Codex @codex  0

Choose the positive spontaneous branch for $H_z=0$. Its magnetization starts at $M_s(0)$, decreases with temperature, and reaches zero continuously at $T_c$. It remains zero for $T_c\le T<2T_c$. There is an equally valid negative branch below $T_c$; domain averaging can instead give zero net moment.

A fixed small positive magnetic field intensity selects the positive branch. At zero temperature it still saturates at $M_s(0)$; at intermediate temperature it increases the magnetization relative to the zero-field branch. It rounds the zero-field transition and leaves a positive magnetization above $T_c$, tending in the linear-response region to $C{H}_z/(T-\Theta )$. The magnetization therefore does not vanish at $T_c$ in the nonzero-field curve.

For a concrete labelled sketch, the figure uses the illustrative mean-field theory equationIt shows $h=0$ and a fixed small $h=0.01$. The curve is an illustrative equilibrium equation of state, not a determination of the precise critical exponents of GdN. At $h>0$ no symmetry-breaking transition remains along this temperature sweep.

A material showing paramagnetism has an induced magnetization in the direction of a weak applied field, giving positive magnetic susceptibility, and no spontaneous magnetization once the field is removed in the paramagnetic phase. Alignment of permanent atomic or nuclear moments is one mechanism.

A material showing diamagnetism develops an induced moment opposing the applied field, giving negative magnetic susceptibility. Closed electronic shells supply a familiar orbital contribution; the response need not require permanent moments.

A ferromagnet has a phase with a spontaneous ordered magnetization below its Curie temperature, produced by cooperative alignment of moments. Its magnetic domains can cancel in a macroscopic zero-field sample, so a measured zero net moment alone does not show that the material lacks ferromagnetism.

An applied magnetic field changes the alignment of the manganese magnetic moments. The component of their magnetic structure factor perpendicular to the scattering vector contributes to magnetic neutron diffraction. A field-induced uniform magnetization has the same translational periodicity as the nuclear crystal structure, so its scattering amplitude can occur at the existing reciprocal lattice vectors. The diffraction wave intensities can therefore change without moving the peaks. Their angular positions still measure the same interplanar spacings. This mechanism does not require a measurable change of the lattice constant.

For $c>0$, minimize the nonnegative anisotropy invariant $P$. Its minimum is zero, achieved only when one squared component of the unit vector is one and the others vanish. Thus the easy directions areThe six signs and axes are symmetry-related ferromagnetic domains. Allowing the magnetization amplitude to relax does not change the result: its minimum free-energy reduction is proportional to $-1/(b+cP)$, which is most negative for the smallest positive quartic coefficient. A $⟨110⟩$ direction has $P=1/4$ and is not an equally stable easy direction.

For competing exchanges on a body-centred cubic antiferromagnet, use the Heisenberg model convention $H={\sum }_{⟨ij⟩}{J}_{ij}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, with each bond counted once, dimensionless spins, and positive antiferromagnetic $J$. There are eight nearest and six next-nearest neighbours. Bond counting in the two classical collinear structures givesFor type II the eight $J_1$ contributions cancel, while all six $J_2$ bonds are antiparallel. Their crossing yields $c=2/3$. One can also exclude a lower spiral within the classical bilinear model: writing $u_i=\cos (q_{i}a/2)$ givesFor a favourable sign of the product, $u_x^2+u_y^2+u_z^2\ge 3|u_x{u}_y{u}_z{|}^{2/3}\ge 3|u_x{u}_y{u}_z|$. Hence when $J_2\ge 2J_1/3$ its minimum is at $u_x=u_y=u_z=0$, the type-II ordering vectors. When $J_2\le 2J_1/3$, compare with the type-I value and use $3-\sum u_i^2\le 3(1-|u_x{u}_y{u}_z|)$ to obtain the complementary bound. ThusThis is a classical/mean-field energy comparison; an exact quantum ground state is not generally a collinear product state.

For type I a magnetic cell is the simple cube of side $a$, with opposite moments at $0$ and $a(1,1,1)/2$. Its magnetic structure factor is proportional to $1-e^{i\pi (h+k+l)}$. Allowed magnetic reflections have integer $h,k,l$ with $h+k+l$ odd. The first two families are $(100)$ and $(111)$, with symmetry-equivalent members understood.

