Landau Collision Operator

The fact that two-particle Coulomb collisions are dominated by small-angle scattering events allows some simplification of the Boltzmann collision operator in a plasma. According to Equations (3.26) and (3.70), the Boltzmann collision operator for two-body Coulomb collisions between species-$s$ particles (with mass $m_s$ and charge $e_s$) and species-$s'$ particles (with mass $m_{s'}$ and charge $e_{s'}$) can be written

$\displaystyle C_{ss'}= \int\!\!\int\!\!\int u_{ss'}\,\frac{d\sigma}{d{\mit\Omega}}\,(f_s'\,f_{s'}'-f_s\,f_{s'})\,
d^3{\bf v}_{s'}\,d{\mit\Omega},$ (3.71)

where

$\displaystyle \frac{d\sigma}{d{\mit\Omega}} = \frac{1}{4}\left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,\mu_{ss'}\,u_{ss'}^{2}}\right)^2 \frac{1}{\sin^4(\chi/2)}.$ (3.72)

Here, ${\bf u}_{ss'}$ is the relative velocity prior to a collision, and $d{\mit\Omega}= \sin\chi\,d\chi\,d\phi$, where $\chi$ is the angle of deflection, and $\phi$ is an azimuthal angle that determines the orientation of the plane in which a given two-body collision occurs. Recall that $f_s'$, $f_{s'}'$, $f_s$, and $f_{s'}$ are short-hand for $f_s({\bf r}, {\bf v}_s',t)$, $f_{s'}({\bf r}, {\bf v}_{s'}',t)$, $f_s({\bf r}, {\bf v}_s,t)$, and $f_{s'}({\bf r}, {\bf v}_{s'},t)$, respectively.

The species-$s$ and species-$s'$ particle velocities prior to the collision are ${\bf v}_s$ and ${\bf v}_{s'}$, respectively, so that ${\bf u}_{ss'}={\bf v}_s-{\bf v}_{s'}$. Let us write the corresponding velocities after the collision as (see Section 3.3)

$\displaystyle {\bf v}_s'$ $\displaystyle = {\bf v}_s + \frac{\mu_{ss'}}{m_s}\,{\bf g}_{ss'},$ (3.73)
$\displaystyle {\bf v}_{s'}'$ $\displaystyle = {\bf v}_{s'} -\frac{\mu_{ss'}}{m_{s'}}\,{\bf g}_{ss'}.$ (3.74)

Here, ${\bf g}_{ss'}= {\bf u}_{ss'}'-{\bf u}_{ss'}$ is assumed to be small, which implies that the angle of deflection is also small. Expanding $f_s'\equiv f_s({\bf r},{\bf v}_s', t)$ to second order in ${\bf g}_{ss'}$, we obtain

$\displaystyle f_s({\bf v}_s') \simeq f_s({\bf v}_s) + \frac{\mu_{ss'}}{m_s}\,{\...
...}_{ss'}: \frac{\partial^2 f_s({\bf v}_s)}{\partial {\bf v}_s\partial{\bf v}_s}.$ (3.75)

Likewise, expanding $f_{s'}'\equiv f_{s'}({\bf r},{\bf v}_{s'}', t)$, we get

$\displaystyle f_{s'}({\bf v}_{s'}') \simeq f_{s'}({\bf v}_s) - \frac{\mu_{ss'}}...
...ac{\partial^2 f_{s'}({\bf v}_{s'})}{\partial {\bf v}_{s'}\partial{\bf v}_{s'}}.$ (3.76)

Note that, in writing the previous two equations, we have neglected the ${\bf r}$ and $t$ dependence of $f_s({\bf r}, {\bf v}_s',t)$, et cetera, for ease of notation. Hence,

