Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 41 (Astrophysical Fluid Dynamics)

Integrating the given expansion equation gives $R^{3/2}=t/t_0$, with the integration constant fixed by $R(t_0)=1$. Thus the scale factor and Hubble parameter are $R=(t/t_0{)}^{2/3}$ and $H=2/(3t)$. The Newtonian derivation of the flat Friedmann equation gives $H^2=8\pi G\rho /3$, soUse the comoving coordinate $\mathbf{q}=\mathbf{x}/R$ and the density contrast $\delta ={\rho }_1/\rho$. Linearized mass conservation, the peculiar-velocity Euler equation for fluid motion and the Poisson equation for the peculiar gravitational potential areDifferentiate the first equation, substitute the second and use the third. A Fourier component of comoving wavenumber $k$ obeysSince $c^2{k}^2/R^2=k^2{c}^2(t_0)t_0^{2\gamma -2/3}{t}^{-2\gamma +2/3}$, the normalization is $\Lambda =kc(t_0)t_0^{\gamma -1/3}$, as printed in the PDF. The resulting linear cosmological density perturbation equation isTo reduce it to a Bessel equation, put $\nu =\gamma -4/3$, $z=\Lambda /(\nu t^{\nu })$ and $\epsilon =t^{a}y(z)$. As $\dot{z}=-\nu z/t$, multiplying the substituted equation by $t^{2-a}$ givesTaking $a=-1/6$ makes the first two coefficients proportional to those in the Bessel equation, and the remaining constant is $-25/36$. HenceFor noninteger $\lambda$, the Bessel density modes of a polytropic expanding universe areThere is an exceptional case within the allowed interval: if $\lambda =n$ is an integer, $J_{-n}=(-1{)}^n{J}_n$, so these are not two independent solutions. The full solution then uses $t^{-1/6}[A{Y}_n(z)+B{J}_n(z)]$, with the Bessel function of the second kind providing the missing branch. For example $\gamma =29/18$ gives $\lambda =3$. A basis using $J_{\lambda },Y_{\lambda }$ is valid for every allowed exponent.

For early times, $z\to \infty$. The large-argument asymptotic expansion of the Bessel function of the first kind has oscillatory phase $z-\pi a/2-\pi /4$ and envelope $(2/(\pi z){)}^{1/2}$ for order $a$; the Bessel function of the second kind has the same envelope. ThusThis decreases because $3\gamma <5$. At late times $z\to 0$, the leading powers are $J_{\pm \lambda }(z)\propto z^{\pm \lambda }$ for noninteger order, giving $t^{-1}$ and $t^{2/3}$. The $Y_n$ branch gives the same growing power in the integer case. Therefore a generic perturbation grows as $t^{2/3}$; an initial condition with zero growing-mode coefficient instead has the decaying $t^{-1}$ mode.

To find the growth scale including its dependence on $\nu$, set $\epsilon =t^{2/3}h(t)$. The equation becomesSubstitute $h=1+a{\Lambda }^2{t}^{-2\nu }+\cdots$. The leading coefficient equation is $-2\nu (5/3-2\nu )a+1=0$. The pressure correction to a growing polytropic cosmological mode is consequentlySince $ck/R=\Lambda t^{-\nu -1}$ and $t^{-2}=6\pi G\rho$, the correction becomes small when $(ck/R{)}^2{t}^2$ is small compared with $2\nu (5/3-2\nu )$. The factor $2(5/3-2\nu )$ is between two and $10/3$, giving the stated order-of-magnitude onset conditionFor a well-developed asymptotic growing mode, replace $\lesssim$ by $\ll$. This is a Jeans instability interpretation: the physical acoustic frequency must be sufficiently low compared with the gravitational expansion scale, or equivalently the physical wavelength must be sufficiently long for self-gravity to overcome pressure support. It is an approximate condition for the dust-like growing branch, not an exact initial-condition-independent instant at which any perturbation begins increasing. The instantaneous sign change of the coefficient in the original equation instead occurs at $c^2{k}^2/R^2=4\pi G\rho$.

