Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The exact constant drift solution in subcritical crossed fields

Physical scope

A symbolic specialization of the subcritical crossed-field theorem. The chosen field parameters describe an exact model solution and are not measured values; general initial data still give inverse-boosted magnetic orbits.

Example

Let B0>0 be in tesla and set E0=cB0/2 in newtons per coulomb, so E=E0e1, B=B0e3 is subcritical. In Subcritical crossed fields and exact drift-frame orbits, the drift velocity and boost factor are vd=−(c/2)e2 in metres per second and γd=2/3. The constant-velocity orbit x(t)=x0+vdt in a chart with x0=ct solves the adopted model for any m>0 and constant charge q: its Lorentz force vanishes. This is one exact solution; other initial data yield boosted magnetic orbits.

Facts & Assumptions

Given: The chosen fields, positive B0, m>0, constant q, and the subcritical external-field model of the theorem.

[F1]

Subcritical crossed fields and exact drift-frame orbits: for ∣E0∣<c∣B0∣, vd=−(E0/B0)e2, γd=(1−∣vd∣2/c2)−1/2, and E+vd×B=0; general solutions are inverse boosts of magnetic orbits.

Verification

1.1givenF1algebra

The ratio E0/(cB0)=1/2 verifies the strict subcritical hypothesis. Substituting E0=cB0/2 in [F1] gives vd=−(c/2)e2 and γd=(1−1/4)−1/2=2/3.

2.1step 1.1F1algebra∎

Directly, vd×B=−(cB0/2)(e2×e3)=−(cB0/2)e1=−E. Thus [F1] makes the Lorentz force zero, momentum stays constant, and the straight worldline is an exact orbit. The theorem's inverse-boost construction applies to general initial data, so this example does not assert that every orbit has this constant velocity.

Depends on

Dependency tree · 2 levels

Sources