This question is inspired by the page at NED and references it throughout the question.

On the linked page there is a method for calculating the location of the first acoustic peak of the CMB. It involves calculating the distance to the surface of last scattering.

The expression they have for calculating this is

$$d_{sls} = \frac{c}{H_0 \sqrt{\Omega_m} (1+z_{sls})}\int_{0}^{z_{sls}} [\Omega_m(1+z)^3+\Omega_\Lambda]^{-1/2} dz$$

Where $\Omega_m = .3$ is the current mass density, $\Omega_\Delta = .7$ is the current dark energy density, and $z_{sls} = 1100$ is the redshift at the surface of last scattering.

They then calculate the integral by estimating it as a binomial expansion. As a result, they calculate (and I verified) that

$$\int_{0}^{z_{sls}} [\Omega_m(1+z)^3+\Omega_\Lambda]^{-1/2} dz \approx 1.21$$

And as a result, they calculate $l \approx 200$.

However, if I plug this integral into Wolfram Alpha, I get

$$\int_{0}^{z_{sls}} [\Omega_m(1+z)^3+\Omega_\Lambda]^{-1/2} dz \approx 3.19$$

And as a result, I calculate $l \approx 500$.

Given that $l \approx 200$ has been measured by the Planck satellite, this is a huge discrepancy! I have two questions:

  1. Is there something about Wolfram Alpha that would given a bad approximation to this integral? Is it possible that the binomial approximation method gives a more accurate result?
  2. If not, then why is such a bad approximation used in the calculation and accepted as accurate?

Apologies if this is not clear. At the end of the day, I'm trying to understand this calculation better and anything that helps me do so would be appreciated.

Thank you!


1 Answer 1


The webpage you linked to is remarkable, to say the least. It contains a whole chain of errors, producing a wrong result that the authors mistakenly interpret as accurate. Based on what I've seen on that page alone, I wouldn't trust anything else from that article.

So, first of all, Wolfram Alpha is correct, and I applaud you for checking their calculations. I should point out that $$ d_{sls} = \frac{c}{H_0\,(1+z_L)}\int_0^{z_L}\left[\Omega_{m,0}(1+z)^3 + \Omega_{\Lambda,0}\right]^{-1/2}\,\text{d}z, $$ so in your question you've put an extra $\sqrt{\Omega_{m,0}}$ in the denominator, which shouldn't be there.

Let's see where things go wrong:

The most glaring mistake is the step from (55) to (56): $$ \begin{align} d_{sls} &= \frac{2c}{7H_0(1+z_L)}\left[7\,\Omega_{m,0}^{-1/2} - 2\,\Omega^{\vphantom{1/2}}_{\Lambda,0}\,\Omega_{m,0}^{-3/2} + \mathcal{O}\left((1+z_L)^{-1/2}\right)\right]\\ &\approx \frac{2c}{7H_0(1+z_L)}\left[9\,\Omega_{m,0}^{-1/2} -2\,\Omega_{m,0}^{-3/2}\right] = \frac{2c\,\Omega_{m,0}^{-1/2}}{7H_0(1+z_L)}\left[9 -2\,\Omega_{m,0}^{-1}\right]. \end{align} $$ So there's a $\Omega_{m,0}^{-1}$ term instead of $\Omega_{m,0}^3$. As a result, $d_{sls}/r_s \approx 60$ instead of $221$.

The second mistake is that (55) itself is wrong. The binomial expansion of the integrand in $d_{sls}$ is only accurate for large values of $z$, but the integration is taken over the entire range $[0,z_L]$, and depends crucially on the value of the primitive function in $z=0$. This is why this approximation is so different from the correct numerical calculation with Wolfram Alpha. If you insert the correct value, you should get $d_{sls}/r_s \approx 149$ (and not $300$, because of the $\sqrt{\Omega_{m,0}}$ you added).

But the problems don't stop there. The calculation of the sound horizon $r_s$ is also wrong. Equation (47) should be $$ r_s = a_L\int_0^{t_L}\frac{c_s}{a(t)}\,\text{d}t. $$ Using (49) and the convention $a_0 = 1$, this can be written as $$ r_s = \frac{a_L}{H_0}\int_0^{a_L}\frac{c_s\,\text{d}a}{\sqrt{\Omega_{r,0} + \Omega_{m,0}\,a + \Omega_{k,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}. $$ The authors then make the assumption that the speed of sound $c_s$ remains roughly constant and that the matter term is dominant, such that all other terms can be ignored. This then leads to $$ r_s \approx \frac{c_s\,a_L}{H_0\sqrt{\Omega_{m,0}}}\int_0^{a_L}a^{-1/2}\,\text{d}a = \frac{2\,c_s}{H_0\sqrt{\Omega_{m,0}}}a_L^{3/2} = \frac{2\,c_s}{H_0\sqrt{\Omega_{m,0}}}(1+z_L)^{-3/2}, $$ so there is a factor $3$ wrong in (53). However, this is also a poor approximation, because the radiation term $\Omega_{r,0}$ is quite significant in the early universe. A better approximation is therefore $$ \begin{align} r_s &\approx \frac{c_s\,a_L}{H_0}\int_0^{a_L}\frac{\text{d}a}{\sqrt{\Omega_{r,0} + \Omega_{m,0}\,a}}\\ &= \frac{2\,c_s}{H_0\,\Omega_{m,0}}(1+z_L)^{-1}\left[\sqrt{\Omega_{r,0} + \Omega_{m,0}\,(1+z_L)^{-1}} - \sqrt{\Omega_{r,0}}\,\right]. \end{align} $$ Using the value $\Omega_{r,0}\approx 9\times 10^{-5}$, this leads to $d_{sls}/r_s \approx 86$. A full calculation with non-constant $c_s$, which I won't derive here, gives $d_{sls}/r_s \approx 96$.

Finally, a first-order approximation to the position of the first acoustic peak is not given by $d_{sls}/r_s$, but by $\pi d_{sls}/r_s$, and we find $\pi 96 \approx 300$. As you see, this is still not very close to the actual value of $\approx 220$. The reason for this is that the low-order acoustic peaks are significantly influenced by the propagation of the density fluctuations and by gravitational redshift effects in the early universe. The full treatment of these is very complex, and for specialists only (which I'm not, by the way).

I don't think I've ever seen a website that looks professional, yet is so wrong on so many levels. My advice would be to find a better course. For instance, Weinberg's cosmology textbook has everything you need, but it's no easy read.

  • $\begingroup$ Very interesting! I have the Weinberg textbook, so I'll try to go through it. It's just, as you say, no easy read. $\endgroup$ Jul 20, 2016 at 14:38
  • $\begingroup$ @PaulHlebowitsh Excellent. You'll find the relevant formulae on pages 143-145, and 355. $\endgroup$
    – Pulsar
    Jul 20, 2016 at 16:21

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