= Independent spin Barnett response
{c}
{title2=$\langle J_z\rangle=(\hbar/2)\tanh(\hbar\Omega/(2k_BT))$}
Consider an explicitly stipulated dilute <independent spin-one-half two-level system> with zero applied <magnetic field>. Each constituent has $J_z=\pm\hbar/2$ 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> $E_\pm=\mp\hbar\Omega/2$. Define the dimensionless variable $x=\hbar\Omega/(2k_BT)$, where $\hbar$ is the <reduced Planck constant> and $k_B$ is the <Boltzmann constant>. The <partition function> and <Boltzmann distribution> give
$$
Z=e^x+e^{-x}=2\cosh x,\qquad
p_+=\frac{e^x}{Z},\qquad p_-=\frac{e^{-x}}{Z}.
$$
Taking the <expected value> gives
$$
\langle J_z\rangle=\frac\hbar2\tanh x,\qquad
\langle m_z\rangle=\gamma\frac\hbar2\tanh x,\qquad
M_z=n\gamma\frac\hbar2\tanh x,
$$
where $n$ 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 $|x|\ll1$, expand the <hyperbolic tangent> to obtain
$$
M_z\simeq\frac{n\gamma\hbar^2}{4k_BT}\Omega
=\frac{n\gamma^2\hbar^2}{4k_BT}B_\Omega.
$$
The 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 $T\to0^+$ limit saturates $|\langle J_z\rangle|$ at $\hbar/2$; 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.
\Image[/barnett-independent-spin-rotation-response.png]
{title=Signed <magnetic moment> and <angular momentum> response to rotation}
{description=Original calculated <Independent spin Barnett response>, with $\gamma<0$. The left panel reaches the formal saturation limits; the right panel is the extremely small-polarization room-<temperature> regime. The stipulated inputs describe a toy collection of independent <spins>; neither panel represents measurements of a <ferromagnet> or a fit to the gadolinium experiment.}
For a numerical example, stipulate $n=10^{28}\,\mathrm{m^{-3}}$, $T=300\,\mathrm K$, $\gamma=-1.76\times10^{11}\,\mathrm{s^{-1}\,T^{-1}}$, and rotation <frequency> $f=1000\,\mathrm{Hz}$. With $\Omega=2\pi f$, $\hbar=1.054571817\times10^{-34}\,\mathrm{J\,s}$, and $k_B=1.380649\times10^{-23}\,\mathrm{J/K}$, the model gives
$$
B_\Omega=-35.7\,\mathrm{nT},\qquad
x=8.00\times10^{-11},\qquad
M_z=-7.42\times10^{-6}\,\mathrm{A/m}.
$$
These are teaching inputs and calculated outputs, not measured material values. The corresponding internal <angular momentum> density is $M_z/\gamma=+4.22\times10^{-17}\,\mathrm{J\,s/m^3}$. 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.
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