= Secular tidal torque
{title2=$\langle\tau\rangle$}
A <secular tidal torque> is the <torque> remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned <tidal deformation> lets the companion's <gravity> change the spin <angular momentum>; the orbit receives the opposite <torque>. <Tidal dissipation> converts <mechanical energy> into <heat>, while total <angular momentum> is conserved in an isolated system. Thus the spin-orbit coupling exchanges <angular momentum> rather than destroying it. https://link.springer.com/article/10.1007/s11214-024-01055-4[Fuller, Guillot, Mathis and Murray (2024): tidal dissipation in giant planets]{ref}.
For intuition near <synchronous rotation> on a <circular orbit>, hold <mean motion> $n$ and <moment of inertia> $I$ fixed and adopt the illustrative linear dissipative law
$$
I\dot\omega=\langle\tau\rangle=-K(\omega-n),\qquad K>0,\qquad \omega=\dot\theta.
$$
Here $\omega$ is spin <angular velocity>, $\theta$ is a marked surface direction in an <inertial frame>, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss are
$$
\omega(t)-n=[\omega(0)-n]e^{-Kt/I},\qquad
\dot E_{\rm mech}=\langle\tau\rangle(\omega-n)=-K(\omega-n)^2\le0.
$$
The orbital contribution to this power is $-n\langle\tau\rangle$, and the spin contribution is $\omega\langle\tau\rangle$; their sum becomes <heat>. Holding $n$ fixed is a local approximation to the coupled evolution. Actual <secular tidal torque> depends on the deformation response and <orbital eccentricity>, so this simple law does not establish which <spin-orbit resonance> an eccentric body captures into.
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