Larmor radius in magnetic bottle/increasing magnetic field

Question

I am confused about how Larmor radius changes in an increasing magnetic field.

The Larmor radius means the radius of a charged particle’s circular motion in a magnetic field, its formula is

$r_{L} = {{m v_{\perp}}\over{q B}}$

where $v_{\perp}$ is the velocity component perpendicular to the magnetic field lines.

In my opinion, since a static magnetic field does no work, and the convering field lines has a decelerating effect on the parallel velocity component of the particle. From conservation of kinetic energy, as particle moving in an increasing magnetic field, $v_{\parallel}$ should decrease and $v_{\perp}$ should increase. Thus $r_{L}$ should increase.

However, I googled about this and got this graph, showing that in a magnetic bottle(field is stonger at both ends), the Larmor radius decreases where field is stronger:

Please explain to me how the Larmor radius would change in an increasing magnetic field, and why my derivation or the graph is different(or wrong).

Thank you for your time and discussion

Answer 1

In the situation of an axially symmetric magnetic field with $r$ and $z$ components that vary on length scales longer than the gyroradius $r_L$ then you can assume that the magnetic moment, $\mu$, of the particle is an adiabatic invariant, where $$ \mu =\frac{mv_\perp^2}{2B}\ , \tag*{(1)} $$ and $$ \frac{d\mu}{dt} = \frac{mv_\perp}{B}\frac{dv_\perp}{dt} - \frac{mv_\perp^2}{B^2} \frac{dB}{dt}= 0\ \tag*{(2)} $$ where $dB/dt$ is the changing B-field experienced by the particle as it moves along the z-axis.

Note this assumes non-relativistic motion, but the generalisation to relativistic motion does not change anything because the Lorentz factor, along with the total kinetic energy is an invariant, since static magnetic fields do no work on the charged particle. Thus $$\frac{dK}{dt} = mv_\parallel \frac{dv_\parallel}{dt} + mv_{\perp}\frac{dv_\perp}{dt} = 0\ . \tag*{(3)}$$

Equation (2) can be manipulated by setting $dB/dt = v_\parallel \partial B/\partial z$ which is just saying that if $v_\parallel$ is postive and the B-field increases as we move along the z-axis then the particle experiences an increasing magnetic field with time. Then from (2) we can write $$v_\perp \frac{dv_\perp}{dt} = \frac{v_\perp^2}{B} \frac{\partial B}{\partial z} v_\parallel $$ and hence, substituting this into (3) and using (1), that $$ \frac{dv_\parallel}{dt} = -\frac{2\mu}{m} \frac{\partial B}{\partial z}\ .$$ In this way, the parallel velocity decreases and may become negative - resulting in a reflection.

In terms of what happens to the speeds and gyroradii, we see that as $v_\parallel$ decreases, then $v_\perp$ must increase (reaching a maximum when $v_\parallel = 0$), but at the same time (from equation (1)), the B-field experienced by the particle must be increasing as $v_\perp^2$ in order that $\mu$ is an invariant. Then since the gyroradius is proportional to $v_\perp/B$, then $r_L$ must decrease.

Answer 2

This is a small supplement to ProfRob's answer.

As ProfRob correctly noted, a static magnetic field does not net work on charged particles, so the total kinetic energy must be conserved. One can state this as: $$ V_{o}^{2} = V_{\parallel, o}^{2} + V_{\perp, o}^{2} = V_{\parallel, f}^{2} + V_{\perp, f}^{2} = \text{constant} \tag{0} $$ where the subscript $o$($f$) corresponds to the initial(final) point of interest, $\parallel$($\perp$) is the direction parallel(perpendicular) with respect to the local quasi-static magnetic field vector, $\mathbf{B}$, and $V_{j}$ is the j^th component of the particle velocity.

If the magnetic field gradient is slow/gradual compared to the time necessary for the particle to gyrate about the magnetic field, then one can assume that the first adiabatic invariant of particle species $s$, $\mu_{s}$, is conserved (e.g., see https://physics.stackexchange.com/a/670591/59023) for details. Mathematically, this is written as: $$ \mu_{s} = \frac{ \gamma \ m_{s} \ V_{\perp}^{2} }{ 2 \ B } = \text{constant} \tag{1} $$ where $\gamma$ is the Lorentz factor, $m_{s}$ is the rest mass of particle species $s$, and $B$ is the local quasi-static magnetic field magnitude.

We know that the particle gyroradius is given by: $$ \rho_{s} = \frac{ \gamma \ m_{s} \ V_{\perp} }{ q_{s} \ B } \tag{2} $$ where $q_{s}$ is the charge of particle species $s$.

So if Equations 0 and 1 are true, then we can show that: $$ \frac{ V_{\perp, o}^{2} }{ B_{o} } = \frac{ V_{\perp, f}^{2} }{ B_{f} } = \text{constant} \tag{3} $$ because $\gamma$ = constant for constant kinetic energy. Using Equations 2 and 3, we can then show that: $$ \begin{align} V_{\perp} & \propto B^{1/2} \tag{4a} \\ \rightarrow \rho_{s} & \propto B^{-1/2} \tag{4b} \end{align} $$

Therefore, as the magnetic field increases(decreases), $\rho_{s}$ should decrease(increase).