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The Barkhausen effect consists of abrupt magnetization changes during a smooth applied magnetic field sweep. Its inductive signal is Barkhausen noise, often arising from collective magnetic domain wall motion.^[ref]

The distinction is between a smooth drive and an intermittent material response. A smooth macroscopic hysteresis curve can hide smaller jumps. By Faraday's law of induction, a pickup coil responds to the rate of change of linked magnetic flux, so a rapid jump can be conspicuous even when the total change is small. A pulse is not one chemical atom flipping: neither its height nor the number of audible clicks counts individual Electron spins. Qualitative crackling alone does not establish quantum spin; collective switching also admits a classical model with trapped interfaces. Compare the distinct mechanical readout of angular momentum transfer in the Einstein-de Haas effect.

The Barnett effect is the change of a body's magnetization caused by mechanical rotation. Turning the whole specimen can bias the internal angular momentum carrying its magnetic moments, even with zero applied magnetic field. It is the reciprocal physical question to the Einstein-de Haas effect: impose rotation and measure magnetization, rather than change magnetization and observe mechanical motion. The relation between the two becomes especially transparent when the same signed gyromagnetic ratio connects magnetic moment and internal angular momentum.

Samuel Jackson Barnett published Magnetization by Rotation on 1 October 1915 in Physical Review. The Barnett rotation magnetometry experiment below provides an inspected modern example. The important distinction is between an equivalent term in a Hamiltonian and the actual external magnetic field measured by a magnetometer: the first biases the material; its resulting magnetization produces the second.

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 ${\tau }_p(t)$: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 $m_z(t)=\mathrm{Re}(\tilde{m}{e}^{\mathrm{i}\omega t})$ and similarly for $\theta$ and ${\tau }_p$, differentiation multiplies by $\mathrm{i}\omega$. Substitution givesThis is the torque-to-twist transfer function. With ${\tilde{\tau }}_p=0$ and real positive $\tilde{m}=m_1$, the twist amplitude isFor $\gamma <0$, the twist phase relative to $m_z$ is $\pi /2-atan2(b\omega ,\kappa -I{\omega }^2)$. 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 $Q>1/\sqrt{2}$. To verify the first statement, divide the squared denominator of $A_{\mathrm{EdH}}$ 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.

For a fully stipulated example, choose $I=2.00\times 10^{-10}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =8.00\times 10^{-10}\mathrm{N}m/rad$, $b=8.00\times 10^{-11}\mathrm{N}\mathrm{m}s/rad$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and $m_1=1.76\times 10^{-4}\mathrm{A}{\mathrm{m}}^2$. These teaching inputs give ${\omega }_0=2.00rad/s$, $Q=5.00$, reaction torque amplitude $2.00\times 10^{-15}\mathrm{N}\mathrm{m}$ at ${\omega }_0$, andAn optical lever at distance $D=1.00\mathrm{m}$ gives small-displacement spot amplitude $x\simeq 2DA=25.0\mu \mathrm{m}$; 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 $\Delta L_{\mathrm{body}}=-2.00\times 10^{-15}\mathrm{kg}{\mathrm{m}}^2/s$ and initial $\Delta \dot{\theta }=-1.00\times 10^{-5}rad/s$, 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, thenFor $\Delta I=I$ in the stipulated example, $T_a=\pi \mathrm{s}$ and $T_b=\sqrt{2}\pi \mathrm{s}$, 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.

The Einstein-de Haas effect is mechanical rotation or twisting caused by a change in a body's magnetization. The microscopic carriers of magnetic dipole moment also carry angular momentum: changing their net orientation can transfer angular momentum to the body. A freely suspended ferromagnet can therefore turn when its magnetization reverses. A suspension instead stores part of the response as an elastic twist. This makes a microscopic property accessible through a macroscopic angle.

The central measurement compares a change of magnetic dipole moment with transferred angular momentum, giving a gyromagnetic ratio. It complements an ordinary magnetic dipole torque measurement: even with the applied magnetic field parallel to the suspension axis, so that an ideal axial magnetic dipole moment has $\mathbf{m}\times \mathbf{B}=0$, changing magnetization can drive an axial twist. The external drive supplies energy; conservation of angular momentum still requires accounting for the suspension and any external axial torque.