The Einstein–de Haas and Barnett Effects in Biquaternionic Form

Introduction

A magnet and a gyroscope are two faces of one body. Change the magnetization of a freely suspended ferromagnet, and it begins to rotate; set a ferromagnet rotating, and it becomes magnetized. The first is the Einstein–de Haas effect, the second the Barnett effect, and the two are reciprocal readings of a single fact: the magnetization of matter is carried by angular momentum, so a change of one entails a change of the other. This article treats the pair in the biquaternion framework, classically and non-quantum mechanically. The magnetism is intrinsic to the body; the body's rotation is a rotor of the material sector; and the reciprocity is the statement that one bilinear coupling governs both effects.

The physical content of the effects is standard and is imported as such. An isolated body conserves its total angular momentum, and if a part of that angular momentum is magnetic — tied to the magnetic moment by $\boldsymbol{\mu} = \gamma\mathbf{S}$ — then a change of the moment changes the mechanical angular momentum by the same amount and in the opposite sense. Equivalently, in a frame rotating with the body the spin sees an effective magnetic field, and a body at rest in that frame acquires the corresponding magnetization. The measured quantity in both effects is the gyromagnetic ratio $\gamma$, or its dimensionless companion $g = 2m\gamma/q$; the effects are, historically, how $g\approx 2$ was established for the ferromagnets, and therefore how the magnetization of iron was shown to be carried by electron spins rather than by orbital currents.

The biquaternion algebra contributes the description of the rotation. A rigid body's orientation is a unit real quaternion $\tilde{R}\in\mathbb{H}_{\mathbb{B}}$, its angular velocity is a pure real quaternion $\tilde{\boldsymbol{\omega}} = \omega_ke_k$, and its motion is the rotor conjugation $\mathbf{x}(t) = \tilde{R}(t)\mathbf{x}(0)\tilde{R}(t)^{*}$, generated by $\dot{\tilde{R}} = \tfrac{1}{2}\tilde{\boldsymbol{\omega}}\tilde{R}$. The same rotor that describes a rotation of the body describes the rotation of the magnetization it carries, and the half-angle of the rotor is the same half-angle that governs the intrinsic magnetic moment. The rotor conventions needed here are stated in full in the next section, because the companion article on the rigid-body rotor is written in parallel with this one and is not presupposed.

The conventions are those of the foundational articles and of the companion article Larmor Precession and the Classical Magnetic Moment in Biquaternionic Form. The biquaternion algebra is $\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$, with quaternion basis $e_0 = 1, e_1, e_2, e_3$, $e_k^2 = -e_0$, and central imaginary $i$. The material sector is $\mathbb{M}_-$, the informational sector is $\mathbb{M}_+$, and the real-quaternion subspace is $\mathbb{H}_{\mathbb{B}}$. The axial vectors of the material sector — the magnetic moment, the magnetization, the angular momentum, the angular velocity — are pure real quaternions in $\mathbb{H}_{\mathbb{B}}$, and the intrinsic magnetic moment is the same vector slot with the central factor, $\mu_k\,ie_k\in\mathbb{M}_+$.

The Gyromagnetic Coupling

The Moment and the Angular Momentum

An intrinsic magnetic moment is tied to an intrinsic angular momentum by the gyromagnetic relation

$$ \boldsymbol{\mu} = \gamma\,\mathbf{S}, \qquad \gamma = g\,\frac{q}{2m}, $$

with $\mathbf{S}$ the spin angular momentum of the carrier and $\gamma$ its gyromagnetic ratio. The moment and the spin are parallel for a positive $g$ and antiparallel for a negative one; the electron, with $q = -e$ and $g\approx 2$, has its moment antiparallel to its spin. In the biquaternion algebra both are material axial vectors,

$$ \tilde{\boldsymbol{\mu}} = \mu_ke_k, \qquad \tilde{\mathbf{S}} = S_ke_k, \qquad \tilde{\boldsymbol{\mu}} = \gamma\,\tilde{\mathbf{S}}, $$

and the intrinsic (informational-sector) representative of the same moment is $\mu_k\,ie_k\in\mathbb{M}_+$, the representative used when the moment is paired with the spin observable rather than with the material field. The two representatives are exchanged by the central $i$, exactly as the two sectors are, and the factor between them is the sector exchange and not a physical difference.

