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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.

Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

The two-state calculation is not a general formula for a ferromagnet. Exchange interactions, magnetic domains, magnetic domain-wall pinning, magnetic anisotropy, and hysteresis can determine which response is accessible. Both spin angular momentum and orbital angular momentum can contribute; a single constant gyromagnetic ratio may be only a restricted effective description. Rotation sets an internal bias, while the material and its preparation determine magnetization. The Barkhausen effect illustrates why a smooth bias can produce jumps instead of the toy smooth curve.

For a limited, reversible, isotropic linear model, suppose the same intrinsic volume magnetic susceptibility ${\chi }_v$ responds to the applied bias and to $B_{\Omega }/{\mu }_0$. For a uniformly magnetized ellipsoid along a principal axis, include its demagnetizing factor $N_d$:Solving this linear equation yieldsHere $H_{\mathrm{int}}$ and $H_{\mathrm{app}}$ are actual magnetic field intensities; the added rotational bias is an equivalent drive in the constitutive response. In SI, dimensionless ${\chi }_v$ multiplies an intensity in $A/m$, so $M_z={\chi }_v\Omega /\gamma$ would omit ${\mu }_0$ and have the wrong units. A susceptibility defined instead as $\partial M/\partial B$ has different units. State the convention before comparing formulas.

At zero applied magnetic field intensity, a stipulated ${\chi }_v=3$ gives prefactor $3/2$ in a sphere, with $N_d=1/3$, and a prefactor approaching $3$ in a long axial ellipsoid. This is a geometry check of the assumed linear model. A real cylinder does not have an exactly uniform demagnetizing field, and an irreversible ferromagnet cannot in general be represented by this single linear equation. Inferring an intrinsic gyromagnetic ratio requires an independently calibrated material response and a geometry-dependent conversion of the measured stray magnetic field.

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.

In 1915, Albert Einstein and Wander Johannes de Haas suspended a soft-iron cylinder by glass fibre between coaxial coils. Alternating magnetization drove torsional resonance; a mirror reflected lamplight onto a scale. They compensated Earth's magnetic field, tilted the coils, and rotated the suspension attachment to test unwanted magnetic dipole torques. Added known moment of inertia calibrated the mechanical response.^[ref]

Their paper acknowledges a failed phase determination and claims only ten-percent precision for the revised measurement. Its agreement with the orbital gyromagnetic ratio prediction was not a direct discovery of spin.^[ref] Comparing that orbital gyromagnetic ratio prediction with the electron spin g-factor explains why a successful qualitative demonstration of angular momentum transfer can coexist with an inadequate quantitative interpretation. The stated precision is the authors' estimate; it does not establish that all systematic errors were bounded by that amount.

The general control problem follows directly from the model. Unwanted magnetic dipole torque can drive the same mechanical resonance as the desired reaction torque. In the ideal geometry $m_z$ and $B_z$ alone produce no axial cross product, but transverse magnetic dipole moments, stray transverse magnetic fields and imperfect alignment can. Nulling stray magnetic fields and deliberately varying alignment test this contribution. Merely observing a large resonant twist cannot identify its cause.

For a nearly instantaneous magnetic response, a sinusoidal magnetization makes $m_z$ follow the drive while $-{\dot{m}}_z/\gamma$ is in phase quadrature. An ordinary magnetic contribution proportional to $m_z$ is then distinguishable by phase, after removing the mechanical frequency response. Hysteresis or delayed magnetic response can spoil this simple separation. The measured magnetization, rather than the coil electric current alone, must set the reference. Mechanical vibration, magnetostriction coupled through imperfect geometry, and changing suspension properties require separate checks; none is eliminated simply by reversing the coil electric current.

The dimensionless coefficient in the spin part of the electron magnetic moment. It is distinct from a gyromagnetic ratio, which has units: in the angular convention, ${\gamma }_e=-g_{s}e/(2m_e)$. It is also distinct from the Landé g-factor $g_J$ of a combined atomic angular momentum and from a nuclear or hyperfine g-factor.

For spin one-half, the largest magnitude of a spin-component magnetic dipole moment is $g_s{\mu }_B/2$, not $g_s{\mu }_B$. Thus the ideal value $g_s=2$ gives one Bohr magneton. This follows directly from the spin angular momentum projections $S_z=\pm \hslash /2$.

