= Solution
Average the equations in part <1/b> and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf v=\delta\mathbf u$, $m=\delta M$ and $\pi=\delta p$, the <linearized equations> are
$$
\partial_tv_i=-\partial_i\pi+\partial_jm_{ij},\qquad
\partial_tm_{ij}=v_A^2(\partial_jv_i+\partial_iv_j),\qquad
\partial_iv_i=0.
$$
No unmodelled small-scale response has been inserted into this closure.
Introduce the <fluid displacement> $\boldsymbol\xi$ by $\mathbf v=\partial_t\boldsymbol\xi$ and choose $\nabla\cdot\boldsymbol\xi=0$. Integrating the <magnetic second-moment tensor> equation gives
$$
\boxed{m_{ij}
=v_A^2(\partial_i\xi_j+\partial_j\xi_i)+s_{ij}(\mathbf r),}
$$
where $s$ is a time-independent integration tensor set by the initial <stress>. For a pure oscillatory <normal mode>, $s=0$. Its <restoring force> is therefore $v_A^2\nabla^2\boldsymbol\xi$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with <shear modulus> $\rho_0v_A^2$.
For a <plane wave> proportional to $e^{i(\mathbf k\cdot\mathbf r-\omega t)}$, with $\omega\ne0$, <incompressibility> gives $\mathbf k\cdot\widehat{\mathbf v}=0$. The tensor equation and <magnetohydrodynamic momentum equation> become
$$
\widehat m_{ij}
=-\frac{v_A^2}{\omega}
(k_i\widehat v_j+k_j\widehat v_i),\qquad
-i\omega\widehat v_i=-ik_i\widehat\pi+ik_j\widehat m_{ij}.
$$
Dotting the second equation with $\mathbf k$ yields $\widehat\pi=0$ for $k\ne0$, since $k_i k_j\widehat m_{ij}=0$. Substitution then gives
$$
\boxed{\omega^2=v_A^2k^2,\qquad
\widehat m_{ij}=iv_A^2(k_i\widehat\xi_j+k_j\widehat\xi_i),\qquad
\mathbf k\cdot\widehat{\boldsymbol\xi}=0.}
$$
There are two independent <transverse polarizations>. The <phase speed> is $v_A$ in every direction, and the two signs of $\omega$ give <group velocities> $\pm v_A\widehat{\mathbf k}$. These <magnetoelastic waves in an isotropic magnetic tangle> exist because a shear deformation distorts the second moment of the tangled <magnetic field>, producing a restoring <magnetic tension> even though its mean is zero.
For comparison, a uniform <magnetic field> with <Alfvén velocity> $\mathbf B_0$ gives the incompressible <Alfvén wave> relation $\omega^2=(\mathbf k\cdot\mathbf B_0)^2$. Its <frequency> depends on the direction of $\mathbf k$, and <energy> propagates along $\mathbf B_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through <magnetic tension>, rather than through compressive sound.
\Image[/past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2019-magnetoelastic-wave-speeds.png]
{title=Directional phase speeds for a magnetic tangle and a uniform magnetic field with the same mean-square strength}
{description=The radial coordinate is phase speed divided by the tangle wave speed. The circular tangle curve is direction independent; the uniform-field curve is proportional to the absolute cosine of the propagation angle. These are phase-speed curves, not group-velocity trajectories.}
\b[The propagating branch is not the entire perturbation space.] For example, $\mathbf v=0$, $m_{ij}=f(\mathbf r)\delta_{ij}$ and $\pi=f(\mathbf r)$ solve the <linearized equations> without propagation, for any small, slowly varying $f$. Such a <static stress perturbation in magnetoelastic wave equations> balances a <pressure gradient>. Thus the requested <wave> conclusion applies to the oscillatory branch; it must not exclude stationary initial <stress> perturbations.
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