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---
title: "Coulomb logarithm"
sort_title: "Coulomb logarithm"
date: 2021-10-03
categories:
- Physics
- Plasma physics
layout: "concept"
---
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 like so to the cross-section $$\sigma$$,
the total density $$n$$ of charged particles,
and the 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](/know/concept/debye-length/) $$\lambda_D$$,
i.e. the plasma's self-shielding length.
We thus find:
$$\begin{aligned}
\boxed{
\sigma_\mathrm{small}
= 8 \sigma_\mathrm{large} \ln(\Lambda)
= \frac{q_1^2 q_2^2 \ln(\Lambda)}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
}
\end{aligned}$$
Here, $$\ln(\Lambda)$$ is known as the **Coulomb logarithm**,
with the **plasma parameter** $$\Lambda$$ defined below,
equal to $$9/2$$ times the number of particles
in a sphere with radius $$\lambda_D$$:
$$\begin{aligned}
\boxed{
\Lambda
\equiv \frac{\lambda_D}{b_\mathrm{large}}
= 6 \pi n \lambda_D^3
}
\end{aligned}$$
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.
|
