Categories: Electromagnetism, Physics, Plasma physics.

# Lorentz force

The **Lorentz force** is an empirical force used to define
the electric field $\vb{E}$
and magnetic field $\vb{B}$.
For a particle with charge $q$ moving with velocity $\vb{u}$,
the Lorentz force $\vb{F}$ is given by:

## Uniform electric field

Consider the simple case of an electric field $\vb{E}$ that is uniform in all of space. In the absence of a magnetic field $\vb{B} = 0$ and any other forces, Newton’s second law states:

$\begin{aligned} \vb{F} = m \dv{\vb{u}}{t} = q \vb{E} \end{aligned}$This is straightforward to integrate in time, for a given initial velocity vector $\vb{u}_0$:

$\begin{aligned} \vb{u}(t) = \frac{q}{m} \vb{E} t + \vb{u}_0 \end{aligned}$And then the particle’s position $\vb{x}(t)$ is found be integrating once more, with $\vb{x}(0) = \vb{x}_0$:

$\begin{aligned} \boxed{ \vb{x}(t) = \frac{q}{2 m} \vb{E} t^2 + \vb{u}_0 t + \vb{x}_0 } \end{aligned}$In summary, unsurprisingly, a uniform electric field $\vb{E}$ accelerates the particle with a constant force $\vb{F} = q \vb{E}$. Note that the direction depends on the sign of $q$.

## Uniform magnetic field

Consider the simple case of a uniform magnetic field $\vb{B} = (0, 0, B)$ in the $z$-direction, without an electric field $\vb{E} = 0$. If there are no other forces, Newton’s second law states:

$\begin{aligned} \vb{F} = m \dv{\vb{u}}{t} = q \vb{u} \cross \vb{B} \end{aligned}$Evaluating the cross product yields three coupled equations for the components of $\vb{u}$:

$\begin{aligned} \dv{u_x}{t} = \frac{q B}{m} u_y \qquad \quad \dv{u_y}{t} = - \frac{q B}{m} u_x \qquad \quad \dv{u_z}{t} = 0 \end{aligned}$Differentiating the first equation with respect to $t$, and substituting $\idv{u_y}{t}$ from the second, we arrive at the following harmonic oscillator:

$\begin{aligned} \dvn{2}{u_x}{t} = - \omega_c^2 u_x \end{aligned}$Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
which may be negative:

Suppose we choose our initial conditions so that the solution for $u_x(t)$ is given by:

$\begin{aligned} u_x(t) = u_\perp \cos(\omega_c t) \end{aligned}$Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity. Then $u_y(t)$ is found to be:

$\begin{aligned} u_y(t) = \frac{m}{q B} \dv{u_x}{t} = - \frac{m \omega_c}{q B} u_\perp \sin(\omega_c t) = - u_\perp \sin(\omega_c t) \end{aligned}$This means that the particle moves in a circle, in a direction determined by the sign of $\omega_c$.

Integrating the velocity yields the position,
where we refer to the integration constants $x_{gc}$ and $y_{gc}$
as the **guiding center**, around which the particle orbits or **gyrates**:

The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:

Finally, it is easy to integrate the equation for the $z$-axis velocity $u_z$, which is conserved:

$\begin{aligned} z(t) = z_{gc} = u_z t + z_0 \end{aligned}$In conclusion, the particle’s motion parallel to $\vb{B}$ is not affected by the magnetic field, while its motion perpendicular to $\vb{B}$ is circular around an imaginary guiding center. The end result is that particles follow a helical path when moving through a uniform magnetic field:

$\begin{aligned} \boxed{ \vb{x}(t) = \frac{u_\perp}{\omega_c} \begin{pmatrix} \sin(\omega_c t) \\ \cos(\omega_c t) \\ 0 \end{pmatrix} + \vb{x}_{gc}(t) } \end{aligned}$Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$ is the position of the guiding center. For a detailed look at how $\vb{B}$ and $\vb{E}$ can affect the guiding center’s motion, see guiding center theory.

## References

- F.F. Chen,
*Introduction to plasma physics and controlled fusion*, 3rd edition, Springer.