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Finite-size rounding replaces a limiting discontinuity or divergence by a finite crossover or peak. For a spatial system near a continuous phase transition, the bulk description ceases to apply unchanged when the correlation length becomes comparable with the sample size. How the crossover width and location depend on size also depends on the model, geometry and boundary conditions; there is no universal $1/N$ rule for every phase transition.

An exact finite example is the Curie–Weiss model with $s_i=\pm 1$, $J>0$, and a conjugate field $h$ in energy units. Set $K=J/(k_{B}T)$ and $u=h/(k_{B}T)$. Its statistical Hamiltonian and finite partition function areThe binomial coefficient counts the microstates with $k$ positive Ising spins. The diagonal terms in the square add a configuration-independent constant to the energy; retaining them fixes the normalization used here. Divide each summand by $Z_N$ to obtain the probability mass function of $m_k=(2k-N)/N$. In particular,The variance expression makes the finite response nonnegative, while spin inversion symmetry gives $⟨m⟩=0$ exactly at $u=0$ for every plotted $N$.

Every curve is computed from the finite partition function, without selecting one minimum of the Bragg-Williams free energy of the Ising model. The field panel holds $K=1.5$ fixed and shows increasingly rapid switching between opposite spin magnetizations. The temperature panel uses $u=0$: its nonzero $⟨|m|⟩$ measures the typical magnitude, while the mean spin magnetization stays zero. The reference $k_{B}T/J=1$ marks the limiting critical temperature, not a singular point of any finite curve.

A short limiting argument explains the field panel. Combining the binomial coefficient with its leading entropy gives a Helmholtz free energy per spin whose stationarity equation is $m=\tanh (Km+u)$. For $K>1$ and $u=0$ the two ordered minima have $m=\pm m_{\ast }$, where $m_{\ast }=\tanh (K{m}_{\ast })$. For large $N$ well below the critical temperature, their relative Boltzmann weight is approximately $e^{2Nu{m}_{\ast }}$, giving $⟨m⟩\approx m_{\ast }\tanh (Nu{m}_{\ast })$. Thus this two-peak approximation has a field width of order $1/(N{m}_{\ast })$ in $u$. It is not valid at the critical temperature, where the peaks merge; the figure uses the exact finite sum throughout.

The approximate partition function obtained by decoupling each Ising bond has this free energy. The quadratic constant compensates double counting. Its stationary condition is the self-consistency equation; away from stationarity $m$ parametrizes the assumed effective field and need not equal that distribution's mean Ising spin. It agrees at stationary points with the Bragg-Williams free energy of the Ising model, although their off-equilibrium expansions differ.

For a ferromagnetic Ising model with coordination number of a lattice $z$, its mean-field self-consistency equation is $M=\tanh (zJM/(k_{B}T))$. Put $A=T_c/T$. For $A\le 1$, the only solution is zero. For $A>1$, strict concavity of $\tanh (AM)-M$ on $M>0$ gives one positive solution and its spin-reversed partner. Expanding $\tanh (AM)=AM-(AM{)}^3/3+O(M^5)$ gives $M^2=3(A-1)/A^3+O((A-1{)}^2)$, proving a continuous onset with order-parameter critical exponent $\beta =1/2$. The Bragg-Williams free energy of the Ising model has a positive quartic coefficient at $T_c$, so these are the stable ordered minima rather than a first-order jump.

The mean-field approximation replaces each site's fluctuating surroundings by a self-consistent average. On a cubic lattice each spin has coordination number of a lattice $z=2D$, even though summing only over positive basis directions counts each bond once. Write ${\sigma }_n=M+\delta {\sigma }_n$ and drop products of deviations. Each Ising bond becomes ${\sigma }_n{\sigma }_m\simeq M{\sigma }_n+M{\sigma }_m-M^2$, so the approximate statistical Hamiltonian isThe constant corrects counting each bond twice. The resulting independent-spin partition function is $Z_{\mathrm{MF}}=e^{-{\beta }_{T}NzJ{M}^2/2}[2\cosh ({\beta }_{T}zJM){]}^N$, where ${\beta }_T=1/(k_{B}T)$. The one-spin probabilities giveEquating this average with the assumed $M$ givesHere $J>0$ is assumed; a ferromagnetic order parameter is not the appropriate choice for an antiferromagnetic coupling. Put $A=T_c/T$. For $A\le 1$, $\tanh (AM)<AM\le M$ at any $M>0$, so only $M=0$ is possible. For $A>1$, the function $\tanh (AM)-M$ has positive slope at zero, is strictly concave on $M>0$ and is negative at $M=1$. It therefore has exactly one positive zero and a spin-reversed negative zero. Expanding at the bifurcation givesThe transition is continuous, with order-parameter critical exponent $\beta =1/2$. Stability can also be checked from the Bragg-Williams free energy of the Ising model per Ising spin:Its derivative is $-zJM+k_{B}T\mathrm{arctanh}M$, giving the same self-consistency equation. Near zero,The quartic coefficient is positive at $T_c$, the stable minima approach zero continuously, and the minimized free energy has no first-derivative jump. Thus mean-field approximation predicts a second-order transition at $T_c=2DJ/k_B$. This is the mean-field Ising critical bifurcation.

For short-range interactions, large dimension gives many neighbours and suppresses the relative fluctuations of their average. More quantitatively, Gaussian field theory fluctuations averaged over a correlation volume have variance of order ${\xi }^{2-D}$, while the squared ordinary mean-field order parameter is of order $|t|\sim {\xi }^{-2}$. Their ratio is proportional to ${\xi }^{4-D}$. The Ginzburg criterion therefore makes neglect of critical fluctuations asymptotically consistent above four dimensions, inconsistent below four, and marginal at four, where logarithms occur. Large dimension is a route to validity; sufficiently long-range or infinite-range interactions can also produce mean-field behaviour. In low-dimensional short-range systems even the existence and location of a transition can differ from this approximation.

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.