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

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.