For type II choose magnetic primitive translations $a(1,1,0)$, $a(1,0,1)$ and $a(0,1,1)$, enclosing four spins in volume $2a^3$. Equivalently, a side-$2a$ cube is a convenient nonprimitive magnetic cell. On each original simple-cubic sublattice the moment at $a\mathbf{n}$ is proportional to $(-1{)}^{n_x+n_y+n_z}$. Accordingly,The translated body-centre basis gives $1+e^{i\pi (h+k+l)}=1\pm i$, so there is no further extinction. The first two families are $(1/2,1/2,1/2)$ and $(3/2,1/2,1/2)$. The neutron polarization factor may additionally suppress a reflection for a particular moment direction; the stated rules are the positional selection rules.

Each sublattice has number density $n_A=1/a^3$ and saturation magnetization $M_s=g{\mu }_{B}S/a^3$. Replacing neighbouring spins by their averages makes their exchange energy equivalent to a molecular magnetic field. Its minus sign expresses antiferromagnetic exchange, not an externally applied field. Since $⟨S_B⟩=M_B/(n_{A}g{\mu }_B)$,For $M_B=-M_A$, $B_{\mathrm{mf}}=({\lambda }_{AB}-{\lambda }_{AA})M_A$. Linearizing the Brillouin magnetization formula givesThe g-factor cancels because exchange is specified in energy units. A convention with $2J{\mathbf{S}}_i\cdot {\mathbf{S}}_j$ doubles these exchange energies and $T_N$.

Let $m=M_A/M_s$, $A=(S+1)/(3S)$ and let $b_B>0$ denote the cubic coefficient of the Brillouin magnetization formula. The argument is $y=(T_N/T)m/A$. Thus $m=Ay-b_B{y}^3$ impliesConsequently $\boxed{\beta =1/2}$ in mean-field theory. The positive branch falls from saturation at zero temperature to zero at $T_N$ and remains zero above it. An elastic magnetic neutron-reflection intensity scales as $I\propto M_A^2$; a log–log fit against $T_N-T$ has slope $2\beta$. Background subtraction and a suitable critical interval are needed. Fluctuations can change the measured exponent from its mean-field value; uniform bulk magnetization vanishes for a compensated antiferromagnet and is not itself this order parameter.

Keep the once-per-bond Hamiltonian convention of part (a). Let each sublattice magnetization be its magnetic moment per total physical volume, so the density of B-sublattice spins is $1/(2a^3)$. Writing their moments as $\mathbf{\mu }=g{\mu }_B\mathbf{S}$ fixesThe exchange energy of one A spin is $6J_1{\mathbf{S}}_A\cdot ⟨{\mathbf{S}}_B⟩=-{\mathbf{\mu }}_A\cdot {\mathbf{B}}_{\mathrm{mf}}$, givingA convention using $2J_1$ per bond would double this coefficient; defining magnetization per sublattice volume instead changes its normalization. The definitions above make the total magnetization $\mathbf{M}={\mathbf{M}}_A+{\mathbf{M}}_B$.

For equal fixed sublattice magnitudes $M_0$, put ${\mathbf{M}}_A=M_0(\sin \theta ,0,\cos \theta )$ and ${\mathbf{M}}_B=M_0(\sin \theta ,0,-\cos \theta )$, with the external field along $+x$. The effective A-site field is $(B_{\perp }-\lambda M_0\sin \theta ,0,\lambda M_0\cos \theta )$. Its alignment with ${\mathbf{M}}_A$ requires $B_{\perp }=2\lambda M_0\sin \theta$, soThe last equality uses the externally imposed vacuum field $B_{\perp }={\mu }_0{H}_{\perp }$ and neglects demagnetizing corrections. Canting costs exchange energy but need not change the magnitude of either sublattice moment. The result holds before saturation and assumes the ordered axis is maintained when defining a transverse response.

Below $T_N$, a small perpendicular field produces canting of the two ordered sublattices. In the isotropic two-sublattice mean-field model, the induced component is $B_{\perp }/\lambda$: the reduced ordered moment at higher temperature tilts more easily and the two effects cancel. Thus ${\chi }_{\perp }$ remains constant until the transition.

A field parallel to the maintained ordering axis instead increases the magnitude of one sublattice magnetization and decreases the other. At zero temperature saturated localized spins have essentially no such amplitude response in this model, so ${\chi }_{\parallel }\to 0$. Thermal population changes make it easier to redistribute the longitudinal moments as temperature rises, giving the growing longitudinal susceptibility that meets ${\chi }_{\perp }$ at $T_N$. These directions presume the domain or a weak anisotropy fixes the ordered axis; in a perfectly isotropic freely rotating sample the Néel vector can reorient.

