Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2017 / 1

Assume a constant downward gravitational acceleration $g$, constant composition, and an ideal gas law $p=\rho \mathcal{R}T$, where $\mathcal{R}=k_B/(\mu m_u)$ is the gas constant per unit mass. Here $m_u$ is the unified atomic mass unit; this convention makes the specified mean molecular weight dimensionless. Hydrostatic equilibrium givesIntegrating from sea level yieldsDividing by the temperature ratio gives the hydrostatic atmosphere with a constant temperature lapse rate:The model requires $T_0-\alpha z>0$ and is used here only within the troposphere.

Convert the atmospheric lapse rate to $\alpha =6.5\times 10^{-3}\mathrm{K}{\mathrm{m}}^{-1}$. At $11\mathrm{km}$, $T=228.5\mathrm{K}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66054\times 10^{-27}\mathrm{kg}$ and $\mu =28$, the density exponent is $4.0825$. HenceThe mass density is therefore approximately one third of its sea-level value in this model.

The original PDF has absolute-value bars around both gradients; they are missing from the transcription. The stability argument must respect the sign of the upward coordinate.

Displace a small parcel upward by $\xi$. Assume it remains in pressure equilibrium with its surroundings but exchanges negligible heat, so it follows an adiabatic process. For a perfect gas with adiabatic index $\gamma$, $T\propto p^{(\gamma -1)/\gamma }$ along that parcel's adiabat. Its temperature gradient is thereforeThe last step uses hydrostatic equilibrium, and $c_p$ is the specific heat capacity at constant pressure. At its new height, the temperature difference between the parcel and the environment is $(T_{\mathrm{ad}}^{\prime }-T^{\prime })\xi$. At equal pressure, its relative density excess is the negative of its relative temperature excess. The upward buoyancy acceleration is thusThis is the squared Brunt–Väisälä frequency. Convective stability requires $N^2>0$, or, in signed-gradient form,For a temperature decreasing upward, this is exactly the PDF's Schwarzschild criterionThe absolute-value condition is also sufficient more generally, although a sufficiently steep temperature inversion remains stable even when that stronger condition fails. Reversing the signed inequality, as the bar-free transcription does, would instead describe an unstable temperature decrease.

Take incidence from A into B and first specify pressure amplitude. For a plane acoustic wave traveling normally to the interface, the acoustic impedance is $Z=\rho c_s$. A right-moving wave has pressure amplitude $P=ZU$, where $U$ is its fluid-velocity amplitude; a left-moving reflected wave has $P=-ZU$.

Let $P_i,P_r,P_t$ denote incident, reflected, and transmitted pressure amplitudes. Continuity of pressure and normal velocity at the interface givesSolving the two linear equations gives the acoustic reflection and transmission at an interface coefficientsFor the same perfect gas, take a common adiabatic index $\gamma$. The adiabatic sound speed satisfies $c_s^2=\gamma p/\rho$. The equal equilibrium pressure therefore implies $Z_B/Z_A=\sqrt{{\rho }_B/{\rho }_A}=\sqrt{\delta }$, givingThe word amplitude is otherwise ambiguous. For fluid velocity, or displacement at the same frequency, the corresponding answer isThese are consistent because pressure and velocity amplitudes differ by the impedance. As an energy check, the time-averaged acoustic intensity is $|P{|}^2/(2Z)$, so its transmission coefficient is $\mathcal{T}=4\sqrt{\delta }/(1+\sqrt{\delta }{)}^2\le 1$, even when the transmitted pressure amplitude exceeds the incident one.

Equal mean molecular weight alone does not logically guarantee equal $\gamma$. If the gases have different values of the adiabatic index, replace $\sqrt{\delta }$ in these formulas by $\sqrt{{\gamma }_B\delta /{\gamma }_A}$; a density-only answer requires the common-$\gamma$ assumption.