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The baryon asymmetry of the Universe is quantified in terms of $$\eta=\frac{n_B-n_{\bar B}}{n_\gamma}|_0= (6.21\pm 0.16)\times 10^{-10}\tag{1}$$ or $$Y_{\Delta B}=\frac{n_B-n_{\bar B}}{s}|_0=(8.75\pm 0.23)\times 10^{-11}\tag{2}$$ $s$ represents the entropy density and it is related to the number density of photon as $s=7.04\times n_\gamma$. These two relations (1) and (2), in these form, do not seem to clearly express the fact that there are an overwhelmingly large number of baryons over antibaryons. From these relations how can we get an estimate of the number of baryons corresponding to one antibaryon per comoving volume?

I ask "What is the number excess of baryons per antibaryons"? I tried to argue as follows. As I understand, the Eq.(1) says that there are $\sim 10^{-10}\times n_\gamma$ number of baryons corresponding to one antibaryon. However, this turns out to be a ridiculously small number. But I think $\sim 10^{-10}\times n_\gamma$ should be a very large number because there are almost no antibaryons per baryon. But it's not!

What is wrong in my interpretation?

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  • $\begingroup$ this summarizes the experimental evidence for the asymmetry cfpa.berkeley.edu/Classes/Class_Archive/Fall00/Physics_250/… . It is an experimental fact, whence are the formulas that claim to describe the asymmetry, to start with.?. CP violation at such levels is still a research theoretical topic. $\endgroup$
    – anna v
    Commented May 22, 2017 at 6:23
  • $\begingroup$ OK, I see that Bob has given the rational for the gamma part. What is the s in the second formula? Can you give a link for the formulas? $\endgroup$
    – anna v
    Commented May 22, 2017 at 6:33
  • $\begingroup$ Here is a review arxiv.org/pdf/1411.3398.pdf $\endgroup$
    – anna v
    Commented May 22, 2017 at 7:37
  • $\begingroup$ It does not address your question, it gives a background to your formula for physicists in general. You just stated the formulas and I needed further confirmation $\endgroup$
    – anna v
    Commented May 22, 2017 at 8:28
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    $\begingroup$ This paper seems to be closer to clear up your misunderstanding : it talks of unit cells defined to be constant through the expansion, web.stanford.edu/~savas/papers/BaryonNumber-of-the-Universe.pdf . With the usual definition of asymmetry , the formulas do not make sense to me. They imply that the number of baryons is practically the same as the number of antibaryons. $\endgroup$
    – anna v
    Commented May 22, 2017 at 10:17

3 Answers 3

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The universe is in approximate local thermal equilibrium. Today, the temperature of the universe is much smaller than the rest mass energy of the proton (divided by $k_B$). This means that the number of anti-baryons is essentially zero, the asymmetry is almost entirely carried by baryons. There is a non-zero baryon chemical potential, which is very close to the rest mass of the proton.

The density of baryons is indeed $n_B\sim 10^{-10}n_\gamma$. The universe is almost empty.

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  • $\begingroup$ "The universe is in approximate local thermal equilibrium". I would have thought it was out of equilibrium, almost everywhere. $\endgroup$
    – Mitchell Porter
    Commented May 21, 2017 at 22:03
  • $\begingroup$ In our immediate neighborhood baryon number is obviously clumped, matter is not a 3K, etc. The numbers quoted in the question refer to averages on large (larger than a galaxy) scales. On that scale, we have approximate equilibrium. $\endgroup$
    – Thomas
    Commented May 21, 2017 at 22:19
  • $\begingroup$ Would this then refer to the intergalactic medium? Or do we need to think even bigger, an "inter-cluster medium", a population of primordial particles that aren't bound to galactic clusters? $\endgroup$
    – Mitchell Porter
    Commented May 21, 2017 at 23:53
  • $\begingroup$ @Thomas My question is "How can I read off the number excess of baryon over the number of antibaryons from Eq.(1) or (2)?" My question is not quite about the thermal history but about the interpretation of the equation (1) and (2) in the present epoch. I've edited and attempted to clarify my question. Apologies if the question was unclear. $\endgroup$
    – SRS
    Commented May 22, 2017 at 6:27
  • $\begingroup$ The density is indeed $10^{-10}n_\gamma$. The universe is almost empty. The thermal history is relevant, because we determine the baryon number at earlier times (BBN and decoupling). $\endgroup$
    – Thomas
    Commented May 22, 2017 at 12:41
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SRS, look at it this way. The thought is that when baryons and anti baryons existed in about equal numbers, they were in equilibrium with photons, and the numbers of each were about the same. That's because in equilibrium you can think of baryon antibaryin collisions typically producing 1 photon and viceversa.

