Why is it said that even if the baryon asymmetry existed as an initial condition, the asymmetry would have been destroyed by inflation. How does inflation get rid of initial Baryon asymmetry (if any)?

EDIT: For reference see the section 1.1 (last line of second paragraph) of this review.

  • $\begingroup$ there are no baryons ore even quarks at the time of inflation, just the inflaton field. As no baryons, no asymmetry. $\endgroup$
    – anna v
    Commented Mar 19, 2017 at 14:37
  • 3
    $\begingroup$ Actually, your claim is too strong. Assume that there is an initial baryon number density $n_B$, which is not interacting during inflation. Then, by the continuity equation $\dot{n}_B + 3 H n_B = 0$ with $H$ almost constant. This implies that any density decays exponentially in time during inflation. However, the issue is not so trivial since one should also care about the entropy density generated during inflation/reheating and evaluate the ratio between $n_B$ and this quantity. $\endgroup$
    – user47224
    Commented Mar 19, 2017 at 15:46
  • $\begingroup$ @user47224 with a little expansion (haha, get it?) that should be an answer $\endgroup$ Commented Mar 19, 2017 at 18:03
  • $\begingroup$ I found this journals.aps.org/prd/abstract/10.1103/PhysRevD.89.063523 $\endgroup$
    – anna v
    Commented Mar 19, 2017 at 18:42

2 Answers 2


The comment on the review you indicated can be explained in the following way. Inflation is a period of nearly exponential expansion of the Universe that is required to solve the flatness problem, the horizon problem, ecc.. These cosmological constraints, quite independently on the specific model of inflation (single field inflaton, modified gravity and so on), require for the Hubble function to remain almost constant for at least $N=50$ e-folds. In other words the scale factor at the end of inflation $a_f$ compared to the one at the beginning $a_i$ must be such that $$ \frac{a_f}{a_i} \simeq e^{50} $$

Now, lest define $\Delta = n_B - n_{\bar B}$ as the difference between the baryon and anti-baryon number densities. For our purposes we also assume for $\Delta$ to be different from zero at the start of inflation. Additionally, we assume that no baryon-violating processes happens during inflation. From this it follows that the total number of baryons and anti-baryons are conserved separately. Since the universe expands, then the number densities obey the differential equation: $$ \dot n_b + 3 H n_B = 0 $$ with $H\sim const$. The same equation holds also for $\Delta$. The solution is, approximately: $$ \Delta(t_f) = \Delta(t_i) e^{-3 H (t_f - t_i)} = \Delta(t_i) e^{-3 N} \approx \Delta(t_i) e^{-3 \times 50} $$

Therefore, in this picture of non-interacting baryons their asymmetry would be completely washed out at the end of inflation.

  • $\begingroup$ This answers my question. $\endgroup$
    – SRS
    Commented Mar 20, 2017 at 3:54

There can be a relation with inflation and leptogeneis which also can lead to asymmetric baryogenesis. The inflation will indeed smooth out everything, but there were no baryons then, as @Anna correctly states. Could not be, temperature was too high for baryons, though if there had been they would have been smoothed out by many orders of magnitude.

It's a problem and not really likely. Besides the temperature not being right and there being no baryons then (a sort of not too small problem), there is an extreme fine tuning that would have been required. The smoothing was 50 or more e foldings, whereas the observed asymmetry now is about 1 part of about $10^{7}$ would have required some real fine tuning before inflation, Of many orders of magnitude. That mechanism is deemed to have been not likely to have been the cause of the observed remaining asymmetry.

The Phys Rev review said that if there had been baryon asymmetry at the Big Bang (i.e., primordial) it would have been diluted. There were none till reheating, but the statement is not inaccurate, just misleading.

Inflation does however enter in, and there are models of baryogenesis and leptogenesis that depend on the model for the inflation field, the Inflaton. Thus, depending on the inflation field model (and there are many) there are papers that show an asymmetry from the inflation field that then leads to asymmetric leptogenesis including parity breaking neutrinos) and baryogenesis, all after inflation and during or after reheating. Some of the models show how the baryon asymmetry then comes about. But really nothing with it being diluted by inflation.

See also various other papers like https://arxiv.org/abs/hep-ph/0103229, or Google leptogenesis and inflation.

  • $\begingroup$ Even if there are no baryons, there can still be baryon asymmetry- it just means having more quarks than antiquarks. $\endgroup$
    – Chris
    Commented May 2, 2018 at 19:29
  • $\begingroup$ Yes, and after it cools it'll mean more baryons than antibaryons $\endgroup$
    – Bob Bee
    Commented May 25, 2018 at 0:04
  • $\begingroup$ You seem to imply there can't be any primordial baryon asymmetry, though. The fact that there were no baryons in the early universe is irrelevant to whether there was baryon asymmetry or not. $\endgroup$
    – Chris
    Commented May 25, 2018 at 0:11

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