Expand knowledge base

7 files changed, 479 insertions, 6 deletions

diff --git a/content/know/concept/coulomb-logarithm/index.pdc b/content/know/concept/coulomb-logarithm/index.pdc
new file mode 100644
index 0000000..649806b
--- /dev/null
+++ b/content/know/concept/coulomb-logarithm/index.pdc
@@ -0,0 +1,199 @@
+---
+title: "Coulomb logarithm"
+firstLetter: "C"
+publishDate: 2021-10-03
+categories:
+- Physics
+- Plasma physics
+
+date: 2021-09-23T16:22:18+02:00
+draft: false
+markup: pandoc
+---
+
+# Coulomb logarithm
+
+In a plasma, particles often appear to collide,
+although actually it is caused by Coulomb forces,
+i.e. the "collision" is in fact [Rutherford scattering](/know/concept/rutherford-scattering/).
+In any case, the particles' paths are deflected,
+and it would be nice to know
+whether those deflections are usually large or small.
+
+Let us choose $\pi/2$ as an example of a large deflection angle.
+Then Rutherford predicts:
+
+$$\begin{aligned}
+    \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu b_\mathrm{large}}
+    = \tan\!\Big( \frac{\pi}{4} \Big)
+    = 1
+\end{aligned}$$
+
+Isolating this for the impact parameter $b_\mathrm{large}$
+then yields an effective radius of a particle:
+
+$$\begin{aligned}
+    b_\mathrm{large}
+    = \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu}
+\end{aligned}$$
+
+Therefore, the collision cross-section $\sigma_\mathrm{large}$
+for large deflections can be roughly estimated as
+the area of a disc with radius $b_\mathrm{large}$:
+
+$$\begin{aligned}
+    \sigma_\mathrm{large}
+    = \pi b_\mathrm{large}^2
+    = \frac{q_1^2 q_2^2}{16 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
+\end{aligned}$$
+
+Next, we want to find the cross-section for small deflections.
+For sufficiently small angles $\theta$,
+we can Taylor-expand the Rutherford scattering formula to first order:
+
+$$\begin{aligned}
+    \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu b}
+    = \tan\!\Big( \frac{\theta}{2} \Big)
+    \approx \frac{\theta}{2}
+    \quad \implies \quad
+    \theta
+    \approx \frac{q_1 q_2}{2 \pi \varepsilon_0 |\vb{v}|^2 \mu b}
+\end{aligned}$$
+
+Clearly, $\theta$ is inversely proportional to $b$.
+Intuitively, we know that a given particle in a uniform plasma
+always has more "distant" neighbours than "close" neighbours,
+so we expect that small deflections (large $b$)
+are more common than large deflections.
+
+That said, many small deflections can add up to a large total.
+They can also add up to zero,
+so we should use random walk statistics.
+We now ask: how many $N$ small deflections $\theta_n$
+are needed to get a large total of, say, $1$ radian?
+
+$$\begin{aligned}
+    \sum_{n = 1}^N \theta_n^2 \approx 1
+\end{aligned}$$
+
+Traditionally, $1$ is chosen instead of $\pi/2$ for convenience.
+We are only making rough estimates,
+so those two angles are close enough for our purposes.
+Furthermore, the end result will turn out to be logarithmic,
+and is thus barely affected by this inconsistency.
+
+You can easily convince yourself
+that the average time $\tau$ between "collisions"
+is related as follows to the cross-section $\sigma$,
+the density $n$, and relative velocity $|\vb{v}|$:
+
+$$\begin{aligned}
+    \frac{1}{\tau}
+    = n |\vb{v}| \sigma
+    \qquad \implies \qquad
+    1
+    = n |\vb{v}| \tau \sigma
+\end{aligned}$$
+
+Therefore, in a given time interval $t$,
+the expected number of collision $N_b$
+for impact parameters between $b$ and $b\!+\!\dd{b}$
+(imagine a ring with these inner and outer radii)
+is given by:
+
+$$\begin{aligned}
+    N_b
+    = n |\vb{v}| t \: \sigma_b
+    = n |\vb{v}| t \:(2 \pi b \dd{b})
+\end{aligned}$$
+
+In this time interval $t$,
