Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 66 (Galaxies and Dark Matter)

The separated disk modes. Away from the plane of a razor-thin disk, there is no volume source, so the Newtonian gravitational potential obeys the Laplace equation in cylindrical coordinates. Separation of variables givesHere $k>0$ is a radial wavenumber, with units of inverse length. Regularity on the axis selects the Bessel function of the first kind $J_0(kR)$ rather than the singular second solution. Decay away from the disk, continuity across it and reflection symmetry select $Z=e^{-k|z|}$. These conditions apply to each nonzero-wavenumber mode; a uniform sheet is a separate zero-mode problem.

Integrate the Poisson equation for Newtonian gravity across the disk to obtain the jump conditionFor the unit-amplitude separated mode, the jump is $-2k{J}_0(kR)$. ThusAn individual basis density is signed; positive physical surface densities are built by superposition.

An arbitrary axisymmetric surface density. Define its order-zero Hankel transform byMatching the coefficients of the mode densities gives the Hankel representation of thin-disk gravity,The boundary conditions exclude added external fields. For sufficiently decaying surface density, this is the isolated-disk potential; for infinite scale-free disks, only potential differences or regularized forces need be finite.

The Mestel disk. Put $C={\Sigma }_0{R}_0$. The Bessel function integral ${\int }_0^{\infty }{J}_0(k{R}^{\prime })d{R}^{\prime }=1/k$, understood with a decaying regulator, gives $\hat{\Sigma }=C/k$. Differentiate the potential before removing the regulator and use $J_0^{\prime }=-J_1$:The last integral is $1/R$, since ${\int }_0^{\infty }{J}_1(u)du=[-J_0(u){]}_0^{\infty }=1$. Therefore the circular speed and enclosed mass areConsequently $v_c^2=GM(R)/R$. This equality follows from the specific Mestel disk calculation; the spherical shell theorem cannot be applied to general disks. Its total mass and absolute potential referenced to infinity diverge. A convenient finite midplane reference is $\Phi (R,0)=v_0^2\log (R/R_0)$.

The rotating distribution function. Use specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. For the printed positive exponential, take $\epsilon =-E$ and $s:=\sigma >0$; $s$ has units of squared velocity. If instead $\epsilon$ means ordinary total energy and $\sigma >0$, the velocity integral diverges, so that literal convention cannot define the requested galactic distribution function. Equivalently one can retain ordinary energy and choose the printed $\sigma$ negative.

With the binding-energy convention, write the amplitude as $A$ to distinguish it from the function:Integration over both radial velocities and positive azimuthal velocities givesThe half-line integral follows by setting $u=v_{\varphi }^2/(2s)$ in the defining Gamma function integral and requires $q>-1$. Matching its radial power and normalization to the Mestel disk yields the rotating Mestel disk distribution function parametersSince $v_0^2,s>0$, the required integrability condition is automatic. The amplitude uses the stated potential reference; changing the additive potential constant changes $A$ without changing the physical distribution.

First velocity moment. Assume a stationary, spherical, nonstreaming collisionless stellar system with constant mass-to-light ratio ${\rm Y}$. Use a luminosity-weighted galactic distribution function, with luminosity density $\lambda$. Multiplying the Collisionless Boltzmann equation by $v_i$ and integrating over velocities, with its velocity-space boundary terms vanishing, givesSpherical symmetry makes the stress tensorHere ${\sigma }_t^2$ is one tangential component, as in the paper; the total tangential variance is $2{\sigma }_t^2$. Its radial divergence is $q^{\prime }+2(q-q_t)/r$, since the divergence of the radial unit vector is $2/r$. With the velocity-anisotropy parameter $\beta =1-{\sigma }_t^2/{\sigma }_r^2$, and enclosed mass $M(r)={\rm Y}L(r)$, this gives the Spherical Jeans equationThis uses a constant ratio of total gravitating mass to luminosity, rather than an arbitrary independent dark halo.

Projection and elimination of anisotropy. Write $R$ for projected radius and $r$ for spherical radius. The Abel transform of the luminosity density giveswhose inverse is the displayed density inversion in the paper. At a point on this line of sight, ${\sin }^2\theta =R^2/r^2$. The line-of-sight second stellar velocity moment is ${\sigma }_r^2{\cos }^2\theta +{\sigma }_t^2{\sin }^2\theta ={\sigma }_r^2(1-\beta R^2/r^2)$. Therefore the measured quantity $J(R)=\mu (R){\sigma }_p^2(R)$ satisfiesThe Spherical Jeans equation gives $\beta q=-r{q}^{\prime }/2-\gamma \lambda L/(2r)$. Substitution yieldsThe inner $\gamma L$ printed in the PDF's definition of $p$ must therefore be $\lambda L$: the literal expression has an extra gravitational factor and lacks the density. Both this derivation and dimensional consistency require the correction. Define the projected spherical Jeans residual byTo express it entirely in terms of the radial stress, putDifferentiating the nonsingular $z$ representation gives $A^{\prime }(R)=R{\int }_R^{\infty }{q}^{\prime }(r)/\sqrt{r^2-R^2}dr$. HenceIf $R^{2}A$ vanishes at zero and infinity, then ${\int }_0^{\infty }6\pi Rp(R)dR=6\pi [R^{2}A{]}_0^{\infty }=0$. These are the absence-of-boundary-stress conditions needed for the global result.

