Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 40 (Galaxies)

Use the instantaneous recycling approximation and treat the interstellar medium as well mixed. If star formation initially consumes mass $dM$, returning a fraction $\beta$ leaves locked stellar mass $ds=(1-\beta )dM$. The original metal mass removed from the interstellar medium is $Z_{i}dM$. Of this original material, $\beta f{Z}_{i}dM$ returns and remains in the galaxy. In addition, the stellar yield produces fresh metal mass $y_{i}ds$, of which the retained part is $f{y}_{i}ds$. Therefore metal accounting givesThis is the retained-ejecta chemical evolution equation. It assumes that the same retention fraction applies to original and newly synthesized metals. Pristine galactic gas inflow can change $g$ without contributing to $d(g{Z}_i)$; enriched galactic gas inflow would require an additional metal source. Here $0\le \beta <1$, and $0\le f\le 1$.

A star inherits the gas-phase metallicity at its birth. Suppose enrichment is monotone and a fixed initial mass function, with an appropriate surviving tracer selection, produces $\kappa$ observable stars per unit locked mass. For two birth abundances $Z_a<Z_b$, the relative count isThus $s(Z_i)$ is proportional to the cumulative metallicity distribution function; its derivative gives the differential metallicity distribution function. The normalized distribution after final locked mass $s_{\infty }$ is $s_{\infty }^{-1}ds/d{Z}_i$. For logarithmic abundances, the logarithmic metallicity distribution instead has $dN/d{\log }_{10}{Z}_i=(\ln 10)Z_{i}dN/d{Z}_i$.

The fixed initial mass function and survival selection matter: a mass distribution is not automatically the number distribution of every presently observable star. If enrichment is nonmonotone, add the contributions $\kappa |ds/d{Z}_i|$ from all birth-time branches attaining that abundance, rather than using one inverse.

The independent dimensionless variable is $S=s/s_{\infty }$. Since $g{Z}_i=s_{\infty }G{Z}_i$ and $dS/ds=1/s_{\infty }$, the chain rule gives $d(g{Z}_i)/ds=d(G{Z}_i)/dS$. The derivative in the normalized equation must consequently be with respect to $S$. Substituting the gas history in the retained-ejecta chemical evolution equation yieldsThis step allows a varying retention fraction $f(S)$; no derivative of $f$ is needed because $f$ enters the instantaneous metal source and loss terms.

Set $H=G{Z}_i$ and now take $f=S$. The coefficient in this first-order ordinary differential equation decomposes asHence an integrating factor is $\mu (S)=S^{1/(1-\beta )}/(1-S)$. With $q=1+1/(1-\beta )=(2-\beta )/(1-\beta )$, multiplication by this integrating factor givesThe general solution has an additional term $C/S^q$. Finite initial gas-phase metallicity rules it out. Consequently, for $0<S<1$,The upper limit is the dimensionless $S$, not the unnormalized stellar mass. On each compact subinterval of $0\le S<1$, the geometric series $1/(1-u)={\sum }_{k\ge 0}{u}^k$ converges uniformly, permitting integration term by term. ThusIn particular $Z_i\sim y_{i}S/(q+1)$ at small $S$. For positive stellar yield, the series is strictly increasing, so the inverse needed for the metallicity distribution function exists. Differentiating either representation gives $Z_i^{\prime }=y_i/(1-S)-q{Z}_i/S>0$. It diverges logarithmically as gas exhaustion approaches; the trace-metal approximation eventually becomes inadequate.

The quadratic gas-history enrichment model is not a closed gas reservoir: its gas mass initially increases. This can be made consistent with the metal equation by pristine galactic gas inflow. Indeed gas accounting gives $dg/ds=-(1-\beta f)/(1-\beta )+d{M}_{\mathrm{in}}/ds$. Since $dg/ds=1-2S$ and $f=S$, the required inflow is $d{M}_{\mathrm{in}}/ds=q(1-S)\ge 0$. It adds no metals and therefore leaves the preceding derivation intact.

The pressure tensor has Cartesian components $p_{ab}=\int f(v_a-u_a)(v_b-u_b)d^{3}v$. With zero mean velocity and an isotropic galactic distribution function, off-diagonal components vanish by reflection symmetry and diagonal components are equal by rotational symmetry. ThereforeHere $p$ is a tracer number-weighted random-motion stress, rather than thermodynamic pressure per volume. Its tensor divergence is $(\mathrm{div}\mathbf{p}{)}_a={\partial }_b(p{\delta }_{ab})={\partial }_{a}p$. The time-independent, nonstreaming Jeans equation consequently reduces toThe gravitational acceleration convention is $\nabla \psi$: this $\psi$ differs by a sign from the usual Newtonian gravitational potential $\Phi$.

Let $z$ measure distance along the line of sight. Spherical density projection givesBecause isotropy makes the local second stellar velocity moment along every line of sight equal to $p/n$, and because there is no streaming motion, the projected moment isThis is the forward relation for isotropic stellar pressure deprojection.

