Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 64 (Dynamics of Astrophysical Discs)

Take the accretion rate $\dot{M}>0$ to mean inward flow. In a steady state, conservation of mass makes the inward mass flux independent of radius, so $\mathcal{F}=\dot{M}$. Integrating its radial expression givesFor Keplerian rotation, $\Omega =(G{M}_{\ast }/r^3{)}^{1/2}$ and the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma \sqrt{G{M}_{\ast }r}$. The zero-torque inner boundary condition therefore sets $\bar{\nu }\Sigma =0$ at $r_{\ast }$. It fixes $C=-\dot{M}\sqrt{r_{\ast }}/(3\pi )$, giving the steady density profileThis is the Keplerian accretion disk solution on $r_{\ast }<r<r_{\mathrm{out}}$. It determines the kinematic viscosity–surface density product; a separate viscosity closure is needed to turn it into an explicit power law. If the kinematic viscosity is finite and nonzero at the inner edge, the surface density tends to zero there.

The standard thin-disk dissipation flux, summed over both faces, follows by substituting the Keplerian accretion disk profile into the viscous heating rate:The Stefan–Boltzmann law gives $\mathcal{C}=2\sigma T^4$ because there are two emitting faces. Local radiative equilibrium consequently gives the effective-temperature profile of a zero-torque diskHere $\sigma$ is the Stefan-Boltzmann constant. The scale $T_{\mathrm{in}}$ is not the temperature exactly at the inner boundary: the zero-torque inner boundary condition makes that formal temperature zero. Maximizing $T^4$ shows that the maximum effective temperature of a zero-torque disk occurs at $r=49r_{\ast }/36$, with $T_{\max }\simeq 0.488T_{\mathrm{in}}$.

At equal central mass and accretion rate, characteristic effective temperatures scale as $r_{\ast }^{-3/4}$. Thus, comparing corresponding values of $r/r_{\ast }$,The neutron-star disk is about 180 times hotter. The Planck law and Wien displacement law move its characteristic emission to about 180 times higher frequency, or 180 times shorter wavelength. A white dwarf disk commonly emits in optical and ultraviolet bands, while the hotter neutron star disk can emit in X-rays. Absolute bands require an actual accretion rate; the relative shift follows directly from the stated scaling.

For the Rayleigh-Jeans spectrum of a finite blackbody disk, the Rayleigh-Jeans law replaces $(e^{h\nu /(kT)}-1{)}^{-1}$ by $kT/(h\nu )$. Here $h$ is the Planck constant and $k$ the Boltzmann constant. The multitemperature blackbody disk therefore hasFor example, setting $R=r_{\mathrm{out}}/r_{\ast }$ makes its frequency-independent coefficient proportional to the finite dimensionless integralAlthough the exact effective temperature vanishes at the inner edge, the very narrow cold rim where the Rayleigh-Jeans law fails makes a negligible contribution in this limit. More formally, after dividing the integrand by ${\nu }^2$, the inequality $e^x-1\ge x$ bounds it by $(k/h)rT$, so dominated convergence justifies the result even at that edge.

At large radius, $T\propto r^{-3/4}$, and the contribution per logarithmic interval is $r^{2}T\propto r^{5/4}$. The outer disk dominates the low-frequency emission because its much greater area outweighs its lower effective temperature.

For intermediate frequencies use the allowed power-law approximation to the effective temperature and introduceThensoThe lower limit is much smaller than one and the upper limit much larger than one. Extending them to zero and infinity leaves a constant: near zero the integrand behaves as $x^{2/3}$, and at infinity it decays exponentially. Hence the intermediate spectrum isThis is the one-third spectrum of a multitemperature disk. Much of the emission comes from radii where $kT(r)$ is of order $h\nu$, moving inward as the frequency rises. With the exact inner-edge profile, a broad intermediate interval also requires frequency well below $k{T}_{\max }/h$; the supplied approximation captures its slope away from the hottest annuli.

Integrating the standard thin-disk dissipation flux over annular area, with both faces already included in $\mathcal{H}$, givesThus the disk luminosity for $r_{\mathrm{out}}\gg r_{\ast }$ isThe Newtonian gravitational potential decreases by approximately $G{M}_{\ast }/r_{\ast }$ per unit mass from a distant outer edge to the surface, giving a potential-energy release rate $G{M}_{\ast }\dot{M}/r_{\ast }$. The standard thin-disk luminosity is half of this.

