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Define source strength $m$ here as the volume flux through a small sphere around the source. For incompressible potential flow, $\mathbf{u}=\nabla \varphi$ and the governing conditions areEquivalently the Laplace equation holds away from the source, with local behavior $\varphi \sim -m/(4\pi |\mathbf{x}-{\mathbf{x}}_s|)$. The boundary condition expresses impermeability. The method of images uses an equal source below the plane:The image singularity lies outside the fluid, the two vertical derivatives cancel on the wall, and the source has the specified outward flux.

Let $s=\sqrt{x^2+y^2}$. On the wall $u_z=0$ and the tangential radial velocity is $u_s=ms/[2\pi (s^2+a^2{)}^{3/2}]$. With ambient pressure $p_{\infty }$, the steady Bernoulli equation in this irrotational flow givesThe fluid pressure is reduced everywhere on the wall except at its axis and in the infinite-distance limit. The finite hydrodynamic force is the excess over the uniform ambient-pressure force. Integrating its magnitude over the plane yieldsPutting $u=s^2$ gives the last integral as $\frac{1}{2}{\int }_0^{\infty }u(u+a^2{)}^{-3}du=1/(4a^2)$. HenceRelative to ambient pressure on the other side, the boundary is attracted toward the source. This is the hydrodynamic attraction of a plane wall to a point source.

There is an alternative source-strength convention: if $m$ denotes the coefficient in $u_r=m/r^2$ rather than total volume flux, replace the flux above by $4\pi m$. Then $\varphi =-m(1/r_{+}+1/r_{-})$, $p-p_{\infty }=-2\rho m^2{s}^2/(s^2+a^2{)}^3$, and $F=\pi \rho m^2/a^2$. The physical result is identical once the normalization is fixed.

For the independent normal distribution observations, the log likelihood isIts derivative is ${\sum }_i(X_i-\theta )$ and its second derivative is $-n<0$, so the unique maximum-likelihood estimate isFor the prior $\theta \sim N(\mu ,{\tau }^{-1})$ with $\tau >0$, multiplying prior density by likelihood and completing the square givesThus normal-normal conjugacy gives the posterior distributionFor quadratic loss, the conditional expected loss of action $a$ is the posterior variance plus $(a-\mathbb{E}[\theta \mid X]{)}^2$. Its unique minimizer is the posterior mean, so the Bayes estimator isThis weights the sample and prior means by their respective precisions.

For an $n$-dimensional random vector with a density and finite differential entropy in bits, define its entropy power by $N(X)=(2\pi e{)}^{-1}{2}^{2h(X)/n}$. The entropy power inequality states that for independent such random vectors,Here the sum's differential entropy must be defined, with $+\infty$ permitted.

Since $g^{\prime }(x)>0$, the mean value theorem makes $g$ strictly increasing. Its image is an interval $I$, and its inverse on $I$ has derivative $1/g^{\prime }(g^{-1}(y))$. The density change of variables givesThis does not require $g$ to map onto all of $\mathbb{R}$. Taking logarithms at $y=g(X)$ and then expected values proves the differential entropy under an increasing transformation:The two finite expectations in the assumptions justify splitting the expectation and imply that the transformed differential entropy is finite.

To derive the product inequality, assume that $h(Y_i)$ and ${\mu }_i=\mathbb{E}{\log }_2{Y}_i$ are finite and that the differential entropy of the logarithmic sum is defined. Set $Z_i=\ln Y_i$, using natural logarithms for this transformation while measuring differential entropy in bits. On the positive half-line the derivative of $\ln y$ is $1/y$, so the same density argument givesThe $Z_i$ are independent random variables, and the one-dimensional entropy power inequality yieldsSince $Y_1{Y}_2=\exp (Z_1+Z_2)$, another change of variables givesMultiplying the entropy power inequality by $2^{2({\mu }_1+{\mu }_2)}$ proves the multiplicative entropy power inequality:

The printed positivity and density assumptions alone do not ensure that the quantities in this last formula exist. For a concrete example, let $T$ have the standard Cauchy distribution and put $Y=e^T$. Thenis a density, but $\mathbb{E}{\log }_{2}Y=({\log }_{2}e)\mathbb{E}T$ is undefined because its positive and negative parts both diverge. Two independent copies satisfy the printed hypotheses but leave ${\alpha }_1,{\alpha }_2$ undefined. The product inequality therefore needs the finiteness or well-definedness conditions used above.

