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A circular magnetic orbit with zero axial momentum

Physical scope

A symbolic specialization of the conditional constant-magnetic-field test-particle theorem with chosen model parameters. It is not a measurement or empirical claim.

Example

In Exact relativistic helix in a uniform magnetic field, choose qB0>0, p01=p∗>0 with units kilogram metre per second, p02=p03=0, and arbitrary m>0 kilograms. Then S0=m2c2+p∗2 has momentum units, γ0=S0/(mc) is dimensionless and Ω=qB0/(γ0m)>0 has units inverse seconds. The orbit is the circle

x1=x01+cp∗S0Ωsin⁡(Ωt),x2=x02−cp∗S0Ω(1−cos⁡(Ωt)),x3=x03,

of radius p∗/(qB0), with constant energy cS0. At t=π/(2Ω) its momentum is (0,−p∗,0). All values are model parameters, not observations.

Facts & Assumptions

Given: The parameter restrictions above and the adopted magnetic test-particle model in the theorem.

[F1]

Exact relativistic helix in a uniform magnetic field: the momentum and position formulas have signed angular frequency Ω=qB0/(γ0m), constant energy cS0, and transverse radius ∣p⊥∣/∣qB0∣.

Verification

1.1givenF1algebra

With p02=p03=0 and Ω>0, substitution into [F1] gives p1=p∗cos⁡(Ωt), p2=−p∗sin⁡(Ωt) and p3=0. Substitution into the position formulas gives the displayed coordinates and x3=x03.

2.1step 1.1F1algebra∎

The radius clause of [F1] gives ∣p⊥∣/∣qB0∣=p∗/(qB0), and the energy is cS0. At Ωt=π/2, sine is 1 and cosine is 0, so p=(0,−p∗,0). The orbit is planar because p03=0 makes x3 constant.

Depends on

Dependency tree · 2 levels

Sources