Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 58 (Galactic Astronomy and Dynamics)

Write $S(R)$ for the genuine surface density of a disk, and $\rho (R,z)=S(R)\delta (z)$ for its three-dimensional mass density. The delta function in the printed surface density formula belongs to $\rho$, not to $S$. Introduce a reference length $R_{\ast }$ to make logarithms dimensionless. The intended model is the infinite, self-gravitating, scale-free Mestel disk, with no extra source or imposed gravitational field.

The circular speed satisfies $v_c^2=R{\partial }_R\Phi (R,0)$. Thus the flat galaxy rotation curve givesA reflection-symmetric harmonic function with this midplane boundary value isVerify this continuation using the Poisson equation for Newtonian gravity. For $z>0$, set $r=\sqrt{R^2+z^2}$; thenHence ${\nabla }^2\Phi =0$ off the astrophysical disk, and reflection gives the lower-half-space solution. Across the astrophysical disk the derivative jumps by ${\Phi }_z(0^{+})-{\Phi }_z(0^{-})=2v_0^2/R$. The distributional Poisson equation for Newtonian gravity therefore givesAt the origin the enclosed disk mass tends to zero linearly with radius, so there is no additional central point mass. This verifies the Mestel disk potential-density pair. The astrophysical disk has infinite total mass and a logarithmic Newtonian gravitational potential, so it is not an isolated finite-mass model with Newtonian gravitational potential zero at infinity. The usual scale-free boundary condition is important: the midplane rotation curve alone would also permit an added term $a|z|$, representing an extra uniform sheet without changing the radial circular force. That contribution is excluded in the intended Mestel disk model.

Use a mass-weighted planar galactic distribution function, so $F(\mathbf{x},\mathbf{v})d^{2}x{d}^{2}v$ is the stellar mass in a small planar phase space element. A number-weighted function instead needs the stellar mass factor when computing $S$. Assume a steady collisionless stellar system and isotropy in the two in-plane velocity components. Put $u=|\mathbf{v}{|}^2/2$ and write $F=f(\mathbf{x},u)$. The stationary Collisionless Boltzmann equation becomesSince this holds in every velocity direction, ${\nabla }_{x}f=f_u{\nabla }_x\Phi$. In coordinates $(\mathbf{x},E)$ with $E=u+\Phi$, this is exactly ${\nabla }_{x}f{|}_E=0$. Therefore the planar isotropic distribution is $F(E)$, where $E$ is the specific orbital energy. Stationarity is essential; instantaneous isotropy alone would not imply this result.

Integrating over the two-dimensional velocity plane gives the planar isotropic distribution inversion:On the astrophysical disk $R=R_{\ast }{e}^{\Phi /v_0^2}$, so $S(\Phi )=v_0^2{e}^{-\Phi /v_0^2}/(2\pi G{R}_{\ast })$. Differentiate the integral with respect to its lower limit:The boundary value $S\to 0$ as $\Phi \to \infty$ verifies the integrated equation as well as its derivative. Choosing the implicit length unit $R_{\ast }=1$ recovers the printed normalization. Changing the additive energy zero changes this prefactor accordingly.

At a fixed radius, the normalized velocity density isIt is a product of centered Gaussian distributions. Differentiating the supplied Gaussian integral with respect to its coefficient gives the second moments, and odd moments vanish. Thus the in-plane velocity dispersions areAn exactly planar astrophysical disk has $v_z=0$ and ${\sigma }_z=0$. The nonzero circular speed is a property of the force field, not a statement that this hot stellar distribution has net rotation.