Both inversions follow from one Abel transform calculation. Write $F(u)=N(\sqrt{u})$ and $a(v)=n(\sqrt{v})$, so $F(u)={\int }_u^{\infty }a(v)(v-u{)}^{-1/2}dv$. Compose the transforms, using Tonelli's theorem for nonnegative densities, or absolute convergence for signed profiles:Assume sufficient decay to make these integrals finite and sufficient regularity for the last derivative. Then $J^{\prime }(x)=-\pi a(x)$. Setting $x=r^2$ gives the spherical Abel deprojectionReplacing $F(u)$ by $N(\sqrt{u}){\sigma }^2(\sqrt{u})$ and $a(v)$ by $p(\sqrt{v})$ provesThese are the requested expressions, with $u=R^2$. The composed-transform argument avoids an invalid separate differentiation of a divergent lower-end kernel.

For the spherical Abel deprojection, put $A_r=r^2+b^2$ and $s=(R^2+b^2)/A_r$. Then $d{R}^2=A_{r}ds$ and $\sqrt{R^2-r^2}=A_r^{1/2}\sqrt{s-1}$. Consequently the integral appearing in the reconstruction of $n$ isDifferentiate before substituting the numerical value of $I_{5/2}$:For the projected second stellar velocity moment, $N{\sigma }^2=N_0{\sigma }_0^2{b}^6(R^2+b^2{)}^{-3}$. The identical substitution in isotropic stellar pressure deprojection givesFor completeness, setting $t=1/s$ yields $I_{\alpha }=B(1/2,\alpha -1/2)$, where $B$ is the Euler beta function. The Gamma function recurrence gives $I_{5/2}=4/3$ and $I_3=3\pi /8$. This also checks the normalization of the power-law spherical projection kernel.

The radial Jeans equation is $p^{\prime }=n{\psi }^{\prime }$. Since $p^{\prime }=-105N_0{\sigma }_0^2{b}^{6}r/(16A_r^{9/2})$, division by the tracer number density givesWriting $A=315\pi {\sigma }_0^{2}b/128$, the total gravitational acceleration is thereforeTo find all gravitating mass density, apply the Poisson equation for Newtonian gravity, with the same sign convention:Equivalently, the enclosed mass is $M(r)=A{r}^3/[G(r^2+b^2{)}^{3/2}]$, and ${\rho }_{\mathrm{tot}}=M^{\prime }/(4\pi r^2)$. The inferred total mass is $A/G$ and its relative potential is $A/\sqrt{r^2+b^2}$, a Plummer model. The tracer number density has a different radial exponent from this total mass density. One must not multiply $n$ by an arbitrary stellar mass and identify it with all matter: the inferred field can include unobserved stars, gas, and dark matter.

Use gravitational acceleration $\mathbf{g}=\nabla \psi$. Through the plane with upward unit normal, the signed gravitational flux per unit area of the point mass isThus its downward flux magnitude is $Gmb/(R^2+b^2{)}^{3/2}$. Subtracting a constant from the gravitational potential does not change this flux.

The Kuzmin reflection method gives the even relative potential $\psi =Gm/\sqrt{R^2+(|z|+b{)}^2}$. It is harmonic above and below the plane because the corresponding point masses lie outside the respective half-spaces. At the plane the derivative jump isIntegrating the Poisson equation for Newtonian gravity ${\nabla }^2\psi =-4\pi G\Sigma (R)\delta (z)$ through a thin layer, with $\delta$ the Dirac delta, equates this jump to $-4\pi G\Sigma$. ThereforeThis is a Kuzmin disk. As a check, its total mass is $2\pi {\int }_0^{\infty }R\Sigma (R)dR=m$.

Replace the point mass by $dm=\mu (b)db$ and superpose the Kuzmin disks. Linearity of the Poisson equation for Newtonian gravity gives, whenever the force integral converges,Both choices of potential reference give this same surface density. If the unreferenced potential diverges, perform the reference subtraction in each component before taking the integral limit; the force and the surface density may still be finite.

The exponent of the point-mass weight is $\beta$, with $0<\beta <1$. At the origin the unreferenced relative potential has integrand $GB{b}^{\beta -1}$, whose integral diverges at its upper endpoint. In the origin-referenced potential the two terms cancel pointwise there. More generally, first form their difference at finite cutoffs and then pass to the limit. For any fixed nonzero $(R,z)$, the small-$b$ singular term is $-b^{\beta -1}$, which is integrable because $\beta >0$. At large $b$,so the weighted difference is integrable because $\beta <1$. Hence the referenced potential is well defined, and its value at the origin is zero. It is a potential difference for an infinite astrophysical disk, not a potential with zero at infinity. In fact rescaling all lengths shows ${\psi }_{II}(\lambda R,\lambda z)={\lambda }^{\beta }{\psi }_{II}(R,z)$; its finite values approach zero at the origin even though its force is singular there.

For $R>0$, differentiation under the convergent force integral in the plane givesUse $b=R\sqrt{t/(1-t)}$, for which $db=(R/2)t^{-1/2}(1-t{)}^{-3/2}dt$. The powers of $R$, $t$, and $1-t$ then givewhere the Euler beta function evaluates the constant asThe first gamma function argument is $(1+\beta )/2$: it follows from adding one to the exponent of $t$. Both Euler beta function arguments are positive in the stated range.

