The sphaleron interactions in the Standard Model (SM) is $(B-L)$ conserving and $(B+L)$ violating. Each sphaleron transition causes the change in the baryon number $\Delta B$ and the lepton number $\Delta L$ be such that $\Delta(B-L)=0$ but $\Delta(B+L)\neq 0$. These facts can be easily verified from the expressions of baryon and lepton current anomalies.

From this, however, it is not clear to me why is it said that sphaleron interactions will erase any initial baryon asymmetry? Such a statement can be found in the introduction of this paper "Baryogenesis without grand unification by Yanagida and Fukugita". I would like to understand this statement mathematically.

  • $\begingroup$ I have probably misunderstood your question, but if initially you had an asymmetry $B \ne 0$, you can end-up with $B = 0$ provided that $\Delta L=\Delta B$. So no way to track the initial asymmetry, no? $\endgroup$
    – Paganini
    Jan 17, 2015 at 9:58

1 Answer 1


It would help if you would cite the corresponding sentence, argument or explanation of the book. Nevertheless, I will try to answer your question.

I think this paper by Y. Burnier, M. Laine and M. Shaposhnikov might be of interest to you.

The change in baryon number at time $t$ is proportional to the net baryon number at $t$:

$\dot{B}(t)\sim -B(t)+ \eta(t) \sum_{i=1}^{n_G} L_i(t) \ ,$

where $n_G$ is the number of generations and $L_i$ the number of leptons in the $i$-th generation. So the more baryons (and leptons) you have, the faster the changes in those numbers (there is a similar expression for $\dot{L}(t)$).

Furthermore, the sphaleron washout is very rapid at high temperatures. The rate is $\mathcal{A}\sim e^{-E_{sph}/T}$, where the sphaleron energy $E_{sph}=\frac{8\pi v}{9}=\frac{2M_W}{\alpha_W} f\left(\frac{\lambda}{g^2}\right)$ with $f(0)=1.56$, $\ldots$, $f(\infty)=2.72$. (I found this in some notes of mine where I did not cite the source, which I cannot find right now. However, this paper has a similar result for a $SU(2)$+Higgs theory.)

  • $\begingroup$ Something like this is stated in page 45 of the Physics Letters B I linked. @Clever $\endgroup$
    – SRS
    Jul 1, 2017 at 13:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.