# Helical motion on non-uniform magnetic field

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 $$\hat{z}$$!

"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 $$\theta$$.

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 $$\omega_c \gg \frac{1}{B_z}\frac{d}{dz} B_z v_z.$$

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.

• 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) Oct 14 '21 at 0:52
• 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? Oct 14 '21 at 0:54
• 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). Oct 14 '21 at 12:48
• So how does the "non-orthogonal" component of the velocity behaves in a magnetic-mirror? 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 physics.stackexchange.com/a/671056/59023 and you'll see how the velocity components change. Oct 14 '21 at 14:30