The Total Angular Momentum and Its Conservation

A body of mass $m$ and moment of inertia $\hat{I}$, rotating with angular velocity $\boldsymbol{\omega}$, carries a mechanical angular momentum $\mathbf{L} = \hat{I}\boldsymbol{\omega}$, and, if it is magnetized, a spin angular momentum $\mathbf{S} = \boldsymbol{\mu}/\gamma$ distributed over its magnetization. Its total angular momentum is

$$ \mathbf{J} = \mathbf{S} + \mathbf{L}. $$

For an isolated body, with no external torque, the total angular momentum is conserved,

$$ \dot{\mathbf{J}} = 0 \qquad\Longrightarrow\qquad \dot{\mathbf{S}} = -\,\dot{\mathbf{L}} . $$

This single conservation law is the Einstein–de Haas half of the reciprocity. A change of the moment changes $\mathbf{S}$ by $\Delta\mathbf{S} = \Delta\boldsymbol{\mu}/\gamma$, and therefore changes the mechanical angular momentum by

$$ \Delta\mathbf{L} = -\,\frac{1}{\gamma}\,\Delta\boldsymbol{\mu}, $$

which is the Einstein–de Haas effect. The reverse reading — a rotation producing a magnetization — is the reciprocal of the same coupling rather than a second conservation law; it is obtained from the spin-rotation coupling in the rotating frame, and it is treated in the section on the Barnett effect. In the algebra, the conservation reads

$$ \Delta\tilde{\mathbf{S}} + \Delta\tilde{\mathbf{L}} = 0, \qquad \tilde{\mathbf{L}} = \hat{I}\,\tilde{\boldsymbol{\omega}} . $$

The Rotor Algebra of a Spinning Body

The Orientation Rotor

The orientation of a rotating rigid body is a unit real quaternion,

$$ \tilde{R}(t) \in \mathbb{H}_{\mathbb{B}}, \qquad \tilde{R}\,\tilde{R}^{*} = e_0 , $$

and a material vector fixed in the body is carried into the laboratory by rotor conjugation,

$$ \mathbf{x}(t) = \tilde{R}(t)\,\mathbf{x}(0)\,\tilde{R}(t)^{*} . $$

For a rotation at constant rate about a fixed unit axis $\hat{\mathbf{n}}$, the rotor is

$$ \tilde{R}(t) = \exp\!\left(\frac{\omega t}{2}\,\hat{n}_ke_k\right) = \cos\frac{\omega t}{2}\,e_0 + \sin\frac{\omega t}{2}\,\hat{n}_ke_k , $$

and it rotates every body vector through the angle $\omega t$ about $\hat{\mathbf{n}}$ in the right-handed sense. The rotor is normalized, $N(\tilde{R}) = \tilde{R}\tilde{R}^{\natural} = e_0$, and it is a double cover of the rotation: $\tilde{R}(\theta + 2\pi,\hat{\mathbf{n}}) = -\tilde{R}(\theta,\hat{\mathbf{n}})$, so that $\pm\tilde{R}$ describe the same orientation and the group of orientations is $\mathbb{H}_{\mathbb{B}}^1/\{\pm e_0\} = SO(3)$.

