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Loss Cone Angle

\[\sin\alpha_{\rm LC} = \frac{1}{\sqrt{R_m}}, \qquad R_m \equiv \frac{B_{\max}}{B_0}\]

Variables

Symbol Meaning Units
\(\alpha_{\rm LC}\) loss-cone half-angle, measured at the weak-field point rad
\(R_m\) mirror ratio —
\(B_0\) field at the midplane (weakest point) T
\(B_{\max}\) field at the throat (strongest point) T

Assumptions

  • The magnetic moment \(\mu = mv_\perp^2/2B\) is conserved — requires \(r_L \ll L_B\) and slow variation compared with the gyroperiod
  • Energy is conserved (magnetic forces do no work); no electric field along \(\mathbf{B}\)
  • Collisionless over one bounce

Derivation sketch

At the turning point all kinetic energy is perpendicular, so \(v_\perp^2 = v^2\) where \(B = B_{\max}\). Equating \(\mu\) there with \(\mu\) at the midplane, where \(v_\perp = v\sin\alpha\):

\[\frac{v^2\sin^2\alpha}{B_0} = \frac{v^2}{B_{\max}} \quad\Longrightarrow\quad \sin^2\alpha = \frac{B_0}{B_{\max}}\]

Particles with \(\alpha < \alpha_{\rm LC}\) never reach a turning point and escape.

Notes

Mass, charge and speed all cancel — the loss cone is purely geometric. The escaping fraction of an isotropic distribution is \(1 - \cos\alpha_{\rm LC}\) (two cones, normalised to \(4\pi\)). Because \(\sin\alpha_{\rm LC} = R_m^{-1/2}\), reducing losses requires punishingly large mirror ratios: 1% loss needs \(R_m = 50\).