Use finite sets of links $J$, source-sink pairs $S$, and admissible routes $R$, with at least one route for each pair. Route $r$ carries aggregate flow $x_r$. The link-route incidence matrix has $A_{jr}=1$ if route $r$ uses link $j$ and zero otherwise; repeated traversals, if allowed, are counted with their multiplicity. The source-sink route incidence matrix has $H_{sr}=1$ if $r$ serves pair $s$. Thus $y=Ax$ gives link throughputs and $f=Hx$ gives source-sink demands. Write $R_s$ for the routes serving $s$ andA Wardrop equilibrium routes positive flow only on minimum-delay routes:This is a nonatomic equilibrium: an individual infinitesimal user's route change does not alter the aggregate link delays.

For fixed nonnegative demand $f$, the feasible route-flow set $\{x\ge 0:Hx=f\}$ is a nonempty product of finite-dimensional simplexes and hence a compact set. The Beckmann potentialis continuous, so it attains a minimum over this feasible set. Each primitive is a strictly convex function, since its derivative $D_j$ is continuous and strictly increasing. Its route-flow derivative is exactly the corresponding route delay:If a minimizer had positive flow on $r$ and a lower-delay alternative $q\in R_s$, transferring a small amount from $r$ to $q$ would give directional derivative $L_q-L_r<0$, contradicting minimality. Hence every minimizer is a Wardrop equilibrium.

Conversely, let $x$ be a Wardrop equilibrium and $z$ any feasible competitor. Convexity givesFor each $s$, subtract ${\lambda }_s$ inside this sum. The demand constraint makes ${\sum }_{r\in R_s}{\lambda }_s(z_r-x_r)=0$, and the remaining terms are nonnegative: used routes have $L_r-{\lambda }_s=0$, while unused routes have $x_r=0$, $z_r\ge 0$ and $L_r-{\lambda }_s\ge 0$. Thus $V(Az)\ge V(Ax)$. We have proved both existence and the optimization characterization.

Strict convexity in the link vector makes $y$ unique: two different minimizing link vectors would have a feasible midpoint of strictly smaller potential. Thus link delays and the minimum route delays at the Wardrop equilibrium are also unique. The route vector need not be unique, because different route decompositions may have the same $Ax$ and $Hx$. For example, two successive stages with two parallel links at each stage have four routes. With unit demand and link delay $D(u)=u$, the vectorsall give throughput $1/2$ on each of the four links and delay one on every route. This exhibits route-flow nonuniqueness at a Wardrop equilibrium. The equilibrium link throughputs are unique; the individual route flows need not be.

For elastic demand, make the usual physical assumptions explicit: delays are nonnegative on $[0,\infty )$, every route is nonempty, and $B_s:[0,\infty )\to [0,\infty )$ is finite, continuous and strictly decreasing. PutThe limit exists by monotonicity; it need not be zero. The inverse demand function $P_s$ is continuous, strictly decreasing, satisfies $P_s(M_s)=0$, and tends to infinity as $v\downarrow {\ell }_s$.

Choose any reference $b_s\in ({\ell }_s,M_s)$ and use elastic-demand utility with an interior reference:Each utility is strictly concave, with derivative $P_s$. A reference inside the range avoids a potentially divergent integral from zero. For instance, $B(\lambda )=(1+\lambda {)}^{-1/2}$ has inverse $v^{-2}-1$, whose integral from zero is infinite, although the displayed reference-point primitive is finite at every interior demand.

Give $G_s$ its limiting value at ${\ell }_s$, which may be $-\infty$, and extend it by $-\infty$ outside $[{\ell }_s,M_s]$. This encodes the physical demand range in the objective. After substituting $y=Ax$, $f=Hx$, minimizeIts effective feasible set is contained in $\{x\ge 0:{\ell }_s\le (Hx{)}_s\le M_s\}$, a nonempty compact set. The objective is lower semicontinuous, including a possible value $+\infty$ at a lower endpoint, and is finite when all demands equal their interior reference values. It therefore has a finite-valued minimizer.

