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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.

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 $C\ll \Omega$.

A laminar column can contract through nearly hydrostatic equilibrium when its cooling time greatly exceeds its vertical dynamical timescale of a disk. For cooling $A{T}^{1+\beta }$ and fixed column mass, a similarity solution has $\rho \propto {\eta }^{-1}$, $T\propto {\eta }^2$, $P\propto \eta$, and vertical velocity $u_z=\dot{\eta }z/\eta$. The coefficient $C$ also depends on the initial mass density normalization.