Rest-frame spin transport and finite-body size bounds

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ of The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$) for the canonical tangent-bundle and connection suppliers used by the BMT transport and charged-action equivalence. No full Axiom of Choice is assumed.

Adopt the charged-body quantities of Charged and spinning test-body quantities, the SI field convention of The electromagnetic two-form, its observer decomposition and the four-current, and the classical BMT law Classical BMT polarization model. For a trajectory stationary under the local action postulate, the translational force equation is supplied by Stationary charged action and Lorentz force. In flat Minkowski spacetime let $m>0$, $q,g_s\in \mathbb{R}$, $\alpha =q\mathrm{/}m$, $G=g_s\mathrm{/}2$, and let $U,s$ satisfy $D_{\tau }U=\alpha F^{\mathrm{♯}}(U),D_{\tau }s=\alpha [G{F}^{\mathrm{♯}}(s)-(G-1)\frac{U}{c^2}g(s,F^{\mathrm{♯}}(U))],g(U,U)=-c^2,g(U,s)=0.$ The tensor equation is covariant under proper orthochronous Lorentz transformations of the complete field and worldline data. If $F=0$, then $D_{\tau }U=0$ and $D_{\tau }s=0$. For any acceleration $a=D_{\tau }U$ within the Lorentz-force model, the $G=0$ branch is Fermi–Walker transport on the orthogonal spin subspace: $D_{\tau }s=\frac{1}{c^2}(Ug(a,s)-ag(U,s))=\frac{U}{c^2}g(a,s)\mathrm{.}$ For the uniform-field branch, choose an inertial frame $(o;e_0,e_1,e_2,e_3)$ of Inertial frames and inertial charts of the affine Minkowski model with $e_0=U\mathrm{/}c$ for the stationary trajectory and with the positively oriented orthonormal spatial basis $(e_1,e_2,e_3)$. Let $\mathbf{E}=0$ and let the constant spatial field be $\mathbf{B}\mathrm{\ne }0$; its norm and cross product are those of the frame's positive-definite spatial structure. Then $U=(c,\mathbf{0})$ is stationary, $\widehat{\mathbf{B}}:=\mathbf{B}\mathrm{/}\mathrm{\mid }\mathbf{B}\mathrm{\mid }$, and for ${\mathbf{s}}_0:=\mathbf{s}({\tau }_0)$ define ${\mathbf{s}}_{\parallel ,0}:=({\mathbf{s}}_0\cdot \widehat{\mathbf{B}})\widehat{\mathbf{B}}$ and ${\mathbf{s}}_{\perp ,0}:={\mathbf{s}}_0-{\mathbf{s}}_{\parallel ,0}$. The $G$ branch has the exact solution $\mathbf{s}(\tau )={\mathbf{s}}_{\parallel ,0}+\cos (\omega \mathrm{\Delta }\tau ){\mathbf{s}}_{\perp ,0}-\sin (\omega \mathrm{\Delta }\tau )\widehat{\mathbf{B}}\times {\mathbf{s}}_{\perp ,0},\omega =\frac{qG}{m}\mathrm{\mid }\mathbf{B}\mathrm{\mid },\mathrm{\Delta }\tau =\tau -{\tau }_0\mathrm{.}$ The $F=0$, $G=0$, and uniform-field claims are deductions within the adopted model, not experimental statements. For the independent finite-body branch, use the moment and unit convention of Body moments and transport convention; the density-to-spin identity below is stipulated data.

Separately, let a flat body have rest momentum $p$, and define $M:=\sqrt{-g(p,p)}\mathrm{/}c>0$. Choose its momentum-rest inertial frame $(o;e_0,e_1,e_2,e_3)$ of Inertial frames and inertial charts of the affine Minkowski model with $e_0=p\mathrm{/}(Mc)$ and spatial origin at the stipulated body center. Define ${\bar{B}}_{\mathrm{\ell }}(0):=\{\mathbf{x}\in {\mathbb{R}}^3:\mathrm{\mid }\mathbf{x}\mathrm{\mid }\le \mathrm{\ell }\}$ using the Euclidean norm of its positive-definite spatial rest space. In this frame $p$ has components $(Mc,\mathbf{0})$; let $S^{ab}=-S^{ba}$ be its angular-momentum tensor, with $S^{0i}=0$ (hence $S^{ab}{p}_b=0$), and let $e:{\mathbb{R}}^3\to [0,\mathrm{\infty })$ and $j:{\mathbb{R}}^3\to {\mathbb{R}}^3$ be measurable rest densities, with each component of $j$ in $L^1({\mathbb{R}}^3)$; their units are joules per cubic metre and kilogram per square metre per second, respectively. Assume $e$ and $j$ vanish almost everywhere outside ${\bar{B}}_{\mathrm{\ell }}(0)$, $\mathrm{\mid }j\mathrm{\mid }$ is the Euclidean norm and satisfies $\mathrm{\mid }j\mathrm{\mid }\le e\mathrm{/}c$ almost everywhere, $\int e{d}^{3}x=M{c}^2$, and the stipulated componentwise spin-density identity ${\mathbf{s}}_{\ast }:={\int }_{{\mathbb{R}}^3}\mathbf{x}\times j(\mathbf{x})d^{3}x=\frac{1}{2}{ϵ}_{kij}{S}^{ij}{\mathbf{e}}_k,$ where the vector integral is componentwise in the positively oriented spatial basis of this inertial frame. Then $\mathrm{\ell }\ge \frac{\mathrm{\mid }{\mathbf{s}}_{\ast }\mathrm{\mid }}{Mc}\mathrm{.}$ If ${\mathbf{s}}_{\ast }\mathrm{\ne }0$, every observer-induced centroid shift $\mathit{\delta }$ of the flat SSC translation law satisfies $\mathrm{\mid }\mathit{\delta }\mathrm{\mid }<\mathrm{\mid }{\mathbf{s}}_{\ast }\mathrm{\mid }\mathrm{/}(Mc)$, every point of the open disk in ${\mathbf{s}}_{\ast }^{\perp }$ with that radius is attained, and its boundary is a supremum not attained by a timelike observer. If ${\mathbf{s}}_{\ast }=0$, the only shift is $\mathit{\delta }=0$. Under proper orthochronous Lorentz transformations of the complete body data and its momentum-rest hypersurface, the scalars $M$, $\mathrm{\ell }$, $\mathrm{\mid }{\mathbf{s}}_{\ast }\mathrm{\mid }$, and $\mathrm{\mid }{\mathbf{s}}_{\ast }\mathrm{\mid }\mathrm{/}(Mc)$ are unchanged; ${\mathbf{s}}_{\ast }$ itself transforms as a spatial vector, so its components depend on the oriented rest-space basis. No equality between $\mathbf{s}$ and ${\mathbf{s}}_{\ast }$ is assumed, so the size inequalities do not bound the BMT polarization unless such an identification is separately adopted.

The BMT polarization vector belongs to the charged test-body transport model. The body spin ${\mathbf{s}}_{\ast }$ is the dual of an angular-momentum moment in a momentum-rest frame. Their units agree, but their physical identification is additional input. The Lorentz-covariance claim applies to the full tensor model, while the magnetic sine–cosine expression and size assumptions are stated in their specified rest frames.