# How far was the surface of last scattering at the moment of recombination?

If I understand correctly, then the surface of last scattering (the last particles off of which CMB photons scattered before traveling for 13 billion years and reaching us) is currently at a distance of around 42 billion lightyears from us, just before the particle horizon.

Also, when we calculate the curvature of the universe by comparing standard rulers on the CMB to their measured angular sizes, which distance do we use as the "distance to object"? Do we use the distance described in this question, or the distance to the particle horizon, or some other distance?

• As we're made of the stuff that was (in terminology that I believe dates from a hypothetically repetitive combination of matter and antimatter) "recombined", all the material we contain would've been part of it, so, are you asking how far the surface of last scattering was from where we are now? (The "re" syllable in "recombination" is a historical artifact, which I believe traces back to Tolman's "bouncing" cosmology of the 1930's, that ran into a problem with an uncontrolled density of entropy: I guess it's preserved just in case Someone comes up with some kind of an escape valve for it.) Jan 11, 2022 at 4:59

The distance to the surface of last scattering was 0 at the moment of recombination.

The distance of last scattering (the raius of the surface of last scattering) at any time is the distance from which the recombination radiation (which today is the CMB) can have traveled since the time of recombination. Since that time was zero at the time of recombination, that distance must also have been zero.

EDIT:

If this question is to be interpreted as: "If we take the current SLS distance and shrink it along with the Universe as we go back in time, what does that distance become then?", then the calculation is pretty simple. If we assume a standard cosmology with $$H_0=70$$ km/s and $$\Omega_{m,0} = 0.3$$, and a redshift of 1100 for Recombination, then we get a currrent, comoving distance to the surface of last scattering of $$\approx 45$$ Gigalightyear, so we simply divide that number by 1100 to get $$\approx 41$$ million light years.

But that is not really a physically meaningful distance*.

*: Although maybe it is, see @benrg's comment below.

• I think you misunderstood my question. I don't mean how far had the CMB photons traveled at the moment of recombination; I mean how far from us was the surface of last scattering at that moment? The time taken for those photons to reach us was not zero, as you suggest, but rather (the age of the universe) – (time between Big Bang and recombination). May 8, 2019 at 22:31
• But that's what the surface of last scattering is - a sphere with radius determined by how far the CMB has traveled since emission. If you are asking how big the current surface of last scattering was when the CMB was emitted, just divide its current size by the redshift of the CMB, around 1100. But that's not really a physically meaningful quantity. May 9, 2019 at 2:55
• But isn't it meaningful in the sense described by my third paragraph? May 9, 2019 at 16:55
• @cumfy what exactly is it you think I did not understand in the question you just repeated without adding anything? Feb 19, 2020 at 8:37
• I disagree that it's "not really a physically meaningful distance". It's the radius of the intersection of our past light cone with the last-scattering surface. It tells you the physical size of the fluctuations in the CMB that were mapped by WMAP and its successors. To put it another way, it's the angular-size distance to the CMB. Sep 29, 2022 at 3:39

I hope this clarifies things.

If recombination were been an isolated event, i.e. something that happened in a certain place with some finite spatial extent, then it would make sense to ask how far it was from us when t=380kyrs.

However recombination was not an isolated event, it happened everywhere in the universe at the same time, light scattered for the last time in all directions. From our location in earth CMB photons were emitted as well (that's why our distance to the surface of last scattering was cero at that time, because we emitted CMB radiation) but those photons are now very far away from us.

The photons that we do see are those that were far enough at the moment of last scattering so that their trajectories are meeting us today. That far enough distance is the radius of the surface of last scattering, because in every direction, at that radius, we see this last scattered photons.

If another planet looks far enough they will also see their own last scattering surface, with the same radius as ours. They will also observe the CMB because it happened everywhere.