= Magnetomechanical angular momentum balance
{title2=$\tau_{\rm EdH}=-\dot m_z/\gamma$}
Let $m_z=\int_V M_z\,dV$ be the axial <magnetic dipole moment>, where $M_z$ is the axial <magnetization>. Write $J_z$ for the internal <angular momentum> associated with that <magnetic dipole moment>, and $I\dot\theta$ for the body's rotational <angular momentum>, with <moment of inertia> $I$. In a regime admitting a constant signed <gyromagnetic ratio> $m_z=\gamma J_z$, axial <conservation of angular momentum> gives
$$
\frac{d}{dt}(I\dot\theta+J_z)=\tau_{\rm ext},
\qquad I\ddot\theta=\tau_{\rm ext}-\frac{\dot m_z}{\gamma},
\qquad \tau_{\rm EdH}=-\frac{\dot m_z}{\gamma}.
$$
Here $\tau_{\rm EdH}$ names the internal reaction <torque> on the body, not an additional external <torque>. The external term includes the suspension. The model assumes that the changing internal <angular momentum> is transferred to the measured mechanical motion, with other storage or escape channels negligible on the observation timescale. A changing or non-collinear mixture of <orbital angular momentum> and <spin angular momentum> need not admit a single constant $\gamma$.
During a reversal fast compared with the mechanical response, with negligible external angular <impulse>, integrating the balance gives
$$
\Delta L_{\rm body}=-\Delta J_z=-\frac{\Delta m_z}{\gamma}.
$$
For $m_z:+m_s\to-m_s$ and an electronic $\gamma<0$, the body acquires $\Delta L_{\rm body}=2m_s/\gamma<0$: its initial rotation is toward $-z$ by the <right-hand rule>. Reversing the reversal reverses the impulse. Over a complete periodic cycle, $\Delta m_z=0$, so the net reaction angular <impulse> vanishes; a resonant alternating response does not imply steadily accumulating <angular momentum>.
The <orbital angular momentum> prediction can be derived without a material model. A charge $q$ traveling around a circle of <radius> $r$ at <angular frequency> $\omega$ has loop <electric current> $q\omega/(2\pi)$ and <magnetic dipole moment> $m=q\omega r^2/2$. Its <orbital angular momentum> is $L=m_e\omega r^2$, hence, for an <Electron> with <electric charge> $-e$,
$$
\mathbf m_L=-\frac{e}{2m_e}\mathbf L,
\qquad \gamma_L=-\frac{e}{2m_e}.
$$
The <electron magnetic moment> associated with intrinsic <spin> instead obeys
$$
\mathbf m_S=-g_s\frac{e}{2m_e}\mathbf S,
\qquad \gamma_S=-g_s\frac{e}{2m_e},\qquad g_s\simeq2.
$$
Here $e>0$ is the <elementary charge>, $m_e$ is the <electron mass>, and $g_s$ is the <electron spin g-factor>.<The magnetic moment of the electron by Polykarp Kusch>[pp. 299–300: orbital and intrinsic moments and the Dirac value]{ref} Thus the same change of <magnetic dipole moment> corresponds to about half as much transferred <angular momentum> in the pure-<spin> model as in the <orbital angular momentum> model. This factor of two is a relation between <magnetic dipole moment> and <angular momentum>, not evidence that an <Electron> is literally a rotating charged sphere. In a material with both contributions, the inferred coefficient is a material-dependent magnetomechanical <gyromagnetic ratio>.
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