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Rest magnetic spin rotation and independent centroid disk

Physical scope

A worked exact rest-frame BMT rotation in a uniform magnetic field alongside the separate flat-body SSC centroid disk and positive-energy size bound. The polarization and body angular momentum are kept distinct; no experiment or material family is reported.

Example

Assume ACω of The Axiom of Countable Choice (ACω) 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 e0=U/c, positively oriented spatial basis (e1,e2,e3), and constant coordinates for the stationary trajectory, let U=(c,0), E=0, and B=Be3 with constant B>0 tesla. Let the adopted polarization law have q in coulombs, m>0 kilograms, α=q/m, G=gs/2, and choose initial polarization s(τ0)=s0e1 in kilogram metres squared per second. Independently, choose a separate momentum-rest inertial frame (o′;f0,f1,f2,f3) of Inertial frames and inertial charts of the affine Minkowski model with f0=p/(Mc), spatial origin o′ at the stipulated body center, and positive spatial basis (f1,f2,f3). Let a flat material body have rest momentum components p=(Mc,0) in this frame, M>0 kilograms, body angular momentum s∗=S∗f3 with S∗>0 in kilogram metres squared per second, and measurable densities e:R3→[0,∞) and j:R3→R3 with each component of j in L1(R3), expressed in its spatial coordinates x. Define B‾ℓ(0):={x:∣x∣≤ℓ} using this rest-space Euclidean norm. Assume e and j vanish almost everywhere outside this ball for ℓ>0 metres, ∣j∣ is that norm and satisfies ∣j∣≤e/c almost everywhere, ∫e d3x=Mc2, and the componentwise vector integral satisfies s∗=∫x×j d3x. Assume also that its spin tensor is the spatial dual of s∗ and obeys the momentum SSC. Then the polarization and body-centroid results are

s(τ)=s0(cos⁡(ωΔτ)e1−sin⁡(ωΔτ)e2),ω=qGBm,Δτ=τ−τ0,

ℓ≥S∗Mc,δ=(δ1,δ2,0) in the body-frame spatial components,δ12+δ22<(S∗Mc)2.

Every interior shift vector is realized by some future timelike observer. There is no identification between s0 and S∗ in this example.

Facts & Assumptions

[A3]

The Axiom of Countable Choice (ACω): assume only countable choice ACω for the canonical tangent-bundle smooth structure and inherited BMT/action-equivalence supplier hypotheses.

Given: The BMT rest-frame law and exact solution theorem, a constant uniform rest magnetic field, m,M,B,S∗>0, q,gs∈R, and separate material rest densities satisfying the stated energy-current and support hypotheses.

[A2]

The inertial rest field is E=0, B=Be3 with B>0, and the initial polarization is s0e1; independently, in the momentum-rest frame with origin and spatial basis specified in the Example, the body has p=(Mc,0), s∗=S∗f3=∫x×j d3x, the stated spin tensor, and the specified componentwise-integrable rest densities supported in B‾ℓ(0) with the energy conditions.

[A1]

The polarization obeys the adopted BMT model; in a rest frame the magnetic transport law is ds/dτ=αG(s×B). Rest-frame spin transport and finite-body size bounds

[F1]

The component convention is (u×v)i=ϵijkujvk and Fij=ϵijkBk. The cross product in R3; The electromagnetic two-form, its observer decomposition and the four-current

[F2]

The exact constant-field spin solution and the separate body-size and centroid-disk statements hold under their stated assumptions. Rest-frame spin transport and finite-body size bounds; Positive-energy spin–size inequality; Exact flat centroid translation and disk

[F3]

The charge-model polarization s and the flat-body moment spin are distinct model quantities unless an additional identification is supplied. Charged and spinning test-body quantities

Verification

technique · direct
1.1givenA3A1A2F1F2algebra

Under [A2], B^=e3, s∥,0=0, s⊥,0=s0e1, and the angular frequency in the rest-spin theorem is ω=αGB=qGB/m. Substitution in that theorem gives the displayed s(τ); differentiating yields ds/dτ=ωs0[−sin⁡(ωΔτ)e1−cos⁡(ωΔτ)e2], equal to αG(s×Be3). The initial value at τ0 is s0e1.

1.2givenA3A2F2algebra

The independent body data [A2] satisfy the hypotheses of [F2]'s size lemma, so ∣s∗∣=S∗ gives ℓ≥S∗/(Mc). The centroid lemma says the observer shifts fill the open disk in s∗⊥ of radius S∗/(Mc); because s∗=S∗f3, this is exactly the displayed f1–f2 disk in body-frame components, with every strict interior point attainable and the boundary unattained.

2.1F3step 1.1step 1.2∎

The first calculation uses the polarization s and the second uses the body moment s∗. [F3] imposes no equality between them, so the body-size inequality contains S∗ and does not constrain s0 without an extra identification assumption.

Remarks

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.

Depends on

Dependency tree · 2 levels

Sources