For a circular orbit in the plane, $V^2/R=-{\partial }_R{\psi }_{II}$. The prescribed circular speed therefore fixes $C=K^2$ andIndependently, the Kuzmin reflection method givesHere the power of $b$ is one larger than in the force integral, so its Euler beta function arguments differ. Substituting the normalization fixed by the circular speed yieldsThis scale-free Kuzmin superposition has nonnegative surface density, finite central enclosed mass, and infinite total mass. Its force and surface density approach the Mestel disk values as $\beta \downarrow 0$; that limiting disk requires a different potential reference at the origin.

Let $\mathcal{G}$ be the outward viscous torque in an accretion disk. For a declining angular velocity, the vertically integrated viscous shear stress givesIn a steady accretion disk without radiated angular momentum, the net inward angular-momentum flux $Fh-\mathcal{G}$ is constant and equals the swallowed flux $F{h}_0$. Thus conservation of angular momentum gives $F(h-h_0)=\mathcal{G}$. Substituting $F=2\pi r\Sigma u$ cancels the surface density and provesThis uses the stipulated phenomenological viscous shear law; it does not replace it by a different fully relativistic stress prescription.

For the orbital calculation let a dot denote a derivative with respect to proper time. Substituting the geodesic energy and specific angular momentum first integrals into the timelike normalization in Schwarzschild spacetime givesDifferentiate with $h$ and $ϵ$ constant along the geodesic. Where $\dot{r}\ne 0$, division by $2\dot{r}$ givesThis equation also holds at turning points and on circular orbits. To justify that without dividing by zero, put $A=1-2m/r$ in the equatorial geodesic Lagrangian $L=(A{c}^2{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)/2$. Its radial Euler-Lagrange equation is $\ddot{r}=(A^{\prime }/2A){\dot{r}}^2-A{A}^{\prime }{c}^2{\dot{t}}^2/2+Ar{\dot{\varphi }}^2$. Substitution of the timelike normalization cancels the ${\dot{r}}^2$ terms and yields $\ddot{r}=-A^{\prime }{c}^2/2+(A/r-A^{\prime }/2)h^2/r^2$, exactly the displayed radial equation, including when $\dot{r}=0$.

A circular orbit requires $\ddot{r}=\dot{r}=0$. Thereforechoosing the positive sense of rotation. Differentiating the square is simpler than differentiating $h$:It is negative for $3m<r<6m$ and positive for $r>6m$, so the global minimum over timelike circular orbits isAs a stability check, the second radial derivative of $W=(1-2m/r)(c^2+h^2/r^2)$ at a circular orbit, holding $h$ fixed for the perturbation, is $W^{\prime \prime }=2m{c}^2(r-6m)/[r^3(r-3m)]$. Thus the exterior branch is stable, and $6m$ is the innermost stable circular orbit. It is the natural matching radius for the idealized zero-torque inner boundary condition.

Take constant kinematic viscosity $\nu$ and the inner specific angular momentum found above. The Angular velocity of a circular Schwarzschild geodesic gives $-{\Omega }^{\prime }=3c\sqrt{m}/(2r^{5/2})$. Hence the inward radial component of four-velocity in the stated viscous model isTo convert proper time to Schwarzschild time, neglect ${\dot{r}}^2$ in the normalization as prescribed and insert the circular value of $h$. One obtainsThe change in observed Schwarzschild time equals this coordinate-time change when differential light travel time is neglected. Since $dr/d\tau =-u$,This time is nonnegative: $r^2-12m(r-3m)=(r-6m{)}^2\ge 0$ on the integration interval.

Put $x=r/(3m)$ and $X=r_1/(3m)$. The integral becomesPolynomial division gives $x^2/(x-1)=x+1+1/(x-1)$. Separately, writing $y=x-1$ gives $\int 2x/\sqrt{x-1}dx=(4/3)(x-1{)}^{3/2}+4\sqrt{x-1}$. Therefore the antiderivative inside the integral isIts value at $x=2$ is $-4/3$. Subtracting this lower endpoint yields the formal viscous inspiral time in Schwarzschild spacetimeThe logarithm has a positive coefficient. The minus sign in the printed final expression is inconsistent with its preceding integral. Differentiating the corrected bracket gives $6X[X-2\sqrt{X-1}]/(X-1)$, precisely the positive integrand after rescaling. The bracket vanishes at $X=2$; replacing $+6\log (X-1)$ by its negative would give derivative $-12$ there and a negative elapsed time for $X$ just above $2$. This is a sign error, not a choice of potential or time convention.

The result assumes constant kinematic viscosity; if $\nu$ varies with radius it remains inside the time integral. There is also a physical limitation to the stipulated approximation: $h-h_0$ vanishes quadratically at $6m$, so its formal $u$ diverges and eventually violates the neglected-radial-motion condition. The finite integral is the formal extrapolation of the nearly circular viscous model. A real accretion disk must instead match to its plunging region of a black-hole accretion disk; this approximation cannot describe that transition arbitrarily closely.