In this problem you will use the mean field approximation to analyse the behaviour of the Ising model near the critical point.

(a) Prove that, when $x\ll 1,\tanh x\approx x-\frac{1}{3}{x}^3$

(b) Use the result of part (a) to find an expression for the magnetisation of the Ising model, in the mean field approximation, when T is very close to the critical temperature. You should find $M\propto {\left(T_c-T\right)}^{\overline{\beta }},\text{where}\beta$(not to be confused with 1/kT) is a critical exponent, analogous to the f defined for a fluid in Problem 5.55. Onsager's exact solution shows that $\beta =1/8$in two dimensions, while experiments and more sophisticated approximations show that $\beta \approx 1/3$in three dimensions. The mean field approximation, however, predicts a larger value.

(c) The magnetic susceptibility $\chi$is defined as $\chi \equiv (\partial M/\partial B{)}_T$. The behaviour of this quantity near the critical point is conventionally written as $\chi \propto {\left(T-T_c\right)}^{-\gamma }$ , where y is another critical exponent. Find the value of in the mean field approximation, and show that it does not depend on whether T is slightly above or slightly below Te. (The exact value of y in two dimensions turns out to be 7/4, while in three dimensions $\gamma \approx 1.24$.)

In this problem we will use the mean field approximation to analyse the behaviour of the Ising model near the critical point.

a) The function tanh x has the following definition:

$$\begin{gathered} \tanh x=\frac{e^{2x}-1}{e^{2x}+1} \\ {e}^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \\ \tanh x\approx \frac{1+2x+\frac{(2x{)}^2}{2}+\frac{(2x{)}^3}{6}-1}{1+2x+1} \\ \approx \frac{2x+\frac{(2x{)}^2}{2}+\frac{(2x{)}^3}{6}}{2+2x} \\ \approx \frac{x+\frac{2x^2}{2}+\frac{4x^3}{6}}{1+x};(1+x{)}^{-1}\approx 1-x \\ \approx \left(x+x^2+\frac{2x^3}{3}\right)(1-x) \\ =x-x^2+x^2-x^3+\frac{2x^3}{3} \\ =x-\frac{x^3}{3} \end{gathered}$$

(b)This expression will now be used to expand relation 8.50:

role="math" $$\begin{gathered} \overline{s}=\tanh \left(\beta ϵn\overline{s}\right) \\ {T}_c=\frac{nϵ}{k},\beta =\frac{1}{kT},\beta nϵ=\frac{T_c}{T} \\ \approx \frac{T_c}{T}·\overline{s}-\frac{{\left(\overline{s}{T}_c/T\right)}^3}{3} \\ \overline{s}=\frac{T_c}{T}·\overline{s}\left[1-\frac{{\left(T_c/T\right)}^2}{3}·{\overline{s}}^2\right] \\ \frac{T}{T_c}=1-\frac{{\left(T_c/T\right)}^2}{3}·{\overline{s}}^2 \\ \frac{T_c^2}{3T^2}·{\overline{s}}^2=1-\frac{T}{T_c} \\ \frac{T_c^2}{3T^2}·{\overline{s}}^2=1-\frac{T}{T_c}/·\frac{3T^2}{T_c^2} \\ {\overline{s}}^2=3·\frac{T^2}{T_c^2}·\left(1-\frac{T}{T_c}\right) \\ \overline{s}=\sqrt{3·\frac{T^2}{T_c^2}\left(1-\frac{T}{T_c}\right)},T\to T_c^{-} \\ =\sqrt{3·\frac{T^2}{T_c^2}\frac{T_c-T}{T_c}},t=\frac{T-T_c}{T_c} \\ =\sqrt{-3t(1+t{)}^2},\left|t\right|<<1,(1+t{)}^2\approx 1+2r \\ \approx \sqrt{-3t(1+2t)} \\ \approx \sqrt{-3t} \\ =\sqrt{\frac{-3\left(T-T_c\right)}{T_c}} \\ \overline{s}~\sqrt{T_c-T} \\ \beta =0.5 \end{gathered}$$

We may conclude that the same critical exponent applies to magnetisation because it is proportional to $\overline{s}$.

(c) To calculate susceptibility, we must differentiate the non-approximate formula for $\overline{s}$in relation to B, which we shall manually insert:

$$\begin{gathered} \overline{s}=\tanh \left(\beta \right(ϵn\overline{s}+B\left)\right)/\frac{d}{dB} \\ \chi =\frac{\beta (1+ϵn\chi }{{\cosh }^2\left[\beta \right(ϵn\overline{s}+B\left)\right]} \end{gathered}$$

We can get $\chi$by setting B = 0 and rearranging terms:

$$\begin{gathered} \chi =\frac{\beta (1+ϵn\chi )}{{\cosh }^2\left[\beta ϵn\overline{s}\right]} \\ \chi =\frac{\beta }{{\cosh }^2\left[\beta ϵn\overline{s}\right]-\beta ϵn} \\ \beta ϵn=\frac{T_c}{T} \\ \overline{s}\approx \left(3\right|t|{)}^{0.5}\text{for}T\to T_c^{-} \\ \cosh (x)\approx 1+\frac{x^2}{2} \\ {\cosh }^2\left[\frac{T_c}{T}\overline{s}\right]\approx {\left(1+\frac{{\left(T_c/T\right)}^2·3\left|t\right|}{2}\right)}^2 \\ \approx 1+3\frac{T_c^2}{T^2}|t| \\ \chi \approx \frac{\beta }{1+3\frac{T_c^2}{T^2}\left|t\right|-\frac{T_c}{T}} \\ 1+3\frac{T_c^2}{T^2}|t|-\frac{T_c}{T}=3\frac{T_c^2}{T^2}|t|+1-\frac{T_c}{T} \\ =3\frac{T_c^2}{T^2}|t|-|t| \\ =|t|\left(3\frac{T_c^2}{T^2}-1\right) \\ =|t|·\frac{3T_c^2-T^2}{T^2} \\ \chi \approx \frac{\beta }{\left|t\right|·\frac{3T_c^2-T^2}{T^2}} \\ \approx \frac{1}{k_{B}T}·\frac{T^2}{\left|t\right|·\left(3T_c^2-T^2\right)} \\ \approx \frac{1}{k_B}·\frac{T}{\left|t\right|·\left(3T_c^2-T^2\right)}T\approx T_c \\ \approx \frac{1}{k_B}·\frac{T_c}{\left|t\right|·2T_c^2} \\ \chi \approx \frac{1}{2k_B}·\frac{1}{\left|t\right|T_c} \\ \chi \approx {\left(T-T_c\right)}^{-1} \\ \gamma =1 \end{gathered}$$