= Saturated isotropic magnetostrictive strain
{title2=$\varepsilon^{\rm mag}_{ij}=\frac{3\lambda_s}{2}(m_i m_j-\delta_{ij}/3)$}
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 $a+b=\lambda_s$. Hence
$$
\varepsilon^{\rm mag}_{ij}=\frac{3\lambda_s}{2}\left(m_i m_j-\frac{\delta_{ij}}{3}\right),\qquad
\varepsilon_\parallel=\lambda_s,\qquad
\varepsilon_\perp=-\frac{\lambda_s}{2}.
$$
The <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 $a M^2\delta_{ij}+b M_iM_j$; <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 is
$$
\frac{\Delta L_{\mathbf n}}{L}=n_i\varepsilon^{\rm mag}_{ij}n_j
=\frac{\lambda_s}{2}(3\cos^2\theta-1).
$$
For a stipulated $\lambda_s=20\times10^{-6}$ and reference length $L=100\,\mathrm{mm}$, alignment gives $+2\,\mathrm{\mu m}$ and a perpendicular orientation gives $-1\,\mathrm{\mu m}$. Rotating from perpendicular to parallel therefore changes the length by $3\,\mathrm{\mu m}$, even though the saturated longitudinal value relative to the random reference is $2\,\mathrm{\mu m}$. 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 $\Delta V/V=\operatorname{tr}\varepsilon^{\rm mag}=0$ to first order. This is not exact finite <incompressibility>: using the illustrative <deformation gradient> $F=I+\varepsilon^{\rm mag}$ gives
$$
\det F=(1+\lambda_s)(1-\lambda_s/2)^2
=1-\frac34\lambda_s^2+\frac14\lambda_s^3.
$$
A finite-deformation law would need higher-order corrections if exact <volume> conservation were required.
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