# Critical exponent mean field Ising model

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 $$r_0 <0$$):

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

• Who voted to close this as "not about physics"? It's like voting to close a question about Shakespeare as "not about English". – Nathaniel Sep 29 at 8:01
• Shouldn't you get the former condition if you set $v=0$? This does not seem to be the case. – Norbert Schuch Sep 29 at 12:19
• @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"? – Norbert Schuch Sep 29 at 12:19
• 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 $m \sim |T-T_c|^{1/2} \sim |r_0|^{1/2}$. – Dani Sep 29 at 13:00
• @Dani Why don't you continue from the equation you derived? Where are you stuck? – Norbert Schuch Sep 29 at 14:10