The missing half remains as kinetic energy of nearly circular orbital motion: at the inner edge $v_K^2=G{M}_{\ast }/r_{\ast }$, so the specific orbital kinetic energy is $G{M}_{\ast }/(2r_{\ast })$. Equivalently, circular motion has total specific mechanical energy $-G{M}_{\ast }/(2r_{\ast })$. Matter joining a slowly rotating star must shed this orbital motion in an accretion-disk boundary layer, producing approximately another $G{M}_{\ast }\dot{M}/(2r_{\ast })$ of luminosity. A rotating star can retain some energy in spin, so equal disk and accretion-disk boundary layer luminosities assume slow stellar rotation. For a central black hole, there is no material surface, and energy can instead be carried inward.

For a slowly rotating central star, the accretion-disk boundary layer radiates a luminosity $L_{\mathrm{BL}}\sim G{M}_{\ast }\dot{M}/(2r_{\ast })$. A layer of thickness comparable to the disk scale height has emitting area $A_{\mathrm{BL}}\sim r_{\ast }H$ times a geometric constant. Applying the Stefan–Boltzmann law to this blackbody area givesComparing with $\sigma T_{\mathrm{in}}^4=3G{M}_{\ast }\dot{M}/(8\pi r_{\ast }^3)$ gives the boundary-layer temperature scalingOnly the scaling is fixed: for a two-faced annulus of area $4\pi r_{\ast }H$, for example, $T_{\mathrm{BL}}^4/T_{\mathrm{in}}^4=r_{\ast }/(3H)$. A surface belt gives a different order-one coefficient.

For a thin disk, $H/r_{\ast }\ll 1$, so a comparable luminosity emerges from a smaller emitting area at a higher effective temperature. The boundary-layer emission is harder than the disk emission, with its Planck law peak shifted to higher frequency. It is also closer to a single-temperature component than the broad multitemperature blackbody disk spectrum, under the adopted uniform-temperature approximation. Rapid stellar rotation weakens the heating and this conclusion's temperature contrast.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. For any specific quantity $f$, conservation of mass givesDotting the momentum equation with $\Sigma \mathbf{u}$ produces the kinetic energy balanceThe Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Since the shearing-sheet tidal potential is time-independent, its advected potential-energy density obeysFor the isothermal equation of state, define the barotropic energy density $U=c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})$, with fixed positive reference surface density ${\Sigma }_{\mathrm{ref}}$. The continuity equation impliesAdding these balances combines the two pressure terms into $-\nabla \cdot (P\mathbf{u})$. This establishes isothermal shearing-sheet energy conservation. Thus the conserved energy density and its flux areChoosing the density unit so that ${\Sigma }_{\mathrm{ref}}=1$ reproduces the printed logarithm. Changing the reference adds a multiple of the conserved mass to $E$ and its advective energy flux. The barotropic energy density is the mathematical energy of this fixed-isothermal sound speed closure; it is not the microscopic thermal energy of a thermally isolated gas. Maintaining an isothermal process can require heat exchange.

For a local axisymmetric Fourier mode, the Toomre stability criterion balances three contributions to the squared oscillation frequency:The radial epicyclic frequency supplies rotational restoration at long wavelengths; isothermal sound speed and pressure stabilize short wavelengths; disk self-gravity destabilizes intermediate wavelengths. Minimizing over $|k|$ gives $|k|=\pi G{\Sigma }_0/c_s^2$ and ${\omega }_{\min }^2={\kappa }^2(1-Q^{-2})$. Axisymmetric gravitational instability occurs when $Q<1$.

In a centrally dominated Keplerian disk, $\kappa =\Omega$ and vertical hydrostatic equilibrium gives $c_s\simeq H\Omega$. Using ${\Omega }^2=G{M}_{\ast }/r^3$ and a local disk-mass estimate $M_D\sim \pi r^2{\Sigma }_0$,This is the disk mass form of the Toomre criterion. An actual enclosed mass depends on the radial surface density profile and changes an order-one coefficient. In particular, a profile proportional to $r^{-3/2}$ has $M_D(<r)\simeq 4\pi r^2{\Sigma }_0$ when its inner cutoff is negligible.

For the protosolar estimate, take $M_{\ast }\simeq M_{\odot }\simeq 2\times 10^{33}\mathrm{g}$. This solar mass is implicit in identifying the central star with the young Sun. The supplied constant disk aspect ratio givesUsing the printed approximate astronomical unit and gravitational constant, one can also obtain $\Omega (1\mathrm{AU})\simeq 3.7\times 10^{-7}{\mathrm{s}}^{-1}$ and $c_s\simeq 3.7\times 10^5{\mathrm{cm}\mathrm{s}}^{-1}$; they give the same $Q$. The gravitational constant cancels from the mass form.

