Saturated isotropic magnetostrictive strain (${\epsilon }_{ij}^{\mathrm{mag}}=\frac{3{\lambda }_s}{2}(m_i{m}_j-{\delta }_{ij}/3)$)

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.

A full $180^{\circ }$ reversal leaves saturated isotropic magnetostrictive strain unchanged. A $90^{\circ }$ rotation generally changes it. In the simplest unconstrained orientation average over magnetic domains, replace $m_i{m}_j$ by $⟨m_i{m}_j⟩$, not by $⟨m_i⟩⟨m_j⟩$. Equal populations pointing along $+z$ and $-z$ have zero net magnetization but the same preferred strain as a single domain along $z$. An isotropic orientation distribution instead has $⟨m_i{m}_j⟩={\delta }_{ij}/3$ and zero average preferred strain. Elastic compatibility between real magnetic domains can modify the macroscopic response.

Reversing an applied magnetic field does not guarantee that all microscopic orientations retrace the same path. Hysteresis, magnetic domain-wall pinning, and changes in domain populations can alter the measured curve. The reversal identity concerns the same pair of magnetic states with opposite local magnetization; it is not a universal prediction that an arbitrary experimental sweep is a single-valued even function of applied magnetic field.