Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2018 / Section A / A1

Take positive $\dot{Q}$ to mean net removal of thermal energy per unit mass. The isobaric thermal stability of a heated ideal gas depends on the density change accompanying a temperature disturbance. For an ideal gas of fixed composition, $p=\rho k_{B}T/\bar{m}$, so constant pressure gives $\rho T=\text{constant}$. Therefore the net cooling can be writtenwhere $C>0$ is constant for this pressure. At thermal equilibrium $T=T_0$, $C{T}_0^{\alpha -1}=H$.

The first law of thermodynamics at constant pressure includes the expansion work, so the relevant heat capacity is the specific heat capacity at constant pressure, $c_p$, rather than the constant-volume value. Linearizing its temperature equation givesHenceFor optically thin Bremsstrahlung, $\alpha =1/2$, soA hotter parcel expands and becomes less dense. Its cooling per unit mass, proportional to $T^{-1/2}$ at fixed pressure, decreases while its cosmic ray heating stays fixed, making it hotter still. A colder, denser parcel cools faster and runs away in the opposite direction. Here $H$ has units $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$ and $c_p{T}_0$ units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, so $s$ is an inverse time. The conclusion assumes the disturbance remains close to pressure equilibrium and neglects heat conduction and changes of composition. The relevant $\alpha$ is the temperature exponent in the supplied cooling law.

Use the nonrelativistic moving-conductor Ohm law in the formThe Maxwell equations give Faraday's law of induction, ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$. For a nonmagnetic conductor with constant permeability ${\mu }_0$, the quasistatic Ampere-Maxwell law gives $\mathbf{j}\simeq (\nabla \times \mathbf{B})/{\mu }_0$. This step neglects displacement current, requiring it to be small compared with conduction current; nonrelativistic velocity alone is not the complete justification.

For spatially uniform scalar electrical conductivity $\sigma$,Apply the identity $\nabla \times (\nabla \times \mathbf{B})=\nabla (\nabla \cdot \mathbf{B})-{\nabla }^2\mathbf{B}$ and Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$. The resistive induction equation isThe first term transports and deforms the magnetic field with the moving fluid; the second diffuses it. The magnetic diffusivity $\eta$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, so both terms have units tesla per second. No assumption of incompressible flow was needed. If $\sigma$ varies spatially, the conductivity-gradient term in magnetic induction, $-\nabla \eta \times (\nabla \times \mathbf{B})$, must also be retained.

For a plane-parallel slab, all equilibrium quantities depend only on height $z$. The vertical hydrostatic equilibrium equation and the Poisson equation for Newtonian gravity areWith the polytropic equation of state $p=K{\rho }^2$ and constant $K$,In the gas interior, where $\rho >0$, divide by $\rho$ and differentiate once:ConsequentlyThe coefficient $2\pi G/K$ has units inverse square metres. The index-one plane-parallel polytrope illustrates the resulting structure: symmetry about $z=0$ gives $\rho (z)={\rho }_c\cos (kz)$, with $k=\sqrt{2\pi G/K}$. A slab bounded by vacuum ends at its first zero, $|z|=\pi /(2k)$; the negative continuation of the cosine is not physical mass density.