Details

Considering a particle of charge and mass , subjected to an electric field

,

and a magnetic field

,

the resulting force is given by the Lorentz force:

.

In Cartesian coordinates the position vector is

,

then the velocity is

and the acceleration is

.

In this case, Newton's second law,

,

can be written as

.

Substituting the vectors, we obtain

.

Doing the cross products and rearranging terms, we obtain:

;

;

.

These are coupled second-order ordinary differential equations, which can be solved by either analytical or numerical methods. In this system, if , all three equations have the same solutions with constant acceleration. If , all three solutions have constant velocity. Considering , , , , and nonzero, the equation in the direction is the simplest to solve, since there are only constants. Rearranging, we obtain

,

which means that

,

and

,

where and are constants given by the initial conditions and , for velocity and position in the direction, respectively. The other two equations are coupled and thus must be solved together, which means they require a more complicated method to obtain their solutions.

It is possible to solve these equations by multiplying the one in the direction by the imaginary unit and adding it to the equation in the direction. Doing this leads to

.

Dividing this equation by and considering and , it is possible to write this equation as

,

which is a first-order linear differential equation for . By using the integrating factor method, its solution is

,

where is an arbitrary constant that can be determined by the initial condition . In fact, this constant is

,

so the solution is

.

It is possible to decouple the equations in the and directions by introducing the constant that obeys

and

.

This leads to

,

and using the relation , we obtain

and then

.

From the real part of this equation, we have

and from the imaginary part,

.

These equations can be solved by direct integration, which leads to

and

,

where and are arbitrary constants. By the initial conditions and , we obtain

and

.

Using these constants and remembering that , we obtain the final solutions

,

,

and

.

Thus the problem is solved analytically, to give solutions that are plotted in 3D Cartesian coordinates.

Reference

[1] J. D. Jackson, Classical Electrodynamics, 3rd ed., New York: Wiley, 1999.