In the dominant central Newtonian gravitational potential, conservation of energy for gas falling from rest at infinity gives $u^2/2=G{M}_c/R$, hence $u=a{R}^{-1/2}$ with $a=(2G{M}_c{)}^{1/2}$. The characteristic stellar speed has the same radial scaling. For random stellar directions, the cross term in $|\mathbf{u}-\mathbf{v}{|}^2$ averages to zero, so the typical relative speed is also $V=b{R}^{-1/2}$. With the stated kinetic-energy estimate, an rms convention gives $V^2\simeq 4G{M}_c/R$; another convention changes only the constant $b$.

In the steady spherical inflow approximation, mass conservation gives $A=4\pi R^2\rho u$. ThusThis leading calculation treats the inward gas flux as essentially undepleted over the region used. Substantial distributed removal by stars would make the local flux a function of $R$, an important limitation discussed below.

Insert these scalings in the Bondi--Hoyle--Lyttleton accretion rate. The factors $R^{3/2}$ from $V^{-3}$ and $R^{-3/2}$ from $\rho$ cancel:This is quadratic accretion in a central Keplerian potential. Write its fixed coefficient as $\lambda =\Lambda$. Integrating $d(1/M)/dt=-\lambda$ givesThe cancellation of radius means that orbital variation of $R$ alone does not change this coefficient, provided the assumed steady gas profile, central mass and relative-velocity scaling remain fixed.

Conserve the number of stars along these deterministic mass trajectories. Inverting the map and taking its Jacobian determinant givesThe mass-spectrum transport under quadratic accretion is thereforeFor a narrow but nonzero initial mass interval, the mapping stretches the high end especially strongly. If its endpoints are $m_1<m_2$, their evolved ratio is $(m_2/m_1)(1-\lambda m_{1}t)/(1-\lambda m_{2}t)$, which can become large as $t$ approaches $1/(\lambda m_2)$ from below. In an intermediate high-mass range with $\lambda Mt\gg 1$, mapped back into a portion of the initial interval where $F_0$ changes little,Here $M_{\ast }$ is near $1/(\lambda t)$ on the populated side of the initial interval. Finite initial support gives a finite upper mass cutoff at every time before the earliest blow-up; this is an approximate broad-range power law, not a literal infinite tail of a finite pre-blow-up population. A strictly monochromatic initial population remains monochromatic under identical accretion coefficients and does not generate a spectrum.

The limitations of this accretion model are substantial. The star cluster must remain centrally dominated, and both the central mass and gas supply must remain effectively fixed. Stellar capture depletes the gas flux and eventually invalidates the $R^{-3/2}$ profile. Gas pressure, angular momentum, shocks, radiative feedback and winds can reduce capture relative to the cold Bondi–Hoyle accretion estimate; a broad relative-speed distribution is not faithfully represented by one typical speed, especially because capture scales as $V^{-3}$. A growing star can perturb the surrounding potential, overlap capture regions, change its orbit through momentum loading, or compete with its neighbors for the same gas. Finally, number conservation excludes mergers, new star formation and stellar losses. The formal finite-time mass divergence cannot persist under a finite available gas mass, so the inferred mass-spectrum slope applies only while these approximations remain adequate.

Fix the sign convention first: depth $z$ increases downward, while the vertical displacement amplitude $\xi$ is positive upward. Write $\Pi =\delta p$ for the Lagrangian pressure perturbation, and $p_1,{\rho }_1$ for Eulerian fluid perturbations. Hydrostatic equilibrium gives $dp/dz=\rho g$, and the adiabatic relation gives $\delta \rho =\Pi /c^2$. Since the vertical displacement vector has component $-\xi$ along increasing $z$,In the Cowling approximation, horizontal momentum gives ${\mathbf{\xi }}_h=i\mathbf{k}{p}_1/(\rho {\omega }^2)$. The mass conservation equation $\delta \rho =-\rho (i\mathbf{k}\cdot {\mathbf{\xi }}_h-{\xi }^{\prime })$ and vertical momentum ${\omega }^2\rho \xi =-p_1^{\prime }+g{\rho }_1$ then eliminate the horizontal displacement. The plane-parallel Cowling displacement-pressure equations areFor ${\omega }^2=gk$, the second equation admits $\Pi =0$ and the first reduces to ${\xi }^{\prime }=-k\xi$. Thus the plane-parallel stellar f-mode hasIts Eulerian eigenfunctions are $p_1=\rho g\xi$ and ${\rho }_1={\rho }^{\prime }\xi$, while $\delta \rho =0$. Hence it has no Lagrangian compression even in a stratified star. It decays inward but grows upward, explaining the failure of a pure small-amplitude continuation to $z\to -\infty$.

