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While reading about the discovery of quasars and the spectroscopic analysis of 3C 273 in this paper by Maarten Schmidt, I came across the following quote:

At these distances, corrections to luminosities depend on the specific world model. For an approximate determination of the fluxes in the emission lines we use the formulae given by Sandage (1961) based on Mattig (1958). For the energy contained within an emission line, in which we integrate over the entire profile, the correction to the inverse square law is of the form $(1 + z/2)^2$ if the cosmological constant and deceleration parameter are zero. It is $(1 + z)^2$ for the steady-state model.

First, why is a correction to the inverse square law necessary? Is it to do with interstellar extinction?

Second, why are the two models different? I understand what the cosmological constant and deceleration parameter are, but why is the redshift prediction for a steady state model twice as much as when $\Lambda=q=0$?

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  • $\begingroup$ Isn't the correction needed to account for expansion? $\endgroup$ – Alchimista Feb 1 '19 at 9:07
  • $\begingroup$ Ok, to account for expansion. But why? And why is it half of the steady-state? $\endgroup$ – Jim421616 Feb 20 '19 at 21:54
  • $\begingroup$ Expansion must be accounted as for energy flux is considered. So we have the geometrical inverse square and reddening of photons. For why this is more relevant in steady state I can only think of some time dilation due to density. When photons leaves from the source in a steady state world they travel in a denser universe as compared to an expanding one. But it's a guess. I have just asked recently if gravitational shift is included in cosmological shift or is negligible or cancel out at emission & reception places but I did not get an answer. One should really look for the works mentioned. $\endgroup$ – Alchimista Feb 21 '19 at 8:34
  • $\begingroup$ Just related but some links might be useful physics.stackexchange.com/questions/184115/… $\endgroup$ – Alchimista Feb 21 '19 at 8:36

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