The Angular Velocity

Differentiating the rotor gives the angular velocity. For the fixed-axis rotor above,

$$ \dot{\tilde{R}} = \frac{\omega}{2}\,\hat{n}_ke_k\,\tilde{R} = \frac{1}{2}\,\tilde{\boldsymbol{\omega}}\,\tilde{R}, \qquad \tilde{\boldsymbol{\omega}} = \omega\,\hat{n}_ke_k . $$

For a general time-dependent rotor the same relation defines the angular velocity, $\tilde{\boldsymbol{\omega}} = 2\,\dot{\tilde{R}}\,\tilde{R}^{*}$, which is a pure real quaternion because $\tilde{R}\tilde{R}^{*} = e_0$ implies $\dot{\tilde{R}}\tilde{R}^{*} + \tilde{R}\dot{\tilde{R}}^{*} = 0$. The equation $\dot{\tilde{R}} = \tfrac{1}{2}\tilde{\boldsymbol{\omega}}\tilde{R}$ is the biquaternion form of the rigid-body kinematics, and it is the same relation that the Larmor precession uses with $\tilde{\boldsymbol{\omega}}_L = \gamma\mathbf{B}$.

Acting on a material vector, the rotor equation reproduces the elementary velocity field:

$$ \dot{\mathbf{x}} = \tfrac{1}{2}\left[\tilde{\boldsymbol{\omega}},\mathbf{x}\right] = \boldsymbol{\omega}\times\mathbf{x}, $$

the commutator of two pure real quaternions being twice their cross product. Every point of the body therefore moves with the velocity of a rigid rotation, and the mechanical angular momentum is $\tilde{\mathbf{L}} = \hat{I}\tilde{\boldsymbol{\omega}}$.

The Body Frame and the Space Frame

A rotation can be referred to the axes fixed in the laboratory or to the axes fixed in the body, and the two angular velocities are related by the rotor itself. Define the space-frame angular velocity by $\tilde{\boldsymbol{\omega}} = 2\dot{\tilde{R}}\tilde{R}^{*}$ and the body-frame angular velocity by

$$ \tilde{\boldsymbol{\omega}}_{\rm body} = \tilde{R}^{*}\,\tilde{\boldsymbol{\omega}}\,\tilde{R} = 2\,\tilde{R}^{*}\,\dot{\tilde{R}} . $$

The rotor equation then reads in the two frames as

$$ \dot{\tilde{R}} = \frac{1}{2}\,\tilde{\boldsymbol{\omega}}\,\tilde{R} = \frac{1}{2}\,\tilde{R}\,\tilde{\boldsymbol{\omega}}_{\rm body}, $$

the two forms differing by the placement of the rotor. The space-frame generator is the one that acts on laboratory vectors, $\dot{\mathbf{x}} = \tfrac{1}{2}[\tilde{\boldsymbol{\omega}},\mathbf{x}]$, while the body-frame generator is the constant one for a body rotating about a principal axis. For the effects of this article the axis of the magnetization and the axis of rotation coincide, so the two coincide, and the distinction is recorded only because the rotor algebra is stated in full here. The magnetization is carried by the same rotor: its director $\hat{\mathbf{m}}$ is a body vector, $\hat{\mathbf{m}}(t) = \tilde{R}(t)\hat{\mathbf{m}}_0\tilde{R}(t)^{*}$, so a rotation of the body is a rotation of the magnetization, which is the kinematic content of the reciprocal effects.

The Einstein–de Haas Effect

Statement

Let a ferromagnetic body be suspended so that it can turn freely about an axis, with its magnetization along that axis. If the magnetization is changed — reversed by a field pulse, or changed in magnitude — the spin angular momentum of the body changes, and since the total angular momentum is conserved, the body acquires the opposite mechanical angular momentum and begins to rotate. The effect was predicted from the Ampère molecular currents by Richardson and first observed by Einstein and de Haas; its measurement gives the ratio of the magnetic moment to the mechanical angular momentum, that is, the gyromagnetic ratio.