At one astronomical unit the disk is very stable, with $Q\simeq 640$. Formal extrapolation givesThis is far beyond the planetary region and any plausible extent of the minimum-mass solar nebula. Moreover, the extrapolated enclosed disk mass is already a substantial fraction of a solar mass, so the centrally dominated approximation becomes questionable. The formal radius is not a prediction of a real unstable outer nebula. Direct gas fragmentation by gravitational instability is unlikely to have formed Solar System planets in this model. Core accretion is the more natural route; an earlier substantially more massive or colder disk would be a different model. Even $Q<1$ alone does not guarantee fragmentation, because sufficiently rapid cooling is also needed.

For nonzero real wavenumber $k$, seek a Fourier mode of the Newtonian gravitational potential in the form ${\Phi }^{\prime }=f(z)e^{ikx}$. Away from the razor-thin sheet, the Poisson equation for Newtonian gravity becomes $f^{\prime \prime }-k^{2}f=0$. Requiring the perturbation to decay on both sides and remain continuous gives $f=A{e}^{-|k||z|}$.

Integrating the Poisson equation for Newtonian gravity through $z=0$ fixes the derivative jump:Thus the perturbing gravitational potential isAt the midplane this reduces to the razor-thin disk Poisson kernel. The observable real disturbance is the real part. Writing the perturbation amplitude as ${\Sigma }_a$ instead of ${\Sigma }_0$ gives the same formula with ${\Sigma }_a$; linearity makes it independent of the background density. The $k=0$ disturbance is a uniformly changed sheet and has potential proportional to $|z|$, rather than a decaying nonzero-wavenumber solution.

First eliminate the density-dependent definition of the auxiliary $\mathbf{g}$. The Newtonian gravitational stress tensor can be written without derivatives of its normalization:Here ${\delta }_{ij}$ is the Kronecker delta, and repeated indices are summed. Differentiating givesThe first and third terms cancel because mixed partial derivatives commute; the final step uses the Poisson equation for Newtonian gravity. Therefore the gravitational force density is a stress divergence:The identity holds for spatially varying mass density: substituting the definition of $\mathbf{g}$ before differentiating prevents erroneous extra density-gradient terms.

For the gravitational stress contribution to angular-momentum transport, the anisotropic term $\rho g_i{g}_j$ supplies angular momentum transport. In cylindrical components its radial–azimuthal entry isIn the momentum conservation equation, $r{T}_{r\varphi }$ is the radial stress contribution to the angular-momentum flux, whose sign depends on the correlated radial and azimuthal field components. It vanishes for a perfectly axisymmetric potential but can be nonzero for spiral disturbances.

The other term is isotropic and acts as an effective negative pressure, $P_g=-\rho g^2/2=-|\nabla \Phi {|}^2/(8\pi G)$. Its force contribution is $-\nabla P_g=+\nabla (\rho g^2/2)$, modifying normal compression and force balance. It has no off-diagonal shear component and therefore no direct radial angular momentum transport. In an axisymmetric averaged disk it supplies no azimuthal torque. The complete symmetric Newtonian gravitational stress tensor also expresses conservation of angular momentum without an internal couple.

Use perturbations $\mathbf{v},\mathbf{b},p,\vartheta$ for velocity, magnetic field, magnetohydrodynamic total pressure and buoyancy displacement variable. SetThis is advection by the background Keplerian shearing sheet. Subtracting the equilibrium before retaining first-order terms gives the nine linearized equations:The azimuthal coefficient $\Omega /2$ combines the Coriolis acceleration with perturbation advection of the background shear. The $-3\Omega b_x/2$ term is the winding of radial magnetic field by that shear. The tidal force cancels when the equations at a fixed position are subtracted; it does not add a separate Eulerian perturbation force. Magnetic pressure is already included in $p$, leaving only linear magnetic tension. In this normalization $\vartheta$ has dimensions of length because $D_0\vartheta =v_x$.

