The reason for not assuming the Baryon asymmetry of the universe as an initial condition is often attributed to the following fine-tuning problem: If the baryon asymmetry had been an initial condition, then, for every 6,000,000 antiquarks, there should have been 6,000,001 quarks.

How are these numbers estimated? Is there a calculation which will dictate these numbers?

Note: I know there are other arguments for it and I understand them. But I'm particularly interested in understanding these numbers as mentioned in this review (second paragraph, page 4).


2 Answers 2


The calculation is actually reasonably simple. Most of the photons in the Universe are CMB photons, and the number density of CMB photons has been measured to exquisite accuracy. There are about $N_\gamma = 4.11\times10^8\,{\rm m}^{-3}$. The mass density of baryons has also been measured, both "directly" via elemental abundances and the Big Bang nucleosynthesis model, and "indirectly" from CMB analysis. It is $\Omega_{\rm b}\approx0.02$, where $\Omega_{\rm b}$ is the density in units of the critical density for closure $\rho_{\rm crit}=137.6\,{\rm M}_\odot\,{\rm kpc}^3$. I'll spare you all the unit conversion; under sensible assumptions for the composition of the baryons in the Universe, you get an average number density of about $N_{\rm bar} = 0.11\,{\rm m}^{-3}$. Taking the ratio of the two number densities, you find that for every baryon, there are $3.7\times10^9$ photons. $q\bar{q}$ annihilation is complicated, but from what I surmise the result is $\mathcal{O}(1)$ photon per quark, so initially there would have been about $3.7\times10^{9}+1$ baryons worth of quarks for every $3.7\times10^{9}$ anti-baryons worth of quarks (noting that it takes both quarks and anti-quarks to make a typical baryon, or anti-baryon).

I think the numbers 6,000,000 and 6,000,001 in the article you linked must be typos (should rather be ~6,000,000,000), as I can't think of anything that would account for 3 orders of magnitude difference between the (current) ratio $N_{\rm b}/N_\gamma$ and the (initial) ratio $(N_{\rm b} - N_{\bar{\rm b}})/N_{\rm b}$. If you look around the literature, both popular and technical, on the topic, the figure '1 part per billion', rather than ~1 part per million, is the one that gets quoted.

  • 2
    $\begingroup$ I just bounced this off one of my particle physics friends and it's reinforced my conviction that the article has a typo/error. $\endgroup$
    – Kyle Oman
    Commented Jan 12, 2017 at 21:38

To start, it is important to notice that a statement like "For each x antiquarks, there are x+1 quarks" is vacuous unless you specify the instant at which you are looking at the system. To give an example, if you start with 100 antiquarks for every 101 quark, there will be a time when you'll have 50 for every 51, and finally 0 for every 1.

This said, I'd recommend to disregard this unfortunate statement in an otherwise awesome review and turn to Kolb and Turner's The Early Universe, where a similar comment is made (near equation 6.2, page 159):

Although the baryon asymmetry is maximal today, i.e., no antimatter, at high temperatures ($T \gtrsim 1\,\textrm{GeV}$) thermal quark-antiquark pairs were present in great numbers ($n_q \sim n_{\bar q} \sim n_\gamma$), so that the baryon asymmetry observed today corresponds to a tiny quark-antiquark asymmetry at early times ($t \lesssim 10^{-6}\,\textrm{s}$): $$\frac{n_q-n_{\bar q}}{n_q} \simeq 3 \times 10^{-8}\,.$$ That is, for every 30 million antiquarks, there were 30 million and 1 quarks present! A very tiny asymmetry indeed.

Unlike Davidson, Nardi and Nir, these authors refer to a specific time, namely around and before $t \sim 10^{-6}$ seconds, corresponding to temperatures about and above $1\,\textrm{GeV}$.

Lets take a stab at computing their quoted value. At such time in the history of the Universe, particles such as the light quarks move relativistically, and in this limit $n_q \simeq n_{\bar q} \simeq \frac{3}{4} n_\gamma$ (see eq. 3.52 of the same book).

Additionally, the density of photons $n_\gamma$ and the entropy density $s$ are related at all times by (chapter 3.4): $$s = \frac{\pi^4\, g_{*s}}{45\, \zeta(3)} \,n_\gamma\simeq 1.8\, g_{*s}\,n_\gamma\,.$$

Here $g_{*s}$ is a function of time. For temperatures around $1\,\textrm{GeV}$ one has $g_{*s} \simeq 62$ according to this detailed StackExchange answer. Hence,

$$ \frac{n_q - n_{\bar q}}{n_q} \simeq \frac{3 n_B - 3 n_{\bar B}}{\frac{3}{4} n_\gamma} \simeq 4 \frac{n_B - n_{\bar B}}{n_\gamma} \simeq 4 \frac{s}{n_\gamma} \frac{n_B - n_{\bar B}}{s} \simeq 4 \, (1.8 \cdot 62) \, Y_{\Delta B}\,,$$ where $Y_{\Delta B} \equiv (n_B - n_{\bar B})/{s}$ is measured to be $\simeq 9 \times 10^{-11}$ today according to (1.2) of Davidson, Nardi and Nir. The nice thing is that this quantity is unchanged from the moment baryon violating interactions become inactive (way before $t \sim 10^{-6}\,\textrm{s}$ in a log scale), and assuming no entropy production.

So we can plug it in and get: $$ \frac{n_q - n_{\bar q}}{n_q} \simeq 4 \cdot 1.8 \cdot 62 \cdot 9 \times 10^{-11} \simeq 4 \times 10^{-8} \,,$$ which is in good enough agreement with what is quoted in the book :)

  • $\begingroup$ "if you start with 100 antiquarks for every 101 quark, there will be a time when you'll have 50 for every 51, and finally 0 for every 1" I don't follow this. Are you talking about the number in the fixed volume? Because once the production/annihilation reactions freeze out, the number of particles is constant, and the number density of photons and baryons (quarks) scales in the same way. $\endgroup$
    – Kyle Oman
    Commented Jan 13, 2017 at 3:56
  • $\begingroup$ @Kyle Sorry, I was just giving a toy example (particles and antiparticles in, if you wish, a box with annihilations always happening and from which photons are removed) to illustrate the need to specify a time/temperature. In fact, as you point out, interactions freeze out in an expanding Universe :). Kolb and Turner have a nice discussion on this in their chapter 5.2 "Freeze Out: Origin of Species", namely pp. 127-128. $\endgroup$
    – J-T
    Commented Jan 13, 2017 at 12:10

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