= Solution
Set $\Delta p_0=p_{\perp0}-p_{\parallel0}$ and $\mathcal T_0=\Delta p_0+B_0^2/(4\pi)$. Distinguish the scalar <magnetic field> strength perturbation $\delta B$ from the vector perturbation $\delta\mathbf B$. To first order,
$$
\delta\mathbf B=\delta B\,\widehat{\mathbf z}+B_0\delta\mathbf b,
\qquad \delta b_z=0,
\qquad k_\perp\delta b_x=-k_\parallel\frac{\delta B}{B_0},
$$
where the last identity is the <solenoidal magnetic-field constraint>. For a <Fourier mode> with the stated time convention, put
$$
\delta q=\delta p_\perp+\frac{B_0\delta B}{4\pi},\qquad
\delta\mathcal T=\delta p_\perp-\delta p_\parallel+\frac{B_0\delta B}{2\pi}.
$$
The linear <momentum conservation> equation, after division by $i$, is
$$
-\omega\rho_0\mathbf u=-\mathbf k\,\delta q+k_\parallel\widehat{\mathbf z}\,\delta\mathcal T+\mathcal T_0\left[k_\parallel\delta\mathbf b+\widehat{\mathbf z}(\mathbf k\cdot\delta\mathbf b)\right].
$$
Use the <solenoidal magnetic-field constraint> in its $z$ component. The coefficient of $\widehat{\mathbf z}$ reduces to $-k_\parallel[\delta p_\parallel+\Delta p_0\delta B/B_0]$. Thus
$$
\boxed{-\omega\rho_0\mathbf u=-k_\perp\widehat{\mathbf x}\left(\delta p_\perp+\frac{B_0\delta B}{4\pi}\right)-k_\parallel\widehat{\mathbf z}\left(\delta p_\parallel+\Delta p_0\frac{\delta B}{B_0}\right)+k_\parallel\mathcal T_0\delta\mathbf b.}
$$
Both scalar <pressure> perturbations act within the $xz$ plane. The $y$ component therefore decouples, independently of their closure. The <ideal magnetohydrodynamic equations> give
$$
\partial_t\delta\mathbf B=B_0\partial_z\mathbf u-B_0\widehat{\mathbf z}\,\nabla\cdot\mathbf u,
\qquad
-\omega\delta b_y=k_\parallel u_y,
\qquad
-\omega\rho_0u_y=k_\parallel\mathcal T_0\delta b_y.
$$
Eliminating the amplitudes gives the <Alfvén wave> <dispersion relation>
$$
\boxed{\omega^2=\frac{k_\parallel^2\mathcal T_0}{\rho_0}=k_\parallel^2v_A^2\left(1+\frac{4\pi\Delta p_0}{B_0^2}\right),\qquad v_A^2=\frac{B_0^2}{4\pi\rho_0}.}
$$
For this <Shear Alfvén wave>, the <velocity> and fluid displacement are along $y$, perpendicular to both $\mathbf k$ and the equilibrium <magnetic field>. Let $\boldsymbol\xi=\xi_y\widehat{\mathbf y}$; then
$$
\mathbf u=-i\omega\boldsymbol\xi,\qquad
\delta\mathbf B=ik_\parallel B_0\boldsymbol\xi,\qquad
\delta\mathbf b=ik_\parallel\boldsymbol\xi.
$$
A real snapshot $\xi_y=\xi_0\cos(\mathbf k\cdot\mathbf r-\omega t)$ bends the <magnetic field lines> sideways, with $\delta B_y=B_0\partial_z\xi_y$. There is no first-order compression or change of <magnetic field> strength: $\nabla\cdot\boldsymbol\xi=0$, $\delta\rho=0$ and $\delta B=0$. Scalar <pressure> perturbations are unnecessary in this transverse subsystem. For a traveling stable branch, $\delta B_y/B_0=-k_\parallel u_y/\omega$, so the <velocity> and transverse <magnetic field> perturbations are in phase or opposite phase according to the branch. At $k_\parallel=0$ the bending <force> and this <angular frequency> vanish.
\Image[/past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2016-alfven-firehose-displacement.png]
{title=Transverse displacement and the sign of the effective tension}
{description=Original schematic for a Fourier displacement varying along the equilibrium magnetic field. With positive effective tension the force opposes the displacement; with negative effective tension it reinforces it.}
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