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This is a follow up to this question here about the $(1+z)$ factor in the Luminous Distance formula. As the universe expands, I understand why the energy falls as the wavelength is stretched (by a factor of $z+1$), but in the Luminous Distance formula $$L_D=\chi(1+z)$$ we have an additional an additional factor of $z+1$. I've read Dodelson and this article (page 201) and both seem to argue that the frequency of photons crossing the shell is different between emission and observation:

We must take into account the fact that the rate of photon reception is smaller than the rate of emission by a factor of $a = 1/(1+z)$.

Both authors wave their hands and say 'this is so', but I don't understand the process. The time interval, $\Delta t$, is the same from emission to observation (that is, expansion doesn't change the time interval). The number of photons crossing the entire shell doesn't change (you can't add or remove photons from the expanding shell). So how does the rate of photons crossing the shell (or hitting a detector) change?

ɪdɪət strəʊlə
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Quark Soup
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1 Answers1

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The additional $1+z$ factor is cosmological time dilation (e.g. see Zhang et al. 2013). The frequency with which photons are received is reduced by this factor.

That is, if a distant source emits $n$ photons per second in its rest frame, then those are received in the current epoch at a rate of $n/(1+z)$. The reason is precisely the same reason that individual photons have frequencies reduced by the same factor.

ProfRob
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