Are critical exponents below and above the critical point always same?

The scaling relations don't distinguish the the critical exponents below and above the critical value. In the mean field level, I understand these critical exponents are same whatever one approaches the phase transition from the order phase or disorder phase. However, beyond the mean field treatment, are they always same?

Are there examples where they are different below and above the critical value? Or are there some theoretical arguments that the must be same? Can anyone give me some hints or good reference?

• As far as I can tell, the critical exponents are the same for both side of the transition (in the exact calculation say), since the critical exponents label the universality class of the transition. Universality classes do not depend from where you approach the transition from my empirical experience... A good ref. on the subject is John Cardy's "Scaling and Renormalization in Stat. Phys." book. – VanillaSpinIce Oct 10 '14 at 15:23

Critical exponents are properties of the RG fixed point that drives the phase transition. They are computed by linearising the RG flow equations close to the fixed point. The exponents are the derivatives of the beta functions evaluated at the fixed point. They know nothing of the way you approach the fixed point. In particular if you are flowing slightly above or below the critical temperature.

An obvious exception to this is the scaling of the order parameter,

$$m \sim (-\tau)^{\beta} \, .$$

Above the critical temperature, $\tau > 0$, we have $m=0$ and there is no critical exponent.

Edit (15 sept 2015): I recently read this paper. They show, using Renormalisation Group (RG) methods, that critical exponents can be different above and below a phase transition if there is a dangerously irrelevant operator involved. An anisotropic term is added. This term breaks a continuous symmetry and forces the order parameter to take one of $n$ countable values. They consider $n=6$. The value of $n$ depends on the way the symmetry is broken and is not important.

Basically, below $T_c$, the RG flow becomes able to double past the previously attractive Infra-Red (IR) RG fixed point where it terminates in the symmetric case. See the figure in the paper. The closer you are to the phase transition, the closer you come to this fixed point and the larger the susceptibility gets. There is an additive contribution to the critical exponent of the susceptibility because of this. When $T>T_c$, the IR attractive fixed point is Gaussian and its attractiveness is not affected by the anisotropy. Then the critical exponent is not affected.

For what it's worth, it is claimed that the critical exponents differ above and below the critical point for some exactly solvable 2-dimensional model: http://www.ujp.bitp.kiev.ua/files/journals/49/11/491114p.pdf (Ukr. J. Phys., v.49, #11, p.1122 (2004)).

• That looks very interesting. Can you explain it? – Steven Mathey Oct 13 '14 at 14:08
• @Steven Mathey: I worked in the area of exactly solvable problems of statistical physics very long time ago, so I am not sure I can offer an explanation. The authors of the referenced article offer some comments. See also their article arxiv.org/abs/1012.0607v1 , where "The reasons for the violation of the scaling law hypothesis and the universality hypothesis for the models are clarified." – akhmeteli Oct 14 '14 at 10:56
• Yes, I can read the abstract too. I don't have time to get into these papers now. Do you know what their point is? Where does this violation come from? – Steven Mathey Oct 14 '14 at 14:53