The boundary exclusion for elastic Wardrop demand shows that every minimizing demand is interior. If $f_s={\ell }_s$ and the utility there is finite, adding a small flow $h$ on any route serving $s$ costs at most a bounded constant times $h$, while the utility gain is at least $h{P}_s({\ell }_s+h)$, whose ratio to $h$ tends to infinity. This decreases the objective. If that endpoint utility is $-\infty$, a finite minimizer cannot occur there in the first place. At $f_s=M_s>0$, choose a used route. Its delay is strictly positive, since its nonempty links have positive throughput and strictly increasing nonnegative delays. Removing a small amount of its flow saves a positive first-order cost, whereas the utility loss is $o(h)$ because $P_s(M_s)=0$. This also decreases the objective. Hence ${\ell }_s<f_s<M_s$ for every pair.

The interior-demand route derivatives are nowAt a minimizer they are zero on positive route flows and nonnegative on zero route flows, by increasing or decreasing a single route flow. Since $f_s>0$, some route is used, soThese are precisely the elastic-demand Wardrop equilibrium conditions. Conversely, those conditions give the same nonnegative directional derivative for every competitor; convexity then proves global minimality. The equivalent constrained form is the requested minimization of $V(y)-G(f)$ with $Hx=f$, $Ax=y$, $x\ge 0$. Strict convexity determines both $y$ and $f$ uniquely, while their route decomposition can still be nonunique.

Minimizing the residual sum of squares $\Vert Y-Xb{\Vert }^2$ gives the least-squares normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Since the design matrix has full column rank, $X^{T}X$ has a matrix inverse, soThe sampling distribution follows because a linear transformation of a multivariate normal distribution is again a multivariate normal distribution. Its expectation is $\beta$, and its covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}{X}^{T}X(X^{T}X{)}^{-1}={\sigma }^2(X^{T}X{)}^{-1}$.

The hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. It is the orthogonal projection matrix onto the column space of $X$: $H^T=H$, $H^2=H$ and $HX=X$. Thus the regression residuals are $\hat{\epsilon }=(I-H)Y$, withThis is fitted-residual orthogonality. The vectors are jointly normal, so independence of uncorrelated jointly normal variables strengthens both zero-covariance statements to independence.

For the quadratic linear regression, the columns of the design matrix are $1,x,x^2$. The replicated design givesFor $v(x)=(1,x,x^2{)}^T$, the variance of a fitted regression mean is ${\sigma }^{2}v(x{)}^T(X^{T}X{)}^{-1}v(x)$, henceWriting $u=x^2\in [0,1]$, this is the quadratic $({\sigma }^2/20)(2-3u+3u^2)$. Its derivative vanishes at $u=1/2$; it decreases before that point and increases afterwards. Therefore the maximum is ${\sigma }^2/10$ at $x=-1,0,1$, and the minimum is ${\sigma }^2/16$ at $x=\pm 1/\sqrt{2}$.

At $100^{\circ }\mathrm{C}$, $x=0$ and the expected yield is $\alpha$. The unbiased residual variance estimate is $s^2=2.43/(30-3)=0.09$. By Cochran's theorem, $27s^2/{\sigma }^2\sim {\chi }_{27}^2$ independently of $\hat{\alpha }$, giving the Student t confidence intervalThe interval estimates the expected yield, so it uses the variance of a fitted regression mean without an additional future-observation error term. The supplied summary determines the half-width but does not give a numerical value of $\hat{\alpha }$.

At a continuous phase transition the equilibrium order parameter tends continuously to its critical value, there is no latent heat, and the correlation length diverges. For $u>0$ at zero field, minimizing $V(M)=r{M}^2/2+u{M}^4/4$ gives $M=0$ when $r\ge 0$ and $M=\pm \sqrt{-r/u}$ when $r<0$. These minima join continuously at $r=0$. The minimized singular free-energy density is zero above and $-r^2/(4u)$ below, so its first thermal derivative is continuous, while its second derivative has a finite jump in mean-field approximation.

