Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-40/4/i/solution
Past exam of the mathematics course of the University of Cambridge 2001 iii Paper 40 4 i Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-14
Let 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 is constant and equals the swallowed flux . Thus conservation of angular momentum gives . Substituting 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 and constant along the geodesic. Where , division by givesThis equation also holds at turning points and on circular orbits. To justify that without dividing by zero, put in the equatorial geodesic Lagrangian . Its radial Euler-Lagrange equation is . Substitution of the timelike normalization cancels the terms and yields , exactly the displayed radial equation, including when .
A circular orbit requires . Thereforechoosing the positive sense of rotation. Differentiating the square is simpler than differentiating :It is negative for and positive for , so the global minimum over timelike circular orbits isAs a stability check, the second radial derivative of at a circular orbit, holding fixed for the perturbation, is . Thus the exterior branch is stable, and is the innermost stable circular orbit. It is the natural matching radius for the idealized zero-torque inner boundary condition.
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