The Equation of Motion

The conservation $\dot{\mathbf{S}} = -\dot{\mathbf{L}}$ with $\mathbf{S} = \boldsymbol{\mu}/\gamma$ and $\mathbf{L} = I\boldsymbol{\omega}$ (the scalar moment of inertia about the fixed axis) gives

$$ \boxed{\;\dot{\boldsymbol{\omega}} = -\,\frac{1}{\gamma I}\,\dot{\boldsymbol{\mu}}\;.} $$

The angular velocity of the body is proportional to the rate of change of its magnetic moment, with the coefficient $-1/(\gamma I)$. Reversing the magnetization, so that the moment changes by $\Delta\boldsymbol{\mu} = -2\boldsymbol{\mu}_0$, makes the body acquire the angular velocity change $\Delta\boldsymbol{\omega} = 2\boldsymbol{\mu}_0/(\gamma I)$, whose sense is fixed by the sign of $\gamma$. In the biquaternion algebra the same statement is

$$ \Delta\tilde{\boldsymbol{\omega}} = -\,\frac{1}{\gamma I}\,\Delta\tilde{\boldsymbol{\mu}}, \qquad \Delta\tilde{R} = \exp\!\left(\frac{\Delta\theta}{2}\,\hat{\mu}_ke_k\right), \qquad \Delta\theta \propto -\,\frac{\Delta\mu}{\gamma I}, $$

so that a change of the moment is described by a rotation rotor of the body: the body's orientation is the conjugate variable to its magnetic moment, and a change of one is a rotation generated by the change of the other. This is the biquaternion form of the Einstein–de Haas effect, and it exhibits the effect as the statement that the magnetic moment and the body orientation are canonically paired.

The Observed Rotation

In the torsional-oscillation experiment, the body hangs from a fiber of torsional constant $\kappa$, and the angular impulse delivered by the reversal is read from the amplitude of the ensuing oscillation. The impulse is

$$ \int \tau\,dt = \Delta L = -\,\frac{1}{\gamma}\,\Delta\mu , $$

and the amplitude is $\theta_{\rm max} = \Delta L/\sqrt{\kappa I}$. The measurement therefore gives $\gamma$ directly from the ratio of the magnetic moment change to the mechanical angular momentum change, and hence $g = 2m\gamma/q$. For an iron cylinder of magnetization $M_s\approx 1.7\times10^6\,\text{A m}^{-1}$ and volume $10^{-6}\,\text{m}^3$, a full reversal changes the moment by $2M_sV\approx 3.4\,\text{A m}^2$; with $\gamma = g\gamma_e/2$ and $g\approx 2$ this gives $\Delta L\approx 1.9\times10^{-11}\,\text{kg m}^2\,\text{s}^{-1}$, and with $I\approx10^{-7}\,\text{kg m}^2$ an angular velocity change of order $2\times10^{-4}$ rad s$^{-1}$. The rotation is minute, which is why the effect is measured as a torsional resonance and not by watching the body turn.

The historical outcome of the measurement is the part relevant here. The measured ratio for the ferromagnets is close to $q/m$, that is $g\approx 2$, and not to the orbital value $q/2m$. The magnetization of iron is therefore carried by the spins, whose intrinsic gyromagnetic factor is two, and not by orbital currents, whose factor is one. The account of the factor two is the subject of the companion article The Classical Origin of g = 2 in Biquaternionic Form.

The Barnett Effect

The Rotating Frame

The inverse effect is the magnetization of a rotating body. Let the body rotate with angular velocity $\boldsymbol{\omega}$, and let the analysis be performed in the frame that rotates with it. In that frame the spin is at rest, and the only change from the laboratory description is the rotation of the frame itself. The spin's equation of motion in the rotating frame is obtained from the Larmor equation of the companion article Larmor Precession and the Classical Magnetic Moment in Biquaternionic Form by subtracting the frame's rotation,

$$ \dot{\mathbf{S}}_{\rm rot} = \gamma\,\mathbf{S}\times\mathbf{B} - \boldsymbol{\omega}\times\mathbf{S} = \mathbf{S}\times\left(\gamma\mathbf{B} + \boldsymbol{\omega}\right). $$

Equivalently, in the rotating frame the spin behaves as if the field were the effective field

$$ \boxed{\;\mathbf{B}_{\rm eff} = \mathbf{B} + \frac{\boldsymbol{\omega}}{\gamma}\;.} $$

The term $\boldsymbol{\omega}/\gamma$ is the Barnett field, the field equivalent of the rotation. In the absence of an applied field, $\mathbf{B}_{\rm eff} = \boldsymbol{\omega}/\gamma$, and the spin precesses about the rotation axis exactly as it would precess about a magnetic field of that magnitude.