A first-order phase transition instead has competing minima whose free energies cross while their order parameters remain distinct. The first derivatives of the equilibrium free energy generally jump; along a temperature-driven crossing this gives latent heat. For the symmetric sextic potential with $u<0,v>0$, let $y=M^2>0$. At phase coexistence, the stationary points and equality of the two free energies requireEliminating $r$ gives $-u{y}^2/4-v{y}^3/3=0$, henceAt coexistence $V(M)=v{M}^2(M^2+3u/(4v){)}^2/6$, proving that all three degenerate minima are global. The equilibrium order parameter jumps from zero to one of the two nonzero values even though $r_{\mathrm{coex}}>0$. The disordered minimum loses local stability at $r=0$, and the nonzero stationary points first appear at $r=u^2/(4v)$. These spinodal points delimit metastability and are not the equilibrium transition line. Cubic terms, when symmetry permits them, provide another common route to a first-order transition.

Let $t=(T-T_c)/T_c$, and suppose $r=r_{t}t$ with $r_t>0$ while $u>0$ and $\kappa >0$ remain finite at the transition. Critical exponents describe the leading singular powers: $M\sim (-t{)}^{\beta }$ below the transition, $\chi \sim |t{|}^{-\gamma }$, $\xi \sim |t{|}^{-\nu }$, $C_{V,s}\sim |t{|}^{-\alpha }$, $M(t=0)\sim \mathrm{sgn}(h)|h{|}^{1/\delta }$, and $G(R,t=0)\sim R^{-(D-2+\eta )}$.

The quartic equation of state $h=rM+u{M}^3$ immediately gives $M=\sqrt{-r/u}$ at zero field below the transition, so $\beta =1/2$. Differentiating with respect to the conjugate field gives ${\chi }^{-1}=r+3u{M}^2$, which is $r$ above and $2|r|$ below; therefore $\gamma =1$. At $r=0$, $M=(h/u{)}^{1/3}$ with its real sign, giving $\delta =3$. The quadratic correlation function has $\xi \propto |r{|}^{-1/2}$ and critical Ornstein--Zernike correlation function proportional to $q^{-2}$, giving $\nu =1/2$ and $\eta =0$.

Finally the minimized potential below the transition is $V_s=-r^2/(4u)$, while it is zero above. Its second temperature derivative is a finite step, so the mean-field heat-capacity critical exponent is $\alpha =0$. In particular,These are derived mean-field critical exponents. In the fluctuation-dominated critical region of a phase transition, one instead derives exponents from the scaling eigenvalues of the renormalization-group fixed point, as in the next question. A zero heat-capacity power exponent can also accompany logarithms in other theories, so specifying the power alone does not completely describe that singularity.

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 ${\beta }_T=1/(k_{B}T)$ for inverse temperature, to distinguish it from the order-parameter critical exponent $\beta$. In a translation-invariant pure thermodynamic phase, the connected two-point correlation function isFor finite correlation length, at $|r|\gg \xi$ it decays exponentially, possibly with an algebraic prefactor; ${\xi }^{-1}$ is the asymptotic exponential decay rate. Below an ordered transition the pure phase is selected by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A zero-field symmetric mixture of the two ordered phases has a nondecaying connected term and does not describe the usual pure-phase magnetic susceptibility.

Let $\mathcal{F}=-k_{B}T\log Z$ denote extensive physical free energy and $S={\sum }_r{\sigma }_r$. Since the field contributes $-hS$ to the statistical Hamiltonian, differentiation givesExpanding the variance and using translation invariance proves the correlation-function susceptibility sum rule:The sum includes $r=0$. If the source is instead defined as the dimensionless quantity ${\beta }_{T}h$, the corresponding derivative omits the displayed inverse-temperature factor.

A real-space renormalization group step of length factor $b>1$ uses a local blocking kernel $B_b({\sigma }^{\prime },\sigma )\ge 0$ satisfying ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$. For example, a deterministic majority-spin rule assigns one coarse Ising spin to each block; tied blocks can be resolved with equal probabilities. Define the blocked statistical Hamiltonian byThe spacing becomes $a^{\prime }=ba$. All generated interactions, including multispin terms and the field-independent constant, must be retained for an exact transformation. Summing over ${\sigma }^{\prime }$ uses normalization to prove $Z(u^{\prime },C^{\prime },N^{\prime })=Z(u,C,N)$. Iteration gives $N_p=b^{-pD}N$ and $Z(u_p,C_p,N_p)=Z(u_0,C_0,N)$.