For the lower part of the matched chromospheric interfacial mode, put ${\omega }^2=gk(1-\epsilon )$, $\xi =e^{-k(z-z_0+\epsilon \varphi )}$ and $\Pi =\epsilon f\xi$. The pressure amplitude is of order $\epsilon$ because the limiting f-mode has identically zero Lagrangian pressure amplitude; the frequency correction perturbs that zero solution. Expand $g{k}^2/{\omega }^2=k(1+\epsilon )+O({\epsilon }^2)$ and ${\omega }^2-g^2{k}^2/{\omega }^2=-2gk\epsilon +O({\epsilon }^2)$. The leading equations areExclude the lower homogeneous component $f\propto e^{2kz}$, which would make $\Pi$ grow into the star. Integrating the remaining equation with its integrating factor givesThe accompanying correction is $\varphi (z)={\int }_{z_0}^z[1+(f/\rho )(1/g-1/(k{c}^2))]ds$, with the integrand evaluated at $s$. This fixes $\varphi (z_0)=0$ and determines the lower eigenfunctions to the stated order.

For the upper part, distinguish the rescaled sound speed by writing $\rho =\epsilon \hat{\rho }$ and $c^2={\epsilon }^{-1}{\hat{c}}^2$. Set $\Pi =\epsilon \psi$, $\psi =A{e}^{k(z-z_0+\epsilon \varphi )}$ and $\xi =\mu \psi$. To leading order the first displacement equation becomes ${\xi }^{\prime }=-k\xi -k\psi /(g\hat{\rho })$, while the pressure equation gives ${\psi }^{\prime }=k\psi +\epsilon [k\psi +2gk\hat{\rho }\xi ]+O({\epsilon }^2)$. ConsequentlyThe component $\mu \propto e^{-2kz}$ would give $\xi \propto e^{-kz}$, which diverges upward. Remove it by the upper boundary condition. The leading upper solution isIntegrating ${\varphi }^{\prime }$ with $\varphi (z_0)=0$ gives the first correction to the upper exponential. The displayed $\xi =\mu \psi$ satisfies the required decay only for compatible upper density profiles; integral convergence alone is not the full boundary condition.

To match, define the positive weighted integralsThe lower normalization is $\xi (z_0)=1$ and $f(z_0)=-2gk{I}_b$. Continuity of Lagrangian pressure perturbation makes the upper $A=f(z_0)$; continuity of fluid displacement then requires $\mu (z_0)A=1$. Since $1/\hat{\rho }=\epsilon /\rho$, the upper integral gives $\mu (z_0)=-(k/g)\epsilon I_a$. The two matching conditions therefore giveAs a sign and normalization check, two incompressible constant-density layers give $I_a=(2k{\rho }_a{)}^{-1}$ and $I_b={\rho }_b/(2k)$, so $\epsilon \simeq 2{\rho }_a/{\rho }_b$. This agrees with expansion of the exact interfacial gravity-wave dispersion relation ${\omega }^2=gk({\rho }_b-{\rho }_a)/({\rho }_b+{\rho }_a)$.

For validity, require $\epsilon \ll 1$, convergent weighted integrals, a matching level separating the low-density high-sound-speed atmosphere from the denser envelope, and small accumulated corrections $\epsilon k|\varphi |$ over the regions carrying the matching integrals. The upper expansion also needs $g/(k{c}^2)\ll 1$. The resulting upper displacement must tend to zero: for example if $\hat{\rho }\propto e^{a(z-z_0)}$ at great height, its leading amplitude is proportional to $e^{(k-a)(z-z_0)}$, requiring $a<k$; the weaker condition $a<2k$ merely makes the upper integral converge. Plane geometry, constant $g$ and the Cowling approximation must be appropriate on the wavelength scale. The physical mode amplitude must remain small compared with both its wavelength and local background scale heights, so the adiabatic fluid perturbation remains linear. If these requirements fail, neither the matching formula nor continuation of the leading expansions is controlled.

