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If $B_{y2}=0$ but $B_{y1}\ne 0$, tangential momentum continuity in the formforces $A_1^2=1$. Since the normal components are positive,so the upstream velocity equals the upstream Alfvén speed. This is the limiting switch-off shock condition.

Solved by gpt-5.6-sol high.

Uniform ${\rho }_0,p_0,$ and ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$ obey the unperturbed ideal magnetohydrodynamics equations whenIndeed $({\mathbf{v}}_0\cdot \nabla ){\mathbf{v}}_0=0$, while the Coriolis and tidal terms cancel:

For perturbations proportional to $e^{i(\omega t-kz)}$, the horizontal velocity and magnetic perturbations decouple from the compressive variables. Their linear equations areEliminating $B_x,B_y,v_x,v_y$ and writing the Alfvén speed as $v_A^2=B_0^2/(4\pi {\rho }_0)$ givesSince $d{\Omega }_0^2/d\log r=-2q{\Omega }_0^2$, a root has ${\omega }^2<0$ precisely when the constant term is negative:This is the magnetorotational instability criterion.

For a circular Kepler orbit, $q=3/2$. With $K=k^2{v}_A^2$, the unstable branch isMinimizing it givesand henceThe e-folding time is of order the dynamical time, so the growth is rapid: several e-foldings occur in one orbit.

Instability requires $k^2{v}_A^2<3{\Omega }_0^2$, making the critical wavelength of the magnetorotational instabilityFor a thin isothermal disk, $H=c_{s,0}/{\Omega }_0$. Setting ${\lambda }_{\mathrm{crit}}=2H$ gives $v_A^2=3c_{s,0}^2/{\pi }^2$, and thereforeAbove this field strength the shortest unstable vertical MRI wavelength exceeds the full disk thickness, so no such vertical mode fits inside the disk. Magnetic tension then stabilizes this local mode; the result corresponds to a magnetic-to-gas pressure ratio $B^2/(8\pi p)=3/(2{\pi }^2)$.

Solved by gpt-5.6-sol high.