= Solution
Work in the fixed reference domain $\Omega_0$, and write $v_i=\delta x_i$. For the unconstrained <hyperelastic material> considered here, the material-first <nominal stress> is $N_{\alpha i}=W_{,A_{i\alpha}}$. The increment of the <deformation gradient> and the <incremental elastic moduli> are
$$
\delta A_{i\alpha}=v_{i,\alpha},\qquad c_{\alpha i\beta j}=\frac{\partial^2W}{\partial A_{i\alpha}\partial A_{j\beta}},\qquad \delta N_{\alpha i}=c_{\alpha i\beta j}v_{j,\beta}.
$$
All coefficients are evaluated at the base deformation. The <Hessian> symmetry is $c_{\alpha i\beta j}=c_{\beta j\alpha i}$; minor interchange symmetry is unnecessary. Linearizing the momentum equation gives
$$
\boxed{(c_{\alpha i\beta j}v_{j,\beta})_{,\alpha}+\rho_0\delta g_i=\rho_0\ddot v_i\quad\text{in }\Omega_0.}
$$
The reference <mass density> and reference outward unit <normal vector> are fixed, so the incremental boundary conditions are
$$
\boxed{\nu_\alpha c_{\alpha i\beta j}v_{j,\beta}=\delta t_i^0\quad\text{or}\quad v_i=\delta x_i^0.}
$$
In particular, no variation of a current surface normal belongs in this material-coordinate <traction> condition. If a material constraint were imposed, one would additionally linearize its <constraint reaction in hyperelastic stress> and its admissibility equation; the present moduli formula is the unconstrained one.
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