= Calibrated Einstein-de Haas torsion response
{c}
{title2=$H_\tau(\omega)=(\kappa-I\omega^2+\mathrm i b\omega)^{-1}$}
An original reduced model makes the calibration explicit. Let the <torsion pendulum> have restoring <torque> $-\kappa\theta$, with <torsion constant> $\kappa$, and <viscous damping> <torque> $-b\dot\theta$, with $b>0$. Allow an unwanted axial <torque> $\tau_p(t)$:
$$
I\ddot\theta+b\dot\theta+\kappa\theta
=-\frac{\dot m_z}{\gamma}+\tau_p(t),\qquad
\omega_0=\sqrt{\frac{\kappa}{I}},\qquad
Q=\frac{I\omega_0}{b}.
$$
Here $\omega_0$ is the undamped natural <angular frequency> and $Q$ the <quality factor>. The <forced harmonic oscillator> response must be included before a twist can be interpreted as a <torque>.
Using the <complex amplitude> convention $m_z(t)=\operatorname{Re}(\widetilde m e^{\mathrm i\omega t})$ and similarly for $\theta$ and $\tau_p$, differentiation multiplies by $\mathrm i\omega$. Substitution gives
$$
H_\tau(\omega)=\frac{1}{\kappa-I\omega^2+\mathrm i b\omega},
\qquad
\widetilde\theta=H_\tau(\omega)
\left(-\frac{\mathrm i\omega}{\gamma}\widetilde m+\widetilde\tau_p\right).
$$
This is the torque-to-twist <transfer function>. With $\widetilde\tau_p=0$ and real positive $\widetilde m=m_1$, the twist <amplitude> is
$$
A_{\rm EdH}(\omega)=\frac{\omega m_1/|\gamma|}{\sqrt{(\kappa-I\omega^2)^2+(b\omega)^2}}.
$$
For $\gamma<0$, the twist <phase> relative to $m_z$ is $\pi/2-\operatorname{atan2}(b\omega,\kappa-I\omega^2)$. At $\omega=\omega_0$ it is zero, although the reaction <torque> leads $m_z$ by $\pi/2$. A <phase> measured at the mirror is not automatically the <phase> of the applied <torque>.
A fixed-<amplitude> reference <torque> $\tau_c\cos\omega t$ measures $|H_\tau|$ directly. The <Einstein-de Haas effect> instead contains an extra factor $\omega$ when $m_1$ stays fixed. Consequently its twist <amplitude> peaks exactly at $\omega_0$ in this model, whereas the fixed-<torque> peak is at $\omega_0\sqrt{1-1/(2Q^2)}$ when $Q>1/\sqrt2$. To verify the first statement, divide the squared denominator of $A_{\rm EdH}$ by $\omega^2$: the quantity to minimize is $(\kappa/\omega-I\omega)^2+b^2$, minimized at $\omega_0$. Both response curves tend to zero at high <frequency>; only the magnetomechanical curve also tends to zero at zero <frequency>.
\Image[/einstein-de-haas-calibrated-torsion-response.png]
{title=Calibrated forced torsion: distinguish the mechanical response from the magnetic drive}
{description=Original calculated curves, not historical measurements or fitted data. A constant reference <torque> and a constant-<amplitude> <magnetic dipole moment> share the same mechanical <transfer function>. The lower plot uses each input's own cosine as its <phase> reference. Markers at $\omega_0=2\,\mathrm{rad/s}$ show equal twist <amplitudes> but different <phases>.}
For a fully stipulated example, choose $I=2.00\times10^{-10}\,\mathrm{kg\,m^2}$, $\kappa=8.00\times10^{-10}\,\mathrm{N\,m/rad}$, $b=8.00\times10^{-11}\,\mathrm{N\,m\,s/rad}$, $\gamma=-1.76\times10^{11}\,\mathrm{s^{-1}\,T^{-1}}$, and $m_1=1.76\times10^{-4}\,\mathrm{A\,m^2}$. These teaching inputs give $\omega_0=2.00\,\mathrm{rad/s}$, $Q=5.00$, reaction <torque> <amplitude> $2.00\times10^{-15}\,\mathrm{N\,m}$ at $\omega_0$, and
$$
A_{\rm EdH}(\omega_0)=\frac{m_1}{|\gamma|b}=1.25\times10^{-5}\,\mathrm{rad}.
$$
An <optical lever> at distance $D=1.00\,\mathrm m$ gives small-displacement spot <amplitude> $x\simeq2DA=25.0\,\mu\mathrm m$; its full peak-to-peak width is $4DA=50.0\,\mu\mathrm m$. For a fast reversal from $+m_1$ to $-m_1$, the same inputs give $\Delta L_{\rm body}=-2.00\times10^{-15}\,\mathrm{kg\,m^2/s}$ and initial $\Delta\dot\theta=-1.00\times10^{-5}\,\mathrm{rad/s}$, provided the external angular <impulse> is negligible.
One independent mechanical calibration adds known <moment of inertia> $\Delta I$, keeping $\kappa$ unchanged. If $T_a,T_b$ are the undamped <oscillation periods> before and after adding it, then
$$
T_a^2=\frac{4\pi^2I}{\kappa},\qquad
T_b^2=\frac{4\pi^2(I+\Delta I)}{\kappa},\qquad
\kappa=\frac{4\pi^2\Delta I}{T_b^2-T_a^2}.
$$
For $\Delta I=I$ in the stipulated example, $T_a=\pi\,\mathrm s$ and $T_b=\sqrt2\pi\,\mathrm s$, recovering the chosen $I$ and $\kappa$. Real damped periods require a <damping> correction: a free decay has angular envelope $e^{-bt/(2I)}$ and oscillation <angular frequency> $\sqrt{\kappa/I-(b/(2I))^2}$. Measuring that decay determines $b/(2I)$. Calibrating the optical scale, mechanical response and change of <magnetic dipole moment> separately is what makes the inferred <gyromagnetic ratio> meaningful.
Back to article page
