Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 320 (Modern Stellar Dynamics)

Choose intrinsic Cartesian coordinates in which the oblate spheroid isLet $\ell$ be distance along the line of sight and let $y$ be the sky coordinate in the plane containing the line of sight and the symmetry axis. A rotation through the inclination $i$ givesSubstitution into the ellipsoid equation and minimization over $\ell$, equivalently requiring the quadratic in $\ell$ to have zero discriminant on the projected boundary, givesTherefore the projected axis ratio of an oblate spheroid isIt correctly gives $Q=1$ face-on and $Q=q$ edge-on.

Uniform orientation on the unit sphere means that probability is proportional to the solid-angle elementBecause the minor axis is unoriented, it is enough to take $0\le i\le \pi /2$. After integration over the azimuth and normalization,Equivalently, the random variable $\mu =\cos i$ has the uniform distribution on $[0,1]$.

At fixed intrinsic ratio $q$, part a givesSince $\mu$ is uniform, the change-of-variables formula for a probability density yields the conditional probability density functionThe inverse-square-root singularity at $Q=q$ is integrable, and direct integration gives one.

For an intrinsic probability density function $\mathcal{P}(q)$, the law of total probability averages the conditional density from part c. An object observed with ratio $Q$ can only have $q\le Q$, soThis is an Abel transform of $\mathcal{P}(q)/\sqrt{1-q^2}$. Its normalization follows by reversing the order of integration and using ${\int }_q^1\mathcal{P}(Q\mid q)dQ=1$.

For the uniform distribution $\mathcal{P}(q)=1$, substitute $q=Q\sin \theta$ into part d:where $K$ is the complete elliptic integral of the first kind. Both $Q$ and $K(Q)$ increase on $(0,1)$; indeed $\mathcal{P}(Q)\sim \pi Q/2$ near zero and diverges logarithmically as $Q\uparrow 1$. Thus the apparent distribution is strongly skewed toward higher $Q$: projection makes many intrinsically flattened systems look round.

Withthe factor $\sqrt{1-q^2}$ cancels from the Abel transform:HenceThis probability density function is normalized and again favors rounder projections.

For an ordinary intrinsic density, inversion of the Abel transform givesIf $\mathcal{P}(Q)=1$, the integral is $\pi /2$ for every $q>0$, so its derivative vanishes. The missing probability is an endpoint atom: all systems must be infinitely thin,Indeed, putting $q=0$ directly into part c gives $\mathcal{P}(Q\mid 0)=1$. Thus a uniform apparent-axis-ratio distribution corresponds to a Dirac delta distribution of ideal zero-thickness disks, rather than to a regular intrinsic density.

Yes. Even under random orientation, a population concentrated at one nonzero intrinsic ratio $q_0$ haswhich decreases from an integrable divergence at the lowest allowed value $Q=q_0$. It is therefore skewed toward the lower end of its support. More generally, a sufficiently narrow intrinsic distribution can retain such low-$Q$ peaks after the mixture in part d.

If the random-orientation assumption is relaxed, preferentially edge-on selection makes $\cos i$ concentrate near zero and hence makes $Q$ concentrate near $q$. Dust extinction, surface-brightness selection, or alignment by environment can instead bias the sample in either direction. Triaxial or prolate galaxy shapes also invalidate the projected axis ratio of an oblate spheroid formula and can produce other low-$Q$ distributions.

Write $\mu =GM$ and $s=\sqrt{b^2+r^2}$. The gravitational potential of the spherical isochrone model obeysThe spherical Poisson equation therefore givesIts small-radius asymptotic expansion isso the model has a finite-density core. At large radius,The $r^{-4}$ envelope has finite total mass $M$.

For a bound orbit let $E<0$ be the specific energy and $L$ the specific angular momentum. Introduce the dimensionless radius suggested in the question,The radial energy equation becomeswhere $x_{-}$ and $x_{+}$ are the two turning points. Comparing coefficients givesThe time from periapsis to apoapsis isThus the radial orbital period isIt depends on $E$ but not on $L$, which is the defining isochrone property.

During half a radial oscillation,Useand the two root productsThe stated elementary integrals then give the azimuthal advance over one radial period:Consequently the frequency ratio of the spherical isochrone model isor equivalently $T_{\varphi }/T_r=2/(1+L/\sqrt{L^2+4GMb})$.

Deep in the constant-density core, $L^2\ll 4GMb$ and ${\Omega }_{\varphi }/{\Omega }_r\to 1/2$, as for an isotropic harmonic oscillator: the radius completes two oscillations per revolution. Far outside the core, $L^2\gg 4GMb$ for the corresponding circular scale and the ratio tends to one, recovering the closed Kepler orbit. Intermediate orbits undergo apsidal precession because the ratio is generally not rational.

The radial action isWith the variable $x$ from part b this becomesApplying the supplied integral twice, with $c=0$ and $c=2$, and using the root sum and products from parts b and c givesThis is the radial action of the spherical isochrone model. Rearranging,and squaring yields the Hamiltonian in action-angle variables:As a check, differentiating this Hamiltonian gives ${\Omega }_r=(-2E{)}^{3/2}/(GM)$ and the frequency ratio found in part c.

At fixed radius, use spherical coordinates in velocity space with polar angle $\alpha$ measured from the radial direction:The galactic distribution function and volume element give angular weightAll dependence on $v$, $r$, and $g(E)$ cancels from ratios of second moments. Symmetry in the tangential plane givesWriting the angular integrals as beta function integrals and using the Gamma function recurrence,Therefore, for every admissible energy factor $g$,Equivalently, the velocity-anisotropy parameter is the constant $\beta =1-p/2$.

Put $\Psi =-\Phi =GM(r^p+a^p{)}^{-1/p}$. The spherical Poisson equation gives the hypervirial model density

For the proposed hypervirial distribution function, write $\mathcal{E}=-E=\Psi -v^2/2$ and $n=(3p+1)/2$. Direct velocity integration givesMatching this expression to the density fixesThis coefficient is positive for $0<p\le 2$, so the distribution function self-consistently generates the stated density and potential.

The density has the formInsert the ansatz ${\sigma }_r^2=k\Psi$ into the Spherical Jeans equationThe radial-power derivative cancels the anisotropy term because $p-2+2\beta =0$, leaving $k(2p+2)=1$. ThusTheir sum is independent of $p$:Therefore the local kinetic-energy density and gravitational potential-energy density areand they satisfy the local virial relation of the hypervirial modelat every radius. This pointwise identity is not generic. The ordinary virial theorem constrains suitable global integrals, with boundary terms when the system is truncated, but does not normally impose a virial balance shell by shell.