BMT constraint and restricted-ansatz algebra — Examples

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.

Use the charged-body conventions and adopted BMT transport equation of Charged and spinning test-body quantities and Classical BMT polarization model. On a local potential domain, take the trajectory stationary under the action postulate so that Stationary charged action and Lorentz force supplies its translation equation. Let $U$ be future timelike, $g(U,U)=-c^2$, and $g(U,s)=0$. Write $\alpha =q\mathrm{/}m$, $G=g_s\mathrm{/}2$, $\kappa =g(s,F^{\mathrm{♯}}(U))$, and $a=D_{\tau }U=\alpha F^{\mathrm{♯}}(U)$. The BMT law is $D_{\tau }s=\alpha [G{F}^{\mathrm{♯}}(s)-(G-1)\frac{U\kappa }{c^2}]\mathrm{.}$

Here Fermi–Walker transport means $D_{\tau }s=(Ug(a,s)-ag(U,s))\mathrm{/}{c}^2$, where $a:=D_{\tau }U$; on the orthogonal subspace $g(U,s)=0$, this reduces to $D_{\tau }s=Ug(a,s)\mathrm{/}{c}^2$.

(a) If $g_s=0$ ($G=0$), then $D_{\tau }s=Ug(a,s)\mathrm{/}{c}^2$, the Fermi–Walker transport law on $U^{\perp }$.

(b) If $g_s=2$ ($G=1$), then $D_{\tau }s=\alpha F^{\mathrm{♯}}(s)$ while $D_{\tau }U=\alpha F^{\mathrm{♯}}(U)$, so the same linear field operator transports both vectors. Both parameter cases preserve orthogonality and spin norm by BMT orthogonality and conserved spin length.

The $G=0$ branch is Fermi–Walker transport for the prescribed Lorentz-force acceleration. The $G=1$ branch is a shared-generator statement; it does not claim $s=U$ or identify the polarization with a body angular-momentum moment.

Assume the static Coulomb-exterior support-radius hypotheses of Coulomb-exterior support-radius energy bound, so every member obeys $\mathrm{\ell }\ge q^2\mathrm{/}(8\pi {\epsilon }_{0}M{c}^2)$.

(a) For fixed $q\mathrm{\ne }0$ and fixed finite $M>0$, the lower bound is the positive constant $q^2\mathrm{/}(8\pi {\epsilon }_{0}M{c}^2)$. Thus no family satisfying these hypotheses can have $\mathrm{\ell }\to 0$ while keeping both $q$ and $M$ fixed.

(b) For $q(\lambda )=\lambda q_0$, $M(\lambda )=\lambda M_0$, and $\mathrm{\ell }(\lambda )=\lambda {\mathrm{\ell }}_0$ with $\lambda >0$, $q_0\mathrm{\ne }0$ coulombs, $M_0>0$ kilograms, and ${\mathrm{\ell }}_0>0$ metres, the same necessary condition reduces to ${\mathrm{\ell }}_0\ge q_0^2\mathrm{/}(8\pi {\epsilon }_0{M}_0{c}^2)$, independent of $\lambda$. The inequality therefore does not rule out this scaling, although it does not construct a body family or prove that its interior energy remains admissible.

This comparison separates a fixed nonzero charge and fixed mass limit from a scaling in which both charge and mass vanish. It does not establish that any such family solves a closed matter or Maxwell evolution model.

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.

In a charged-body inertial frame of Inertial frames and inertial charts of the affine Minkowski model with $e_0=U\mathrm{/}c$, positively oriented spatial basis $(e_1,e_2,e_3)$, and constant coordinates for the stationary trajectory, let $U=(c,\mathbf{0})$, $\mathbf{E}=0$, and $\mathbf{B}=B{\mathbf{e}}_3$ with constant $B>0$ tesla. Let the adopted polarization law have $q$ in coulombs, $m>0$ kilograms, $\alpha =q\mathrm{/}m$, $G=g_s\mathrm{/}2$, and choose initial polarization $\mathbf{s}({\tau }_0)=s_0{\mathbf{e}}_1$ in kilogram metres squared per second. Independently, choose a separate momentum-rest inertial frame $(o^{\mathrm{\prime }};f_0,f_1,f_2,f_3)$ of Inertial frames and inertial charts of the affine Minkowski model with $f_0=p\mathrm{/}(Mc)$, spatial origin $o^{\mathrm{\prime }}$ at the stipulated body center, and positive spatial basis $(f_1,f_2,f_3)$. Let a flat material body have rest momentum components $p=(Mc,\mathbf{0})$ in this frame, $M>0$ kilograms, body angular momentum ${\mathbf{s}}_{\ast }=S_{\ast }{f}_3$ with $S_{\ast }>0$ in kilogram metres squared per second, and measurable densities $e:{\mathbb{R}}^3\to [0,\mathrm{\infty })$ and $j:{\mathbb{R}}^3\to {\mathbb{R}}^3$ with each component of $j$ in $L^1({\mathbb{R}}^3)$, expressed in its spatial coordinates $\mathbf{x}$. Define ${\bar{B}}_{\mathrm{\ell }}(0):=\{\mathbf{x}:\mathrm{\mid }\mathbf{x}\mathrm{\mid }\le \mathrm{\ell }\}$ using this rest-space Euclidean norm. Assume $e$ and $j$ vanish almost everywhere outside this ball for $\mathrm{\ell }>0$ metres, $\mathrm{\mid }j\mathrm{\mid }$ is that norm and satisfies $\mathrm{\mid }j\mathrm{\mid }\le e\mathrm{/}c$ almost everywhere, $\int e{d}^{3}x=M{c}^2$, and the componentwise vector integral satisfies ${\mathbf{s}}_{\ast }=\int \mathbf{x}\times j{d}^{3}x$. Assume also that its spin tensor is the spatial dual of ${\mathbf{s}}_{\ast }$ and obeys the momentum SSC. Then the polarization and body-centroid results are

$\mathbf{s}(\tau )=s_0(\cos (\omega \mathrm{\Delta }\tau ){\mathbf{e}}_1-\sin (\omega \mathrm{\Delta }\tau ){\mathbf{e}}_2),\omega =\frac{qGB}{m},\mathrm{\Delta }\tau =\tau -{\tau }_0,$

$\mathrm{\ell }\ge \frac{S_{\ast }}{Mc},\mathit{\delta }=({\delta }_1,{\delta }_2,0)\text{ in the body-frame spatial components},{\delta }_1^2+{\delta }_2^2<{\left(\frac{S_{\ast }}{Mc}\right)}^2\mathrm{.}$

Every interior shift vector is realized by some future timelike observer. There is no identification between $s_0$ and $S_{\ast }$ in this example.

This is a model calculation. The field values are stipulated constants, not observations; the inequalities are conditional consequences for a separate body satisfying the stated material hypotheses.

Assume the neutral pole-dipole equations of Pole-dipole and closed multipole test-body models and the TD admissible region of TD spin and momentum norms: $p$ is future timelike, $K$ is invertible, $W=K^{-1}p$ is future timelike, and the initial spin tensor satisfies $S^{ab}=0$. Then throughout the local solution the spin remains zero, $p$ is parallel transported, and $U=\frac{c}{\mu }p,\mu =\sqrt{-g(p,p)}=\text{constant},D_{\tau }U=0.$ Thus the representative curve is a timelike geodesic with constant rest mass $M=\mu \mathrm{/}c$ in this restricted branch.

The result is the zero-spin specialization of the adopted neutral TD model. It does not derive that model from the body-moment definition and does not assert completeness of the geodesic or the full spinning-body system.