+we can thus turn our earlier sum
+into an integral of $N_b$ over $b$:
+
+$$\begin{aligned}
+    1
+    \approx \sum_{n = 1}^N \theta_n^2
+    = \int N_b \:\theta^2 \dd{b}
+    = n |\vb{v}| t \int 2 \pi \theta^2 b \dd{b}
+\end{aligned}$$
+
+Using the formula $n |\vb{v}| \tau \sigma = 1$,
+we thus define $\sigma_{small}$ as the effective cross-section
+needed to get a large deflection (of $1$ radian),
+with an average period $t$:
+
+$$\begin{aligned}
+    \sigma_\mathrm{small}
+    = \int 2 \pi \theta^2 b \dd{b}
+    = \int \frac{2 \pi q_1^2 q_2^2}{4 \pi^2 \varepsilon_0^2 |\vb{v}|^4 \mu^2 b^2} b \dd{b}
+\end{aligned}$$
+
+Where we have replaced $\theta$ with our earlier Taylor expansion.
+Here, we recognize $\sigma_\mathrm{large}$:
+
+$$\begin{aligned}
+    \sigma_\mathrm{small}
+    = \frac{q_1^2 q_2^2}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2} \int \frac{1}{b} \dd{b}
+    = 8 \sigma_\mathrm{large} \int \frac{1}{b} \dd{b}
+\end{aligned}$$
+
+But what are the integration limits?
+We know that the deflection grows for smaller $b$,
+so it would be reasonable to choose $b_\mathrm{large}$ as the lower limit.
+For very large $b$, the plasma shields the particles from each other,
+thereby nullifying the deflection,
+so as upper limit
+we choose the Debye length $\lambda_D$,
+i.e. the plasma's self-shielding length.
+We thus find:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_\mathrm{small}
+        = 8 \ln\!(\Lambda) \sigma_\mathrm{large}
+        = \frac{q_1^2 q_2^2 \ln\!(\Lambda)}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
+    }
+\end{aligned}$$
+
+Here, $\ln\!(\Lambda)$ is known as the **Coulomb logarithm**,
+with $\Lambda$ defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \Lambda
+        \equiv \frac{\lambda_D}{b_\mathrm{large}}
+    }
+\end{aligned}$$
+
+The above relation between $\sigma_\mathrm{small}$ and $\sigma_\mathrm{large}$
+gives us an estimate of how much more often
+small deflections occur, compared to large ones.
+In a typical plasma, $\ln\!(\Lambda)$ is between 6 and 25,
+such that $\sigma_\mathrm{small}$ is 2-3 orders of magnitude larger than $\sigma_\mathrm{large}$.
+
+Note that $t$ is now fixed as the period
+for small deflections to add up to $1$ radian.
+In more useful words, it is the time scale
+for significant energy transfer between partices:
+
+$$\begin{aligned}
+    \frac{1}{t}
+    = n |\vb{v}| \sigma_\mathrm{small}
+    = \frac{q_1^2 q_2^2 \ln\!(\Lambda) \: n}{2 \pi \varepsilon_0^2 \mu^2 |\vb{v}|^3}
+    \sim \frac{n}{T^{3/2}}
+\end{aligned}$$
+
+Where we have used that $|\vb{v}| \propto \sqrt{T}$, for some temperature $T$.
+Consequently, in hotter plasmas, there is less energy transfer,
+meaning that a hot plasma is hard to heat up further.
+
+
+
+## References
+1.  P.M. Bellan,
+    *Fundamentals of plasma physics*,
+    1st edition, Cambridge.
+2.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/elastic-collision/index.pdc b/content/know/concept/elastic-collision/index.pdc
new file mode 100644
index 0000000..9873144
--- /dev/null
+++ b/content/know/concept/elastic-collision/index.pdc
@@ -0,0 +1,160 @@
+---
+title: "Elastic collision"
+firstLetter: "E"
+publishDate: 2021-10-04
+categories:
+- Physics
+- Classical mechanics
+
+date: 2021-09-23T16:22:39+02:00
+draft: false
+markup: pandoc
+---
+
+# Elastic collision
+
+In an **elastic collision**,
+the sum of the colliding objects' kinetic energies
+is the same before and after the collision.
+In contrast, in an **inelastic collision**,
+some of that energy is converted into another form,
+for example heat.
+
+
+## One dimension
+
+In 1D, not only the kinetic energy is conserved, but also the total momentum.
