Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 54 / 2

For a razor-thin disk, integrate the delta function in height to getThe corresponding horizontal force kernel is proportional to $(\mathbf{R}-{\mathbf{R}}^{\prime })/(|\mathbf{R}-{\mathbf{R}}^{\prime }{|}^2+{ϵ}^2{)}^{3/2}$. It removes the short-distance point-force singularity and smooths forces on separations of order $ϵ$. This is height-evaluation gravitational softening.

To represent finite vertical thickness, choose $ϵ$ of order the disc disk scale height $H$. There is no universal exact numerical choice: a true vertical mass density profile produces the Fourier reduction factor ${\Sigma }^{-1}\int \rho (z)e^{-k|z|}dz$, which is not generally $e^{-kϵ}$. Matching the long-wave term gives $ϵ=⟨|z|{⟩}_{\rho }$, the vertical-profile softening match. For an exponential vertical profile this mean is $H$, while its full reduction factor is $(1+kH{)}^{-1}$.

Use the horizontal Fourier convention $\tilde{f}(\mathbf{k})=\int e^{-i\mathbf{k}\cdot \mathbf{R}}f(\mathbf{R})d^{2}R$. For each nonzero $k=|\mathbf{k}|$, the Poisson equation becomesDecay away from the sheet and continuity at it give $\tilde{\Phi }=C{e}^{-k|z|}$. Its derivative jump is $-2kC=4\pi G\tilde{\Sigma }$, hence the off-plane razor-thin Poisson kernel isThe spatially uniform mode has a different vertical solution and an arbitrary additive potential reference; it is not obtained by substituting $k=0$ into this decaying-mode formula.

Let $k=|k_x|$, $c_s^2=(dP/d\Sigma {)}_0$ and ${\kappa }^2=2\Omega (2\Omega -S)$. Linearize the barotropic closure of a razor-thin disk and shearing sheet about the given state. Write the radial and azimuthal velocity perturbations as $u$ and $v$. Axisymmetry removes advection by the background shear, givingThe softened potential is ${\Phi }_1=-2\pi G{\Sigma }_1{e}^{-kϵ}/k$. Eliminating the velocity components for the compressive branch gives the softened Toomre dispersion relationThe Coriolis/shear combination supplies the radial epicyclic restoring term, pressure supplies the short-wave term, and self-gravity lowers the squared frequency. The full three-variable determinant also has a stationary axisymmetric geostrophic mode, with $u=0$ and azimuthal flow balancing the pressure-plus-gravity gradient. The displayed relation describes the density-wave pair; eliminating by division by $\omega$ must not silently deny that stationary mode.

Assume $\kappa ,c_s,{\Sigma }_0>0$, and use the positive Toomre parameter. The dimensionless dispersion relation isAt fixed dimensionless gravitational softening $\delta$, maximum growth corresponds to maximizing $F(s)=2s{e}^{-\delta s}-s^2$. Differentiating gives the most unstable softened disk wavenumberFor $\delta >0$, the right side is strictly decreasing on $0<s<1/\delta$, while the left side increases; there is exactly one positive solution. The equation implies $\delta s_{\ast }<1$ and $s_{\ast }<1-\delta s_{\ast }$, soInstability occurs precisely when $Q^2<F(s_{\ast })$. Thus the critical Toomre parameter with exponential softening isThe printed description of a minimum $Q$ for instability reverses the threshold: $Q_c$ is the upper boundary of unstable values and the lower boundary of stable ones. Because $s_{\ast }<1/(1+\delta )\le 1$ and $2s-s^2$ increases on $[0,1]$,Both strict comparisons become equality in the unsoftened limit $\delta =0$: then $s_{\ast }=1$ and $Q_c=1$. Gravitational softening suppresses short-wave gravity, shifts the most dangerous mode to a longer wavelength and requires stronger self-gravity, or smaller $Q$, for instability.

The plot holds $Q$ fixed and changes gravitational softening. If $Q$ is varied physically at fixed $\kappa ϵ/c_s$, remember $\delta =(\kappa ϵ/c_s)/Q$: the marginal curve must be evaluated at its corresponding $\delta$, rather than treating these two dimensionless parameters as independently fixed along that physical variation.