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I understand the mathematical derivation of magnetic mirroring, which usually starts from the conservation of the magnetic dipole moment (e.g. in a plasma).

But physically: a mirrored particle is effectively coming to a halt and then starting to move again, in the opposite direction.

1) It loses kinetic energy whilst slowing down - where does this energy go?

2) It recovers it when it is fully reflected - where does this energy come from? where was it stored?

ACuriousMind
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SuperCiocia
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  • I think you should take a look at the actual motion of a mirrored particle and calculate the kinetic energy. You will find, that it is a constant. The motion only comes to a stop and reverses in one direction, but not perpendicular to it. A magnetic mirror reverses the momentum of a particle reflected by it, which, of course, is the same as the momentum transfer of a ball bouncing off a wall, except that the force here is generated by the magnetic field, rather than an atomic interaction. Where did the momentum go? It was imparted on the magnet creating the magnetic field. – CuriousOne Aug 18 '14 at 15:24
  • What makes a particle that is brought to a stop by a magnetic mirror different from a particle that is simply sitting stationary in a magnetic mirror? – SuperCiocia Sep 08 '14 at 22:50
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    @Harold - CuriousOne is correct, the kinetic energy, T, does not change here. The entire interaction is completely reversible and is considered to be much like an elastic collision. A somewhat careless (it's careless because energy is a scalar, not a vector) approach is to break up the T into parallel and perpendicular components with respect to the magnetic field. The total T = constant, but $T_{\parallel}$ and $T_{\perp}$ are not. Regarding your second question, a particle at rest with respect to a magnetic field experiences no force/acceleration. – honeste_vivere Oct 03 '14 at 16:12
  • @SuperCiocia - I wrote a detailed answer at http://physics.stackexchange.com/a/252885/59023 that explains mirroring in the context of Fermi acceleration. – honeste_vivere May 22 '16 at 13:24

1 Answers1

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To answer this question, we should have a look at the motion of a charged particle in a magnetic field: it basically gyrates around the magnetic field line where the gyration frequency $\omega$ depends on the magnetic field strength $B$: $$ \omega = q B / m, $$ with $q$ the particle's charge and $m$ its mass. Larger magnetic field strength results in a higher gyration frequency.

A simple answer to your question therefore is that with increasing magnetic field strength, the perpendicular (perpendicular to $\mathbf{B}$) velocity of the particle increases. To conserve energy, the parallel velocity has to decrease. And that might end up in the particle being reflected.

(In a bit more detail, one can explain it with Faraday's law being responsible for the parallel deceleration and perpendicular acceleration: the change of magnetic flux over the surface given by the particle's gyration induces an electric field and thus a force on the particle leading to the described behaviour.)

What puzzled you is that you described the particle's motion without taking into account the gyromotion, you only looked at its guiding center (something quite common in plasma physics).

Alf
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  • Sorry, I didn't get this: parallel velocity has to decrease. And that might end up in the particle being reflected.

    If I am not mistaken, this reflection is parallel to B. If parallel velocity reduces and becomes zero, then how is that the particle gets reflected?

    Your explanation invoking Faraday's law makes sense. This electric field will be present throughout the motion of that particle, correct?

     – sreeraj t Sep 08 '23 at 02:41