Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 65 / 4

Let ${\rho }_0$ denote the midplane mass density and choose the symmetry plane at $z=0$. Isothermal hydrostatic equilibrium and the Poisson equation for Newtonian gravity giveso $c_s^2(d^2/d{z}^2)\ln \rho =-4\pi G\rho$. For $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, direct differentiation yieldsThus the profile is a self-gravitating isothermal slab when $H^2=c_s^2/(2\pi G{\rho }_0)$. Integrating over both sides gives $\Sigma =2{\rho }_{0}H$, henceThe corresponding potential is ${\Phi }_0(z)-{\Phi }_0(0)=2c_s^2\ln \cosh (z/H)$, whose gradient vanishes at the midplane and tends to $\pm 2\pi G\Sigma$ far from the slab. The profile parameter $H$ is not its asymptotic exponential mass density scale height: that is $H/2$.

To distinguish Eulerian and Lagrangian fluid perturbations, use $\mathbf{\xi }=({\xi }_x,{\xi }_y,{\xi }_z)$ for the fluid displacement, $\varphi ={\Phi }^{\prime }$ for the Eulerian potential perturbation, and $\delta \rho$ for the Eulerian mass density perturbation. The equilibrium mass density $\rho (z)$ is the profile from part (a); subscript $z$ below denotes differentiation. Linearized mass conservation and the isothermal equation of state giveThe velocity perturbation is $-i\omega \mathbf{\xi }$. Linearized momentum initially has the formSince $\nabla {\Phi }_0=-c_s^2\nabla \ln \rho$, the pressure and perturbed mass density gravity terms combine into $-\rho \nabla (c_s^2\delta \rho /\rho )$. Define $W=c_s^2\delta \rho /\rho +\varphi$. The closed linearized system isThe $y$ component is a neutral transverse sector; it vanishes for nonzero-frequency modes.

To prove reality rather than assume it, multiply momentum by $\rho {\mathbf{\xi }}^{\ast }$ and integrate over $z$. With $I=\int \rho |\mathbf{\xi }{|}^{2}dz>0$, integration by parts and the conjugate continuity equation giveUsing the conjugate Poisson equation for Newtonian gravity in the last integral givesTake disturbances with finite $I$, finite pressure and gravitational integrals, and vanishing pressure work and gravitational boundary terms. For $k\ne 0$, require the potential to decay at both infinities and the fluid displacement to make $\rho {\xi }_z^{\ast }W\to 0$ as well. In a source-free outer region the decaying potential is proportional to $e^{-|k||z|}$. The isothermal displacement energy of a self-gravitating slab then givesSince $I$ is strictly positive for a nonzero displacement, ${\omega }^2$ is real. Positive values describe oscillations; negative values give pure growth and decay. The result is the energy identity for the pressure-gravity displacement operator, a self-adjoint operator in the mass density-weighted inner product. The identity excludes an oscillatory growing eigenmode under these boundary conditions. At $k=0$ the same identity applies when its boundary terms vanish and its integrals remain finite, with the gravitational integral involving only $|{\varphi }_z{|}^2$.

Consider the nontrivial gravitational mode with $k\ne 0$. At ${\omega }^2=0$, horizontal momentum requires $ikW=0$, hence $W=0$ andSubstituting into the Poisson equation for Newtonian gravity and using $4\pi G{\rho }_0/c_s^2=2/H^2$ givesSet $\tau =\tanh (z/H)$ and $\nu =kH$. Since ${\partial }_z=(1-{\tau }^2){\partial }_{\tau }/H$, this becomes the associated Legendre equation of degree one and order $\nu$:

To verify the two explicit solutions, put $\eta =z/H$ and $T=\tanh \eta$. They arebecause $e^{2\eta }=(1+\tau )/(1-\tau )$. Differentiating $F_{+}$ twice giveswhich cancels identically against $[2(1-T^2)-{\nu }^2]F_{+}$. Reflection $\eta \mapsto -\eta$ proves the same for $F_{-}$. Their Wronskian is $2\nu (1-{\nu }^2)$, evaluated at $\eta =0$; thus they are generally independent but coincide or become dependent at the exceptional orders $0,\pm 1$.

Use $n=|k|H>0$ to analyze decay, which is unaffected by the sign of $k$. The solution decaying at $+\infty$ is proportional to $e^{-n\eta }(n+\tanh \eta )$. At $-\infty$ its leading term is $(n-1)e^{n|\eta |}$, so it also decays there only if $n=1$. At this valueTherefore the marginal fragmentation mode of an isothermal slab hasAt $n=1$ the second independent solution obtained by reduction of order grows at infinity, so the coincidence of the two explicit formulas does not supply an extra admissible mode. At $k=0$ a uniform vertical translation gives another neutral disturbance, and mass-preserving relabeling displacements can have $\delta \rho =\varphi =0$. Neither is the nonzero-wavenumber fragmentation threshold just calculated.

The unstable fragmentation modes have wavelengths greater than $2\pi H$. Gas-pressure support becomes less effective relative to self-gravity at long wavelength. The side of the threshold can also be checked from the energy identity: at $|k|=1/H$ the marginal mass density disturbance has zero quadratic energy. Keep that mass density disturbance and decrease $|k|$; the pressure integral stays fixed, while the gravitational term becomes more negative because the positive inverse operator $(-{\partial }_z^2+k^2{)}^{-1}$ increases as $k^2$ decreases. Thus the quadratic energy becomes negative, giving an unstable mode on the longer-wavelength side. This argument concerns the gravitational sector and its stated decay conditions.