Work at fixed enclosed mass $m$, so the perturbations are fluid Lagrangian perturbations. The equilibrium acceleration vanishes. Varying the factor $r^{-4}$ in the gravitational force contributes $4Gm\delta r/(4\pi r^5)$, while varying the factor multiplying the acceleration gives no first-order term because the background acceleration is zero. With $\delta r=r\xi e^{-i\omega t}$, the linearized momentum equation isConservation of the mass of each displaced shell gives its fractional density change as minus the fractional volume change. ThusThe adiabatic relation is $\delta p=\gamma p\delta \rho /\rho =-\gamma p\chi$. Use $dm/dr=4\pi r^2\rho$ and the stellar hydrostatic equation $p^{\prime }=-Gm\rho /r^2$ to turn the mass-coordinate momentum equation into $(\delta p{)}^{\prime }-{\omega }^2\rho r\xi +4p^{\prime }\xi =0$. It follows that the radial stellar pulsation equation isThe stellar adiabatic exponent may vary with radius, so it must remain inside the derivative.

Expand the derivative and multiply by $r^3$. The self-adjoint differential equation becomesHence $f(r)=\gamma p{r}^4$ and $g(r)=r^3[(3\gamma -4)p{]}^{\prime }+{\omega }^2\rho r^4$ in the requested form. Multiplying by a real eigenfunction and integrating gives the boundary term $[\gamma p{r}^4\xi {\xi }^{\prime }{]}_0^R$. It vanishes for a regular centre and bounded admissible surface displacement and derivative, since $p(R)=0$. Therefore the weighted stellar pulsation Rayleigh quotient isReal coefficients and the symmetric endpoint conditions give the Sturm-Liouville problem; the Rayleigh-Ritz variational principle identifies its minimum with the lowest squared frequency. For complex amplitudes the corresponding quadratic forms use modulus squares, and the real eigenfunctions may be chosen for this calculation.

Integrate the second term of $K$ by parts. Its boundary term $-[(3\gamma -4)p{r}^3{\xi }^2{]}_0^R$ again vanishes. The remaining integrand isSince $\chi =a+3\xi$, one has $3{\xi }^2+2a\xi =({\chi }^2-a^2)/3$. Substitution proves the positive-square radial pulsation energy identityIf $\gamma \ge 4/3$ throughout the positive-pressure interior, both terms are nonnegative for every admissible displacement. No negative squared frequency is then possible. Hence radial dynamical instability requires $\gamma <4/3$ somewhere inside the star.

For the opposite necessary condition, test the quotient with a nonzero constant $\xi =C$. It has $\chi =3C$ andIf $\gamma \le 4/3$ everywhere and is strictly below $4/3$ on a positive-pressure interval, this is negative. The lowest eigenvalue is at most this negative trial quotient, proving instability. Strict dynamical stability requires the stronger weighted conditionand in particular $\gamma >4/3$ somewhere. This is the pressure-weighted radial instability criterion viewed as a necessary test for strict stability.

The word “stability” needs a marginal qualification: if $\gamma =4/3$ throughout, $\xi =C$ solves the pulsation equation with $\omega =0$. The star is marginally stable in this radial sense, so the strict $\gamma >4/3$ requirement applies to positive fundamental squared frequency, not to a definition that includes a zero-frequency neutral mode. There is no exponentially growing mode in this equality case, but the zero restoring force also permits a secular solution linear in time.

The conditions are useful as rapid tests, but local threshold crossings are not a complete criterion for a mixed exponent profile. If $\gamma \ge 4/3$ everywhere and exceeds it on a positive-pressure interval, the square identity forces $K>0$ for every nonzero eigenfunction: zero energy would first force constant $\xi$ and then force that constant to vanish. If $\gamma \le 4/3$ everywhere with strict inequality on an interval, the constant trial proves instability. When both types of region occur, $\gamma <4/3$ somewhere does not alone prove instability, and $\gamma >4/3$ somewhere does not alone prove stability. A negative pressure-weighted average proves instability, while a positive one tests only the homologous displacement; another radial shape may still have negative energy. Determining the mixed case requires minimizing the full weighted stellar pulsation Rayleigh quotient or solving the radial stellar pulsation equation. These are radial adiabatic statements and do not rule out nonradial, convective or nonadiabatic instabilities.