Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 321 / 2

For $|z|\ll r$, the gravitational potential has the Taylor expansionHence the vertical gravity is $-{\partial }_z\Phi \simeq -{\Omega }_K^{2}z$, where ${\Omega }_K^2=G{M}_{\star }/r^3$. Vertical hydrostatic equilibrium balances a pressure gradient of order $P/H$ against $\rho {\Omega }_K^{2}H$. Since $c_s^2\sim P/\rho$,The thin disk condition $H/r\ll 1$ is therefore equivalent to a highly supersonic orbital speed.

Put $K(r)=e^{s(r)}$. The perfect gas relation in the question becomes the polytropic equation of stateAt fixed $r$, vertical hydrostatic balance isAfter one integration,Defining the midplane adiabatic sound speed by $c_{s0}^2=\gamma P_0/{\rho }_0=\gamma K{\rho }_0^{\gamma -1}$ givesThis is a vertically truncated polytropic atmosphere; the physical perfect-gas case has $\gamma >1$.

At the midplane, radial hydrostatic equilibrium givesIf $d{P}_0/dr<0$, then $\Omega <{\Omega }_K$: the pressure gradient provides part of the inward force and the gas is Sub-Keplerian. For radial pressure scale comparable to $r$,Using $H\sim c_s/{\Omega }_K$ and ${\Omega }_K-\Omega \simeq ({\Omega }_K^2-{\Omega }^2)/(2{\Omega }_K)$,

The linear term $-(3/2){\Omega }_{K}x{\mathbf{e}}_y$ is the local Keplerian shear in the shearing sheet. The remaining constantis the gas's azimuthal velocity relative to the local Keplerian frame. The outwardly decreasing midplane pressure found in part (a)(iii) makes the gas Sub-Keplerian, so

Write $\Delta \mathbf{u}=(U,W,0)$. The background shear givesThe constant radial and azimuthal components of the dust equation are thereforeWith the Stokes number $\mathrm{St}=\tau {\Omega }_K$, their solution is

The gas is slower than a Keplerian particle, so the particle feels an aerodynamic headwind. Drag force removes its angular momentum, and it drifts radially inward because $U<0$. Its radial speed isDifferentiation with respect to $\mathrm{St}>0$ shows that the unique maximum occurs atwhere $|U|=|V|$.

For the axisymmetric perturbation, defineLinearization about the uniform dust density and drifting equilibrium givesThe factors $2{\Omega }_K$ and ${\Omega }_K/2$ are the Coriolis acceleration and Keplerian-shear couplings. Applying ${\mathcal{D}}_{\tau }({\mathcal{D}}_{\tau }^2+{\Omega }_K^2)$ to the continuity equation and using the three momentum equations eliminates ${\mathbf{u}}^{\prime }$. All remaining forcing terms are proportional to $e^{-i\omega t}$, sowhere $F$ is independent of time.

On the forcing $e^{-i\omega t}$,The other three factors have roots with real decay rate $-1/\tau$, so a real-frequency gas wave can resonate only with the undamped factor $\mathcal{D}$. The resonance condition is thereforeAt resonance, $\mathcal{D}$ annihilates the forcing, so a particular solution acquires one power of time:This is a streaming instability resonance: the gas wave's phase velocity along $\mathbf{k}$ matches the dust drift projected along the same wavevector.