I am given the following expression for the free energy:

$$f = \frac{1}{2}r_0 m^2+um^4+vm^6,$$

where $r_0=k_B (T-T_c)$ with $T_c$ the critical temperature and $u=\frac{1}{12}k_B T$ and $v=\frac{1}{30}k_B T$. I can prove from the self-consistency equation: $m = \tanh(m \tau)$ where $\tau=T_c/T$ that $$m \approx m\tau -\frac{1}{3}(m\tau)^3 + \frac{2}{15}(m\tau)^5,$$ so that neglecting the $m^5$-order terms and higher that $m^2 = 3(\frac{T_c-T}{T_c})$ and hence that the critical exponent of $m$ is $\beta = 1/2$. But does this change when I include a term with $m^6$ in the free energy? In order words does a critical exponent in this model depend on the order of $f$ you consider? I calculated the minima to be at (for $r_0 <0$):

$$m^2_{\pm} = \frac{-4u+ \sqrt{16u^2-24r_0 v}}{12v}.$$

  • $\begingroup$ Who voted to close this as "not about physics"? It's like voting to close a question about Shakespeare as "not about English". $\endgroup$ – Nathaniel Sep 29 at 8:01
  • $\begingroup$ Shouldn't you get the former condition if you set $v=0$? This does not seem to be the case. $\endgroup$ – Norbert Schuch Sep 29 at 12:19
  • $\begingroup$ @Nathaniel That's where the "Homework" flag is categorized. Maybe the text should be changed to "This question does not appear to be on-topic on Physics.SE"? $\endgroup$ – Norbert Schuch Sep 29 at 12:19
  • $\begingroup$ No since the abc-equation for quadratic polynomials only applies if $a \neq 0$. I already solved it, you can just take $r_0 \approx 0$ and Taylor expand the square root to find $m \sim |T-T_c|^{1/2} \sim |r_0|^{1/2}$. $\endgroup$ – Dani Sep 29 at 13:00
  • $\begingroup$ @Dani Why don't you continue from the equation you derived? Where are you stuck? $\endgroup$ – Norbert Schuch Sep 29 at 14:10

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