Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 40 / 4

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.