= Solution
Linearized <magnetic flux freezing> gives the Eulerian <magnetic field> perturbation
$$
\delta\mathbf B=\nabla\times(\boldsymbol\xi\times\mathbf B_0)=(\mathbf B_0\cdot\nabla)\boldsymbol\xi-\mathbf B_0\nabla\cdot\boldsymbol\xi.
$$
The background is uniform, and <incompressibility> gives $\nabla\cdot\boldsymbol\xi=0$. For a horizontal displacement proportional to $e^{ikz}$, $\delta\mathbf B=ikB_0\boldsymbol\xi$ and $\delta B_z=0$. Thus the first-order <magnetic pressure> perturbation vanishes, and the <magnetic-tension response to an incompressible displacement> is
$$
\boxed{\delta\mathbf f=\frac{B_0}{\mu_0}\partial_z\delta\mathbf B=-\frac{k^2B_0^2}{\mu_0}\boldsymbol\xi.}
$$
Bending a field line produces a restoring <force> proportional to displacement, like a spring obeying <Hooke's law>. Per unit <mass>, its squared spring <frequency> is $k^2v_A^2$, where $v_A=B_0/\sqrt{\mu_0\rho}$ is the <Alfvén speed>. The equilibrium can be independent of $z$ while the perturbation varies with $z$; neglect of background vertical structure does not set this perturbation's <wavenumber> to zero.
Back to article page