Consider an explicitly stipulated dilute independent spin-one-half two-level system with zero applied magnetic field. Each constituent has $J_z=\pm \hslash /2$ and $m_z=\gamma J_z$, with constant signed gyromagnetic ratio $\gamma$. Ignore interactions and the specimen's own magnetic field. At positive temperature $T$, thermal equilibrium with the rotating lattice gives energy levels $E_{\pm }=\mp \hslash \Omega /2$. Define the dimensionless variable $x=\hslash \Omega /(2k_{B}T)$, where $\hslash$ is the reduced Planck constant and $k_B$ is the Boltzmann constant. The partition function and Boltzmann distribution giveTaking the expected value giveswhere $n$ is number density. This is the rotational version of the independent-moment magnetic response. The angular momentum aligns with positive rotation even when the magnetic moment points oppositely. The probability ratio $p_{+}/p_{-}=e^{2x}$ provides a direct population check of the sign.

For $|x|\ll 1$, expand the hyperbolic tangent to obtainThe slope against $B_{\Omega }$ is positive for either sign of $\gamma$, while the slope against $\Omega$ has the sign of $\gamma$. At fixed $T>0$, reversing $\Omega$ reverses both means, and $\Omega =0$ gives zero. At fixed nonzero $\Omega$, the formal $T\to 0^{+}$ limit saturates $|⟨J_z⟩|$ at $\hslash /2$; the linear approximation must not be extrapolated there. At $T\to \infty$, the response vanishes. These limits describe the toy canonical ensemble, whose equilibration is assumed.

For a numerical example, stipulate $n=10^{28}{\mathrm{m}}^{-3}$, $T=300\mathrm{K}$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and rotation frequency $f=1000\mathrm{Hz}$. With $\Omega =2\pi f$, $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}J/K$, the model givesThese are teaching inputs and calculated outputs, not measured material values. The corresponding internal angular momentum density is $M_z/\gamma =+4.22\times 10^{-17}\mathrm{J}s/{\mathrm{m}}^3$. Keeping a rotor at prescribed $\Omega$ while its internal angular momentum changes requires the drive to supply the appropriate angular impulse. In an isolated body the mechanical part instead changes oppositely, as in magnetomechanical angular momentum balance. There is no creation of net angular momentum by choosing a rotating frame.

The angular frequency of free Larmor precession in a uniform magnetic field of magnitude $B$, with gyromagnetic ratio $\gamma$. Interactions can shift a collective resonance away from this free-spin scale.

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 givesHere ${\tau }_{\mathrm{EdH}}$ 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 givesFor $m_z:+m_s\to -m_s$ and an electronic $\gamma <0$, the body acquires $\Delta L_{\mathrm{body}}=2m_s/\gamma <0$: 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 $-e$,The electron magnetic moment associated with intrinsic spin instead obeysHere $e>0$ is the elementary charge, $m_e$ is the electron mass, and $g_s$ is the electron spin g-factor.^[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.

An electrically neutral neutron still has a magnetic moment $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A collimated beam passing through an inhomogeneous magnetic field experiences the force derived from $H=-\mathbf{\mu }\cdot \mathbf{B}$. In a region whose field axis is approximately fixed, a spin one-half neutron has two field-axis spin eigenstates, producing two allowed transverse forces and two separated beam components after a drift. A homogeneous field can cause Larmor precession but cannot separate positions through this force.

The neutron's negative gyromagnetic ratio interchanges which spin projection moves toward increasing field; it does not change the number of branches. An unpolarized beam gives equal ideal populations. A beam prepared in one field-axis eigenstate populates only one branch, so splitting into two observed spots presupposes both components and sufficient spatial resolution.

The magnetic dipole moment of an atomic nucleus. Its relation to nuclear spin is expressed by the gyromagnetic ratio; its interaction with an applied magnetic field underlies nuclear magnetic resonance.

Nuclear magnetic resonance detects the response of atomic nuclei with a magnetic moment to a radio frequency magnetic field in a steady magnetic field. Its striking possibility is to measure a nuclear property through an ordinary electrical signal from a bulk sample: an enormous number of tiny magnetic moments can produce a measurable change in a receiving circuit.