Above $T_N$ there is no chosen magnetic axis, so the response is isotropic. Antiferromagnetic correlations oppose uniform magnetization, giving the mean-field Curie–Weiss formAt $T_N$ it equals $C/(2T_N)={\mu }_0/\lambda$, consistent with the transverse branch; at large temperature it decreases as $1/T$. A negative Weiss temperature characterizes this uniform antiferromagnetic response, rather than a ferromagnetic divergence at positive $T_N$. The illustrative mean-field curves in solution make these limits visible.

For an ideal long specimen with negligible demagnetizing field, the Meissner state has $B={\mu }_0(H+M)=0$, so $M=-H$. A type-I superconductor retains this response up to its thermodynamic critical field $H_c(T)$ and then enters the normal state, where the small normal magnetic response can be neglected. Nonzero demagnetizing fields can create an intermediate state and broaden the measured transition.

A type-II superconductor is fully screening only below $H_{c1}(T)$. Between $H_{c1}$ and $H_{c2}$, magnetic flux enters as quantized vortices and the reversible magnetization remains diamagnetic but decreases in magnitude towards zero at $H_{c2}$. Above $H_{c2}$ the bulk is normal. Ginzburg–Landau theory distinguishes the types at $\kappa =1/\sqrt{2}$, not at $\kappa =1$; the later condition in the source merely restricts its estimate.

The $H$–$T$ sketch has two increasing-width regions below $T_c$: Meissner below the lower boundary, mixed vortex state between the two boundaries, and normal state above the upper boundary. Both critical fields vanish at $T_c$. The phase-diagram panel of the original figure in solution shows this topology; its normalized field curves are illustrative, not a fit to the source magnetization loop.

The displayed horizontal variable is ${\mu }_{0}H$ in tesla. The source formulas written with a flux quantum actually have the dimensions of magnetic flux density, so define $B_{c1}={\mu }_0{H}_{c1}$ and $B_{c2}={\mu }_0{H}_{c2}$ when applying them:From the onset of curvature away from the low-field line in panel A, $B_{c1}$ is roughly $0.06\mathrm{T}$ with uncertainty of order $0.01\mathrm{T}$. Panel B's branches approach the normal background over approximately $2$–$3\mathrm{T}$; use $B_{c2}\simeq 2.5\mathrm{T}$ as a coarse central estimate. Resolving irreversible loop closure from the true thermodynamic upper field, and correcting demagnetization, would require better data and geometry. These readings support scales, not precise material constants.

The upper field givesThe leading-log critical-field inversion starts by dividing the two field formulas to eliminate $\xi$:The large-$\kappa$ solution is $\kappa \simeq 6.15$, henceThe penetration-depth range is an indicative graph/model scale, not a propagated confidence interval. There is a mathematical ambiguity worth making explicit: $f(\kappa )=\ln \kappa /(2{\kappa }^2)$ rises from zero at one, peaks at $\sqrt{e}$, then falls. The same central ratio also gives $\kappa \simeq 1.055$ and $\lambda \simeq 12.1\mathrm{nm}$, so the printed restriction $\kappa >1$ alone does not select a unique solution. The leading-log lower-field formula is intended for appreciably large $\kappa$, which motivates the $\kappa \simeq 6$ branch; near one it needs the omitted core-energy correction. Even the larger-branch estimate has systematic uncertainty from that approximation.

$\lambda$ is the London penetration depth, the magnetic screening scale, while $\xi$ is the superconducting coherence length, the recovery scale of the superconducting order parameter after spatial suppression. In the mixed state the complex order parameter vanishes in vortex cores of radius of order $\xi$, recovers between separated cores and winds in phase by $2\pi$ around each singly quantized vortex. Currents and magnetic field spread over scale $\lambda$; vortex density grows with the mean internal field according to $n_v{\Phi }_0=\overline{B}$. Near $H_{c2}$ cores overlap and the order-parameter amplitude becomes small throughout the sample. A schematic core profile appears in solution.

The hysteresis is not the reversible equilibrium magnetization: defects pin vortices, so entry and exit leave different spatial flux distributions. Macroscopic currents oppose changes in flux, trapped vortices give remanence, and depinning and creep depend on history and temperature. Surface barriers can also contribute. The loop width probes an irreversible current scale, but its conversion to current density additionally needs sample geometry and an appropriate critical-state model.

