= Solution
The <displacement field> is $u(x)=y(x)-x$, and its <gradient> is $Du=Dy-I$. <Linear elasticity> expands about an undeformed stress-free reference state, taking $y=x+\epsilon v$ with $|Du|\ll1$ and external loads of the same small order. A rigid translation need not itself be small; it is the <deformation gradient> that controls the <infinitesimal strain tensor>.
For $W\in C^2$ with $DW(I)=0$, a <Taylor series> gives
$$
W(I+H)=W(I)+\frac12\mathbb A_{iA jB}H_{iA}H_{jB}+o(|H|^2),\qquad
\mathbb A_{iA jB}=\frac{\partial^2W}{\partial F_{iA}\partial F_{jB}}(I).
$$
The constant $W(I)$ can be set to zero. The <tensor> $\mathbb A$ is the <elasticity tensor>. Commutation of continuous <mixed partial derivatives> proves its major <symmetry>:
$$
\boxed{\mathbb A_{iA jB}=\mathbb A_{jB iA}.}
$$
<Material frame indifference> at a stress-free reference also makes the quadratic energy depend only on the <infinitesimal strain tensor> $e=(H+H^T)/2$, leading to the usual minor <symmetries> after spatial and reference indices are identified.
In this scaling, an external <force> of order $\epsilon$ may be evaluated on the reference geometry: its change due to the order-$\epsilon$ displacement contributes only at order $\epsilon^2$. Thus the leading forces are fixed <dead loads>. Likewise $F=I+O(\epsilon)$, $J=1+O(\epsilon)$ and $S=O(\epsilon)$, so the Piola transformation gives $T=J S F^{-T}=S+O(\epsilon^2)$. Hence \b[Cauchy and first Piola stresses agree to linear order]. This reasoning uses the usual small-load, stress-free linearization; an order-one prestress or leading follower-load effect requires a different expansion.
Back to article page
