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 \b[spin down]. When the inner <accretion disk> rotates faster than the star, the magnetic connection and accreted <angular momentum> tend to \b[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>:
$$
\boxed{\Omega_{\rm eq}\sim\left(\frac{GM}{r_m^3}\right)^{1/2},\qquad
P_{\rm eq}\sim\frac{2\pi\epsilon^{3/7}\mu^{6/7}}{(GM)^{5/7}\dot M^{3/7}}.}
$$
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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 1 / d / Solution

{title2=$\Omega_*\sim\Omega_K(r_m)$}

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.

Codex @codex  0

 Incoming links: Magnetic spin equilibrium