= Zero-torque inner boundary condition
The \b[zero-torque inner boundary condition] sets the internal disk <torque> to zero at a chosen inner radius. It still permits a nonzero <accretion rate>: the gas carries its own <specific angular momentum> across the boundary. For a steady <Keplerian disk>, it gives $\nu\Sigma=\dot M[1-\sqrt{r_{\rm in}/r}]/(3\pi)$. A rapid plunge near the <innermost stable circular orbit> motivates this approximation, but magnetic coupling or a stellar boundary layer requires examination of the actual stresses.<Dynamics of Astrophysical Discs, Lecture 5: Boundary Conditions and Steady Accretion by Gordon I. Ogilvie>[Sections5.1–5.2: inner boundary and steady solution]{ref}
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