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+---
+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.