Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2021 / Section B / 6

For a steady axisymmetric flow ${\mathbf{u}}_0=r\Omega (r)\hat{\mathbf{\varphi }}$, incompressible flow is automatic. The only material acceleration is centripetal, $-r{\Omega }^2\hat{\mathbf{r}}$. The radial Euler equations therefore requireWith no viscosity there is no tangential shear stress to redistribute angular momentum. Any sufficiently regular $\Omega (r)$ can therefore be a steady inviscid profile, with the radial pressure adjusted by integrating this equation. This is a statement of equilibrium, not of stability: some such profiles are centrifugally unstable.

For a homogeneous Newtonian fluid with constant kinematic viscosity $\nu >0$, assume the same steady purely azimuthal flow. In cylindrical coordinates, the azimuthal Navier-Stokes equation has no pressure forcing or convective term in this state, and becomesEquivalently, ${\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r=0$, or $(r^3{\Omega }^{\prime }{)}^{\prime }=0$. Two integrations give the steady circular Couette profile,This is the basic Taylor-Couette flow between rotating cylinders. The constants depend on boundary angular velocities; for radii $R_{<},R_{>}$ and prescribed angular velocities ${\Omega }_{<},{\Omega }_{>}$,If the domain includes $r=0$, regularity requires $B=0$, leaving solid-body rotation. On an annulus a nonzero $B$ is allowed and wall torques maintain the shear.

The restriction concerns a steady profile. An arbitrary initial angular velocity does not cease to exist when viscosity is introduced; within the purely azimuthal ansatz it evolves according to ${\partial }_t\Omega =\nu ({\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r)$ until compatible boundary conditions and angular-momentum constraints determine its long-time state.

Expand the axisymmetric Euler equations about the rotating equilibrium, retaining only terms linear in $u_r,u_{\varphi },u_z$ and in the pressure perturbation. The symbol in this question is the kinematic pressure perturbation,The linearized centripetal term in the radial equation is $-2\Omega u_{\varphi }$, after subtracting the equilibrium term $-r{\Omega }^2$. Radial advection of the background azimuthal velocity, together with the cylindrical-coordinate term $u_r{v}_{\varphi }/r$, givesThis is the azimuthal coupling in the second equation; it also expresses conservation of specific angular momentum $j=r^2\Omega$ for a radially displaced element. The axial equation is acceleration balanced by $-{\partial }_z\pi$. Finally, the axisymmetric incompressible flow condition in cylindrical coordinates iswhich expands to the last supplied equation. Background azimuthal advection contributes no $\Omega {\partial }_{\varphi }$ term because the perturbations are axisymmetric.

Use the thin-annulus approximation stated in the source, so $u_r^{\prime }+ik{u}_z=0$. LetHere ${\kappa }_{\mathrm{ep}}^2$ is the conventional Rayleigh discriminant, also the squared local radial epicyclic frequency when positive. For $k\ne 0$ and $\omega \ne 0$, the azimuthal and axial mode equations giveThe thin-gap incompressible flow condition then gives $\pi =i\omega u_r^{\prime }/k^2$. Substitute these into the radial equation:ThusTo put this into exactly the form printed in the paper, defineThe narrow-gap centrifugal stability equation is consequentlyThe two rigid cylinders are impermeable, so the no-penetration boundary condition givesThere is no inviscid no-slip condition on the tangential or axial components.

The sign deserves emphasis: the source's printed plus sign makes its $\Phi$ the negative of the conventional Rayleigh discriminant. Using the conventional positive discriminant instead requires a minus sign in front of that potential term. These are equivalent conventions; replacing one without adjusting the other would falsely predict instability of solid-body rotation. The $k=0$ and marginal $\omega =0$ cases, excluded by these eliminations, must be treated separately or by limits.

Let $u_r$ be a nonzero admissible mode, multiply the equation of part (d) by its complex conjugate and integrate between the cylinders. The endpoint conditions remove the boundary term in integration by parts, giving the Rayleigh stability integral identityRearrange to obtain a real Rayleigh quotient:The denominator is strictly positive. Consequently, with the paper's convention,The weak inequality $\Phi \le 0$ excludes exponentially growing axisymmetric modes, but permits marginal cases and possible algebraic growth when the restoring effect vanishes. Strict negativity avoids that degeneracy. Equivalently, the conventional Rayleigh discriminant is positive, or the squared specific angular momentum increases outward:This is the Rayleigh centrifugal stability criterion. In the suggested Time-independent Schrödinger equation analogy, a growing mode has ${\omega }^2<0$; if $\Phi \le 0$, its effective potential $k^2\Phi /{\omega }^2$ is nonnegative. The nonnegative kinetic term and that potential cannot have the required negative eigenvalue $-k^2$, which gives the same exclusion.

As a sign check, for solid-body rotation ${\kappa }_{\mathrm{ep}}^2=4{\Omega }^2$ and $\Phi =-4{\Omega }^2$. In a gap of width $d=R_{>}-R_{<}$, the radial modes $u_r\propto \sin [n\pi (r-R_{<})/d]$ haveas expected for inertial waves. The argument establishes axisymmetric inviscid stability within the stated thin-gap model, rather than a claim about all perturbations of a viscous flow.