advanced34 min studyLesson 295 in Physics

Magnetic mirrors and loss cones

Use energy and magnetic-moment conservation to derive reflection conditions and the correct velocity-space loss fraction.

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What you’ll understand

  • Calculate a mirror field from an initial pitch angle.
  • Derive the two-ended loss cone and its solid-angle fraction.
  • Identify electrostatic, collisional, and adiabatic limits of simple mirror confinement.

A stronger magnetic field can reflect a magnetized particle even though the magnetic force does no work. The field converts parallel kinetic energy into perpendicular kinetic energy while the total stays fixed. Reflection depends on pitch angle as well as field strength. A mirror therefore selects directions in velocity space; it cannot confine every particle merely by having a large maximum magnetic field.

Two invariants locate a turning point

Consider a static mirror field with minimum magnitude B0B_0 at the midplane and equal maxima Bmax⁡B_{\max} at its two ends. Neglect electric fields, collisions, and the small drift kinetic energy. Total kinetic energy KK and adiabatic scalar moment μ\mu are then conserved. Parallel kinetic energy at any sampled field is K−μBK-\mu B. A turning point occurs when that expression reaches zero; stronger fields would require an impossible negative parallel energy.

At the midplane let pitch angle α0\alpha_0 measure velocity direction from the local field, so perpendicular energy is Ksin⁡2α0K\sin^2\alpha_0. Substitution gives the turning field Bm=B0/sin⁡2α0B_m=B_0/\sin^2\alpha_0. A particle launched almost parallel to the field has a small perpendicular fraction and needs an enormous turning field. A perpendicular launch is already at zero parallel velocity. These limiting cases are useful checks on the direction of the inequality.

K∥(B)=K−μB,μ=Ksin⁡2α0B0,Bm=B0sin⁡2α0K_\parallel(B)=K-\mu B,\qquad \mu=\frac{K\sin^2\alpha_0}{B_0},\qquad B_m=\frac{B_0}{\sin^2\alpha_0}
K and its parallel component are energies in J; μ is J/T and B is T. This leading guiding-centre result assumes a static smooth field with adiabatic moment conservation, no electric potential variation, and negligible collisional scattering during the bounce.↔ Scroll wide equations horizontally.

The loss cone is a cone in direction space

Define mirror ratio Rm=Bmax⁡/B0>1R_m=B_{\max}/B_0>1. Reflection before the maximum requires sin⁡2α0>1/Rm\sin^2\alpha_0>1/R_m. Directions closer to either parallel axis fall inside a loss cone and escape toward the corresponding end. The equality is the separatrix between passing and trapped orbits; a trajectory can asymptotically approach a smooth maximum there, so it should not be counted as a robustly trapped finite-period orbit.

For isotropically distributed velocity directions at the midplane, the population fraction in a cone is its solid angle divided by 4π4\pi. Two opposite cones together occupy fraction 1−cos⁡αc1-\cos\alpha_c, not simply 2αc/π2\alpha_c/\pi. The sine weighting of spherical area matters. This is the fraction of an initially isotropic population in loss directions; the particle flux through an aperture also weights parallel speed and is therefore a different quantity.

sin⁡2αc=1Rm,floss=1−cos⁡αc=1−1−1Rm\sin^2\alpha_c=\frac1{R_m},\qquad f_{\rm loss}=1-\cos\alpha_c=1-\sqrt{1-\frac1{R_m}}
α_c is the half-angle of each midplane loss cone. The fraction assumes isotropic velocity directions and two equal open ends with no electrostatic barrier. It is dimensionless and is not a steady loss rate or an angular fraction measured uniformly in pitch angle.↔ Scroll wide equations horizontally.

A mirror is conditional confinement

The parallel force in the averaged description is −μ ∂B/∂s-\mu\,\partial B/\partial s, directed away from stronger field. It acts on either charge sign in this approximation because scalar moment is positive. Real divergence-free mirror geometry also changes transverse field components; a purely axial field whose magnitude varies along its own straight direction everywhere would violate the magnetic divergence constraint. The one-dimensional formula follows motion along a field line, rather than specifying a complete field configuration.

