From e28d2a982d0c65fcad9a2d2a4c20d06a9848fa8f Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Tue, 5 Oct 2021 19:31:12 +0200
Subject: Expand knowledge base

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+---
+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.
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