When the temperature of the universe was lowered by expansion, baryon anti baryons were still annihilating and producing photons, but the photons didn't have enough energy to produce baryon anti-baryon pairs. So what is left is the balance (mostly) of baryons over anti baryons.

So that equation in @Thomas's answer is that there's about $10^{10}$ photons for every baryon, and that has not changed much since photons became unable, on a cosmological scale, to produce baryon/anti-baryon pairs. That is more or less what it was then and what it is now.

Of course they still can in high enough energy regions in the universe,
but at the cosmological scales of megaparsecs and more, and even in most of intergalactic space, tHey can't. The average photons temperature now is that of the cosmic microwave background, or about 3 degrees Kelvin. They Are too weak to produce baryons anti-baryon pairs.

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  • $\begingroup$ "there's about $10^{10}$ photons for every baryon"-How is that supposed to represent a baryon asymmetry? Shouldn't baryon asymmetry be a measure of how many baryons are there compared to the number of antibaryons? The Eq.(1) in my question says that there are $\sim10^{-10}\times n_\gamma$ number of baryons corresponding to one antibaryon per comoving volume. That amounts to a very small number. Since both the numerator and denominator represent densities the ratio can also be thought of as comoving numbers (instead of number densities). Where am I making mistake in this estimate? @Bob bee $\endgroup$
    – SRS
    Commented May 22, 2017 at 5:51
  • $\begingroup$ Well, I thought I explained this in my answer. Let's do a simple example: at the time when the temperature was hot enough that photons, in a rough equilibrium at that temperature, were energetic enough that they could create baryon antibaryon (call them bab) pairs. Because baba pair could annihilate and create photons, there was an equilibrium with about the same number of photons and bab pairs. Let's say that was 1000000 (a million). Now let's say the b and ab asymmetry was one in that million, i.e., there were 1000000 pairs, plus one extra baryon. Continued $\endgroup$
    – Bob Bee
    Commented May 22, 2017 at 23:21
  • $\begingroup$ When the universe cooled and photons were unable to create pairs, all those pair annihilated into photons. Now there were 2000000 photons, plus 1 baryon. So the number of baryons to photons was 1 in 2000000. That's that's the same as 0.5 in 1000000, or approximately 1 in 1000000 (we can't measure the cosmological numbers that accurately, there's always other pairs and photons getting formed from other reactions, so that's an order of magnitude estimate. That is a ratio of $10^{-6}$ of leftover baryons to photons. That ratio is also approximately the ratio of extra baryons to anti baryons. Next $\endgroup$
    – Bob Bee
    Commented May 22, 2017 at 23:33
  • $\begingroup$ The other million annihilated. That is why the baryon asymmetry, then and now is the same number as the ratio of baryons to photons now. We measure it and it's not 1 in a million but one in ten billion, as we measure it now. That is about $10^-{10}$. The two ratios, baryons to photons, and extra baryons to the anti baryons in the early universe, is approx. the same. That's the kind of approximations you have to make in cosmology. Hope this is clearer. $\endgroup$
    – Bob Bee
    Commented May 23, 2017 at 1:28
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I tried to understand the discrepancy of language with numbers. This is the baryon asymmetry for a hypothetical universe.

$$\eta=\frac{n_B-n_{\bar B}}{n_\gamma}|_0= (6.21\pm 0.16)\times 10^{-10}\tag{1}$$

If the number of baryons is equal to the number of antibaryons it is zero,but the number of primordial antibaryons is found to be zero.

Since at the present there is no antimatter $(n_{\bar{B}}= 0)$, this ratio is actually $\eta=n_{B}/n_\gamma$.

So both numbers in the numerator are defined experimentally. The formula would give exact numbers in comparing asymmetry only by introducing the experimental values. The formula itself is derivable from a cosmological model, but to get at numbers one needs the observational value of the antibaryons, which at present is estimated as zero.

The smallness of the number reflects the small contribution of matter (masses) with respect to radiation/photons to the disorder of the universe.

Suppose that you have 1000 baryons left over from the initial soup, and 10 antibaryons, it will be a very small measured difference in this definition of asymmetry. The division just changes units appropriate to the cosmological model.

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