The coarse probability law is the exact marginal law of these block variables. Indeed, for any coarse observable $A$, its expected value is the microscopic expected value of ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )A({\sigma }^{\prime })$. A blocking rule that retains the long-wavelength order-parameter mode, with source and observable transformations tracked, therefore retains long-distance response and correlation functions. Their field normalization may change, while their physical length scales and critical singularities do not. Partition-function equality alone would not justify this assertion for an arbitrary kernel that discards all slow variables. Truncating the generated operator family is likewise an approximation.

For the free-energy recursion use dimensionless free energy per site,in the thermodynamic limit. Exact partition-function invariance gives $F(u_0,C_0)=b^{-pD}F(u_p,C_p)$. If one step sends $C$ to $C^{\prime }=b^{D}C+c(u)$, the coefficient $c(u)$ records the identity-operator contribution from eliminated fluctuations and normalization. ConsequentlyRepeated substitution derives the additive free-energy recursion under blocking:In physical energy units, divide this convention by ${\beta }_T$. The additive term is essential for absolute free energies even when it does not affect the leading critical powers.

A renormalization-group fixed point satisfies $R_b(u_{\ast })=u_{\ast }$. Linearize in scaling coordinates $v_i$ so that $v_i^{\prime }=b^{y_i}{v}_i+O(v^2)$. A relevant operator has $y_i>0$, an irrelevant operator has $y_i<0$, and a marginal operator has $y_i=0$ with behaviour decided by nonlinear terms. Thus the eigenvalues of the discrete linear map are $b^{y_i}$; the sign criterion applies to the generator exponents $y_i$, not to the discrete multipliers. The critical surface is the stable manifold on which every relevant scaling field has been tuned away. Irrelevant coordinates flow to the renormalization-group fixed point along it. A repulsive renormalization-group trajectory leaves the renormalization-group fixed point along an unstable relevant direction, ultimately entering a noncritical phase or another scaling regime.

To separate the singular free-energy density from its regular background, suppose the eliminated-mode contribution $g$ is analytic near the renormalization-group fixed point. Look for an analytic background $a(u)$ satisfying $a(u)=b^{-D}a(R_{b}u)+g(u)$. Then $F_s=f-a$ satisfies $F_s(u)=b^{-D}{F}_s(R_{b}u)$. In linear thermal and field coordinates, an analytic monomial $g_{mn}{t}^m{h}^n$ has background coefficientprovided the denominator is nonzero. This explicitly explains when analytic subtraction of an inhomogeneous renormalization recursion works. The printed inhomogeneous equation may describe the free-energy function before this subtraction, or a singular part defined only up to analytic terms. The inhomogeneous sum need not be small compared with the full free energy: an analytic constant can dominate a vanishing singular contribution. It can be omitted in deriving leading singular powers after regular subtraction, provided no renormalization-group free-energy resonance, marginal logarithm or dangerously irrelevant coupling obstructs the homogeneous critical asymptotics. At a renormalization-group free-energy resonance $m{\lambda }_t+n{\lambda }_h=D$, additive logarithms can remain, so simply dropping the term would miss them.

Suppose there are exactly two relevant scaling fields, thermal $t$ and magnetic $h$, with positive exponents ${\lambda }_t,{\lambda }_h$, and the remaining fields approach a regular limit. Physical reduced temperature and magnetic field agree with these coordinates to leading order, up to nonuniversal metric factors. The homogeneous scaling hypothesis for critical phenomena isas an asymptotic scaling equation. Choose $\ell =|t{|}^{-1/{\lambda }_t}$ to make the thermal argument $+1$ or $-1$. Defining $f_{\pm }(z)=F_s(\pm 1,z)$ givesThe two functions describe respectively $t>0$ and $t<0$: approaching the transition from disordered and ordered sides need not give the same function. In particular the ordered zero-field function has one-sided derivatives associated with spontaneous magnetization. Metric factors multiply this formula in different microscopic normalizations.