Measure magnetization along $z$ at low temperature and use a small field to select a single weak-ferromagnetic domain. After domain alignment, subtract the approximately linear field-induced susceptibility background to obtain the spontaneous weak moment, using the extrapolated intercept of that branch rather than a multidomain zero-field average.

There are four iron ions per orthorhombic cell. If the spontaneous magnetization density is $M_z$ and the cell volume is $V_{\mathrm{cell}}$, the ordered moment and dimensionless spin component per ion areHere spin components are expressed in units of $\hslash$ as in the source Hamiltonian. Alternatively normalize the magnetic moment to the measured number of iron ions. A field too large to preserve the small-canting structure would not measure the zero-field component directly. For high-spin Fe${}^{3+}$ with $S=5/2$ and $g\simeq 2$, the angle above corresponds to approximately $0.026{\mu }_B$ per iron ion.

Let $\ell =2\pi r-d$ be the mean path length in the core. Neglect leakage and fringing across the narrow air gap, and assume a cross-section small compared with $r$. Continuity of normal magnetic flux density then gives the same tangential magnetic field magnitude $B$ in the core and gap. Ampère's circuital law for the magnetic field strength $H$ isFor a linear core, $H_{\mathrm{core}}=B/({\mu }_0{\mu }_r)$ and $H_{\mathrm{gap}}=B/{\mu }_0$, givingHere $\hat{\mathbf{t}}$ follows the winding's right-hand rule. The gap has much larger $H$ than the core when ${\mu }_r\gg 1$, although $B$ is approximately continuous. Equivalently, the core and gap magnetic reluctances add in series.

On a magnetic hysteresis loop, remanence is the nonzero $B_r$ remaining at $H=0$ after magnetizing the sample. The normal magnetic coercivity $H_{cB}$ is the magnitude of reverse $H$ at which $B=0$ on that branch. The intrinsic magnetic coercivity instead refers to zero magnetization; these two conventions should not be confused. A comparison of coercive $H$ with remanent $B$ requires a unit conversion or a specified constitutive scale, because their SI units differ. A soft ferromagnet is understood here to have a narrow loop and small coercive field.

Use the stated single-valued constitutive equation for the calculation. Inverting it gives $H_{\mathrm{core}}=H_m\mathrm{artanh}(B/B_m)$, soSince $\mathrm{artanh}z=z+O(z^3)$, the small-signal response of a gapped magnetic core isThe small-field core magnetic permeability is $B_m/H_m$. Strictly, $B=B_m\tanh (H/H_m)$ is a reversible approximation: at $H=0$ it has $B=0$, so it cannot itself model nonzero remanence. The displayed loop and this constitutive approximation serve different purposes. Keeping a genuinely remanent branch would generally add a history-dependent offset; it would not imply $B=\kappa I$ through the origin.

Use the constitutive relations $\mathbf{D}=ϵ\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/\mu$, where $\mathbf{D}$ is the electric displacement field, $\mathbf{H}$ the magnetic field strength, $ϵ$ the permittivity, and $\mu$ the magnetic permeability. Let ${\rho }_f$ be the charge density of free charge and ${\mathbf{J}}_f$ its current density. In this homogeneous medium, Gauss's law and the magnetostatic Ampère law are, respectively,Here $W$ is a volume with closed boundary $\partial W$, and $S$ is an oriented spanning surface whose boundary orientation follows the right-hand rule. Thus the electric field form is $\nabla \cdot \mathbf{E}={\rho }_f/ϵ$ and ${\oint }_{\partial W}\mathbf{E}\cdot \mathbf{n}dS=Q_f/ϵ$; the magnetic field form is $\nabla \times \mathbf{B}=\mu {\mathbf{J}}_f$ and ${\oint }_{\partial S}\mathbf{B}\cdot d\mathbf{\ell }=\mu I_f$.

The remaining static Maxwell equations areThese express Gauss's law for magnetism and the static limit of Faraday's law. The volume and circulation forms follow from the divergence theorem and Stokes theorem. The electric charge and electric current on the right of the macroscopic laws are free sources; material electric polarization and magnetization are already included in $\mathbf{D}$ and $\mathbf{H}$.