+Let $v_1$ and $v_2$ be the initial velocities of objects 1 and 2,
+and $v_1'$ and $v_2'$ their velocities afterwards:
+
+$$\begin{aligned}
+    \begin{cases}
+        \quad\! m_1 v_1 +\quad m_2 v_2
+        = \quad\, m_1 v_1' +\quad m_2 v_2'
+        \\
+        \displaystyle\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2
+        = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2
+    \end{cases}
+\end{aligned}$$
+
+After some rearranging,
+these two equations can be written as follows:
+
+$$\begin{aligned}
+    \begin{cases}
+        m_1 (v_1 - v_1')
+        \qquad\quad\:\;\; = m_2 (v_2' - v_2)
+        \\
+        m_1 (v_1 - v_1') (v_1 + v_1')
+        = m_2 (v_2' - v_2) (v_2 + v_2')
+    \end{cases}
+\end{aligned}$$
+
+Using the first equation to replace $m_1 (v_1 \!-\! v_1')$
+with $m_2 (v_2 \!-\! v_2')$ in the second:
+
+$$\begin{aligned}
+    m_2 (v_1 + v_1') (v_2' - v_2)
+    = m_2 (v_2 + v_2') (v_2' - v_2)
+\end{aligned}$$
+
+Dividing out the common factors
+then leads us to a simplified system of equations:
+
+$$\begin{aligned}
+    \begin{cases}
+        \qquad\;\; v_1 + v_1'
+        = v_2 + v_2'
+        \\
+        m_1 v_1 + m_2 v_2
+        = m_1 v_1' + m_2 v_2'
+    \end{cases}
+\end{aligned}$$
+
+Note that the first relation is equivalent to $v_1 - v_2 = v_2' - v_1'$,
+meaning that the objects' relative velocity
+is reversed by the collision.
+Moving on, we replace $v_1'$ in the second equation:
+
+$$\begin{aligned}
+    m_1 v_1 + m_2 v_2
+    &= m_1 (v_2 + v_2' - v_1) + m_2 v_2'
+    \\
+    (m_1 + m_2) v_2'
+    &= 2 m_1 v_1 + (m_2 - m_1) v_2
+\end{aligned}$$
+
+Dividing by $m_1 + m_2$,
+and going through the same process for $v_1'$,
+we arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            v_1'
+            &= \frac{(m_1 - m_2) v_1 + 2 m_2 v_2}{m_1 + m_2}
+            \\
+            v_2'
+            &= \frac{2 m_1 v_1 + (m_2 - m_1) v_2}{m_1 + m_2}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+To analyze this result,
+for practicality, we simplify it by setting $v_2 = 0$.
+In that case:
+
+$$\begin{aligned}
+    v_1'
+    = \frac{(m_1 - m_2) v_1}{m_1 + m_2}
+    \qquad \quad
+    v_2'
+    = \frac{2 m_1 v_1}{m_1 + m_2}
+\end{aligned}$$
+
+How much of its energy and momentum does object 1 transfer to object 2?
+The following ratios compare $v_1$ and $v_2'$ to quantify the transfer:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    = \frac{2 m_2}{m_1 + m_2}
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    = \frac{4 m_1 m_2}{(m_1 + m_2)^2}
+\end{aligned}$$
+
+If $m_1 = m_2$, both ratios reduce to $1$,
+meaning that all energy and momentum is transferred,
+and object 1 is at rest after the collision.
+Newton's cradle is an example of this.
+
+If $m_1 \ll m_2$, object 1 simply bounces off object 2,
+barely transferring any energy.
+Object 2 ends up with twice object 1's momentum,
+but $v_2'$ is very small and thus negligible:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    \approx 2
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    \approx \frac{4 m_1}{m_2}
+\end{aligned}$$
+
+If $m_1 \gg m_2$, object 1 barely notices the collision,
+so not much is transferred to object 2:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    \approx \frac{2 m_2}{m_1}
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    \approx \frac{4 m_2}{m_1}
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/hookes-law/index.pdc b/content/know/concept/hookes-law/index.pdc
index 94ceb1b..6a7ccb5 100644
--- a/content/know/concept/hookes-law/index.pdc
+++ b/content/know/concept/hookes-law/index.pdc
@@ -43,7 +43,7 @@ In light of this fact, we replace the traditional spring
 with a rod of length $L$ and cross-section $A$.
 