For a nucleus with spin angular momentum $\mathbf{J}$, define its gyromagnetic ratio by $\mathbf{\mu }=\gamma \mathbf{J}$. Throughout this article $\gamma$ is angular, measured in radians per second per tesla; $\gamma /(2\pi )$ is the corresponding value in cycles per second per tesla. With ${\mathbf{B}}_0=B_0\hat{\mathbf{z}}$, the Hamiltonian is $H=-\gamma B_0{J}_z$. Adjacent energy levels differ byThus a measured resonance frequency and a calibrated magnetic field determine $|\gamma |$. Converting that into a nuclear magnetic moment also requires the spin quantum number: the maximum projection has magnitude $|\gamma |\hslash I$. This simple experiment does not by itself determine the sign of $\gamma$.

The Rabi molecular beam magnetic resonance experiment detects transitions through changes in a selected molecular beam. The bulk experiments below instead detect electrical consequences of nuclear magnetization in matter at ordinary density. A small Boltzmann distribution population imbalance is multiplied by a very large number of nuclei. For noninteracting spin one-half nuclei of number density $n$ and positive $\gamma$, their equilibrium magnetization isThis follows by weighting the two projections $\pm \gamma \hslash /2$ with their Boltzmann distribution probabilities. It explains both the smallness of the bulk polarization at ordinary temperature and the value of many nuclei contributing to one receiving circuit.

NMR spectroscopy measures resonant responses of atomic nuclei to applied electromagnetic radiation in a magnetic field. In residual-group monitoring of polyurethane formation, distinguishable signal integrals supply a relative residual-group readout; overlap or very weak signals can limit that inference.

The nuclear magneton is the unit of magnetic momentwhere $m_p$ is the proton mass. For a nuclear spin operator $\mathbf{I}$, the nuclear magnetic moment is $\mathbf{\mu }=g_I{\mu }_N\mathbf{I}/\hslash$. Consequently its gyromagnetic ratio is ${\gamma }_I=g_I{\mu }_N/\hslash$, and its Larmor frequency in a uniform magnetic field of magnitude $B$ is $|{\gamma }_I|B$. The nuclear magneton is smaller than the Bohr magneton by the electron-to-proton mass ratio.

Use the isolated nuclear spin Zeeman effect implied by the supplied nuclear magnetic moment, with hyperfine structure neglected. The proton gyromagnetic ratio isThe two spin levels are separated by $g_p{\mu }_N{B}_0$. Resonant nuclear magnetic resonance therefore requires the Larmor frequency

The duration is a Rabi pi pulse: in a frame rotating with the resonant transverse magnetic field, the spin must rotate through $\pi$. If $B_1$ denotes the magnitude of a circularly rotating field with the correct sense of rotation, the Rabi angular frequency is ${\Omega }_R={\gamma }_p{B}_1$. The shortest pulse is then

There is an amplitude convention to specify. If the applied transverse magnetic field is instead the linearly oscillating $B_1\cos (\omega t)\hat{\mathbf{x}}$, its resonant circular component has magnitude $B_1/2$. The rotating-wave approximation gives ${\Omega }_R={\gamma }_p{B}_1/2$ and henceThe wording does not specify this polarization convention, so the two pulse durations should be distinguished. Odd multiples of either shortest Rabi pi pulse also invert the spin within the corresponding ideal model.

A Stern-Gerlach experiment sends a collimated beam through an inhomogeneous magnetic field and then measures its transverse position after a drift region. The neutron is electrically neutral but has a nonzero magnetic moment proportional to its spin, $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A predominantly $z$-directed field gives the interaction Hamiltonian operator and transverse forceSince a neutron has spin one-half, its spin component along the local field has only two eigenvalues, $S_z=\pm \hslash /2$. Hence there are two opposite force branches, corresponding to ${\mu }_z=\pm {\gamma }_n\hslash /2$, and a beam containing both spin components gives two displaced detector spots. This is neutron Stern-Gerlach splitting, not an electric-charge Lorentz force. A uniform magnetic field can cause Larmor precession but does not provide the required separating magnetic field gradient.

The neutron's gyromagnetic ratio is negative, so its magnetic moment points opposite to its spin; this reverses which spin label goes to which spot without changing the two-way splitting. Equal populations occur for an initially unpolarized beam. A beam prepared entirely in one field-axis spin eigenstate gives only its corresponding branch, explaining the qualification “in general”.