In the Wikipedia, it says that, when calculating the Baryon Acoustic Oscillations (sound horizon), we measure $150\text{ Mpc}$, saying that the sound horizon is the "Physical Length of sound horizon at today". Then it says that, when correcting for expansion, the physical length must be calculated by:

x(physical length of sound horizon) = a(scale factor) * X(comoving distance of sound horizon at drag epoch) 150Mpc(today)=1*150

(I just copied Wikipedia's text, that's why I don't format it). I have two doubts. The first one is, which os those distances is the one you measure, the $150\text{ Mpc}$?

The second one, why is the comoving distance of sound horizon set to $1$? I see it multiplies three terms. Is that right? So if I were to calculate the angular distance in the sky those oscillations have, I should do $\theta = \frac{150}{d_A}$, being $d_A$ the angular distance of the redshift at which we see it? $d_A=r/(1+z)$, being $r$ the distance.

Am I right there?

EDIT: In view of the poor quality of that wikipedia section (mentioned by @ChrisWhite), I will rephrase the question independently of that section.

We know the BAO oscillations for the sound horizon happen at about $150\text{ Mpc}$. Given that, if I want to calculate the angle in the sky that covers, and if those oscillations are known to be given at a redshift $z$, then, which would be the proper way to calculate that angle?

I know it must be:

$$\theta \approx = \frac{d}{d_A}$$

Where $d_A$ is the angular distance, that can be calculated via $d_A=\frac{1}{1+z}\int_0^z \frac{dz'}{H(z')}$, begin $H(z')$ the inverse Hubble radio at that given redshift. My doubt is whether I should use $d=150Mpc$, or if I should multiply that with the scale factor $a$ to transform it into a physical distance.

  • 3
    $\begingroup$ Wow that's an awful wikipedia article. Just look at the formatting, the grammar, the unprofessional tone, not to mention the lack of capitalization of sentences. Probably better to delete the whole section than try to make sense of it. $\endgroup$
    – user10851
    Commented Jun 15, 2015 at 22:04
  • $\begingroup$ @ChrisWhite I know, I was really surprised. I rephrased the question to make it independent of that wikipedia page. $\endgroup$ Commented Jun 16, 2015 at 7:34

1 Answer 1


The BAOs became imprinted in the background density of the Universe at the time of decoupling, which happened at the same time as recombination, i.e. when the Universe was a factor $z \sim 1100$ smaller. At this time, they had a characteristing wavelength given by the speed of sound in the plasma at that time. This happened to be $\lambda_{\mathrm{BAO},\,z=1100} \sim 0.14 \,\mathrm{Mpc}$ in physical length, i.e. what you would measure if you were there. At this point, the waves got "frozen" in the gas so that it was compressed and rarefied on these scales, but as the Universe expanded, these compressions grew proportionally.

Now when we, today, measure the BAOs, we observe them at various redshifts and with correspondingly different wavelengths. But multiplying the physical length by $(1+z)$, where $z$ is the redshift at which we observe them, we can convert to comoving lengths so that the expansion is factored out, yielding the same result at all $z$, namely ~150 Mpc. Because by definition comoving and physical coordinates coincide today, this is the wavelength that the BAOs have today.

The angular diameter distance $d_\mathrm{A}$ is defined as the distance that relates the physical, not comoving, extend $D$ of an object and the angle $\theta$ covered on the sky by that object: $\theta = D/d_\mathrm{A}$.

For instance, the angle covered by a BAO at a redshift of, say, $z=2$ can be calculated in Python as:

from numpy import pi
from astropy.cosmology import Planck15        # Use a Planck 2016 cosmology
from astropy import units as u
z     = 2
dA    = Planck15.angular_diameter_distance(z) # Angular diameter distance
d_com = 150 * u.Mpc                           # Comoving size of the BAOs
d_phy = d_com / (1+z)                         # Physical size of BAO at z
theta = d_phy / dA * u.rad                    # Angle covered by BAO in radians
print('BAO size in degrees at z = '+str(z)+': ', theta.to(u.deg))

giving $1.6^\circ$ (at $z=2$).

  • $\begingroup$ The velocity of sound in the plasma approximated by $c_s=\frac{c}{\sqrt{3}}$. If recombination occurred at 372,000 years, you get $r_s=c_s t=\frac{c}{\sqrt{3}}\times 1.17\times 10^{13}\space s$, or $r_s=0.066 Mpc$. How did you get $0.14 Mpc$? $\endgroup$
    – user32023
    Commented Dec 28, 2018 at 1:28
  • $\begingroup$ @DonaldAirey Calculating the size by assuming a wave traveling at $c_s$ for 372,000 yr can give an order-of-magnitude size, but this neglects the expansion of the Universe. I didn't calculate the size, though, I just took the observationally inferred value of $\sim100/h$ comoving Mpc. $\endgroup$
    – pela
    Commented Jan 1, 2019 at 11:33
  • $\begingroup$ I answered my own question above. The distance sound moves up to 372,000 yr is 0.07 Mpc. But that is just $\frac{1}{2}$ of the wavelength. The full wavelength is peak-to-trough-to-peak which would be $0.07\times 2 = 0.14\space Mpc$. $\endgroup$
    – user32023
    Commented Jan 4, 2019 at 13:47
  • $\begingroup$ Why are you using the angular diameter distance to the SLS instead of the comoving distance? We're not measuring the size of a galaxy, we're measuring the projections of the sound horizon onto a surface. Seems to me like you're mixing apples and oranges here. $\endgroup$
    – user32023
    Commented Jan 4, 2019 at 15:06
  • 1
    $\begingroup$ @DonaldAirey Those 144 Mpc are in comoving coordinates. Physical coords are a factor (1+z) smaller, so physically the imprint in the CMB — the actual size of the waves at that time — are only 0.14 Mpc. These waves were then frozen in the sense that they stopped moving through space, but they did expand along with the Universe until today when they're a factor (1+z) larger, i.e. 144 Mpc in both physical and comoving coords. $\endgroup$
    – pela
    Commented Jan 24, 2019 at 20:52

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