 The constant $k$ depends on, among several things,
-the spring's length $L$ and cross section $A$,
+the spring's length $L$ and cross-section $A$,
 so for our generalization, we want a new parameter
 to describe the proportionality independently of the rod's dimensions.
 To achieve this, we realize that the force $F$ is spread across $A$,
diff --git a/content/know/concept/maxwell-bloch-equations/index.pdc b/content/know/concept/maxwell-bloch-equations/index.pdc
index 3a0df1b..ae7d119 100644
--- a/content/know/concept/maxwell-bloch-equations/index.pdc
+++ b/content/know/concept/maxwell-bloch-equations/index.pdc
@@ -14,7 +14,7 @@ markup: pandoc
 
 # Maxwell-Bloch equations
 
-For an electron in a two level system with time-independent states
+For an electron in a two-level system with time-independent states
 $\ket{g}$ (ground) and $\ket{e}$ (excited),
 consider the following general solution
 to the full Schrödinger equation:
@@ -391,10 +391,10 @@ $$\begin{aligned}
 
 It is trivial to show that $\vb{E}$ and $\vb{P}$
 can be replaced by $\vb{E}^{+}$ and $\vb{P}^{+}$.
-It is equally trivial to convert
+It is also simple to convert
 the dipole $\vb{p}^{+}$ and inversion $d$
 into their macroscopic versions $\vb{P}^{+}$ and $D$,
-simply by summing over all atoms in the medium.
+simply by averaging over the atoms per unit of volume.
 We thus arrive at the **Maxwell-Bloch equations**:
 
 $$\begin{aligned}
diff --git a/content/know/concept/maxwell-boltzmann-distribution/index.pdc b/content/know/concept/maxwell-boltzmann-distribution/index.pdc
index 38b56fd..3328eaf 100644
--- a/content/know/concept/maxwell-boltzmann-distribution/index.pdc
+++ b/content/know/concept/maxwell-boltzmann-distribution/index.pdc
@@ -20,7 +20,7 @@ probability distributions with applications in classical statistical physics.
 
 ## Velocity vector distribution
 
-In the canonical ensemble
+In the [canonical ensemble](/know/concept/canonical-ensemble/)
 (where a fixed-size system can exchange energy with its environment),
 the probability of a microstate with energy $E$ is given by the Boltzmann distribution:
 
@@ -31,7 +31,7 @@ $$\begin{aligned}
 
 Where $\beta = 1 / k_B T$.
 We split $E = K + U$,
-where $K$ and $U$ are the total kinetic and potential energy contributions.
+with $K$ and $U$ the total kinetic and potential energy contributions.
 If there are $N$ particles in the system,
 with positions $\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$
 and momenta $\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$,
diff --git a/content/know/concept/rutherford-scattering/index.pdc b/content/know/concept/rutherford-scattering/index.pdc
index 481e4d1..81bb133 100644
--- a/content/know/concept/rutherford-scattering/index.pdc
+++ b/content/know/concept/rutherford-scattering/index.pdc
@@ -4,6 +4,7 @@ firstLetter: "R"
 publishDate: 2021-10-02
 categories:
 - Physics
+- Plasma physics
 
 date: 2021-09-23T16:22:07+02:00
 draft: false
@@ -231,6 +232,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
+In fact, this formula is also valid if $q_1$ and $q_2$ have opposite signs;
+in that case particle 2 is simply located on the other side
+of particle 1's trajectory.
+
 
