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BMT G=0 Fermi–Walker and G=1 shared-generator cases

Physical scope

Exact parameter specializations of the adopted classical BMT transport model: $g_{s} = 0$ gives Fermi–Walker transport on the orthogonal spin space, and $g_{s} = 2$ makes the same field generator act on $U$ and $s$ . These are model deductions, not measured cases.

Example

Assume ${A C}_{ω}$ of The Axiom of Countable Choice ( ${A C}_{ω}$ ) 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 $α = q / m$ , $G = g_{s} / 2$ , $κ = g (s, F^{♯} (U))$ , and $a = D_{τ} U = α F^{♯} (U)$ . The BMT law is $D_{τ} s = α [G F^{♯} (s) - (G - 1) \frac{U κ}{c^{2}}] .$

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

(a) If $g_{s} = 0$ ( $G = 0$ ), then $D_{τ} s = U g (a, s) / c^{2}$ , the Fermi–Walker transport law on $U^{⊥}$ .

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

Facts & Assumptions

[A3]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): assume only countable choice ${A C}_{ω}$ for the canonical tangent-bundle smooth structure and inherited BMT/action-equivalence supplier hypotheses.

Given: The adopted BMT spin law, a trajectory whose finite segments in a local potential domain are stationary for the adopted action, $m > 0$ , constant $q, g_{s}$ , and an initially orthogonal spin vector.

[A2]

The example selects the model parameters $G = 0$ and $G = 1$ in separate cases and assumes the initial spin is orthogonal to $U$ .

[A1]

Adopt the BMT spin transport law of Classical BMT polarization model. As stipulated in the Example, assume each finite trajectory segment in the local potential domain is stationary for the adopted action, under its $C^{2}$ fixed-endpoint timelike variations. Translation is then supplied by [F3].

[F1]

The spin remains orthogonal to $U$ and its norm is constant within this model. BMT orthogonality and conserved spin length

[F2]

The model spin is a classical polarization vector and $g_{s}$ is a primitive dimensionless parameter; no measured value or quantum derivation is supplied. Charged and spinning test-body quantities

[F3]

Under [A3] and the finite-segment stationarity hypothesis in [A1], the local charged-action equivalence gives $a = D_{τ} U = (q / m) F^{♯} (U)$ . Stationary charged action and Lorentz force

[F4]

At each event $x$ , the model metric $g_{x} : T_{x} M \times T_{x} M \to R$ is symmetric: $g_{x} (v, w) = g_{x} (w, v)$ for all $v, w \in T_{x} M$ .

Verification

technique · direct

1.1givenA3A1A2F1F3F4algebra

In the $G = 0$ case of [A2], [A1] gives $D_{τ} s = α U κ / c^{2}$ . By [F3] and [F4], $g (a, s) = α g (F^{♯} (U), s) = α g (s, F^{♯} (U)) = α κ$ . Hence $D_{τ} s = U g (a, s) / c^{2}$ . The general Fermi–Walker expression subtracts $a g (U, s) / c^{2}$ , which is zero by [A2] and [F1], proving (a).

1.2givenA3A1A2F2F3algebra

In the $G = 1$ case of [A2], [A1] has zero term proportional to $U$ and $D_{τ} s = α F^{♯} (s)$ , while [F3] gives $D_{τ} U = α F^{♯} (U)$ . Thus both vectors are acted on by the same linear map $α F^{♯}$ along the worldline.

2.1F1F2step 1.1step 1.2∎

Applying [F1] to both parameter choices confirms that the initial constraint and spin norm remain unchanged; $G = 0$ and $G = 1$ are selected model parameters only, with no empirical inference.

Remarks

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.

BMT G=0 Fermi–Walker and G=1 shared-generator cases

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources