# Can a wave propagate in an elastic fluid in the absence of volume forces?

A motion (wave) $\mathbf{x}: \mathcal{B}_0 \times [t_0,t_1] \to \mathcal{E}:$ such that
$q-o = \mathbf{x}(p,t)=(p-o)+\mathbf{a}_0 cos(\mathbf{k}_0\cdot(p-o) - \omega_0 t)$

can propagate in an elastic fluid with constitutive equation:
$\mathbf{T}=-\pi(\rho)\mathbf{I}$

not subject to any volume force? What restrictions on polarization vector $\mathbf{a}_0$ and the vector $\mathbf{k}_0$ are needed? $\mathbf{a}_0$ and $\mathbf{k}_0$ should be parallel? What restrictions on $\pi(\rho)$ are needed?

I can write the balance equation of linear momentum, in an eulerian description I suppose: $\rho \mathbf{a} = \mathrm{div}_q \mathbf{T} = -\nabla_q \pi = - \pi' \nabla_q \rho$

But it is easier to compute the acceleration in a lagrangian description:
$\ddot{\mathbf{x}}(p,t) = -\omega_0^2 \mathbf{a}_0cos(\mathbf{k}_0\cdot(p-o) - \omega_0 t)$

At this point computing $\nabla_q\rho$ is not easy I think. There is some mistake here? There is a better and simpler way to prove the results?

$q-o = \mathbf{x}(p,t)$, so $q$ it's the position of $p$ after a time $t$, it's the Lagrangian description of the motion. – unlikely Jan 31 '14 at 14:25