 
 ## References
diff --git a/content/know/concept/spitzer-resistivity/index.pdc b/content/know/concept/spitzer-resistivity/index.pdc
new file mode 100644
index 0000000..6ceed8d
--- /dev/null
+++ b/content/know/concept/spitzer-resistivity/index.pdc
@@ -0,0 +1,109 @@
+---
+title: "Spitzer resistivity"
+firstLetter: "S"
+publishDate: 2021-10-05
+categories:
+- Physics
+- Plasma physics
+
+date: 2021-10-04T14:47:44+02:00
+draft: false
+markup: pandoc
+---
+
+# Spitzer resistivity
+
+If an [electric field](/know/concept/electric-field/)
+with magnitude $E$ is applied to the plasma, the electrons experience
+a [Lorentz force](/know/concept/lorentz-force/) $q_e E$
+(we neglect the ions due to their mass),
+where $q_e$ is the electron charge.
+
+However, collisions slow them down while they travel through the plasma.,
+This can be modelled as a drag force $f_{ei} m_e v_e$,
+where $f_{ei}$ is the electron-ion collision frequency
+(we neglect $f_{ee}$ since all electrons are moving together),
+$m_e$ is their mass,
+and $v_e$ their typical velocity.
+Balancing the two forces yields the following relation:
+
+$$\begin{aligned}
+    q_e E
+    = f_{ei} m_e v_e
+\end{aligned}$$
+
+Using that the current density $J = q_e n_e v_e$,
+we can rearrange this like so:
+
+$$\begin{aligned}
+    E
+    = f_{ei} m_e \frac{J}{n_e q_e^2}
+    = \frac{m_e f_{ei}}{n_e q_e^2} J
+    = \eta J
+\end{aligned}$$
+
+This is Ohm's law, where $\eta$ is the resistivity.
+From our derivation of the [Coulomb logarithm](/know/concept/coulomb-logarithm/) $\ln\!(\Lambda)$,
+we estimate $f_{ei}$ to be as follows,
+where $n_i$ is the ion density,
+$\sigma$ is the collision cross-section,
+and $\mu$ is the [reduced mass](/know/concept/reduced-mass/)
+of the electron-ion system:
+
+$$\begin{aligned}
+    f_{ei}
+    = n_i \sigma v_e
+    = \frac{1}{2 \pi} \Big( \frac{q_e q_i}{\varepsilon_0 \mu} \Big)^2 \frac{n_i}{v_e^3} \ln\!(\Lambda)
+    \approx \frac{1}{2 \pi} \frac{Z q_e^4}{\varepsilon_0^2 m_e^2} \frac{n_e}{v_e^3} \ln\!(\Lambda)
+\end{aligned}$$
+
+Where we used that $\mu \approx m_e$,
+and $q_i = -Z q_e$ for some ionization $Z$,
+and as a result $n_e \approx Z n_i$ due to the plasma's quasi-neutrality.
+Beware: authors disagree about the constant factors in $f_{ei}$;
+recall that it was derived from fairly rough estimates.
+This article follows Bellan.
+
+Inserting this expression for $f_{ei}$ into
+the so-called **Spitzer resistivity** $\eta$ then yields:
+
+$$\begin{aligned}
+    \boxed{
+        \eta
+        = \frac{m_e f_{ei}}{n_e q_e^2}
+        = \frac{1}{2 \pi} \frac{Z q_e^2}{\varepsilon_0^2 m_e} \frac{1}{v_e^3} \ln\!(\Lambda)
+    }
+\end{aligned}$$
+
+A reasonable estimate for the typical velocity $v_e$
+at thermal equilibrium is as follows,
+where $k_B$ is Boltzmann's constant,
+and $T_e$ is the electron temperature:
+
+$$\begin{aligned}
+    \frac{1}{2} m_e v_e^2
+    = \frac{3}{2} k_B T_e
+    \quad \implies \quad
+    v_e
+    = \sqrt{\frac{3 k_B T_e}{m_e}}
+\end{aligned}$$
+
+Other choices exist,
+see e.g. the [Maxwell-Boltzmann distribution](/know/concept/maxwell-boltzmann-distribution/),
+but always $v_e \propto \sqrt{T_e/m_e}$.
+Inserting this $v_e$ into $\eta$ then gives:
+
+$$\begin{aligned}
+    \eta
+    = \frac{1}{6 \pi \sqrt{3}} \frac{Z q_e^2 \sqrt{m_e}}{\varepsilon_0^2 (k_B T_e)^{3/2}} \ln\!(\Lambda)
+\end{aligned}$$
+
+
+
+## References
+1.  P.M. Bellan,
+    *Fundamentals of plasma physics*,
+    1st edition, Cambridge.
+2.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.