Integrate each conservative <ideal magnetohydrodynamics> law through a thin pillbox around the stationary shock. Mass conservation gives
$$
[\rho u_x]=0.
$$
The MHD momentum-flux tensor is
$$
\rho\mathbf u\mathbf u
+\left(p+\frac{B^2}{2\mu_0}\right)I
-\frac{\mathbf B\mathbf B}{\mu_0}.
$$
Its normal and tangential components give
$$
\left[\rho u_x^2+\Pi-\frac{B_x^2}{\mu_0}\right]=0,
\qquad
\left[\rho u_xu_y-\frac{B_xB_y}{\mu_0}\right]=0,
$$
where $\Pi=p+B^2/(2\mu_0)$ is total pressure. The solenoidal condition gives $[B_x]=0$, while steady <Faraday's law> gives continuity of tangential electric field,
$$
[-E_z]=[u_xB_y-u_yB_x]=0.
$$
Finally, material energy flux plus the normal <Poynting vector> component gives
$$
\left[\rho u_x\left(\frac12u^2+h\right)
-\frac{E_zB_y}{\mu_0}\right]=0,
$$
where $h$ is <Enthalpy>[specific enthalpy] per unit mass.

Gravity is a bounded volume force, so its integral across a shock whose thickness tends to zero vanishes. Viscosity and resistivity inside the layer may produce entropy; consequently entropy flux need not be equal on both sides, although the second law requires nonnegative net entropy production.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 3 / a / Solution

Uniform $\rho_0,p_0,$ and $\mathbf B_0=B_0\widehat{\mathbf z}$ obey the unperturbed <ideal magnetohydrodynamics> equations when
$$
\boxed{\mathbf v_0=-q\Omega_0x\widehat{\mathbf y}}.
$$
Indeed $(\mathbf v_0\mathbin\cdot\nabla)\mathbf v_0=0$, while the Coriolis and tidal terms cancel:
$$
-2\Omega_0\widehat{\mathbf z}\times\mathbf v_0
+2q\Omega_0^2x\widehat{\mathbf x}=0.
$$

For perturbations proportional to $e^{i(\omega t-kz)}$, the horizontal velocity and magnetic perturbations decouple from the compressive variables. Their linear equations are
$$
i\omega v_x-2\Omega_0v_y
=-ik\frac{B_0}{4\pi\rho_0}B_x,
$$
$$
i\omega v_y+(2-q)\Omega_0v_x
=-ik\frac{B_0}{4\pi\rho_0}B_y,
$$
$$
i\omega B_x=-ikB_0v_x,
\qquad
i\omega B_y=-q\Omega_0B_x-ikB_0v_y.
$$
Eliminating $B_x,B_y,v_x,v_y$ and writing the <Alfvén speed> as $v_A^2=B_0^2/(4\pi\rho_0)$ gives
$$
\boxed{
\omega^4-omega^2\left[2k^2v_A^2+2(2-q)\Omega_0^2\right]
+k^2v_A^2\left(k^2v_A^2-2q\Omega_0^2\right)=0}.
$$
Since $d\Omega_0^2/d\log r=-2q\Omega_0^2$, a root has $\omega^2<0$ precisely when the constant term is negative:
$$
\boxed{k^2v_A^2+\frac{d\Omega_0^2}{d\log r}<0}.
$$
This is the <magnetorotational instability> criterion.

For a <circular Kepler orbit>, $q=3/2$. With $K=k^2v_A^2$, the unstable branch is
$$
\omega_-^2=K+\frac12\Omega_0^2
-\Omega_0\sqrt{4K+\frac14\Omega_0^2}.
$$
Minimizing it gives
$$
K_{\max}=\frac{15}{16}\Omega_0^2,
\qquad
\omega_-^2=-\frac9{16}\Omega_0^2,
$$
and hence
$$
\boxed{|\omega_{\max}|=\frac34\Omega_0}.
$$
The e-folding time is of order the dynamical time, so the growth is rapid: several e-foldings occur in one orbit.

Instability requires $k^2v_A^2<3\Omega_0^2$, making the <critical wavelength of the magnetorotational instability>
$$
\lambda_{\rm crit}=\frac{2\pi v_A}{\sqrt3\Omega_0}.
$$
For a thin isothermal disk, $H=c_{s,0}/\Omega_0$. Setting $\lambda_{\rm crit}=2H$ gives $v_A^2=3c_{s,0}^2/\pi^2$, and therefore
$$
\boxed{B_{0,\rm crit}^2=4\pi\rho_0v_A^2
=\frac{12}{\pi}\rho_0c_{s,0}^2}.
$$
Above 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.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 347 / 1 / b / Solution

{title2=ideal MHD}
{wiki}

Ideal magnetohydrodynamics assumes vanishing resistivity, so magnetic flux is frozen into the fluid and the electric field in the fluid rest frame vanishes.


Codex @codex  0

 Incoming links: Ideal magnetohydrodynamics