Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 / 3

For a circular mid-plane orbit, radial force balance in the Newtonian gravitational potential is $r{\Omega }^2={\Phi }_{,r}$. HenceAxisymmetry conserves the specific angular momentum $j=r^2\dot{\varphi }$. At fixed $j$, radial motion obeys $\ddot{r}=-{\partial }_r{\Phi }_{\mathrm{eff}}$, where the effective potential is ${\Phi }_{\mathrm{eff}}=\Phi +j^2/(2r^2)$. Linearize around a circular radius $r_0$, writing $r=r_0+\xi$. Since $j^2=r_0^3{\Phi }_{,r}(r_0,0)$ there,This is the radial epicyclic frequency; equivalently ${\kappa }^2=r^{-3}d(r^4{\Omega }^2)/dr$.

Reflection symmetry gives ${\Phi }_{,z}(r,0)=0$ and ${\Phi }_{,rz}(r,0)=0$, so vertical and radial displacements decouple at linear order. The vertical equation is $\ddot{z}=-{\Phi }_{,zz}(r_0,0)z$, giving the vertical epicyclic frequencyPositive squared frequencies give oscillatory stability; a negative squared frequency gives exponential instability rather than a real oscillation frequency.

Write $s=(r^2+z^2{)}^{1/2}$. The Paczyński-Wiita potential is a spherical Newtonian gravitational potential with ${\Phi }^{\prime }(s)=GM/(s-a{)}^2$ and ${\Phi }^{\prime \prime }(s)=-2GM/(s-a{)}^3$. At $z=0$, $s=r$, soSubstituting into the orbital formulas givesChoosing positive orbital and vertical frequencies, $\Omega ={\Omega }_z=\sqrt{GM}/[\sqrt{r}(r-a)]$. For $r>3a$, $\kappa =\Omega \sqrt{(r-3a)/(r-a)}$. For $a<r<3a$, ${\kappa }^2<0$ and radial disturbances grow at rate $\sqrt{-{\kappa }^2}$, even though the vertical motion remains stable. Thus circular orbits are radially unstable for $a<r<3a$, and $r=3a$ is the marginal orbit. This recovers the innermost stable circular orbit of the Paczyński-Wiita circular orbit model.

The marginal circular orbit has conserved specific angular momentum and specific orbital energyAt fixed $j_0$, the derivative of the effective potential factors asAn inward displacement therefore produces inward radial acceleration; no inner circular equilibrium or centrifugal barrier halts the plunge. In the infinitesimal-displacement limit, energy conservation givesSelecting the inward branch and the original positive sense of rotation yields the Paczyński-Wiita ballistic plungeThe azimuthal speed remains positive while the radius decreases, so the trajectory spirals inward. From any fixed starting radius below $3a$, the time to reach $a$ is finite: $1/|v_r|$ behaves as $(r-a{)}^{1/2}$ near $a$, which is integrable.

The exactly marginal circular state with exactly zero displacement remains a circular solution. A finite inward release from rest changes the energy by a term cubic in the displacement; the displayed expression is its limiting separatrix with $E=E_0$. On that separatrix $|v_r|\propto (3a-r{)}^{3/2}$ near $3a$, so departure from exactly $3a$ takes an infinite limiting time. This distinguishes the infinitesimal-release formula from spontaneous motion of the unperturbed marginal orbit.

Use the leading vertically averaged horizontal momentum equations of a thin disk. For an isothermal ideal gas, write $P=\int pdz=c_s^2\Sigma$, with constant isothermal sound speed $c_s$. The disk scale height is of order $H=c_s/{\Omega }_z$; this is exact for local isothermal vertical hydrostatic balance and remains the thickness scale for the estimate. Put $ϵ=H/r\ll 1$. The alpha disk prescription givesAt fixed radii of order $a$ inside the plunge and away from the marginal orbit, $|v_r|,v_{\varphi }$ and $r{\Omega }_z$ are fixed numerical multiples of $(GM/a{)}^{1/2}$. Hence the inflow time is dynamical, while the viscous timescale is longer by $1/(\alpha {ϵ}^2)$.

The Paczyński-Wiita ballistic plunge obeys $v_r{v}_r^{\prime }-v_{\varphi }^2/r=-{\Phi }^{\prime }$ and $v_r{j}_0^{\prime }=0$ exactly. For smoothly varying disk quantities on scale $r$, the pressure correction to radial acceleration has magnitude $c_s^2/r$, and the viscous correction is of order $\nu v/r^2$. Relative to the orbital acceleration scale $r{\Omega }_z^2$, these are respectivelyFinite-height corrections to the mid-plane gravitational force are also $O({ϵ}^2)$ by reflection symmetry. Thus the leading horizontal flow is ${\mathbf{u}}_h=\mathbf{v}+O({ϵ}^{2}v)$, with specific angular momentum differing from $j_0$ by a fraction $O(\alpha {ϵ}^2)$ over a dynamical transit, for $\alpha \lesssim 1$. Vertical structure and the smaller vertical velocity must be supplied by the thin-disk expansion; they are not specified by the planar ballistic expression.

This estimate is not uniform at the marginal orbit. In the ballistic solution, steady mass conservation givesConsequently $d\ln \Sigma /dr=[2(r-a){]}^{-1}+3[2(3a-r){]}^{-1}$. Writing $\delta =3a-r\ll a$, the pressure acceleration scales as $c_s^2/\delta$, while the net ballistic radial force scales as $(GM/a^2)(\delta /a{)}^2$. Their ratio is $O[{ϵ}^2(a/\delta {)}^3]$, evaluated with the characteristic thickness near $3a$. A transonic matching layer of width $\delta /a=O({ϵ}^{2/3})$ is therefore needed. The approximation concerns the bulk plunging flow outside that layer, not a uniform relative approximation arbitrarily close to $3a$.

For the torque comparison, the steady angular-momentum flux is $\mathcal{G}-\dot{M}j=\mathrm{constant}$, where $\mathcal{G}=-2\pi \nu \Sigma r^3{\Omega }^{\prime }$ is the viscous torque in an accretion disk. Inside the plunge, $\Omega =j_0/r^2$, soOutside, the approximately circular disk has $\mathcal{G}=\dot{M}[j(r)-j_0]+O(\alpha {ϵ}^2\dot{M}{j}_0)$, since the inner plunge torque is small. At $4a$,a fixed number independent of disk thickness. At $2a$, radial infall is dynamical and the inner torque is $O(\alpha {ϵ}^2\dot{M}{j}_0)$. Therefore the weak viscous torque in a thin-disk plunging region estimate givesThe numerical prefactor need not be small; the suppression is an asymptotic statement in the thin-disk parameter. Rapid infall and its low surface density explain why a nearly zero inner torque is a useful leading boundary condition for the outer accretion disk.