The correlation length in Bravais lattice units obeys $\xi (t,h)=\ell \xi ({\ell }^{{\lambda }_t}t,{\ell }^{{\lambda }_h}h)$. At $h=0$, the same choice of $\ell$ gives $\nu =1/{\lambda }_t$. Two thermal derivatives of $F_s(t,0)$ give $2-\alpha =D/{\lambda }_t$. One and two field derivatives give, respectively, $\beta =(D-{\lambda }_h)/{\lambda }_t$ and $\gamma =(2{\lambda }_h-D)/{\lambda }_t$, assuming the relevant branch amplitudes are nonzero. At $t=0$, choose $\ell =|h{|}^{-1/{\lambda }_h}$; then $M\propto \mathrm{sgn}(h)|h{|}^{(D-{\lambda }_h)/{\lambda }_h}$, so $\delta ={\lambda }_h/(D-{\lambda }_h)$. HenceSimple algebra now gives the Rushbrooke scaling relation $\alpha +2\beta +\gamma =2$ and the hyperscaling relation $\alpha =2-D\nu$. Also $\beta \delta ={\lambda }_h/{\lambda }_t=\beta +\gamma$, giving the Widom scaling relation. These statements refer to leading singular contributions with the homogeneous two-field scaling assumptions. Above the upper critical dimension, the quartic coupling is dangerously irrelevant because it is still needed to stabilize the ordered phase; ordinary mean-field critical exponents then satisfy Rushbrooke and Widom but generally not hyperscaling. Marginal logarithms also require qualifications at the upper critical dimension.

For the connected-correlation scaling form $G(r)=|r{|}^{-(D-2+\eta )}{f}_G(|r|/\xi )$, the function $f_G(x)$ tends to a finite nonzero value as $x\to 0$ and decays exponentially at large $x$, up to powers of $x$, in a massive short-range phase. The scaling ansatz alone does not fix that prefactor. If the usual massive Gaussian field correlation tail $G(r)\propto r^{-(D-1)/2}{e}^{-r/\xi }$ applies, matching the powers gives $f_G(x)\sim B{x}^{(D-3)/2+\eta }{e}^{-x}$, with a correlation-length-dependent amplitude in $G$ fixed by matching the critical scaling form.

In physical coordinates the lattice sum has site-density factor $a^{-D}$. Its long-distance singular part therefore giveswhere ${\Omega }_{D-1}$ is the area of the unit sphere in $D$ dimensions. For $\eta <2$, finite $f_G(0)$, an integrable exponential tail and a nonzero integral, the last factor tends to a finite nonzero constant as $\xi \to \infty$. Microscopic distances contribute a regular background. Thus ${\chi }_s\propto {\xi }^{2-\eta }\propto |t{|}^{-(2-\eta )\nu }$, proving the Fisher scaling relationEquivalently the field has scaling dimension $(D-2+\eta )/2$, so its conjugate field has ${\lambda }_h=(D+2-\eta )/2$, which gives the same identity from the exponent formulas above.

A partition function sums the Boltzmann weights of all states: $Z={\sum }_s{e}^{-{\beta }_{T}H(s)}$ for discrete states, or the corresponding integral for continuous fields. It normalizes equilibrium probabilities. Its physical Helmholtz free energy is $\mathcal{F}=-k_{B}T\log Z$, and its free-energy density is $\mathcal{F}/V$. With a momentum ultraviolet cutoff ${\Lambda }_0$, a scalar-field example isThe normalization of the functional measure and field-independent energy terms must be tracked when discussing absolute free energy.

Choose a smaller cutoff $\Lambda$ and split the field into retained modes ${\varphi }_{<}$ with $|q|<\Lambda$ and eliminated modes ${\varphi }_{>}$ with $\Lambda <|q|<{\Lambda }_0$. Define the Wilsonian coarse-grained statistical Hamiltonian bySubstitution into the partition function proves $Z=\int \mathcal{D}{\varphi }_{<}{e}^{-{\beta }_T{H}_{\Lambda }[{\varphi }_{<}]}$ for every intermediate cutoff. A source retained in this definition similarly preserves long-wavelength connected correlation functions and response. Thus the coefficients in a Hamiltonian density must run with $\Lambda$: the dependence induced by eliminated modes compensates the dependence of the remaining field integral. Holding the original bare coefficients fixed while changing the cutoff would generally change the system. Exact integration generates all interactions allowed by symmetry, including higher powers, derivative terms and an additive constant; a local derivative expansion or a truncation to a few couplings is an extra approximation.

