Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 / 4

Choose a reference circular orbit at $r_0$, and use local coordinates $x=r-r_0$, $y=r_0[\varphi -{\Omega }_{0}t]$, with the frame rotating at $\Omega ={\Omega }_0=\Omega (r_0)$. Linearizing differential rotation in the shearing sheet gives the background flowThus the velocity perturbation is measured relative to this background. The local orbital shear parameter is $q=2A/\Omega =-d\ln \Omega /d\ln r$; Keplerian rotation has $q=3/2$ and $A=3\Omega /4$. Rotation supplies the Coriolis force, while the linear shear supplies the terms proportional to $A$.

For constant density, $\psi$ is the total-pressure potential after background force balance has been subtracted:up to a function of time and any prescribed perturbation potential that is absorbed into it. The magnetic-pressure gradient has been combined with the gas-pressure gradient, leaving magnetic tension as $\mathbf{B}\cdot \nabla \mathbf{B}/({\mu }_0\rho )$. The pressure also enforces incompressibility. In the unstratified shearing box here, the tidal/background terms have already been removed in deriving the perturbation equations; no independent vertical stratification is being evolved.

At fixed time write the azimuthal shift as $\delta =2A{L}_{x}t$ modulo $L_y$. Integrating a radial derivative over the box gives a boundary difference proportional toThe two integrals agree by translation invariance of an integral over a full azimuthal period. Hence $⟨{\partial }_{x}Q⟩=0$. The ordinary periodic boundary conditions make the $y$ and $z$ boundary differences zero as well. Finally,for every $x,z$. Thus the volume averages in a shearing box satisfyFor smooth shearing-periodic extensions, spatial derivatives and products inherit the same shearing-periodic boundary condition. Their divergence therefore has zero volume average, and integration by parts is available without a residual boundary flux. Time differentiation of an average over this fixed physical box commutes with integration, even though the radial identification shifts with time.

The nonlinear velocity term is a divergence componentwise: $v_j{\partial }_j{v}_i={\partial }_j(v_j{v}_i)$ because $\nabla \cdot \mathbf{v}=0$. Likewise $B_j{\partial }_j{B}_i={\partial }_j(B_j{B}_i)$; their volume averages vanish. Pressure gradients, spatial Laplacians and the background-advection term $x{\partial }_y$ also average to zero by the boundary identities. The mean momentum equation is thereforeIn induction, the two nonlinear terms are divergences, $v_j{\partial }_j{B}_i={\partial }_j(v_j{B}_i)$ and $B_j{\partial }_j{v}_i={\partial }_j(B_j{v}_i)$. Averaging consequently givesThe radial and vertical mean magnetic fields stay constant, and $⟨B_y⟩(t)=⟨B_y⟩(0)-2At⟨B_x⟩(0)$. In particular an initially zero mean magnetic field stays zero.

Writing $V_i=⟨v_i⟩$, the horizontal mean motion satisfiesThis is the local radial epicyclic frequency. When the background is centrifugally stable, $\Omega (\Omega -A)>0$, the whole box supports epicyclic mean-velocity oscillations. For example,while $V_z$ is constant. A Keplerian shearing sheet has $\kappa =\Omega$. For a centrifugally unstable rotation law the same mean equations instead give exponential growth, so an oscillation interpretation requires the positive squared frequency.

No divergence of turbulent stress survives the box average, and a zero initial mean radial velocity cannot acquire a secular accretion drift from internal stresses. More generally, a steady horizontal mean flow has $V_x=V_y=0$ for the usual nondegenerate rotating disk. The epicycle has zero time-averaged radial velocity. Opposite radial faces are identified rather than connected to a central sink and an outer mass reservoir: material crossing a face re-enters the opposite one after an azimuthal shift. Thus the shearing box can model angular-momentum transport and its local stresses, but not develop a global disk accretion flow. Such accretion needs radial gradients, boundaries and mass exchange omitted from the homogeneous local model.

Define the perturbation energy per unit mass by $E=⟨v^2/2+B^2/(2{\mu }_0\rho )⟩$. Dot the momentum equation with $\mathbf{v}$ and the induction equation with $\mathbf{B}/({\mu }_0\rho )$, then take volume averages. The Coriolis force does no work. Pressure work and nonlinear advection are divergences and integrate to zero, as does the background $x{\partial }_y$ transport. The two magnetic exchange terms add toThe velocity shear term supplies $2A⟨v_x{v}_y⟩$, while magnetic shear supplies $-2A⟨B_x{B}_y⟩/({\mu }_0\rho )$. Integration by parts gives the viscous and resistive losses as $-\nu ⟨|\nabla \mathbf{v}{|}^2⟩$ and $-\eta ⟨|\nabla \mathbf{B}{|}^2⟩/({\mu }_0\rho )$. For a solenoidal field $\mathbf{F}$, another integration by parts showsAll boundary terms cancel by the shearing-periodic identities. The resulting balance isThe source is work extracted from maintained differential rotation by Reynolds stress and the magnetic part of the Maxwell stress tensor; internal magnetic/kinetic exchange has canceled.

For the decay bound, the elementary inequality $2|ab|\le a^2+b^2$ givesA uniform spectral gap of a shearing-periodic box is needed; ordinary radial periodicity cannot simply be assumed. At each time the compatible Fourier modes areThey are orthogonal over the box. Their wavevectors areFor every nonconstant mode, $|\mathbf{k}|\ge 2\pi /L$, where $L=\max (L_x,L_y,L_z)$: if $m$ or $p$ is nonzero, its corresponding unsheared component supplies the bound; otherwise $n\ne 0$ and the radial component does. Zero mean excludes precisely the constant mode. The Parseval identity therefore proves the time-uniform Poincaré inequalityApply it componentwise to $\mathbf{v}$ and $\mathbf{B}$. With $\nu =\eta$, the energy decay criterion for a viscous-resistive shearing box becomesHenceThis excludes sustained finite-amplitude turbulence as well as infinitesimal growth. It is a sufficient stability condition, not a claim that every box above the threshold must become turbulent. The zero-mean restrictions matter: an imposed uniform magnetic field or bulk epicycle is not controlled by this spatial spectral-gap inequality.