The observable universe is $4.4 \times 10^{26} m $ in every direction. I told this to my daughter and then added, "And guess what? It's a little bigger today than it was yesterday."


But by how much? My first thought would be: one light-day, in each direction. But that can't be right. The boundary is receding faster than the speed of light.

Yesterday, a lonely photon ended its 13.8-billion-year journey with the collapse of its wave function (or whatever, let's not start on this) on a rock orbiting a smallish star toward the edge of a spiral galaxy in a relatively sparse region of the Universe.

Today, yet another photon met a similar fate.

How much further away was the 2nd photon when it began its trek?


1 Answer 1


How fast does the observable Universe grow?

The observable Universe is $r = 46.3\,\mathrm{Glyr}$ (billion light-years) in radius, so by Hubble's law, galaxies at that distance recede from us at a speed $$ v_\mathrm{rec} = H_0\,r = 9.6\times10^5\,\mathrm{km}\,\mathrm{s}^{-1}, $$ or $3.2$ times the speed of light, $c$.

At the same time, light from ever-increasing distances will have had the time to reach us. Since light travels at the speed of light, this adds another factor to the edge of the observable Universe. That is, its distance recedes at $v_\mathrm{rec}+c = 4.2c$.

Hence, today the radius of the observable Universe is $$ \begin{array}{rcl} dr & = & (v_\mathrm{rec} + c)\,dt\\ & \simeq & 4.2\,\mathrm{light\text{-}days}\\ & \simeq & 10^{11}\,\mathrm{km}\\ & \simeq & 730\,\mathrm{AU} \end{array} $$ bigger than it was yesterday ($dt=1\,\mathrm{day}$).

But remember that most of this growth does not mean that our observable Universe encompasses more galaxies; only that they're farther apart.

How far apart were two photons emitted, if they arrive one day apart?

In reality, we don't see photons from the Big Bang. The oldest photons we can detect are from the cosmic microwave background, but humans don't see those. But for the sake of the argument, consider two photons, both emitted right after cosmic inflation, but at slightly different distances from the place you will once stand.

Taking inflation to end when the Universe was $t\sim10^{-32}\,\mathrm{s}$ old, and the Universe has expanded by a factor of $a^{-1} \simeq 2.3\times10^{26}$ since then. Hence, what we today call the observable Universe was, at the end of inflation, only $$ r_\mathrm{then} = ar\simeq10\,\mathrm{m}. $$

That is, the photon you saw yesterday, was emitted 13.8 Gyr ago, only 10 meters from you.

To find the distance at which the photon you see today was emitted, we can calculate the fractional distance between the two regions emitting the photons — but now we shouldn't use physical coordinates, but instead comoving coordinates, because the part of the extra distance that is due to expansion shouldn't be taken into account. In comoving coordinates, the observable Universe increases a distance of $r_\mathrm{com} =$ one light-day per day, or a fraction $dr_\mathrm{com}/r_\mathrm{com} = dr_\mathrm{com} / r = 6\times10^{-14}$.

In other words, the region emitting photon #2 is, and always has been, a fraction $6\times10^{-14}$ farther away than the region emitting photon #1. If region 1 was 10 meters away, then region 2 was $$ \begin{array}{rcl} \frac{dr_\mathrm{com}}{r_\mathrm{com}}\times 10\,\mathrm{m} & \simeq & 6\times10^{-11}\,\mathrm{cm}\\ & \simeq & 0.6\,\mathrm{picometers}. \end{array} $$

(I originally forgot to calculate in coming coordinates, resulting in a slightly too large value.)

A note on the calculations

In the above calculations, I've used cosmological parameter data from Planck Collaboration et al. (2016), because those are implemented in Python's astropy module. The newer data from Planck Collaboration et al. (2018) would change the result only a little bit. If instead of $10^{-32}\,\mathrm{s}$ you assume $10^{-33}\,\mathrm{s}$ for the end of inflation, as some models predict, you'd get slightly smaller Universe at the end of inflation (a factor $\sqrt{10^{-32}/10^{-33}}\simeq3$).

  • $\begingroup$ The factor of ~3x that I used in my previous blueshift comment came from here. In fact that number should have been either 3.2x or 4.2x :) $\endgroup$ Jul 26, 2020 at 1:49
  • $\begingroup$ @KeithKnauber Ah okay, now I see. But the redshift $z$ is only proportional to the velocity $v$ for small velocities, $v\lesssim0.1c$. For larger distances, the relation becomes a bit more complicated, depending on your model of the Universe, and for distances approaching the edge of the observable Universe (the "particle horizon"), $z\rightarrow\infty$. Galaxies with $z\sim3$ are routinely observed, and the current redshift record is $z=11.1$. $\endgroup$
    – pela
    Jul 27, 2020 at 7:10
  • $\begingroup$ Wikipedia has a plot of the relation here (though it's flipped to give velocity as a function of redshift). You see that the cosmological redshift ("General relativity"), which is what we're discussing here, has a rather large shaded area giving the uncertainty, because it depends on your model (though the errors bars are not this huge anymore). $\endgroup$
    – pela
    Jul 27, 2020 at 7:15

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