In the context of the Landau Theory of phase transitions, applying the mean field theory in an attempt to describe transitions such as the Nematic - Isotropic, the Landau energy density is given by

$f = f_0 + \frac12Am^2 - \frac13Bm^3 + \frac14Cm^4 + hm$

where $m$ is the order parameter $A = a(T-T_c)$ and $B$ and $C$ are independent of the temperature. In this case we find that $m = 0$ minimizes $f$ for $T>T^*:=T_c+\frac{2B^2}{9aC}$ and $m=\frac{B}{2C}+\sqrt{(\frac{B^2}{2C})-\frac{a(T-T_c)}{C}}$ minimizes $f$ for $T<T^*$.

Calculating the entropy I find that $S=\frac12am^2$. I am not sure about this last result but if it is true there is a discontinuity in $T^*$ which implies that this is the temperature at which the phase transition takes place. We calculate all the critical exponents at this temperature, except of β.

Most texts on the subject that I found do not mention critical exponents in this phase transition, but my notes of the lectures I have written that we calculate $β$ at $T_c$ and that $β=\frac12$. I do not understand why this is so. Even if I calculate $β$ in $T_c$ I find it to be 1. I would really appreciate the details of this calculation.


I'm a little rusty on this, but I opened up the Chaikin & Lubensky textbook (Principles of Condensed Matter Physics) because I knew this topic was on section 4.5. I think in the first paragraph, the last thing you wrote was meant to be $T<T^*$, right? Well, from what I am reading in Chaikin you get the entropy as you calculated it but you have to do it for $T=T_c$.

For the exponent $\beta$ you have to analyze the behavior of the order parameter as you approach the critical temperature. From the result you are showing you already have $m \sim \sqrt{T-T_c}$ in the nematic phase, right? Though I think the $B^2/2C$ usually isn't there. How are you calculating the exponent $\beta$?

As you said, this textbook doesn't mention the critical exponents for this transition. Section 2 of this paper does have it and explains it further. Hope that helps.

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  • $\begingroup$ The $B^2/2C$ is most of my problem, because if it is there and if $T$ is close enough to $T_c$ or $T^*$, the order parameter $m$ changes linearly with $T-T_c( or T^*)$. Since I also found the same result for $m$ in my professor's solutions (but they still say that β=1/2) and the algebra involved in minimizing $f$ is very light I believe my derivation is correct. Also, why do the calculation near the critical point even though the phase transition happens at a higher temperature? I think I am missing something. $\endgroup$ – T.T. Sep 5 '18 at 16:39
  • $\begingroup$ I am getting a bit lost in the paper you sent, so I am saving it for later. $\endgroup$ – T.T. Sep 5 '18 at 16:42
  • $\begingroup$ Yes, sorry about that paper, it wasn't the prettiest thing to look at. Excuse me if I'm wrong but I think you say $m$ behaves linearly because you are expanding the square root out. You don't need to do that because you are not trying to arrive at a limit value but rather the overall behavior near that special temperature - a horizontally shifted square root still behaves like a square root. If those results are in your professor's solutions do insist for more explanation, profs sometimes skip things they know but us students are just learning. $\endgroup$ – Daniel Jordán Sep 5 '18 at 18:52
  • $\begingroup$ It is true that the square root still behaves more like a square root than a linear function near any point, but as I have understood the critical exponents, the interesting situation of $m~t^2$ is that the order parameter vanishes (or appears) in a non-trivial fashion -the derivative approaches infinity-, there is some kind of anomaly. This is not the case for the problem in hand, m vanishes approximately linearly and so $m~t$ seems to contain more information about the problem. Thank you for taking the time to reply, discussion on this topic helps. $\endgroup$ – T.T. Sep 6 '18 at 13:18
  • $\begingroup$ No problem, glad to help. For first-order transitions such as this one the parameter vanishes but its derivative doesn't have to go to $\infty$ (I think that's more of a $2^{nd}$-order thing, or for some response functions) but the parameter and its derivative are discontinuous. The linked paper's Fig. 8 has a graph of the parameter. $\endgroup$ – Daniel Jordán Sep 7 '18 at 19:32

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