= Solution
For a homogeneous isotropic solid with small displacement $\mathbf u$, <linear elasticity> gives $\sigma_{ij}=\lambda_L(\nabla\cdot\mathbf u)\delta_{ij}+2\mu e_{ij}$. <Momentum> balance becomes
$$
\rho\,\partial_t^2\mathbf u=(\lambda_L+\mu)\nabla(\nabla\cdot\mathbf u)+\mu\nabla^2\mathbf u.
$$
A <plane wave> with displacement parallel to its <wavevector> is longitudinal, whereas perpendicular displacement is transverse. Writing $K=\lambda_L+2\mu/3$ for the <bulk modulus> and $\mu$ for the <shear modulus>, their bulk speeds are
$$
\boxed{c_P=\sqrt{\frac{K+4\mu/3}{\rho}},\qquad c_S=\sqrt{\frac{\mu}{\rho}}.}
$$
There are two independent transverse polarizations. The <bulk modulus> supplies resistance to compression; shear rigidity supplies transverse restoring <stress>. An ordinary fluid has no static shear rigidity and cannot sustain these propagating bulk shear waves. A thin rod with freely contracting sides instead has longitudinal <speed> $\sqrt{Y/\rho}$, because its lateral <boundary conditions> differ from a bulk <plane wave>. <Reflection>, <refraction> and mode conversion occur at interfaces; a free surface also supports <Rayleigh waves>.
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