The virial relation, term by term. Interchange the order in the residual integral and useIt follows thatThe first term is $2K/{\rm Y}$: spherical averaging assigns one third of the integrated velocity-square to any fixed viewing direction, giving $K=3\pi {\rm Y}\int RJdR$. The gravitational potential energy isThus the integrated residual is exactly $(2K+W)/{\rm Y}$, and its vanishing is the virial theorem. Solving for $\gamma$ gives the global projected virial estimate of a mass-to-light ratio,It is independent of the radial anisotropy profile when the full aperture and the stated boundary conditions are used. A finite aperture retains a boundary term.

The factor called $N$ in the formula must be the stellar number density $n$, not a dimensionless count. Indeed ${\sigma }^3/(G^2{m}^2)$ has dimensions time per volume, so a density in the denominator is essential. If $N_{\ast }$ is the population in a representative volume $V$, then $n=N_{\ast }/V$.

For a weak encounter between equal-mass stars, let $u$ be the asymptotic relative speed and $b$ the impact parameter. In the straight-line impulse approximation, one star's transverse kick isThe other receives the opposite kick, so the relative-velocity change is $\Delta u_{\perp }=4Gm/(bu)$. Encounters in $(b,b+db)$ occur at rate $2\pi bdbnu$. Independent random kicks add in squared magnitude, givingThis is the random-walk derivation of stellar relaxation. The Coulomb logarithm in stellar dynamics is $\ln \Lambda =\ln (b_{\max }/b_{\min })$. The upper cutoff is the size or inhomogeneity scale beyond which independent local encounters are inappropriate. The lower cutoff is of order the strong-deflection scale $G(2m)/u^2$, or a larger stellar-size/softening scale if applicable. A self-gravitating system commonly has $\Lambda$ of order its total number of stars; that population count is distinct from $n$ in the diffusion rate.

To state the numerical convention explicitly, use representative encounter speed $u=\sqrt{2}\sigma$ and define the relaxation estimate from the relative-velocity diffusion rate with reference squared velocity $3{\sigma }^2$. This givesThis recovers the printed coefficient under a specified convention. There is no convention-independent exact numerical prefactor determined by a representative one-dimensional velocity alone. Using the single-star diffusion rate with the same $3{\sigma }^2$ reference instead gives four times this time; averaging over a full velocity distribution also changes the prefactor. The relaxation time conventions in stellar dynamics must therefore distinguish single-star and relative kicks. The physical scaling is $T_R\propto {\sigma }^3/(n{G}^2{m}^2\ln \Lambda )$, and the printed $N$ must be interpreted as number density.

Use $\mu =\gamma m$ for the gravitational parameter, to distinguish it from the Delaunay action called $G$. The displayed actions are per unit reduced mass. Accordingly use specific canonical momenta; with physical relative momenta, all three actions acquire a common reduced-mass factor, which leaves the eccentricity fraction below unchanged.

The canonical transformation. In spherical coordinates, separation of the Kepler Hamilton-Jacobi equation givesA local type-two generating function, continued across turning points with the appropriate momentum branches, isThe old momenta are $\partial W/\partial (r,\theta ,\varphi )$ and the new angles are $l={\partial }_{L}W$, $g={\partial }_{G}W$, $h={\partial }_{H}W$. This defines a canonical transformation, since its generating differential is $dW=p_{r}dr+p_{\theta }d\theta +p_{\varphi }d\varphi +ldL+gdG+hdH$. With standard choices of angle origins, the resulting Delaunay variables arewith $l$ the mean anomaly, $g$ the argument of periapsis, and $h$ the longitude of ascending node. In particular, the signed $H$ includes retrograde orbits; the squared equation in the PDF must not be read as $H\ge 0$.

For bound noncollision Kepler orbits, the independent ranges areThe limiting circular, radial and coplanar cases sit on boundaries where some angle charts degenerate. They have zero measure in the generic phase-volume calculation. The ranges are constrained as above, rather than a Cartesian box with three unconstrained actions.

The eccentricity count. A canonical transformation preserves phase space volume, soIntegrating the three angles contributes $(2\pi {)}^3$, and integrating $H$ gives $2G$. At fixed $L$, the condition that the orbital eccentricity be less than $e_0$ is $G>L\sqrt{1-e_0^2}$. ThusAny energy weight cancels between these integrals. If the total population is finite, or an explicit finite action band is used,This is the Delaunay phase-volume proof of the thermal eccentricity distribution. The source's $e^2$ law is a cumulative count, not a differential eccentricity density. Also, pointwise finite $F(E)$ does not alone imply finite $N_{\mathrm{tot}}$: $F=1$ gives ${\int }_0^{\infty }{L}^{2}dL=\infty$. The finite normalization of an energy-only Kepler distribution is the additional condition needed for a global probability statement; the fraction at every fixed energy is unaffected.