The Magnetization from the Rotation

Thermal equilibrium in the rotating frame is equilibrium in the effective field, so the magnetization of the body is the magnetization of the material in the field $\mathbf{B}_{\rm eff}$,

$$ \mathbf{M} = \chi\,\mathbf{B}_{\rm eff} = \chi\!\left(\mathbf{B} + \frac{\boldsymbol{\omega}}{\gamma}\right), $$

with $\chi$ the magnetic susceptibility, in the constitutive convention in which the factors of $\mu$ are absorbed into $\chi$. In zero applied field the rotation alone produces

$$ \mathbf{M} = \frac{\chi}{\gamma}\,\boldsymbol{\omega}, $$

the Barnett magnetization, parallel to the effective field $\mathbf{B}_{\rm eff}$ and therefore along the rotation axis when $\gamma$ is positive and opposed to it when $\gamma$ is negative. The sense is fixed by the sign of $\gamma$, which is what the Einstein–de Haas measurement of the same material returns. The effect is linear in the rotation rate, and the rotation enters the spin's equation exactly as the additional field $\boldsymbol{\omega}/\gamma$: the frame that is at rest for a moment in the applied field $\mathbf{B}$ rotates at $\boldsymbol{\Omega} = -\gamma\mathbf{B}$, and the applied field that would cancel the Barnett field is $-\boldsymbol{\omega}/\gamma$.

The magnitude of the effect is small. A body rotating at $100$ Hz, $\omega = 628$ rad s$^{-1}$, presents a Barnett field for protons of $|\boldsymbol{\omega}/\gamma_p| = 628/(2.675\times10^8)\approx 2.3\times10^{-6}$ T, a few microtesla. The field is minute, and its observation requires a sensitive magnetometer; the effect was nevertheless detected by Barnett, who observed the magnetization of rotating iron, and it is now observed in rotating paramagnetic salts and in ultracold gases.

The Rotor Form of the Barnett Effect

In the algebra the Barnett effect is the statement that the same rotor that carries the body into the laboratory frame is the rotor whose logarithmic derivative enters the spin's equation of motion. Writing the co-rotating frame by its rotor $\tilde{R}_\omega(t)$ with $\tilde{\boldsymbol{\omega}} = 2\dot{\tilde{R}}_\omega\tilde{R}_\omega^{*}$, the spin's equation in that frame is

$$ \dot{\mathbf{S}}_{\rm rot} = \mathbf{S}\times\left(\gamma\mathbf{B} + \boldsymbol{\omega}\right), $$

and the rotation enters exactly as the additional field $\boldsymbol{\omega}/\gamma$ — the Barnett field. Three apparently different statements are therefore one generator written in three ways: the fictitious field $\boldsymbol{\omega}/\gamma$, the Larmor frame $\boldsymbol{\Omega} = -\gamma\mathbf{B}$ of the applied field, and the spin-rotation coupling $-\boldsymbol{\omega}\cdot\mathbf{S}$.

Reciprocity of the Two Effects

One Coupling, Two Readings

The two effects are not merely inverse in spirit; they are the two derivatives of one coupling term. In the rotating frame the coupling of the moment to the rotation is the bilinear

$$ U_{\rm rot} = -\,\boldsymbol{\Omega}\cdot\mathbf{S} = -\,\frac{1}{\gamma}\,\boldsymbol{\mu}\cdot\boldsymbol{\Omega}, $$

the spin-rotation coupling, which is the exact analogue of the magnetic coupling $U_{\rm mag} = -\boldsymbol{\mu}\cdot\mathbf{B}$. Two opposite rotations are in play here and they must not be conflated: the rotation whose coupling mimics a given field is $\boldsymbol{\Omega} = \gamma\mathbf{B}$, while the rotation that cancels it — the Larmor frame of the companion article — is $\boldsymbol{\Omega} = -\gamma\mathbf{B}$; the two differ by the sign of the coupling. The two effects are the two sides of this one term:

