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Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

A magnetic domain wall separates differently oriented magnetic domains. Its motion changes their volumes and average magnetization. It is a magnetic domain wall.

An original two-magnetic domain model makes the collective change explicit. Suppose opposite regions have magnetizations $+M_s$ and $-M_s$ along the measuring axis, where $M_s$ is the local saturation magnetization. If a magnetic domain wall sweeps volume $\Delta V$ from the negative region into the positive one, in a specimen of fixed volume $V$, thenThe factor two comes from replacing $-M_s$ by $+M_s$. The local magnitude $M_s$ stays fixed while the average magnetization changes; the event size depends on the swept volume, rather than a universal single-spin increment.

Magnetic susceptibility describes how readily a magnetic field changes a material's magnetization. The field variable and normalization matter: volume magnetic susceptibility in the International System of Units is dimensionless, whereas a derivative of total magnetic moment with respect to $B$ has units of $\mathrm{J}{\mathrm{T}}^{-2}$. A slope measured against the applied magnetic field intensity also depends on the sample's demagnetizing field.

For an isotropic, reversible linear response through the origin, writeHere $\mathbf{M}$ is magnetization, or magnetic moment per volume; ${\mathbf{H}}_{\mathrm{int}}$ is the internal magnetic field intensity; ${\mathbf{B}}_{\mathrm{int}}$ is the internal magnetic flux density; and ${\mu }_0$ is the vacuum permeability. Both $M$ and $H$ have units $\mathrm{A}{\mathrm{m}}^{-1}$. The magnetic permeability is ${\mu }_0(1+{\chi }_v)$. ^[ref]

The independent-moment magnetic response below shows why one number cannot describe every field: magnetization approaches a saturation magnetization, while its differential magnetic susceptibility approaches zero. Anisotropy requires a magnetic susceptibility tensor; hysteresis requires specifying the preparation and sweep branch as well as the bias magnetic field, temperature, and timescale. The Gouy balance turns a small response into a measurable force without presuming that its readout is already the intrinsic volume magnetic susceptibility.

A bulk ferromagnet generally separates into magnetic domains. Within each domain many moments align and the local magnetization can be large, but different domain orientations partially or almost completely cancel the vector sum measured for the entire specimen. This reduces its external stray field and magnetostatic energy. Formation of magnetic domain walls costs exchange and anisotropy energy, so the actual pattern balances these costs with the reduction in stray-field energy.

A strong applied field can move domain walls and rotate moments toward a nearly saturation magnetization. When the field is removed, pinning and the sample's magnetic history determine the remanent moment. Thus a small net moment does not show that local ferromagnetic order has disappeared, or that moments within each domain are all randomly oriented. At finite temperature, thermal spin fluctuations also reduce the local ordered moment below its zero-temperature maximum even in a single-domain specimen. These two reductions—between-domain cancellation and within-domain thermal reduction—should be distinguished.

Take two localized spin orientations with moments $\pm {\mu }_B$ along the applied magnetic field. This assumes an electron-like spin $g$ factor of two. Their energies are $\mp {\mu }_{B}B$, so the one-moment partition function is $Z_1=2\cosh x$, with $x={\mu }_{B}B/(k_{B}T)$. The thermal mean moment is $k_{B}T{\partial }_B\ln Z_1={\mu }_B\tanh x$, and the spin-one-half paramagnetic magnetization isFor $|x|\ll 1$, $M\simeq n{\mu }_B^{2}B/(k_{B}T)$. With the same SI convention $\chi =\partial M/\partial H$ and $B\simeq {\mu }_{0}H$,This Curie law is a zero-field or weak-field susceptibility. At arbitrary field the differential response is $(C/T){\mathrm{sech}}^{2}x$, and $M$ saturates at $n{\mu }_B$; extending the linear law to unlimited field would contradict that finite saturation magnetization.

A macroscopic nickel specimen can split into magnetic domains, each with nonzero local spontaneous magnetization but with different orientations. Their vector contributions can cancel, giving an almost zero measured net magnetization below the Curie temperature. This reduces the external stray-field energy, at the expense of magnetic domain wall energy; the balance favours a multidomain state in many ordinary specimens.

