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Use inverse temperature ${\beta }_T=(k_{B}T{)}^{-1}$, reserving $\beta$ without a subscript for the critical exponent. Write $c=qJ$; each undirected bond occurs once, so the number of bonds is $Nq/2$. In the mean-field theory of the Ising model, write each Ising spin as ${\sigma }_i=m+\delta {\sigma }_i$ and neglect the product of fluctuations:The resulting independent-spin statistical Hamiltonian isThe constant corrects the double counting of interaction energy. The Ising spin sums now factorize and can all be evaluated:This is the approximate partition function at an assumed mean field; equilibrium fixes $m$ self-consistently. The corresponding Ising auxiliary mean-field free energy isIts equilibrium value gives the mean-field Helmholtz free energy in the imposed field. Away from a stationary point its parameter $m$ is an assumed field variable, not necessarily the actual mean Ising spin of that independent-spin distribution.

For the full small-$m$ expansion at fixed $h$, set $t_h=\tanh ({\beta }_{T}h)$ and $s_h={\mathrm{sech}}^2({\beta }_{T}h)$. Differentiating $\log \cosh z$ gives successive derivatives $t_h$, $s_h$, $-2t_h{s}_h$, and $2s_h(2-3s_h)$ at $z={\beta }_{T}h$. ThereforeAt zero field, spin inversion symmetry eliminates odd powers and this simplifies toThe quadratic coefficient changes sign and the quartic coefficient is positive atThus the zero-field mean-field prediction is a continuous, continuous phase transition: the stable zero spin magnetization develops two symmetry-related nonzero minima continuously below $T_c$.

Both requested routes give the same mean-field self-consistency equation. First, the one-spin expectation in the effective field isSecond, differentiating the Ising auxiliary mean-field free energy giveswhose stationary condition is precisely that mean-field self-consistency equation. Choose its stable, lowest-free-energy branch rather than every algebraic solution.

An equally useful mean-field approximation parametrizes the trial distribution by its actual mean Ising spin. Its probabilities are $(1\pm m)/2$, giving the Bragg-Williams free energy of the Ising modelThis energy-minus-entropy function has expansionIts stationary equation is $h=k_{B}T\mathrm{arctanh}m-cm$, again equivalent to the mean-field self-consistency equation. The two functions differ away from equilibrium, but agree on stationary branches: for $z={\beta }_T(h+cm)$ and $m=\tanh z$, the entropy bracket equals $mz-\log (2\cosh z)$. Their small-$m$ coefficients therefore need not agree at arbitrary temperature; at $T_c$ they have the same leading critical quartic coefficient $c/12$. This distinction prevents confusing the auxiliary-field expansion with the physical-magnetization variational expansion.

For the order-parameter critical exponent, expand the equation of state at $h=0$:On a nonzero stable branch,Hence $\beta =1/2$. The spontaneous magnetization is understood by selecting a branch with an infinitesimal field after the thermodynamic limit; a finite symmetric sample has zero exact zero-field mean Ising spin.

For the magnetic susceptibility per site, define $\chi =\partial m/\partial h$ with $h$ in energy units. Implicit differentiation givesAbove $T_c$, $m=0$ and $\chi =1/[k_B(T-T_c)]$. Below $T_c$, evaluated on a selected ordered branch, the expansion of $m^2$ gives $\chi \sim 1/[2k_B(T_c-T)]$. Thus the magnetic-susceptibility critical exponent is $\gamma =1$ on both sides, with different amplitudes. At $T=T_c$, the equation of state becomesso the critical-isotherm exponent is $\delta =3$. These are mean-field critical exponents, not a claim that neglecting fluctuations gives the exact Ising transition in every dimension.