$$ \frac{\delta U_{\rm rot}}{\delta\boldsymbol{\mu}} = -\,\frac{\boldsymbol{\Omega}}{\gamma} \quad\text{(rotation produces a magnetization)}, \qquad \frac{\delta U_{\rm rot}}{\delta\boldsymbol{\Omega}} = -\,\frac{\boldsymbol{\mu}}{\gamma} \quad\text{(magnetization produces a rotation)} . $$

The coefficient $1/\gamma$ is the same in both, so a measurement of either effect determines the other. This is the sense in which the pair is reciprocal: the coupling is a single bilinear rather than two independent interactions, and for a medium the mixed second derivatives of the free energy that carries it are the two responses,

$$ \frac{\partial^2 F}{\partial\mu_i\,\partial\Omega_j} = -\,\frac{1}{\gamma}\,\delta_{ij}, $$

so that the magnetization produced by a rotation and the mechanical angular momentum produced by a magnetization change are the two symmetric derivatives of one free energy.

The Gyromagnetic Ratio in Both Effects

Because the reciprocity is exact for the coupling, both effects measure the same $\gamma$, and the measured value is the intrinsic one. For orbital magnetism the convective value is $g = 1$; for the intrinsic magnetism of a ferromagnet the value is $g\approx 2$, and the Einstein–de Haas and Barnett measurements return the second. The two effects therefore form a single measurement of $g$, performed in two directions, and each direction controls a different systematic error: the Einstein–de Haas effect is limited by the mechanical detection of a small angular momentum, the Barnett effect by the magnetic detection of a small moment. The agreement of the two is the classical evidence that the ferromagnetic moment is spin-carried.

Relation to the Quantum Treatment

The spin is treated here as an intrinsic angular momentum of the body, precessing classically in the effective field and coupling to the body by conservation of total angular momentum. The quantum treatment of the same spin is given in the companion articles Spin-1/2 Quantum Mechanics in Biquaternionic Form and Exercise: Spin Precession in a Magnetic Field, where the spin is an $\mathbb{M}_+$ observable and the precession is the evolution of an idempotent. The spin-rotation coupling $-\boldsymbol{\Omega}\cdot\mathbf{S}$ is the same operator in both descriptions; what the classical treatment adds is that the rotation of the body is a material rotor, not a state, and what the quantum treatment adds is the state and the Born rule. The reciprocal pair of effects is classical in its kinematics — conservation of angular momentum and the rotating frame suffice — and it is the same in the two descriptions.

Summary

The magnetization of a body is carried by angular momentum, and the relation between the two is the gyromagnetic coupling $\boldsymbol{\mu} = \gamma\mathbf{S}$. For an isolated body the total angular momentum $\mathbf{J} = \mathbf{S} + \hat{I}\boldsymbol{\omega}$ is conserved, and this one law contains both effects of the pair.

The Einstein–de Haas effect is the change of the body's rotation produced by a change of its magnetization:

$$ \dot{\boldsymbol{\omega}} = -\,\frac{1}{\gamma I}\,\dot{\boldsymbol{\mu}}, \qquad \Delta\mathbf{L} = -\,\frac{1}{\gamma}\,\Delta\boldsymbol{\mu}. $$

A change of the moment is compensated by a mechanical angular momentum, and in the biquaternion algebra the compensation is a rotation rotor $\Delta\tilde{R}$ generated by $-\Delta\tilde{\boldsymbol{\mu}}/(\gamma I)$.

