91影视

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:

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

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

role="math" $$\begin{gathered} \stackrel{炉}{s}=\tanh \left(尾系n\stackrel{炉}{s}\right) \\ {T}_c=\frac{n系}{k},尾=\frac{1}{kT},尾n系=\frac{T_c}{T} \\ 鈮�\frac{T_c}{T}路\stackrel{炉}{s}-\frac{{\left(\stackrel{炉}{s}{T}_c/T\right)}^3}{3} \\ \stackrel{炉}{s}=\frac{T_c}{T}路\stackrel{炉}{s}\left[1-\frac{{\left(T_c/T\right)}^2}{3}路{\stackrel{炉}{s}}^2\right] \\ \frac{T}{T_c}=1-\frac{{\left(T_c/T\right)}^2}{3}路{\stackrel{炉}{s}}^2 \\ \frac{T_c^2}{3T^2}路{\stackrel{炉}{s}}^2=1-\frac{T}{T_c} \\ \frac{T_c^2}{3T^2}路{\stackrel{炉}{s}}^2=1-\frac{T}{T_c}/路\frac{3T^2}{T_c^2} \\ {\stackrel{炉}{s}}^2=3路\frac{T^2}{T_c^2}路\left(1-\frac{T}{T_c}\right) \\ \stackrel{炉}{s}=\sqrt{3路\frac{T^2}{T_c^2}\left(1-\frac{T}{T_c}\right)},T鈫�T_c^{-} \\ =\sqrt{3路\frac{T^2}{T_c^2}\frac{T_c-T}{T_c}},t=\frac{T-T_c}{T_c} \\ =\sqrt{-3t(1+t{)}^2},\left|t\right|<<1,(1+t{)}^2鈮�1+2r \\ 鈮�\sqrt{-3t(1+2t)} \\ 鈮�\sqrt{-3t} \\ =\sqrt{\frac{-3\left(T-T_c\right)}{T_c}} \\ \stackrel{炉}{s}~\sqrt{T_c-T} \\ 尾=0.5 \end{gathered}$$

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:

$$\begin{gathered} \stackrel{炉}{s}=\tanh \left(尾\right(系n\stackrel{炉}{s}+B\left)\right)/\frac{d}{dB} \\ 蠂=\frac{尾(1+系n蠂}{{\cosh }^2\left[尾\right(系n\stackrel{炉}{s}+B\left)\right]} \end{gathered}$$

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

$$\begin{gathered} 蠂=\frac{尾(1+系n蠂)}{{\cosh }^2\left[尾系n\stackrel{炉}{s}\right]} \\ 蠂=\frac{尾}{{\cosh }^2\left[尾系n\stackrel{炉}{s}\right]-尾系n} \\ 尾系n=\frac{T_c}{T} \\ \stackrel{炉}{s}鈮�\left(3\right|t|{)}^{0.5}\text{for}T鈫�T_c^{-} \\ \cosh (x)鈮�1+\frac{x^2}{2} \\ {\cosh }^2\left[\frac{T_c}{T}\stackrel{炉}{s}\right]鈮�{\left(1+\frac{{\left(T_c/T\right)}^2路3\left|t\right|}{2}\right)}^2 \\ 鈮�1+3\frac{T_c^2}{T^2}|t| \\ 蠂鈮�\frac{尾}{1+3\frac{T_c^2}{T^2}\left|t\right|-\frac{T_c}{T}} \\ 1+3\frac{T_c^2}{T^2}|t|-\frac{T_c}{T}=3\frac{T_c^2}{T^2}|t|+1-\frac{T_c}{T} \\ =3\frac{T_c^2}{T^2}|t|-|t| \\ =|t|\left(3\frac{T_c^2}{T^2}-1\right) \\ =|t|路\frac{3T_c^2-T^2}{T^2} \\ 蠂鈮�\frac{尾}{\left|t\right|路\frac{3T_c^2-T^2}{T^2}} \\ 鈮�\frac{1}{k_{B}T}路\frac{T^2}{\left|t\right|路\left(3T_c^2-T^2\right)} \\ 鈮�\frac{1}{k_B}路\frac{T}{\left|t\right|路\left(3T_c^2-T^2\right)}T鈮�T_c \\ 鈮�\frac{1}{k_B}路\frac{T_c}{\left|t\right|路2T_c^2} \\ 蠂鈮�\frac{1}{2k_B}路\frac{1}{\left|t\right|T_c} \\ 蠂鈮�{\left(T-T_c\right)}^{-1} \\ 纬=1 \end{gathered}$$