= Solution
<Magnetic flux freezing> ties a transverse fluid displacement to a bend in the <magnetic field lines>. For the uniform background, the transverse displacement equation is
$$
\rho_0\partial_t^2\boldsymbol\xi_\perp=\mathcal T_0\partial_z^2\boldsymbol\xi_\perp.
$$
For a sinusoidal bend, $\partial_z^2\boldsymbol\xi_\perp=-k_\parallel^2\boldsymbol\xi_\perp$. Positive <pressure-anisotropy-modified magnetic tension> therefore gives a restoring <force>, and <inertia> produces oscillation. In an isotropic <plasma>, this is precisely the positive <magnetic tension> $B_0^2/(4\pi)$.
Particles moving predominantly along bent <magnetic field lines> contribute a transverse <momentum flux>. Their larger parallel <pressure> opposes the restoring curvature <force> supplied by the <magnetic field> and perpendicular <pressure>. Once $p_{\parallel0}-p_{\perp0}>B_0^2/(4\pi)$, the net <force> points along the original displacement: a larger bend produces a larger outward <force>, giving positive feedback. Equivalently, the bending contribution to the quadratic <potential energy> is $\frac12\mathcal T_0\lvert\partial_z\boldsymbol\xi_\perp\rvert^2$ and has become negative.
In this static, homogeneous fluid model the unstable roots are
$$
\boxed{\omega=\pm i\gamma,\qquad \gamma=|k_\parallel|\sqrt{\frac{p_{\parallel0}-p_{\perp0}-B_0^2/(4\pi)}{\rho_0}}.}
$$
With $e^{-i\omega t}$ the root $+i\gamma$ grows. Its real <angular frequency> is zero, so the unstable pattern grows without propagation in the equilibrium rest frame; its partner decays. Advection by a background flow would change that frame-dependent statement. The growth increasing without bound as $|k_\parallel|$ increases is a limitation of this long-<wavelength> model: finite <gyroradius> and other kinetic corrections cannot be neglected at sufficiently short scales.
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