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I understand that when a charged particle enters a magnetic field with a velocity $v$, with a angle $\theta$ within the field lines, its perpendicular and parallel velocity components generate, respectively, a circular and transverse motion, causing the particle to have a helical motion. However, in this development, it is often assumed that the magnetic field is uniform, and, having this premise established, it is not difficult to mathematically demonstrate why the particle describes a helical motion, one very nice and simple way is the first answer on this post: Helical motion of charged particle in external magnetic field.

But let's take the magnetic mirror as an example. In it, the magnetic field is variable, and yet the particle describes a helical movement along $\hat{z}$!

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"Geometrically", it's not difficult to understand, the particle still have the components of the velocity producing the respective motions, after all. But I've not been able to rigorously demonstrate this fact, mathematically.

Summarizing, I want to demonstrate, mathematically, why particles, in non-uniform magnetic field regions, continue to describe helical movements, when entering the field with a angle $\theta$.

honeste_vivere
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Johann Wagner
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2 Answers2

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It doesn't do exactly helical motion, since the idea of the magnetic mirror is that the particle eventually turns around and bounces between the ends. The statement is that if the B field is approximately constant, then we will get approximately helical motion. If it is uniform clearly we get helical motion, then the only part left is to see what the small force is that makes it not exactly helical. This if from a force along mirror axis, or $F_{z}$. The Lorentz force is $$\vec F=q \vec v \times \vec B$$ Assume $v_z \ll v_\perp$, where $v_\perp$ is the gyroscopic motion $v_\perp=r \omega_c$, for $\omega_c=qB/m$. Then if the $B$ tilts radially inward (it has to since $\nabla \cdot B=0$), or $\vec B = -B_r \hat r + B_z \hat z$, we get a force backwards $$\vec F = -q v_\perp B_r \hat z.$$ So there is a force away from the ends, but if this is very small, then the motion is helical.

If it is of interest: from a many step derivation, I get the orbit area, A, changes as $$\dot A = -3 \frac{1}{B_z}\frac{d}{dz} B_z v_z A,$$ so roughly speaking this analysis is OK as long as the area changes slowly compared to the orbital period, or $$\omega_c \gg \frac{1}{B_z}\frac{d}{dz} B_z v_z. $$

Zach Johnson
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Summarizing, I want to demonstrate, mathematically, why particles, in non-uniform magnetic field regions, continue to describe helical movements, when entering the field with a angle $\theta$.

As I showed/stated in https://physics.stackexchange.com/a/670591/59023 and https://physics.stackexchange.com/a/671056/59023, the critical piece of information is how fast the field changes relative to one of the periodic motions of the particle. If the gradient scale length is short and the field gradient strong, the particles can deviate from the usual circular gyration as illustrated in the movies at https://svs.gsfc.nasa.gov/4513. These are charged particles incident on a collisionless, magnetized shock wave that are undergoing something called shock drift acceleration or SDA. Under the right conditions, the particle orbits can actually trace out a cycloid. The particles are undergoing a gradient drift (e.g., see https://physics.stackexchange.com/a/556682/59023) while also being accelerated by an electric field due to a Lorentz transformation.

Regardless, the short answer to your question is just the Lorentz force. In the absence of an electric field, the particle only experiences a force orthogonal to both the magnetic field, $\mathbf{B}$, and the particle velocity, $\mathbf{v}$.

Now suppose there is a static electric field (and static magnetic field) present but that $\lvert \mathbf{B} \rvert > \lvert \mathbf{E} \rvert$. Under this scenario, the particle would merely ExB-drift (e.g., see https://physics.stackexchange.com/a/448523/59023) in a direction orthogonal to both $\mathbf{E}$ and $\mathbf{B}$. Conversely, if $\lvert \mathbf{B} \rvert < \lvert \mathbf{E} \rvert$ then there will be a reference frame where the particle is acted upon by a purely electrostatic field and the particle trajectory would be a hyperbolic path.

honeste_vivere
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  • Thank you! One question: In the specific case of magnetic mirrors, we have the grad-B drift. By the definition of guiding center, we can approximate the movement of a charged particle in this field as the superposition of fast circular motions around a point, having a "drift" associated with that point (continuation below) – Johann Wagner Oct 14 '21 at 00:52
  • The way I interpreted this in the situation of the mirror's variable magnetic field was as follows: When the field was uniform, we simply had the parallel and perpendicular components. Being variable, that's a curvature associated with the field lines. We still have a perpendicular component of the velocity, since the ortogonal planes where the circular motions occurs just start changing it's directions. But we do not have just a parallel component of the velocity, we have a grad-B drift associated with it, and it's velocity has a well-known formula. Is that a fair statement to make? – Johann Wagner Oct 14 '21 at 00:54
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    There is no grad-B drift here as the $\mathbf{B} \times \nabla B$ term is zero, i.e., $\mathbf{B}$ is parallel to the gradient in the magnetic field (both are along the $\hat{z}$ direction in your picture). – honeste_vivere Oct 14 '21 at 12:48
  • So how does the "non-orthogonal" component of the velocity behaves in a magnetic-mirror? – Johann Wagner Oct 14 '21 at 14:20
  • @JohannWagner - In a static magnetic field, the total kinetic energy is conserved as static magnetic fields cannot do work. So look at https://physics.stackexchange.com/a/671056/59023 and you'll see how the velocity components change. – honeste_vivere Oct 14 '21 at 14:30
  • @honeste_vivere is incorrect. There is grad-B drift in magnetic mirrors. In the middle, grad-B is radially inwards, and at the coils it is outwards, so there will be azimuthal drift. There is also curvature drift in the same direction. The plasma tries to straighten out the lines; in the center of the mirror, ions/electrons will move the exact same direction as their counterpart in the coils, and in the opposite direction at the ends of the mirror (at the coil). – Zach Johnson Oct 21 '21 at 17:43
  • To add more, what I said above assumes the most basic magnetic mirror. More realistic mirrors include extra currents (Ioffe bars) and increase the B field radially outward. This is to fix instabilities in the original configuration. – Zach Johnson Oct 21 '21 at 17:49
  • @ZachJohnson - In the center of an ideal magnetic mirror $\nabla B = 0$, so how would there be a gradient drift? In a real lab scenario, yes there may be a finite $\nabla B$ near the boundaries, but generally the experimenter wants there to be no $\nabla B$ except for very near the edge of the chamber. – honeste_vivere Oct 21 '21 at 18:10
  • I'm not sure where you get that. If there are just two coils then there will be non-zero gradient and gradient drift, though it will be proportional to the radial distance from center (obviously something dead center cannot drift azimuthally). Much more complicated currents are used in general, but I don't think the question is about this. – Zach Johnson Oct 21 '21 at 21:26
  • @ZachJohnson - I get this from thought experiments and actual experiments. In the IDEAL thought experiment, the only place a finite $\nabla B$ exists is along the axis of symmetry near the magnetic coils or mirror points. In a real lab setup, albeit a very good one, the magnetic field is held very constant up until very near the chamber walls. This is done precisely to avoid finite $\nabla B$ in the center of the chamber. – honeste_vivere Oct 25 '21 at 14:25