An applied magnetic field can move magnetic domain walls and rotate the domain moments, revealing a large response and eventually saturation magnetization. Thus a zero average does not mean that each local moment has become thermally disordered or that the material is above its Curie temperature. Preparation, shape and domain-wall pinning influence remanence.

Ferromagnetic interactions enhance the response above the ordering transition. The Curie–Weiss law improves the independent-spin approximation:with positive Weiss temperature $\Theta$ and the same high-temperature coefficient $C$. In a simple mean-field theory, take $\Theta =T_c=70\mathrm{K}$. More generally $\Theta$ is fitted and need not equal $T_c$, and the simple pole does not give exact critical behaviour arbitrarily close to the transition.

At zero temperature, a selected ferromagnetic orientation has all magnetic moments aligned. Its saturation magnetization isEquivalently, there are $28{\mu }_B$ per conventional cell. The moment points along either allowed easy-axis direction at zero magnetic field. This is the local or single-domain spontaneous magnetization; a macroscopic multidomain sample can have smaller, even zero, net magnetization in the absence of a selecting field.

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.

Consider a mechanically free isotropic model of a saturated ferromagnet, at fixed temperature and fixed magnitude of magnetization. Let $\mathbf{m}=\mathbf{M}/M_s$ be a unit vector, where $M_s$ is the saturation magnetization. The magnetostrictive strain is an infinitesimal strain tensor, measured relative to a reference with randomly oriented magnetic domains. This reference convention matters: it sets the orientational average of the preferred strain to zero. Define ${\lambda }_s$, the saturation magnetostriction, as the longitudinal relative length change in this model.

Rotational symmetry permits a symmetric second-rank tensor made from $\mathbf{m}$ of the form $a{\delta }_{ij}+b{m}_i{m}_j$, where ${\delta }_{ij}$ is the Kronecker delta. If we restrict attention to a change of shape with zero first-order volume change, its trace gives $3a+b=0$. Requiring the longitudinal component to equal ${\lambda }_s$ gives $a+b={\lambda }_s$. HenceThe tensor product $\mathbf{m}\otimes \mathbf{m}$ is symmetric and unchanged under $\mathbf{m}\mapsto -\mathbf{m}$. It uses two copies of the magnetic direction, so the third-rank isotropic tensor obstruction in isotropic linear magnetostriction vanishes does not apply. More generally, a small-magnetization expansion can contain $a{M}^2{\delta }_{ij}+b{M}_i{M}_j$; isotropy alone does not require its trace to vanish. The displayed saturated formula isolates the volume-preserving orientational contribution, rather than asserting that every material has zero volume magnetostriction.

Along a measuring unit vector $\mathbf{n}$, with $\mathbf{n}\cdot \mathbf{m}=\cos \theta$, the fractional length change isFor a stipulated ${\lambda }_s=20\times 10^{-6}$ and reference length $L=100\mathrm{mm}$, alignment gives $+2\mu \mathrm{m}$ and a perpendicular orientation gives $-1\mu \mathrm{m}$. Rotating from perpendicular to parallel therefore changes the length by $3\mu \mathrm{m}$, even though the saturated longitudinal value relative to the random reference is $2\mu \mathrm{m}$. A negative ${\lambda }_s$ reverses these elongation and contraction signs.

The eigenvalues of this infinitesimal strain tensor are ${\lambda }_s,-{\lambda }_s/2,-{\lambda }_s/2$, so $\Delta V/V=\mathrm{tr}{\epsilon }^{\mathrm{mag}}=0$ to first order. This is not exact finite incompressibility: using the illustrative deformation gradient $F=I+{\epsilon }^{\mathrm{mag}}$ givesA finite-deformation law would need higher-order corrections if exact volume conservation were required.

Saturation magnetization is the limiting magnetization when the responding magnetic moments are fully aligned within the stated model or experimental regime. In independent-moment magnetic response, $M_s=n\mu$. Reaching a large magnetization does not imply a large differential magnetic susceptibility: the remaining slope can be almost zero.