Why the observation does not imply relaxation or equilibrium. No Boltzmann distribution was used: every normalizable energy-only phase-space distribution function has the same eccentricity marginal, irrespective of collisional thermalization. An even stronger counterexample is a genuinely time-dependent collisionless distributionKepler motion has $\dot{l}=\nu (L)$ and constant actions and other angles, so ${\partial }_{t}f+\nu {\partial }_{l}f=0$: this is an exact nonstationary solution of the Collisionless Boltzmann equation. Integrating $l$ removes the cosine and leaves precisely the same $e^2$ cumulative count at every time. An eccentricity marginal alone cannot establish thermal relaxation, a relaxed age, or even a stationary distribution.

An isolating integral is a single-valued integral of motion $I(\mathbf{x},\mathbf{v})$ whose fixed value confines an orbit to a lower-dimensional level set of phase space. Along the flow, $dI/dt=0$; for an autonomous Hamiltonian, this means $\{I,H\}=0$. Several isolating integrals are independent when their gradients are linearly independent on the regular region considered. Counting them requires a specified domain: merely local orbit labels in a flow box do not establish global single-valued integrals. The following cases concern regular globally meaningful orbit constraints.

One integral can occur in an autonomous system with a fully chaotic region and no additional continuous orbit constraints. The specific orbital energy is conserved in a time-independent gravitational potential. If a chaotic stellar orbit is dense in a connected region of its energy hypersurface, every continuous integral of motion is constant on that region. Thus it cannot supply a new independent constraint there, and energy is the only independent isolating integral. A generic nonaxisymmetric potential can contain such regions; not every orbit in such a potential is necessarily chaotic.

No global isolating integral is possible in a sufficiently general time-dependent system without surviving symmetries or other invariants. Foreven energy is then not conserved, and nonsymmetric forcing can destroy all global constraints on the orbit. Time dependence alone does not imply this result: a central time-dependent force still conserves angular momentum, and a rigidly rotating potential can conserve a Jacobi integral of a rigidly rotating potential. The case of no integrals requires absence of such additional structure. A fixed point or locally defined flow labels are not the global orbit family considered here.

More than three integrals occur when additional symmetry or degeneracy restricts motion beyond a generic three-frequency invariant torus. In a time-independent central gravitational potential, energy and the three components of specific angular momentum are four generically independent isolating integrals. They fix the orbital plane as well as the radial-motion parameters. A nonclosed planar rosette typically has two-dimensional orbit closure, rather than a three-dimensional invariant torus.

The Kepler orbit is a stronger example. In addition to $E$ and $\mathbf{L}$, the conserved Laplace-Runge-Lenz vectorfixes the apsidal direction. Its constraints $\mathbf{A}\cdot \mathbf{L}=0$ and $A^2={\mu }^2+2E{L}^2$ leave five independent integrals among $E,\mathbf{L},\mathbf{A}$. It is a maximally superintegrable Hamiltonian system, with closed nondegenerate bound orbits.

More than three integrals need not form a set of Poisson-commuting functions; the three-integral limit applies to an independent commuting set, rather than to every conserved orbit constraint. At a nonstationary point in six-dimensional phase space, integral gradients annihilate the nonzero flow vector, so at most five are independent. This explains the isolating integrals and orbital closure dimension count. A resonance on one isolated orbit does not by itself prove an additional smooth global integral on a neighborhood.

Use $M$ for the mass parameter throughout; the PDF uses $m$ in its formulas and $M$ in the subsequent prose for the same quantity. Set $a=r_j>0$. The Jaffe model density simplifies toIts enclosed mass isIt tends to $M$ as $r\to \infty$, verifying the total mass. It tends to zero at the center, so the cusp carries no extra point mass.

Spherical symmetry gives ${\Phi }^{\prime }(r)=GM(r)/r^2=GM/[r(r+a)]$. Integrating with $\Phi (\infty )=0$ givesThis also directly satisfies the Poisson equation for Newtonian gravity: differentiating $r^2{\Phi }^{\prime }=GMr/(r+a)$ gives $4\pi G{r}^2\rho (r)$. The circular speed isFor $r\ll a$, $v_c=\sqrt{GM/a}[1-r/(2a)+O((r/a{)}^2)]$ is approximately constant. For $r\gg a$, $v_c=\sqrt{GM/r}[1-a/(2r)+O((a/r{)}^2)]\propto r^{-1/2}$. Thus the finite-mass model has an inner nearly flat galaxy rotation curve and an outer Keplerian decline.