Cutoff independence makes arbitrary intermediate coarse-graining scales unphysical. It allows calculations at the scale of the observable, avoids repeatedly resolving irrelevant short-distance details and exposes the renormalization-group fixed points responsible for universality. A microscopic cutoff, such as a Bravais lattice spacing, can still label a physical microscopic model; taking it to infinity as a continuum limit requires tuning appropriate bare couplings. This is distinct from lowering an intermediate Wilsonian cutoff in a fixed model.

To interpret the zero-cutoff limit, define $M=V^{-1}\int \varphi (x)d^{D}x$, the spatially uniform mode of the order parameter. Keep this mode fixed and integrate the others, rather than integrating over $M$ too. Write $\varphi =M+\eta$, with $\int \eta =0$, and define the constrained order-parameter free energy byIn a finite periodic box, once $\Lambda$ is below the smallest nonzero momentum, all nonconstant modes have been eliminated. The remaining effective energy is $H_{\Lambda }[M]=V{U}_{\Lambda }(M)$, so its defining integral is exactly $Z_M$. Therefore, with consistent normalization,When $\mathcal{H}(\Lambda ,M)$ denotes this uniform effective Hamiltonian density, this is the stated identification of $F(M)$ with its zero-cutoff limit. It requires a free-energy density constrained at fixed $M$, not the unconstrained extensive free energy; all integrated fluctuations and additive constants are included, the zero mode in field theory is retained, and a density for the uniform configuration is well defined. The finite-volume argument avoids an interchange of the thermodynamic and cutoff limits without justification. On a homogeneous pure-phase branch one may use a local potential description, while a thermodynamic limit allowing phase separation can convexify the constrained potential. A bare nonconvex Landau polynomial is accordingly a homogeneous-branch approximation, not a claim that the exact equilibrium Legendre potential stays nonconvex.

In a field $h$, the remaining integral is $Z(h)=\int dM{e}^{-{\beta }_{T}V[F_c(M)-hM]}$, up to the consistently chosen zero-mode measure. The thermodynamic saddle then gives $f(h)={\inf }_M\{F_c(M)-hM\}$ and selects the equilibrium magnetization. Landau theory assumes that the relevant uniform potential can be approximated by a regular analytic expansion in $M$, with symmetry-constrained coefficients, and minimizes it. At tree level the integrated fluctuations are neglected, recovering the familiar bare polynomial. With fluctuations retained, the coefficients run and can become singular near criticality; the zero-cutoff identification of constrained free energy alone does not justify mean-field critical exponents.

For an ordinary short-range scalar thermodynamic critical point, the leading interaction is quartic. Absorb inverse temperature and the gradient coefficient into field normalization and writeScale $x=b{x}^{\prime }$ and $\varphi (x)=b^{-(D-2)/2}{\varphi }^{\prime }(x^{\prime })$ to leave the gradient term invariant. The quartic term changes to $g{b}^{4-D}\int {\varphi }^{\prime 4}/4!$, so its engineering scaling exponent is $4-D$. It is irrelevant at the Gaussian fixed point for $D>4$, relevant for $D<4$, and marginal for $D=4$. HenceThe loop expansion makes the breakdown concrete. A quartic-vertex correction is proportional to $g^2{I}_2(r)$, whereFor $0<D<4$, rescaling $q=\sqrt{r}k$ gives $I_2(r)\sim C_D{r}^{(D-4)/2}$, diverging as $r\to 0$. At $D=4$, radial integration giveswhich is logarithmically divergent. Above four dimensions its infrared limit is finite; the leading cutoff-dependent constant can be absorbed into a coupling renormalization. The effective expansion parameter is thus of order $g{\xi }^{4-D}$ below four, or $g\log (\Lambda \xi )$ at four, since $\xi \sim r^{-1/2}$. It eventually becomes large for any fixed bare interaction in the lower-dimensional critical limit, defeating the uncorrected mean-field expansion. The Ginzburg criterion gives the same threshold by comparing correlation-volume fluctuations with the squared mean order parameter.