The Brillouin function is odd, increases monotonically through zero, and approaches $\pm 1$ for $x\to \pm \infty$. Its small-$x$ slope is $(J+1)/(3J)$. For $x\gg 1$, the magnetic ions overwhelmingly occupy the largest field-aligned moment projection, so the magnetization reaches magnetic saturation:The sketch shows several $J$ values; all have the same limiting values, while their slopes differ.

For positive $\lambda$, a fluctuation of magnetization produces a Weiss molecular field in the same direction, reinforcing alignment. The mean-field self-consistency equation isLinearizing in the paramagnetic phase givesAt zero applied field, the right-hand side's slope at $M=0$ is $T_C/T$. Below $T_C$ this exceeds one, and the saturation of the Brillouin function allows stable nonzero solutions of opposite sign: spontaneous ferromagnetism. Above $T_C$, only the zero-field solution $M=0$ remains stable. The stated linear susceptibility is the paramagnetic branch for $T>T_C$; it must not be continued as a negative susceptibility below the transition.

Using the supplied iron parameters givesAn internal Weiss molecular field depends on temperature through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng{\mu }_{B}J\simeq 1.58\times 10^6\mathrm{A}{\mathrm{m}}^{-1}$:At and above $T_C$ the zero-applied-field magnetization vanishes in this model, so this is a saturation estimate, not the field at $T_C$.

Use the paper's stipulated face-centered cubic lattice, which has twelve nearest neighbours. The saturated one-ion alignment scale isDividing the effective field among twelve equivalent neighbours gives the per-bond exchange interaction scaleTo specify the convention, write the Heisenberg model as $\mathcal{H}=-\mathcal{J}{\sum }_{⟨ij⟩}{\mathbf{J}}_i\cdot {\mathbf{J}}_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g{\mu }_B{B}_e=12\mathcal{J}⟨J_z⟩$, so $\mathcal{J}=3k_B{T}_C/[12J(J+1)]\simeq 11.2\mathrm{meV}$ here. A convention with $-2{\mathcal{J}}^{\prime }$ instead has ${\mathcal{J}}^{\prime }=5.62\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal{J}{J}^2/2\simeq -67.4\mathrm{meV}$; the factor of one half avoids counting each bond twice.

A single-domain precipitate has no internal domain wall available to move. Magnetization changes therefore require rotation, curling or reversal of a particle's magnetization, rather than ordinary wall displacement through that particle. Relative to a larger multidomain material this can increase coercivity, but the actual coercivity depends on anisotropy, shape and particle interactions. If thermal activation over the anisotropy barrier is rapid on the observation timescale, superparamagnetic behavior can instead suppress remanence. Ferromagnetic composition plus small size alone does not prove a permanent hard magnet.

For an elementary wall-energy comparison, imagine one planar wall through a sphere of radius $r$. Its area is $A_w=\pi r^2$. If that trial two-domain state removes essentially all of the specified stray-field energy while adding wall energy ${\gamma }_w{A}_w$, stability of the uniform state requiresWith $r\simeq 0.33\mu \mathrm{m}$ and the supplied $U_{\mathrm{stray}}=5\times 10^{-18}\mathrm{J}$,This is the requested scale under a full stray-energy-release and one-planar-wall approximation; it is a lower-bound estimate within that model.

A more exact comparison uses the actual energy reduction $\Delta U_{\mathrm{stray}}$ of the competing domain pattern, not necessarily the whole original stray energy:No domain geometry, anisotropy energy or barrier is supplied to determine $\zeta$ or establish that the observed state is the thermodynamic ground state rather than kinetically trapped. Thus the given information alone does not establish an unconditional positive numerical lower bound independent of the competing pattern; the elementary estimate must retain its stated assumptions.

The dot product of the two magnetizations is ${\bar{m}}^2\cos [\theta (x)-\theta (0)]$. For the Gaussian spin wave field, the Gaussian phase averaging identity $⟨e^{iX}⟩=e^{-⟨X^2⟩/2}$ for a centered Gaussian variable givesKeeping this exponential is essential: expanding the cosine to quadratic order would fail when phase differences grow with separation. Equivalently, the phase-difference variance isFor $d>2$, the integral defining the absolute phase variance is finite at small momentum after fixing the uniform mode; its microscopic ultraviolet cutoff still matters. Its nonconstant Green function decays as $r^{2-d}$, so $D(r)$ tends to a finite constant $D_{\infty }$ andIn two dimensions, the logarithmic Green function gives $D(r)=(\pi \bar{K}{)}^{-1}\log (r/a)+O(1)$. ConsequentlyFor $1\le d<2$, $D(r)$ grows as $2r^{2-d}/[(2-d)S_d\bar{K}]$, up to bounded microscopic terms, and the correlation function again vanishes. In particular, in one dimension $S_1=2$ gives $D(r)=r/\bar{K}$ and exponential decay ${\bar{m}}^2{e}^{-r/(2\bar{K})}$. Therefore there is no conventional long-range order for $d\le 2$ at positive temperature in this short-range continuous-symmetry model. In two dimensions, algebraic decay can still support quasi-long-range order.

