Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 2

Distinguish the cylindrical radius $R$ from the spherical radius $r=\sqrt{R^2+z^2}$. A Taylor expansion of the central gravitational potential for $|z|/R\ll 1$ givesThus the usual thin-disk expansion of the stellar gravitational potential isThe vertical acceleration is $-{\partial }_z\Phi =-{\Omega }_K^{2}z$. The first term can be omitted when considering only vertical forces at fixed $R$, but must be retained for radial dynamics.

Take $\gamma >1$ for the ideal gas. At fixed cylindrical radius, put $\mathcal{R}=k/(\mu m_p)$, where $k$ is the Boltzmann constant, $\mu$ the mean molecular weight and $m_p$ the proton mass. Also write $a=9\alpha \Omega /4$. Vertical hydrostatic equilibrium, thermal equilibrium and the ideal gas law requireConsequently, with $K=A/a$,Integrating from the midplane yieldsThe polytropic vertical structure in stellar gravity therefore has, for $|z|\le H$,whereThis is a polytropic equation of state $P\propto {\rho }^{1+1/\beta }$ with polytropic index $\beta$. For $|z|>H$, set $\rho =P=0$; temperature has no physical gas value in that vacuum. Here $H$ is the finite semi-thickness, rather than an exponential disk scale height.

The vertical acceleration in the thin disk is of order ${\Omega }^{2}H$. Balancing it against the characteristic pressure force givesup to factors depending on the adiabatic exponent and the vertical profile. A displacement of size $\delta z$ has a restoring acceleration of order ${\Omega }^2\delta z$, so $\delta \ddot{z}\sim -{\Omega }^2\delta z$. Equivalently the sound-crossing time is $H/c_s$. Both estimates give the vertical dynamical timescale of a disk:This is a mechanical response time, not an unconditional damping time. An inviscid stable layer can exhibit a vertical breathing mode of an astrophysical disk without settling; its adiabatic frequency satisfies ${\omega }_H^2=(\gamma +1){\Omega }^2$. Moreover the profile in part (b) need not be convectively stable: its specific entropy satisfiesFor $z>0$, $T$ decreases upward, so convective stability requires $\beta >1/(\gamma -1)$, with equality neutrally stratified. Thus interpreting the timescale as re-establishment of a stable equilibrium presupposes suitable stability and damping; the stated equations alone do not guarantee either.

For slow cooling of an astrophysical disk, let $t_{\mathrm{cool}}\gg {\Omega }^{-1}$ and assume a vertically symmetric laminar column with no significant radial mass exchange. Its vertical velocity is of order $H/t_{\mathrm{cool}}$, so the vertical material derivative of that velocity is of order $H/t_{\mathrm{cool}}^2$. Relative to gravity or pressure acceleration $\sim {\Omega }^{2}H$, the inertial correction is $O[(\Omega t_{\mathrm{cool}}{)}^{-2}]$.

The hydrostatic approximation therefore holds to leading order while the column slowly cools and contracts. Non-turbulent evolution removes the prescribed turbulent heating, but not the compressional work in the ideal gas energy equation:The continuity equation must still determine the slow vertical flow; $D_t$ is not generally ${\partial }_t$ at fixed height. This is a quasi-static approximation for a well-prepared, mechanically stable column, with fast free oscillations or growing convection excluded from the assumed slow solution.

At fixed cylindrical radius, let $w=u_z$ and use the continuity equation ${\partial }_t\rho +{\partial }_z(\rho w)=0$. For the similarity solution $z=\xi \eta (t)$ and $\rho =\tilde{\rho }(\xi )/\eta$, the continuity equation reduces toHere $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:The hydrostatic approximation and ideal gas law becomeThe material derivatives are $D_{t}P=\tilde{P}\dot{\eta }$ and $D_t\rho =-\tilde{\rho }\dot{\eta }/{\eta }^2$, so the energy equation givesSeparation of variables requires a positive constant $K$ such thatIntegrating with $\eta (0)=1$ gives the slow cooling of an astrophysical disk similarity factorA closed set of profile equations is thereforeThese have the same polytropic vertical structure in stellar gravity as part (b). Writingwith $P_0=\mathcal{R}{\rho }_0{T}_0$, fixesReturning to physical height, the explicit profiles and evolving semi-thickness for $|z|\le H(t)$ areandOutside this moving surface $\rho =P=0$. The surface density of a disk is conserved because $\rho dz=\tilde{\rho }d\xi$. The column cools and contracts without changing its scaled profile: its midplane temperature falls as ${\eta }^2$, its mass density rises as ${\eta }^{-1}$ and its pressure falls as $\eta$.

The value $T_0$ determines $H_0$, but the mass density normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating givesOtherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is $\ddot{\eta }\xi$, with $\ddot{\eta }/(\eta {\Omega }^2)=2(1+\beta )C^2/[(1+2\beta {)}^2{\Omega }^2(1+Ct{)}^2]$, which remains small and decreases if $C\ll \Omega$.