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A long-wavelength transverse disturbance of an initially isotropic magnetic second-moment tensor behaves as an elastic shear wave when small-scale rearrangements of the tangle are omitted. For $v_A^2=⟨B^2⟩/3$, its fluid displacement obeys ${\partial }_t^2\mathbf{\xi }=v_A^2{\nabla }^2\mathbf{\xi }$ and $\nabla \cdot \mathbf{\xi }=0$ on the oscillatory branch. The tensor perturbation is $\delta M_{ij}=v_A^2({\partial }_i{\xi }_j+{\partial }_j{\xi }_i)$. A zero mean magnetic field does not imply zero restoring magnetic tension.

The magnetic second-moment tensor is $M_{ij}=B_i{B}_j$. This is the magnetic dyadic called the Maxwell tensor in the paper; the isotropic part of the usual Maxwell stress tensor is already contained in the magnetohydrodynamic total pressure. The solenoidal magnetic-field constraint givesConsequently the magnetohydrodynamic momentum equation in part a becomes $D_t{u}_i=-{\partial }_{i}p+{\partial }_j{M}_{ij}$, where $D_t={\partial }_t+u_n{\partial }_n$ is the material derivative. Applying its product rule and the ideal magnetohydrodynamic induction equation givesThus the evolution uses only $\mathbf{u},M,p$, together with ${\partial }_i{u}_i=0$. To determine $p$, take the divergence of the magnetohydrodynamic momentum equation. Since ${\partial }_t{\partial }_i{u}_i=0$ and $u_j{\partial }_j{\partial }_i{u}_i=0$,This is the pressure Poisson equation for incompressible flow, with boundary conditions supplied by the imposed normal acceleration and normal stress, or by periodicity or decay at infinity as appropriate. For example, the normal component of the magnetohydrodynamic momentum equation requires $\mathbf{n}\cdot \nabla p=\mathbf{n}\cdot (\nabla \cdot M-D_t\mathbf{u})$ at a prescribed boundary. The pressure is determined up to a spatially constant function of time.

For equivalence to one resolved magnetic field, the initial $M$ must actually be $\mathbf{B}{\mathbf{B}}^T$ for a divergence-free vector field. In particular it is a symmetric positive semidefinite matrix of matrix rank at most one. The tensor equations also admit more general stresses; the averaged $M$ in part d need not have matrix rank one.

With zero velocity and no time dependence, the magnetic second-moment tensor equation holds automatically, but the magnetohydrodynamic momentum equation still requiresThe magnetic tension must therefore be a pressure gradient. Locally, its necessary integrability condition ison a simply connected region this is also sufficient for a scalar $p_0$, subject to the global boundary conditions. In physical units the same balance is $\nabla P_0={\mathbf{J}}_0\times {\mathbf{B}}_{\mathrm{phys},0}$: the Lorentz force density is balanced by mechanical pressure, rather than necessarily vanishing. A zero mean magnetic field does not make the local magnetic tension zero, so an arbitrary tangle is not automatically a static equilibrium.

Take the trace of the statistically isotropic magnetic second-moment tensor. There are three spatial directions, soChoosing the nonnegative speed,Thus $v_A$ is the root-mean-square value of any one Cartesian magnetic field component in Alfvén velocity units. It is not the magnitude of the mean Alfvén velocity, which is zero here. Statistical isotropy fixes a second moment even when the first moment vanishes.

Average the equations in part 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 areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. 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.

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.

For constant magnetic diffusivity, apply the product rule to $M_{ij}=B_i{B}_j$ and the incompressible resistive induction equation. The identity ${\nabla }^2(B_i{B}_j)=B_i{\nabla }^2{B}_j+B_j{\nabla }^2{B}_i+2{\partial }_n{B}_i{\partial }_n{B}_j$ proves the displayed formula. Omitting the gradient-product term gives the wrong tensor evolution. Its average generally requires information beyond the averaged magnetic second-moment tensor.

The linear system ${\partial }_t{v}_i=-{\partial }_i\pi +{\partial }_j{m}_{ij}$, ${\partial }_t{m}_{ij}=v_A^2({\partial }_i{v}_j+{\partial }_j{v}_i)$ and $\nabla \cdot \mathbf{v}=0$ admits stationary stress as well as propagating waves. For instance, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve it for any sufficiently small smooth $f$. Thus arbitrary initial magnetic second-moment tensor perturbations need not be oscillatory.