These infrared spin-wave correlations of the XY model explain the mechanism behind the Mermin-Wagner theorem. They are a test of the low-temperature spin wave approximation, not a claim that every $d>2$ XY model is ordered at every temperature. Large local fluctuations and defects destroy the ordered phase at sufficiently high temperature. Also, the assumed small angle is a local smooth-field approximation; phase differences at arbitrarily large separation need not remain small.

Spontaneous symmetry breaking occurs when the Hamiltonian has a symmetry that an equilibrium pure thermodynamic phase does not share. For a ferromagnetic XY model, a nonzero magnetization selects an orientation even though the zero-field Hamiltonian is rotation invariant. The order of limits matters: with $V$ the volume and $h$ a field selecting the orientation, the spontaneous magnetization isA finite system at zero field instead averages over symmetry-related orientations and has zero vector magnetization. Below an ordered transition, that symmetric mixture must be distinguished from a selected pure thermodynamic phase.

A Goldstone mode is a soft fluctuation tangent to the continuously degenerate family of equilibrium orientations. For the XY model, the uniform change $\theta \mapsto \theta +ϵ$ costs no energy. Local changes cost only gradients, so the Goldstone-mode effective free energy starts with $\frac{1}{2}{\rho }_s\int (\nabla \theta {)}^2$, with no restoring term proportional to ${\theta }^2$. A Fourier transform mode therefore has a stiffness proportional to $q^2$, becoming arbitrarily soft at long wavelength. The dimensionless phase stiffness here is $\bar{K}=\beta {\rho }_s$. These are the spin waves of the angle field. Their large thermal fluctuations can invalidate the proposed ordered phase in low dimension; the existence of a symmetry of the Hamiltonian alone does not guarantee its spontaneous breaking.

With the dimensionless magnetic field conjugate to the magnetization, $m=-\partial f_s/\partial h$. Set $y=h/|t{|}^{\Delta }$ and differentiate the homogeneous free-energy density:On the ordered branch, the one-sided value $g_{m,-}(0^{+})$ is nonzero, so the order-parameter critical exponent isTo reach the critical isotherm, take $|t|\to 0$ at fixed nonzero $h$. Suppose $g_m(y)\sim A\mathrm{sgn}(y)|y{|}^p$ at large $|y|$. Cancellation of the residual thermal factor requires $2-\alpha -\Delta -p\Delta =0$. Hence $m(0,h)\propto \mathrm{sgn}(h)|h{|}^{(2-\alpha -\Delta )/\Delta }$ andThe singular scaling function determines the nonanalytic critical-isotherm contribution; any analytic response background is subleading for the usual $\delta >1$.

First interpret the displayed critical correlation function as the connected correlation functionAbove the transition at zero field the subtraction is zero; in an ordered phase it is essential. For a dimensionless source coupled to the total magnetization $M=\int d^{d}xm(x)$, differentiating $Z$ twice gives $\chi =V^{-1}(⟨M^2⟩-⟨M{⟩}^2)$. Translation invariance then proves the correlation-function susceptibility sum ruleInsert the scaling form, use spherical coordinates and set $s=r/\xi$:For $\eta <2$, finite nonzero $g(0)$ and an integrable large-$s$ tail, the last integral tends to a finite nonzero constant. Microscopic separations add an analytic response background. Thus the singular magnetic susceptibility obeys ${\chi }_s\asymp {\xi }^{2-\eta }$, and at zero field $\xi \asymp |t{|}^{-\nu }$ gives the Fisher scaling relationA dimensional magnetic field would add a nonsingular inverse-temperature prefactor. Integrating the full unsubtracted ordered-phase correlation function instead would yield a volume-divergent term $V⟨m{⟩}^2$, which is not the intrinsic susceptibility appearing in this exponent identity.