Reversing every retrograde star folds the azimuthal Gaussian to a half-normal distribution. Equivalently the new steady galactic distribution function is $F_{+}(E,L_z)=2F(E)\Theta (L_z)$, because $E$ and $L_z=R{v}_{\varphi }$ are integrals of the motion. Its density and even velocity moments are unchanged. Its streaming velocity isThis maximally prograde stellar distribution still has radial motion and a spread of azimuthal speeds; it does not place every star on a circular orbit. In particular $⟨v_{\varphi }^2{⟩}_{+}=v_0^2$ while ${\sigma }_{\varphi ,+}^2=v_0^2(1-2/\pi )$. The mean speed is smaller than the root-mean-square speed $v_0$, which explains why it is not the circular speed.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

Use the paper's positive binding-energy convention $U>0$; the physical Newtonian gravitational potential energy is $-U$. Assume positive masses, an isolated system, and no collision during the time interval under consideration. Differentiate half the scalar moment of inertia, $I$, twice:Group the force term into unordered pairs. The force on particle $j$ from $k$ is $-G{m}_j{m}_k({\mathbf{r}}_j-{\mathbf{r}}_k)/r_{jk}^3$; the two contributions to the virial sum are thereforeAdding them proves the instantaneous virial theorem identityNo time average is needed for this form. A vanishing average of $\ddot{I}$ would require additional boundedness assumptions.

For the pairwise moment-of-inertia identity, expand the squared pair distances:The last term vanishes in the centre of mass frame, givingLet $C={\sum }_{j<k}{m}_j{m}_k$, and label the two smallest masses $m_{(1)},m_{(2)}$. If $r=\min r_{jk}$, every term in $U$ is at most $G{m}_j{m}_k/r$, so $U\le GC/r$. One pair attains $r$, and its mass product is at least $m_{(1)}{m}_{(2)}$, so $U\ge G{m}_{(1)}{m}_{(2)}/r$. HenceThese constants show the minimum-separation binding-energy bounds: inverse binding energy is comparable to the nearest-pair distance, independently of configuration.

Similarly, every separation is at most $R=\max r_{jk}$ and at least one pair attains $R$. The pairwise identity gives the maximum-separation inertia boundsBoth lower bounds rely on actual extremal separations, rather than an arbitrary numerical lower or upper estimate for all the pair distances.

Negative energy does not imply a positive lower bound on the minimum separation. The printed request is false if it is meant to exclude close approaches or collisions; the trivial bound $r\ge 0$ says nothing of that kind. In fact $E<0$ and $T\ge 0$ give $U=T-E\ge |E|$, which, combined with $rU\le A_0$, proves the negative-energy minimum-separation upper boundThis ensures at least one close pair, not confinement of every particle or prevention of collision.

An explicit negative-energy gravitational collision is a pair initially at rest with separation $a$. Its total energy is $E=-G{m}_1{m}_2/a<0$. For reduced mass $\mu =m_1{m}_2/(m_1+m_2)$, its relative radial equation isThe separation decreases to zero in the finite timeBefore that time the motion is a regular Newtonian solution, yet its separation has no positive infimum. Even at fixed negative energy, bound Kepler orbits with eccentricity approaching one have arbitrarily small pericentre distance. Thus a missing angular momentum or collision-exclusion hypothesis cannot be supplied by the energy sign alone.

For $E>0$, the instantaneous virial theorem gives $\ddot{I}=T+E\ge E$. Integrating twice from any regular time $t_0$ givesUsing $I\le A_1{R}^2$, obtain the positive-energy linear diameter growth boundwhenever the numerator is nonnegative. Consequently, for a solution existing for arbitrarily large future times,This is the precise at-least-linear expansion statement. It does not assert that $R$ is monotone at every instant, nor that all individual particles escape. The large-time conclusion presupposes continued existence of the trajectory; the inequality itself holds on every nonsingular time interval.

Put $\mu =GM$ and use energy and angular momentum per unit particle mass. The Kepler orbit equation is $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$. HenceThus $E$ and $\mathbf{L}=\mathbf{r}\times \dot{\mathbf{r}}$ are integrals of motion. Splitting the velocity into radial and transverse parts, $|\dot{\mathbf{r}}{|}^2={\dot{r}}^2+L^2/r^2$, gives the Kepler radial energy equationHere $\dot{r}$ denotes the derivative of the scalar radius, unlike $\dot{\mathbf{r}}$.

