---
title: "Lorentz force"
firstLetter: "L"
publishDate: 2021-09-08
categories:
- Physics
- Electromagnetism

date: 2021-09-08T17:00:32+02:00
draft: false
markup: pandoc
---

# Lorentz force

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

$$\begin{aligned}
    \boxed{
        \vb{F}
        = q (\vb{E} + \vb{u} \cross \vb{B})
    }
\end{aligned}$$


## 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 $\dv*{u_y}{t}$ from the second,
we arrive at the following harmonic oscillator:

$$\begin{aligned}
    \dv[2]{u_x}{t} = - \omega_c^2 u_x
\end{aligned}$$

Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
which is always positive:

$$\begin{aligned}
    \boxed{
        \omega_c
        \equiv \frac{|q| B}{m}
    }
\end{aligned}$$

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

$$\begin{aligned}
    u_x(t)
    = - u_\perp \sin\!(\omega_c t)
\end{aligned}$$

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:

$$\begin{aligned}
    u_y(t)
    = \frac{m}{q B} \dv{u_x}{t}
    = - \frac{m \omega_c}{q B} u_\perp \cos\!(\omega_c t)
    = - \mathrm{sgn}(q) \: u_\perp \cos\!(\omega_c t)
\end{aligned}$$

Where $\mathrm{sgn}$ is the signum function.
This tells us that the particle moves in a circular orbit,
and that the direction of rotation is determined by $q$.

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**:

$$\begin{aligned}
    x(t)
    = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + x_{gc}
    \qquad \quad
    y(t)
    = - \mathrm{sgn}(q) \: \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + y_{gc}
\end{aligned}$$

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

$$\begin{aligned}
    \boxed{
        r_L
        \equiv \frac{u_\perp}{\omega_c}
        = \frac{m u_\perp}{|q| B}
    }
\end{aligned}$$

Finally, it is trivial to integrate the equation for the $z$-direction velocity $u_z$:

$$\begin{aligned}
    z(t)
    = u_z t + z_{gc}
\end{aligned}$$

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.


## Uniform electric and magnetic field

Let us now consider a more general case,
with constant uniform electric and magnetic fields $\vb{E}$ and $\vb{B}$,
which may or may not be perpendicular.
The equation of motion is then:

$$\begin{aligned}
    \vb{F}
    = m \dv{\vb{u}}{t}
    = q (\vb{E} + \vb{u} \cross \vb{B})
\end{aligned}$$

If we take the dot product with the unit vector $\vu{B}$,
the cross product vanishes, leaving:

$$\begin{aligned}
    \dv{\vb{u}_\parallel}{t}
    = \frac{q}{m} \vb{E}_\parallel
\end{aligned}$$

Where $\vb{u}_\parallel$ and $\vb{E}_\parallel$ are
the components of $\vb{u}$ and $\vb{E}$
that are parallel to $\vb{B}$.
This equation is easy to integrate:
the guiding center accelerates according to $(q/m) \vb{E}_\parallel$.

Next, let us define the perpendicular component $\vb{u}_\perp$
such that $\vb{u} = \vb{u}_\parallel \vu{B} + \vb{u}_\perp$.
Its equation of motion is found by
subtracting $\vb{u}_\parallel$'s equation from the original:

$$\begin{aligned}
    m \dv{\vb{u}_\perp}{t}
    = q (\vb{E} + \vb{u} \cross \vb{B}) - q \vb{E}_\parallel
    = q (\vb{E}_\perp + \vb{u}_\perp \cross \vb{B})
\end{aligned}$$

To solve this, we go to a moving coordinate system
by defining $\vb{u}_\perp = \vb{v}_\perp + \vb{w}_\perp$,
where $\vb{v}_\perp$ is a constant of our choice.
The equation is now as follows:

$$\begin{aligned}
    m \dv{t} (\vb{v}_\perp + \vb{w}_\perp)
    = m \dv{\vb{w}_\perp}{t}
    = q (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B} + \vb{w}_\perp \cross \vb{B})
\end{aligned}$$

We want to choose $\vb{v}_\perp$ such that the first two terms vanish,
or in other words:

$$\begin{aligned}
    0
    = \vb{E}_\perp + \vb{v}_\perp \cross \vb{B}
\end{aligned}$$

To find $\vb{v}_\perp$, we take the cross product with $\vb{B}$,
and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:

$$\begin{aligned}
    0
    = \vb{B} \cross (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B})
    = \vb{B} \cross \vb{E} + \vb{v}_\perp B^2
    \quad \implies \quad
    \boxed{
        \vb{v}_\perp
        = \frac{\vb{E} \cross \vb{B}}{B^2}
    }
\end{aligned}$$

When $\vb{v}_\perp$ is chosen like this,
the perpendicular equation of motion is reduced to:

$$\begin{aligned}
    m \dv{\vb{w}_\perp}{t}
    = q \vb{w}_\perp \cross \vb{B}
\end{aligned}$$

Which is simply the case we treated previously with $\vb{E} = 0$,
with a known solution
(assuming $\vb{B}$ still points in the positive $z$-direction):

$$\begin{aligned}
    w_x(t)
    = - w_\perp \sin\!(\omega_c t)
    \qquad
    w_y(t)
    = - \mathrm{sgn}(q) \: w_\perp \cos\!(\omega_c t)
\end{aligned}$$

However, this result is shifted by a constant $\vb{v}_\perp$,
often called the **drift velocity** $\vb{v}_d$,
at which the guiding center moves transversely.
Curiously, $\vb{v}_d$ is independent of $q$.

Such a drift is not specific to an electric field.
In the equations above, $\vb{E}$ can be replaced
by a general force $\vb{F}/q$ (e.g. gravity) without issues.
In that case, $\vb{v}_d$ does depend on $q$.

$$\begin{aligned}
    \boxed{
        \vb{v}_d
        = \frac{\vb{F} \cross \vb{B}}{q B^2}
    }
\end{aligned}$$



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