 Saturated isotropic magnetostrictive strain

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.