# If a CMB photon traveled for 13.7 billion years to reach me, how far away was the source of that CMB photon when it first emitted it?

If a CMB photon traveled for 13.7 billion years (- 374,000 years) to reach me. How far away was the source of that CMB photon when it first emitted it?

My attempt to solve this question was to use the following assumptions:

1. Temperature of CMB photon today is 2.725 K (will use value of 3 K here)

2. Temperature of CMB photon when it was first emitted is 3000 K

3. A factor of x1000 in temperature decrease results in a factor of x1000 in wavelength increase. (According to Wien's displacement law)

Does this mean that the source of the CMB photon that just reached me today, was actually 13.7 billion light years / 1000 = 13.7 million light years away from me when it first emitted the photon?

• Distance between you and the source of photon at the present day, or distance between you and the source of the photon at the time of emission? Mar 15, 2021 at 1:47
• @joshua lin distance between me and the photon at time of emission is the question. Mar 15, 2021 at 2:00
• Here is a video from FermiLab - If the universe is only 14 billion years old, how can it be 92 billion light years wide? Mar 15, 2021 at 4:01

The comoving distance traveled by light in vacuum between cosmological times $$t_i$$ and $$t_f$$ is $$\displaystyle \int_{t_i}^{t_f} \frac{c\,dt}{a(t)}$$. The metric distance at cosmological time $$t$$ is $$a(t)$$ times the comoving distance. The distance you're looking for is therefore $$\displaystyle \int_{t_i}^{t_f} \frac{a(t_i)}{a(t)} c\,dt$$.
This value depends on $$a(t)$$ over the whole time interval from $$t_i$$ to $$t_f$$, not just at the endpoints. The redshift factor is equal to $$\displaystyle\frac{a(t_f)}{a(t_i)}$$, but you can't just divide $$cΔt$$ by that factor to get the distance.
You can, however, divide the usually quoted comoving distance to the CMBR (around $$46\text{ Gly}$$ or $$14\text{ Gpc}$$) by the CMBR redshift (around $$1100$$) to get the correct distance (around $$42\text{ Mly}$$ or $$13\text{ Mpc}$$). This is because the $$46\text{ Gly}$$ distance is calculated using that first integral, without multiplying by $$a(t_i)$$, and $$a(t_f)=a(t_0)=1$$ by convention.