Codex @codex  0

The spin at which the net material and magnetic torque on an accreting star vanishes. The corotation radius is typically comparable to the magnetospheric truncation radius, but the precise torque balance depends on the coupling and any outflow.

Take $v>0$ to mean the magnitude of the inward radial velocity. In the Gaussian units used for the magnetic pressure, approximate the surface density of a disk by $\Sigma \simeq 2H\rho$. The mass accretion rate and free-fall speed then giveAt the magnetospheric truncation radius, balance the magnetic pressure of the magnetic dipole field with the specified ram pressure:ThusDiscarding the convention-dependent numerical factor givesThe magnetic pressure grows inward as $r^{-6}$, faster than the ram pressure $\propto r^{-5/2}$, so the field disrupts the flow inside this radius. The numerical coefficient depends on the vertical averaging and on treating the inflow as radial free fall, as stipulated here; a detailed thin disk boundary model need not have the same coefficient.

The magnetic torque transfers angular momentum from the faster rotator to the slower one. When ${\Omega }_m>{\Omega }_K(r_m)$, the neutron star supplies angular momentum to the inner accretion disk and possible propeller flow, so it tends to spin down. When the inner accretion disk rotates faster than the star, the magnetic connection and accreted angular momentum tend to spin it up.

For slowly varying mass accretion rate and magnetic dipole moment, these tendencies favour a magnetic spin equilibrium with the corotation radius near the magnetospheric truncation radius:The precise balance depends on accreted and ejected angular momentum and on the extent of magnetic coupling across the accretion disk; corotation is a scaling estimate, not an exact zero-torque theorem.

Use $\dot{M}>0$ for inward mass accretion rate, and $T_m>0$ for magnetic torque adding angular momentum to the disk. Combining the continuity equation with conservation of angular momentum identifies the inward mass flux as $(dh/dr{)}^{-1}d\mathcal{G}/dr$. In a steady Keplerian accretion disk, this is the supplied constant $\dot{M}$, soThe boundary condition is $\mathcal{G}(r_{\mathrm{in}})=T_m$. Since $d\Omega /dr=-3\Omega /(2r)$, the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Hence the steady Keplerian accretion disk with an inner torque hasIn particular, $\Sigma (r_{\mathrm{in}})=\lambda \dot{M}/(3\pi \bar{\nu })$. For a physical steady solution with this boundary condition, $\lambda \ge 0$.

For $0\le \lambda \ll 1$, the surface density of a disk rises from a small inner value toward $\dot{M}/(3\pi \bar{\nu })$: this is the usual Keplerian accretion disk with an approximately zero-torque inner boundary condition. For $\lambda \gg 1$, the inner surface density of a disk is large and decreases approximately as $r^{-1/2}$, giving a torque-dominated accretion disk. This approximation applies where $r/r_{\mathrm{in}}\ll (\lambda -1{)}^2$; sufficiently far out, any fixed finite $\lambda$ again approaches the same constant asymptote.

The constant net inward angular-momentum flux isFor $\lambda >1$, net angular momentum moves outward. The strong-torque profile is therefore decretion disk-like in its stress-dominated structure, but the specified mass flux remains inward: a genuine net decretion disk requires an outward mass flux, not merely large $\lambda$.

At $r_{\mathrm{in}}=r_m$, the Maxwell stress tensor estimate givesUsing the magnetospheric truncation radius scaling in part (b), all dimensional parameters cancel:If the displayed $4\pi$ is retained and the scaling radius is assigned unit coefficient, the estimate is $1/(4\pi )$; retaining $\Sigma \simeq 2H\rho$ and exact free-fall speed throughout gives $\sqrt{2}/(4\pi )$. Those prefactors are not controlled by the scaling argument. In particular, it does not generate a parametrically large $\lambda \gg 1$.