The Barnett effect is the magnetization produced by a rotation:

$$ \mathbf{B}_{\rm eff} = \mathbf{B} + \frac{\boldsymbol{\omega}}{\gamma}, \qquad \mathbf{M} = \chi\,\mathbf{B}_{\rm eff} = \chi\!\left(\mathbf{B} + \frac{\boldsymbol{\omega}}{\gamma}\right), $$

so that a body rotating in zero field is magnetized as if by the field $\boldsymbol{\omega}/\gamma$, the Barnett field. It is the inverse of the Larmor theorem: the frame that is at rest for a moment in the field $\mathbf{B}$ rotates at $\boldsymbol{\Omega} = -\gamma\mathbf{B}$, and the rotation of the body enters the spin's equation exactly as the Barnett field $\boldsymbol{\omega}/\gamma$ does; the applied field that would cancel it is $-\boldsymbol{\omega}/\gamma$.

The two effects are reciprocal, being the two derivatives of the single spin-rotation coupling $U_{\rm rot} = -\boldsymbol{\mu}\cdot\boldsymbol{\Omega}/\gamma$; the same coefficient $1/\gamma$ governs both, so the pair is one measurement of the gyromagnetic factor performed in two directions. The measured value for the ferromagnets is $g\approx 2$, the intrinsic value, and the classical account of that factor is given in the companion article The Classical Origin of g = 2 in Biquaternionic Form. The rotor algebra used throughout — the unit real quaternion $\tilde{R}$, the conjugation $\mathbf{x}\mapsto\tilde{R}\mathbf{x}\tilde{R}^{*}$, the generator $\dot{\tilde{R}} = \tfrac{1}{2}\tilde{\boldsymbol{\omega}}\tilde{R}$, and the double cover $\tilde{R}(\theta+2\pi) = -\tilde{R}$ — is the classical rigid-body kinematics of the material sector.

Summary of Notation

Symbol	Meaning
$\mathbb{B} = \mathbb{C}\otimes_\mathbb{R}\mathbb{H}$	Biquaternion algebra
$e_0 = 1, e_1, e_2, e_3$	Quaternion basis, $e_k^2 = -e_0$
$\mathbb{M}_-$, $\mathbb{M}_+$	Material (anti-Hermitian) and informational (Hermitian) sectors
$\mathbb{H}_{\mathbb{B}}$	Real-quaternion subspace
$\boldsymbol{\mu}$, $\tilde{\boldsymbol{\mu}} = \mu_ke_k$	Magnetic moment (material axial vector)
$\mu_k\,ie_k$	Intrinsic moment's informational-sector representative
$\mathbf{M}$	Magnetization
$\mathbf{S}$, $\tilde{\mathbf{S}} = S_ke_k$	Spin angular momentum
$\mathbf{L} = \hat{I}\boldsymbol{\omega}$	Mechanical angular momentum
$\hat{I}$, $I$	Inertia tensor; its scalar value about the fixed axis
$\mathbf{J} = \mathbf{S} + \mathbf{L}$	Total angular momentum
$\gamma = g\,q/2m$	Gyromagnetic ratio
$g$	Gyromagnetic factor
$\tilde{R}(\theta,\hat{\mathbf{n}})$	Orientation rotor, unit real quaternion
$\mathbf{x}(t) = \tilde{R}\mathbf{x}(0)\tilde{R}^{*}$	Rotor conjugation (body vector into the lab)
$\boldsymbol{\omega} = \omega_ke_k$	Angular velocity (pure real quaternion)
$\tilde{\boldsymbol{\omega}} = 2\dot{\tilde{R}}\tilde{R}^{*}$	Angular velocity biquaternion
$\dot{\tilde{R}} = \tfrac{1}{2}\tilde{\boldsymbol{\omega}}\tilde{R}$	Rotor kinematics
$\mathbf{B}_{\rm eff} = \mathbf{B} + \boldsymbol{\omega}/\gamma$	Barnett effective field
$\boldsymbol{\Omega}$	Rotation in the spin-rotation coupling: $+\gamma\mathbf{B}$ mimics a field, $-\gamma\mathbf{B}$ cancels it
$U_{\rm rot} = -\boldsymbol{\mu}\cdot\boldsymbol{\Omega}/\gamma$	Spin-rotation coupling
$\chi$	Magnetic susceptibility