= Solution
{c}
Restore the particle label suppressed in the question, writing the static family as $\mathbf x_0(\mathbf X;\tau)$. At each fixed $\tau$, its <Cauchy stress tensor> satisfies
$$
\operatorname{div}\mathbf T_0(\tau)+\rho\mathbf b=\mathbf0.
$$
This <dynamic continuation of a static incompressible elastic deformation> sets $\tau=t$ to form a motion. At every time its <deformation gradient> is the same as that of the corresponding static member. For an elastic material the elastic part of the <Cauchy stress tensor> depends on that <deformation gradient>, while <incompressibility> leaves an arbitrary scalar <constraint reaction in hyperelastic stress>. Therefore write
$$
\mathbf T_0=\mathbf S(\mathbf F)-p_0\mathbf1,\qquad
\mathbf T=\mathbf S(\mathbf F)-p\mathbf1,
\qquad
\mathbf T-\mathbf T_0=(p_0-p)\mathbf1.
$$
Here $\mathbf1$ is the <identity matrix>. Subtracting static <balance of linear momentum> from dynamic <balance of linear momentum> gives
$$
\rho\ddot{\mathbf x}_0=\nabla(p_0-p).
$$
Since the homogeneous incompressible material has constant <mass density>, define $\xi=(p-p_0)/\rho$. Necessarily
$$
\boxed{\ddot{\mathbf x}_0(\mathbf X,t)
=-\nabla\xi(\mathbf x_0(\mathbf X,t),t).}
$$
Conversely, suppose this <acceleration potential> exists. Choose $p=p_0+\rho\xi$. Then
$$
\operatorname{div}\mathbf T+\rho\mathbf b
=\operatorname{div}\mathbf T_0+\rho\mathbf b-\rho\nabla\xi
=\rho\ddot{\mathbf x}_0.
$$
The added isotropic <stress> is symmetric, so <angular momentum conservation> remains satisfied. This proves both directions for the field equations and the elastic <constitutive equation>. The surface <tractions> for the motion are those of the adjusted <stress>; identical prescribed static <tractions> would impose an additional boundary restriction on $\xi$.
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