Physics Asked by Timofey Chernyshev on November 30, 2020

Potential is defined as ${phi,, 0,, A,, 0 }$; fields are static and depend only on the axial coordinate $x$: $E_x=-partial_xphi$, $B_z=partial_x A$. Charged particle moves in the ${x,y}$-plane, where $y$ is a direction of drift-motion.

There are two constants of motion:

$frac{m}{2} left(v_{x}^{2} + v_{y}^{2} – v_{x0}^{2} – v_{y0}^{2}right) -ephi $

and

$m left(v_{y} – v_{y0}right) – frac{A e}{c} $, where $v_{x0}$ and $v_{y0}$ are velocities at given initial point $x_0$ (where $phiequiv0$ and $Aequiv0$).

We assume that $phi$ and $A$ are monotonic. Thus, a solution for velocities can be expressed as

$$v_x = pmsqrt{- frac{A^{2} e^{2}}{c^{2} m^{2}} – frac{2 A e v_{y0}}{c m} + frac{2 e phi}{m} + v_{x0}^{2}}\v_y = frac{A e}{c m} + v_{y0}$$

And it works fine in the case of uniform fields. However, it gives multi-valued loops in some cases of nonuniform fields. Figures below show both the correct and incorrect examples. The only difference here is a selection of $x_0$.

So, the question is how to determine the incorrect branch of the solution without step-by-step calculation from $x_0$ to $x$?

In these examples: $phi(x) = {rm erf}(exp(10x))-phi_0$ and $A(x)=164int (1-x^2)^2dx – A_0$, $v_{x0}=v_{y0}=0$, position of incorrect solution is around $xapprox-0.335$.

The phase-loops are shown below for both solutions:

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