We will build upon the mean-field theory for the Ising model by introducing a more efficient notation
ποΈ Note: for now the external magnetic field will be set to zero ie $h=0$
We define the effective free energy $\tilde F(T,P,m)$
ποΈ Note: it is different to the free energy in thermodynamics and statistical mechanics
Starting from the definition of the partition function we write
$$ \begin{aligned} Z(T)&=\sum_{\{s_i\}} \exp \left (- \frac{E[\{s_i\}]}{k_B T}\right ) \\ &=\sum_{\tilde m} \sum_{\{s_i \}_{ \tilde m}} \exp \left ( -\frac{E[\{s_i \}]}{k_B T} \right )=\sum _{\tilde m} \exp \left ( \frac{\tilde F(T,\tilde m)}{k_B T} \right )
\end{aligned} $$
where $\{ s_i \}_{ \tilde m }$ is the set of all spin configurations such that the average magnetisation per lattice is $-1 \le \tilde m \le 1$ ie
$$ \frac 1 N \sum _i s_i = \tilde m $$
ποΈ Notes: $\tilde m$ is not $m=M/N$ instead it is a variable that can assume any value that is tied to a subset of spin configurations. $\tilde m$ is quantised in units of $2/N$ and in the limit of $N\to \infin$ we can treat it continuously as
$$ \sum_{\tilde m} \to \frac{N}{2} \int _{-1}^1 \,\text d\tilde m $$
We start by defining the free energy per lattice site
$$ \tilde f (T, \tilde m ) = \frac{\tilde F (T, \tilde m )}{N} $$
The partition function now becomes
$$ Z(T)=\sum_{\tilde m} \exp \left ( -\frac{N \tilde f(T, \tilde m)}{k_B T} \right )=\frac N2 \int ^1 _{-1} \,\text d \tilde m \exp \left (-\frac{N \tilde f (T,\tilde m )}{k_B T} \right ) $$
For large $N$ the integral above is dominate by the value of $\tilde m$ that minimises the exponent.
Therefore calling $m$ the value of $\tilde m$ that minimises the free energy
$$ \left . \frac{\partial \tilde f(T,\tilde m )}{\partial \tilde m} \right | _{\tilde m=m}=0 $$
we have that
$$ Z(T)=\sum_{\tilde m} \exp \left (-\frac{N\tilde f (T,\tilde m)}{k_B T} \right )\simeq \exp \left (-\frac{N \tilde f (T,m)}{k_B T} \right ) $$
Putting this together we have
$$ F(T)=N \tilde f(T,m) $$
π Conclusions:
If we write out $\tilde f(T,\tilde m)$ we get
$$ {\tilde f(T,\tilde m)=-\frac 12Jz \tilde m^2 -h \tilde m -k_B T \left [ \ln(2) - \frac{1+\tilde m}{2} \ln (1+ \tilde m) -\frac{1-\tilde m}{2} \ln(1-\tilde m) \right ]} $$
If we minimise this expression we get
$$ \left . \frac{\partial \tilde f(T,\tilde m)}{\partial \tilde m} \right |_{\tilde m = m}=0 \quad \Rightarrow Jzm +h=\frac{k_BT}{2}\ln \left ( \frac{1+m}{1-m} \right ) $$
which rearranging gives
$$ m=\tanh \left ( \frac{Jzm+h}{k_B T}\right ) $$
π Conclusion: neglecting the fluctuations of the spin with respect to their average value, provides the same result as assuming that they provide an effective magnetic field
Finally we can find the explicit value for
$$ \frac{F(T)}{N}=\tilde f(T,m)=\frac 12 Jzm^2-k_B T \ln \left [ 2\cosh \left ( \frac{h_\text{eff}}{k_B T} \right ) \right ] $$
𧽠Assumptions of Landauβs theory
- Smooth polynomial expansion: The effective free energy $\tilde f(T,\tilde m)$ can be expressed as a polynomial in the order parameter $\tilde m$ where unphysical or symmetry-violating terms are excluded
- Analytical Coefficients: the coefficients of the polynomial expansion are assumed to be smooth function of the temperature $T$
We can rewrite the the the expression $\tilde f(T,\tilde m )$ for $T\simeq T_C$ where $m\simeq 0$