An electrostatic potential adds qϕq\phi to the conserved energy and can either improve or weaken confinement depending on charge. Collisions change pitch angle and can refill a loss cone after its initial particles escape. Rapid field variation, large gyroradius, field nulls, and collective instabilities further limit the ideal picture. A calculated trapped fraction therefore describes accessible single-particle orbit classes under stated assumptions, not a demonstration of indefinitely stable plasma confinement or a predicted fusion performance.

Test an orbit against a fourfold mirror

A symmetric mirror has midplane field 0.20 T and maximum field 0.80 T. A particle begins at the midplane with 100 eV total kinetic energy and pitch angle 45°. Under the static adiabatic approximation, determine its turning field and whether it reflects before the maximum.

  1. The mirror ratio is Rm=0.80/0.20=4R_m=0.80/0.20=4. Its loss-cone boundary obeys sin⁡2αc=1/4\sin^2\alpha_c=1/4, giving αc=30∘\alpha_c=30^\circ. A 45° launch lies outside both loss cones.

  2. At launch, perpendicular energy is 100sin⁡245∘=50 eV100\sin^2 45^\circ=50\,\mathrm{eV}. Dividing by 0.20 T gives μ=250 eV/T\mu=250\,\mathrm{eV/T}, with energy units used consistently on both sides of the conservation equation.

  3. The turning field is K/μ=100/250=0.40 TK/\mu=100/250=0.40\,\mathrm T. At that location perpendicular energy has risen to 100 eV and parallel energy has fallen to zero. Since 0.40 T is below 0.80 T, reflection occurs before the end maximum.

The result

The particle mirrors at 0.40 T and belongs to the trapped orbit class. The result fixes a field value, not a position; locating the turning point also requires the field profile along its path.

Explore the model

Interactive lab

Interactive lab · vectors

Compare pitch directions at fixed speed

Treat x as parallel and y as perpendicular velocity. Reset and set both controls to angle 20° and magnitude 2; predict a perpendicular fraction below one quarter. Reset and set both controls to angle 45° and magnitude 2; predict a fraction of one half. Inspect projections for the R_m = 4 criterion. The lab has no mirror geometry, bounce motion, or loss-rate readout.

Compare pitch directions at fixed speedPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAt fixed speed the squared perpendicular projection determines the initial moment; a vector projection alone does not simulate confinement.

Check your understanding

For two identical open mirror ends and an isotropic midplane population, which expression gives the fraction initially inside either loss cone?

Try it yourself

A symmetric mirror has ratio 9. Find the loss-cone half-angle through its sine, the approximate initial isotropic population fraction in the two cones, and whether a midplane particle with perpendicular energy one fifth of its total energy is trapped.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the energy fraction directly as sine squared; no inverse trigonometric function is needed for the trapping test.

Compare with the solution

Here sin⁡αc=1/3\sin\alpha_c=1/3, so αc≃19.5∘\alpha_c\simeq19.5^\circ. The combined initial loss fraction is 1−8/9≃0.05721-\sqrt{8/9}\simeq0.0572, or 5.72%. For the proposed particle, sin⁡2α0=1/5\sin^2\alpha_0=1/5 exceeds 1/91/9. Its turning field is 5B05B_0, below the available 9B09B_0, so it is trapped in the stated orbit approximation. The fraction is an initial directional population measure; collisions, electrostatic potentials, and unequal ends would change the relevant physical loss problem.

Take it with you

  • Static mirror reflection exchanges parallel and perpendicular energy at fixed total energy.
  • Mirror ratio fixes a loss cone through sine squared of the pitch angle.
  • A directional trapped fraction is not an indefinite confinement guarantee or a steady escape rate.

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