For an axisymmetric vertical mode of a shearing sheet with real $k\ne 0$, the two divergence constraints give $ik{v}_z=ik{b}_z=0$. The vertical momentum equation then yields $ikp=0$. Thus all three amplitudes vanish:Define the magnetic amplitudes $a_i=b_i/\sqrt{4\pi {\rho }_0}$ and the signed vertical Alfvén velocity $v_A=B_0/\sqrt{4\pi {\rho }_0}$. Write $A=k^2{v}_A^2$. The remaining equations for the normal mode areFor a growing or oscillatory mode with $\sigma \ne 0$, eliminating $a_x,a_y,\vartheta$ givesA nonzero velocity requires the determinant to vanish. After removing its factor $\sigma$, the dynamical dispersion relation isThis is the radially stratified magnetorotational dispersion relation, with $A=k^2{B}_0^2/(4\pi {\rho }_0)$. Keeping the original five-amplitude system instead gives characteristic polynomial $\sigma$ times this quartic. There is also a stationary balanced normal mode; division by $\sigma$ excludes it but loses no exponentially growing mode. At $k=0$ the divergence argument for vanishing vertical components does not apply, so that spatially uniform case must be treated separately.

When the imposed magnetic field vanishes, the dynamical dispersion relation becomesThe nonstationary hydrodynamic branch therefore has ${\sigma }^2=-(N^2+{\Omega }^2)$. The radial Solberg–Høiland instability criterion in this local model isThe radial buoyancy frequency supplies either restoration or a driving force, while the radial epicyclic frequency $\kappa =\Omega$ supplies rotational restoration. Zero sum is marginal, and positive sum gives stable oscillations. The neutral roots do not themselves signify exponential growth.

Because both equilibrium pressure and dimensionless specific entropy decrease outward, their radial gradients have the same sign. The leading minus sign in the expression for $N^2$ therefore makes $N^2<0$: the radial stratification is adverse. For smooth profiles varying over radius $r$, $|{\partial }_{r}P|\sim P/r$ and $|{\partial }_{r}S|\sim 1/r$, soFor a thin disk, the magnitude of this negative radial buoyancy frequency squared is much less than ${\Omega }^2$. Rotation stabilizes the hydrodynamic mode despite the adverse entropy gradient. This estimate assumes gradients on the global radial scale; a sharp thermal feature may be different. It addresses the ideal nondiffusive equations here. Processes such as convective overstability require additional thermal relaxation and are not ruled out by this particular criterion.

Let $A=k^2{v}_A^2>0$ and $C=N^2+{\Omega }^2$. Solving the radially stratified magnetorotational dispersion relation as a quadratic in $s={\sigma }^2$ givesBoth roots are real. If $A+N^2-3{\Omega }^2<0$, their product is negative, so exactly one root is positive and gives exponential growth. Conversely, when $A+N^2-3{\Omega }^2\ge 0$, the coefficient $2A+N^2+{\Omega }^2\ge A+4{\Omega }^2>0$, so both roots are nonpositive. Thus the instability criterion for a specified nonzero wavenumber isIn a local continuum of allowed wavenumbers, some unstable mode exists exactly when $N^2<3{\Omega }^2$. A finite disk only permits wavelengths fitting its vertical boundaries, so this existence statement also requires an allowed mode in the interval.

The growing root is $s_{+}$, whose derivative and curvature areSetting the derivative to zero gives the fastest-growing interior mode:For a real, nonzero wavenumber, this expression requires $|N^2+{\Omega }^2|<4{\Omega }^2$, equivalentlyNonnegative right-hand side permits the endpoints, but there the stationary point is at $k=0$, outside the nonzero-wavenumber mode used above. Substituting the interior maximizing value givesThis is the maximum growth rate of radially stratified magnetorotational instability, taking $\Omega >0$. For $N^2=0$ it reduces to the usual $3\Omega /4$, at $k^2{v}_A^2=15{\Omega }^2/16$. Negative $N^2$ enhances growth and extends the unstable band; positive $N^2$ suppresses growth and eventually eliminates it at $N^2\ge 3{\Omega }^2$. Magnetic tension enables angular-momentum exchange between displaced parcels, weakening the rotational stabilization that protected the adverse hydrodynamic stratification.

If $N^2\le -5{\Omega }^2$, the interior formula is no longer the physical maximum: $s_{+}$ decreases for $A>0$, and its supremum as $k\to 0$ is $-N^2-{\Omega }^2$. The corresponding limiting growth rate is $\sqrt{-N^2-{\Omega }^2}$, dominated by the already unstable hydrodynamic branch. At $N^2=-5{\Omega }^2$ this joins continuously to $2\Omega$. For the smooth thin disk estimate $N^2\sim -(H/r{)}^2{\Omega }^2$, the ordinary interior maximum applies and its enhancement over $3\Omega /4$ is only of fractional order $(H/r{)}^2$.