Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 321 / 3

The razor-thin disk approximation uses vertically integrated pressure and surface density of a disk. To close the printed equations, interpret the barotropic equation of state as $P=P(\Sigma )$, or supply an equivalent fixed vertical conversion from the printed $\rho$ to $\Sigma$. Without that conversion, the literal $P=P(\rho )$ does not close this two-dimensional system. Set $c_s^2=(dP/d\Sigma {)}_0>0$, and write $\Sigma ={\Sigma }_0+\sigma$, $\mathbf{u}={\mathbf{u}}_0+{\mathbf{u}}^{\prime }$, $h^{\prime }=P^{\prime }/{\Sigma }_0=c_s^2\sigma /{\Sigma }_0$. Axisymmetric flow here means ${\partial }_y=0$, so background advection vanishes. The background shear enters through ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}_0=-3\Omega u_x^{\prime }{\mathbf{e}}_y/2$. The complete linearized equations areThese are the compressible Keplerian shearing sheet equations for an inertial-acoustic wave. No perturbed self-gravity is present in the given force law, so no gravitational potential perturbation is to be added. This closure is the barotropic closure of a razor-thin disk.

For a normal mode proportional to $e^{ikx-i\omega t}$, the linear equations giveFor $\omega \ne 0$, eliminate $U_y=-i\Omega U_x/(2\omega )$ and $\hat{\sigma }=k{\Sigma }_0{U}_x/\omega$. The density-wave dispersion relation in a non-self-gravitating disk isThe ${\Omega }^2$ term is the radial epicyclic frequency squared for Keplerian rotation: Coriolis force and background shear restore an orbital displacement. The $c_s^2{k}^2$ term is acoustic restoration by a pressure gradient. Long wavelengths are mainly epicyclic motion; short wavelengths are mainly acoustic waves. The given model contains no disk self-gravity term.

For completeness, the full three-variable determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. Its additional $\omega =0$ branch has $U_x=0$ and $2\Omega U_y=ik{c}_s^2\hat{\sigma }/{\Sigma }_0$. This is a stationary pressure/Coriolis balance, the zero-frequency balanced mode of an axisymmetric disk, rather than a propagating density wave. At $k=0$ it reduces to a uniform surface-density change. Dividing by $\omega$ is legitimate only for the wave branches just derived.

Choose the positive-frequency branch and real radial amplitude $u$. The azimuthal amplitude of the inertial-acoustic wave is $U_y=-i\Omega u/(2\omega )$, so taking real parts givesFor $|k|c_s/\Omega \ll 1$, $\omega =\Omega [1+O((k{c}_s/\Omega {)}^2)]$. Thus, to leading long-wavelength order,The quarter-cycle phase shift and the factor $1/2$ are the epicyclic motion relation in a Keplerian shearing sheet. This approximation drops the small frequency detuning; over times $|\omega -\Omega |t\gtrsim 1$ that phase approximation is not uniform. The associated fractional surface density of a disk perturbation is $\sigma /{\Sigma }_0=(ku/\omega )\cos (kx-\omega t)$, so linearization also requires $|ku/\omega |\ll 1$. The source's arbitrary real $u$ sets the wave phase/amplitude within that small-perturbation regime.

The pressureless dust fluid in a shearing sheet has no pressure restoration and is driven by drag force toward the prescribed gas velocity; gas backreaction is omitted in the given model. After linearizing about the common background shear, axisymmetry givesApply $D_{ϵ}={\partial }_t+ϵ\Omega$ to the first equation and use the second to eliminate $v_y^{\prime }$. The exact linear scalar equation isFor the supplied leading-order epicycle, set $\theta =kx-\Omega t$. Since ${\partial }_t{u}_x^{\prime }=\Omega u\sin \theta$ and $u_y^{\prime }=u\sin \theta /2$, this becomesThe final cosine term is missing from the printed equation. Dropping it is justified at leading order in $0<ϵ\ll 1$, recovering the intended forcing, but the printed equation retains an order-${ϵ}^2$ frequency term on the left and is not an exact derivation at that order. A direct countercheck is ${\mathbf{v}}^{\prime }={\mathbf{u}}^{\prime }$: the drag vanishes and the pair solves both dust equations exactly, while $v_x^{\prime }=u\cos \theta$ leaves residual ${ϵ}^2{\Omega }^{2}u\cos \theta$ in the printed scalar equation. This is the forced dust epicycle with aerodynamic drag; its damping time is $(ϵ\Omega {)}^{-1}$.

For either scalar equation, the complementary solution isso initial velocity transients decay when $t\gg (ϵ\Omega {)}^{-1}$. In the exact corrected forced dust epicycle with aerodynamic drag, $v_{x,p}^{\prime }=u\cos \theta$ is a particular solution, and the original radial equation then gives $v_{y,p}^{\prime }=u\sin \theta /2$. ConsequentlyThis is resonant dust entrainment by an epicyclic gas wave: in the adopted long-wavelength gas approximation, the forcing is a free epicycle of the dust too, so matching the gas makes the drag zero. The convergence holds for every $ϵ>0$ in this linear model; at $ϵ=0$ transients never decay, so the zero-drag and late-time limits do not commute.

If instead the printed scalar equation is treated literally for finite $ϵ$, its particular solution isIt has amplitude $|u|/\sqrt{1+{ϵ}^2/4}$ and phase offset $\arctan (ϵ/2)$, not exact phase/amplitude agreement. Agreement in that truncated equation means only leading order as $ϵ\to 0$. This distinguishes the intended approximation from a false exact reading of the printed assertion.

There is a further order-of-limits condition: a real finite-wavelength gas wave has $\omega -\Omega \ne 0$. Leading resonant entrainment over the damping time requires $|\omega -\Omega |\ll ϵ\Omega$, or $(k{c}_s/\Omega {)}^2\ll ϵ$. Otherwise solve the dust transfer equations at the actual $\omega$; detuning can reduce the dust response even though $|k|c_s/\Omega \ll 1$.

Let ${\Sigma }_{d0}$ be the uniform dust background and put ${\delta }_d={\Sigma }_d^{\prime }/{\Sigma }_{d0}$. The linearized continuity equation is ${\partial }_t{\delta }_d=-{\partial }_x{v}_x^{\prime }$ because background shear has no divergence or axisymmetric advection. Using the late entrained velocity $v_x^{\prime }=u\cos \theta$ givesThus the coherent travelling wave has maximum fractional excess $|ku|/\Omega$, and its crest density is ${\Sigma }_{d0}[1+|ku|/\Omega ]$ when the stationary part is absent. This is the density contrast of an entrained dust wave, and requires $|ku|/\Omega \ll 1$ in the present linear approximation. Its fractional oscillating density amplitude is the same as the gas's long-wavelength amplitude, so the coherent wave produces no additional dust-to-gas concentration ratio at this order.

Velocity damping alone does not erase the time-independent $C(x)$; initial density and the history of the forcing are needed to fix it. The printed data therefore determine the travelling-wave contrast, rather than a unique total late-time concentration for arbitrary initial histories. As an explicit example, switch on the prescribed gas epicycle at $t=0$ with initially uniform dust and ${\mathbf{v}}^{\prime }=0$. The corrected velocity pair gives $v_x^{\prime }=u(1-e^{-ϵ\Omega t})\cos \theta$. Integrating continuity then leavesat late times. It is smaller by order $ϵ$ in the stipulated weak-coupling regime, and confirms the stated leading wave-crest estimate while showing why arbitrary stationary density structure is not removed by the velocity transient.