For the Laplace-Runge-Lenz vector, differentiate directly:The vector triple-product identity yieldsThe terms cancel, proving every component of $\mathbf{A}$ is conserved.

Write $q=\bar{r}(r)$ and $\tau =\bar{t}$ for a monotone radial orbit time transformation. The chain rule givesImpose $q^{\prime }=(r/q)\dot{r}$, equivalentlyMultiplying the radial energy equation by $r^2/q^2$ then givesThis is a transformation along each trajectory; the new clock is obtained by integrating its radius-dependent rate.

For the Kepler–harmonic radial duality, take the old inverse-radius term as the new constant energy, $\bar{E}=\mu r/q^2$. On a positive-radius branch $\bar{E}>0$, soThe remaining term is minus the new Newtonian gravitational potential:For $E<0$ this is an attractive isotropic harmonic oscillator with ${\bar{\omega }}^2=-2E(\bar{E}/\mu {)}^2$. For $E>0$ it is an inverted oscillator, and for $E=0$ the Newtonian gravitational potential vanishes. Calling it a confining oscillator therefore requires a bound original Kepler orbit.

For the angular reconstruction of a radial orbit transformation, choose the new angle by $d\bar{\theta }/d\tau =L/q^2$. The same $L$ then supplies its centrifugal term. From $d\theta /dt=L/r^2$, the transformation is $d\bar{\theta }/d\theta =(r/q)(dq/dr)$; in the harmonic case this gives $\bar{\theta }=\theta /2$ up to a constant. Keeping the old angle unchanged would not in general preserve the angular momentum used in the displayed transformed radial equation.

For the second transformation use the explicit displayed split of the energy terms. It assigns weight $\lambda$ to $E$ and $1-\lambda$ to $\mu /r$; the surrounding printed prose interchanges these weights. Define the transformed quantities consistently byTheir difference is exactly $(r^2/q^2)(E+\mu /r)$, so they give the same transformed radial energy equation.

Let $a=(1-\lambda )\mu$, $h=\lambda E$, and $\epsilon =\bar{E}$. Constancy of the new energy imposesFor the standard regular isochrone branch take $h>0$, $\epsilon >0$, and $a\ge 0$. The positive-radius root isFor $q>0$ it is monotone. The corresponding clock is also regular there:With $\alpha =\sqrt{\epsilon /h}$ and $s=\sqrt{q^2+b^2}$, rationalization gives $r/q^2=\alpha /(s+b)$. Eliminate $r^2/q^2$ using $h{r}^2/q^2=\epsilon -ar/q^2$:where one possible identification isThis is the Kepler–isochrone radial transformation to the spherical isochrone model, up to the additive energy constant $k$.

The attractive branch of the Kepler–isochrone transformation requires a suitable parameter domain. The displayed square root needs $\lambda E\bar{E}>0$, and the positive-root branch used above needs $\lambda E>0$, $\bar{E}>0$. An attractive regular isochrone also needs $b\ge 0$ and $\bar{M}>0$. For bound original orbits a convenient domain is $E<0$, $\lambda <0$, $\bar{E}>0$: it satisfies all these requirements and gives $\bar{E}-k=\bar{E}/\lambda <0$. For $E>0$ one may instead take $1/2<\lambda \le 1$, with $\bar{E}>0$; these map to unbound attractive isochrone orbits. Other parameter choices can produce a repulsive Newtonian gravitational potential or a different branch, so no unrestricted real-parameter statement is justified. The singular algebraic case $\lambda =0$ is the harmonic transformation already handled separately. The angular reconstruction $d\bar{\theta }/d\tau =L/q^2$ likewise completes the isochrone radial solution into a central-force orbit.