=2. Doesn't this suggest that one could in principle see this source to very extreme redshifts (pending the sensitivity of the instrument, of course). Or have I made a mistake in the scaling?
     
  2. jcsd
  3. Sep 3, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That is only the scaling from redshift, the increased distance has to be taken into account as well to get the luminosity distance.
    Otherwise all sources in our galaxy (including the Sun) would have the same brightness...
     
  4. Sep 4, 2017 #3
    Ah of course, makes sense, thank you!

    So the correct scaling should be:

    $$S \propto \frac{1}{(1+z)D_L^2}$$

    Is that correct?
     
  5. Sep 4, 2017 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The luminosity distance takes redshift into account already.
     
  6. Sep 4, 2017 #5
    Ok, I thought I might be double counting somewhere. So the flux is just inversely proportional to the square of the luminosity distance (makes sense!)? And the (1+z) factor is just from the luminosity distance.
     
  7. Sep 5, 2017 #6

    kimbyd

    User Avatar
    Science Advisor
    Gold Member

    Yes. The luminosity distance is defined as:

    $$D_L = (1+z)^2 D_A$$

    Here ##D_A## is known as the "angular diameter distance", which is the distance that we would measure from looking at the apparent size of the object (see here if you want the gory details, as distance in a curved space-time is a tricky concept). The factor of ##1+z## multiplies the distance because the redshift dims the light coming from the source. You might have expected the factor of ##1+z## to have a square root in front of it given your earlier posts, rather than squared. One way to understand this is to think about how the expansion impacts black body radiation. The temperature of black body radiation in an expanding universe is proportional to ##1/(1+z)##. This can be seen from looking just at the behavior of the peak of Planck's Law, which is directly proportional to temperature. But as the energy density of black body radiation scales as the fourth power of temperature, so does the flux. Thus the energy scaling of incoming radiation due to the expansion itself scales as ##1/(1+z)^4##. Hopefully that's clear.

    The flux is then:

    $$S = {L \over 4\pi D_L^2}$$

    Here ##L## is the luminosity.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted