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1 files changed, 65 insertions, 117 deletions

diff --git a/content/know/concept/lorentz-force/index.pdc b/content/know/concept/lorentz-force/index.pdc
index 2362766..83357d2 100644
--- a/content/know/concept/lorentz-force/index.pdc
+++ b/content/know/concept/lorentz-force/index.pdc
@@ -5,6 +5,7 @@ publishDate: 2021-09-08
 categories:
 - Physics
 - Electromagnetism
+- Plasma physics
 
 date: 2021-09-08T17:00:32+02:00
 draft: false
@@ -27,6 +28,44 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+## 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
@@ -64,12 +103,12 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
-which is always positive:
+which may be negative:
 
 $$\begin{aligned}
     \boxed{
         \omega_c
-        \equiv \frac{|q| B}{m}
+        \equiv \frac{q B}{m}
     }
 \end{aligned}$$
 
@@ -78,7 +117,7 @@ the solution for $u_x(t)$ is given by:
 
 $$\begin{aligned}
     u_x(t)
-    = - u_\perp \sin\!(\omega_c t)
+    = u_\perp \cos\!(\omega_c t)
 \end{aligned}$$
 
 Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity.
@@ -87,13 +126,12 @@ Then $u_y(t)$ is found to be:
 $$\begin{aligned}
     u_y(t)
     = \frac{m}{q B} \dv{u_x}{t}
-    = - \frac{m \omega_c}{q B} u_\perp \cos\!(\omega_c t)
-    = - \mathrm{sgn}(q) \: u_\perp \cos\!(\omega_c t)
+    = - \frac{m \omega_c}{q B} u_\perp \sin\!(\omega_c t)
+    = - u_\perp \sin\!(\omega_c t)
 \end{aligned}$$
 
-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 $q$.
+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 $y_{gc}$
@@ -101,10 +139,10 @@ as the **guiding center**, around which the particle orbits or **gyrates**:
 
 $$\begin{aligned}
     x(t)
-    = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + x_{gc}
+    = \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + x_{gc}
     \qquad \quad
     y(t)
-    = - \mathrm{sgn}(q) \: \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + y_{gc}
+    = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + y_{gc}
 \end{aligned}$$
 
 The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:
@@ -112,16 +150,18 @@ The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_
 $$\begin{aligned}
     \boxed{
         r_L
-        \equiv \frac{u_\perp}{\omega_c}
+        \equiv \frac{u_\perp}{|\omega_c|}
         = \frac{m u_\perp}{|q| B}
     }
 \end{aligned}$$
 
-Finally, it is trivial to integrate the equation for the $z$-direction velocity $u_z$:
+Finally, it is easy to integrate the equation
+for the $z$-axis velocity $u_z$, which is conserved:
 
 $$\begin{aligned}
     z(t)
-    = u_z t + z_{gc}
+    = z_{gc}
+    = u_z t + z_0
 \end{aligned}$$
 
 In conclusion, the particle's motion parallel to $\vb{B}$
@@ -129,116 +169,24 @@ 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:
+when moving through a uniform magnetic field:
 
 $$\begin{aligned}
-    \dv{\vb{u}_\parallel}{t}
-    = \frac{q}{m} \vb{E}_\parallel
-\end{aligned}$$
-
-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 $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:
-
-$$\begin{aligned}
-    0
-    = \vb{B} \cross (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B})
-    = \vb{B} \cross \vb{E} + \vb{v}_\perp B^2
-    \quad \implies \quad
     \boxed{
-        \vb{v}_\perp
-        = \frac{\vb{E} \cross \vb{B}}{B^2}
+        \vb{x}(t)
+        = \frac{u_\perp}{\omega_c}
+        \begin{pmatrix}
+            \sin\!(\omega_c t) \\ \cos\!(\omega_c t) \\ 0
+        \end{pmatrix}
+        + \vb{x}_{gc}(t)
     }
 \end{aligned}$$
 
-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 $q$:
-
-$$\begin{aligned}
-    \boxed{
-        \vb{v}_d
-        = \frac{\vb{F} \cross \vb{B}}{q B^2}
-    }
-\end{aligned}$$
+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](/know/concept/guiding-center-theory/).