$\displaystyle f_s'\,f_{s'}' - f_s\,f_{s'}$ $\displaystyle \simeq \mu_{ss'}\,{\bf g}_{ss'}\cdot\left(
\frac{\partial f_s}{\p...
...{m_s} -\frac{f_s}{m_{s'}}\,\frac{\partial f_{s'}}{\partial {\bf v}_{s'}}\right)$    
  $\displaystyle \phantom{=}+\frac{1}{2}\,\mu_{ss'}^{2}\,{\bf g}_{ss'}{\bf g}_{ss'...
...l f_s}{\partial{\bf v}_s}\,\frac{\partial f_{s'}}{\partial{\bf v}_{s'}}\right).$ (3.77)

It follows that

$\displaystyle C_{ss'}$ $\displaystyle \simeq \frac{1}{4}\left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,\mu_...
...}_{ss'}\right]\frac{d^3{\bf v}_{s'}\,d{\mit\Omega}}{u_{ss'}^3\,\sin^4(\chi/2)},$    
    (3.78)

where

$\displaystyle {\bf J}_{ss'} = \frac{\partial f_s}{\partial {\bf v}_s}\,\frac{f_{s'}}{m_s} -\frac{f_s}{m_{s'}}\,\frac{\partial f_{s'}}{\partial {\bf v}_{s'}}.$ (3.79)

Let ${\bf l}$, ${\bf m}$, and ${\bf n}$ be a right-handed set of mutually orthogonal unit vectors. Suppose that ${\bf u}_{ss'}= u_{ss'}\,{\bf l}$. Recall that ${\bf u}_{ss'}'={\bf u}_{ss'}+{\bf g}_{ss'}$. Now, in an elastic collision for which the angle of deviation is $\chi$, we require $\vert{\bf u}_{ss'}'\vert=\vert{\bf u}_{ss'}\vert$, $\vert{\bf u}_{ss'}\times {\bf u}_{ss'}'\vert=u_{ss'}^2\,\sin\chi$, and ${\bf u}_{ss'}' = {\bf u}_{ss'}$ when $\chi=0$. In other words, we need $\vert{\bf g}_{ss'}+u_{ss'}\,{\bf l}\vert=u_{ss'}$, $\vert{\bf l}\times{\bf g}_{ss'}\vert=u_{ss'}\,\sin\chi$, and ${\bf g}_{ss'}={\bf0}$ when $\chi=0$. We deduce that

$\displaystyle {\bf g}_{ss'} \simeq u_{ss'}\left[(\cos\chi-1)\,{\bf l} + \sin\chi\,\cos\phi\,{\bf m} +\sin\chi\,\sin\phi\,{\bf n}\right].$ (3.80)

Thus,

$\displaystyle \int \frac{{\bf g}_{ss'}\,d{\mit\Omega}}{\sin^4(\chi/2)}$ $\displaystyle =\int_0^\pi\oint \frac{{\bf g}_{ss'}\,\sin\chi\,d\chi\,d\phi}{\si...
...}}{\sin^4(\chi/2)}=
-{\bf u}_{ss'} \int\frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)},$ (3.81)

and

$\displaystyle \int \frac{{\bf g}_{ss'}{\bf g}_{ss'}\,d{\mit\Omega}}{\sin^4(\chi/2)}$ $\displaystyle \simeq \frac{u_{ss'}^{2}}{2}\,({\bf m}{\bf m}+{\bf n}{\bf n})\int...
...u_{ss'}^2\,({\bf I}-{\bf l}{\bf l})\int\frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)},$ (3.82)

where use has again been made of the fact that $\chi$ is small.

Now,

$\displaystyle \int \frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)}=\int\frac{4\pi\,\sin...
...)}{\sin(\chi/2)}= 16\pi\,\ln\left(\frac{\chi_{\rm max}}{\chi_{\rm min}}\right),$ (3.83)

where $\chi_{\rm max}$ and $\chi_{\rm min}$ are the maximum and minimum angles of deflection, respectively. However, according to Equation (3.64), small-angle two-body Coulomb collisions are characterized by

$\displaystyle \chi\simeq \frac{e_s\,e_{s'}}{2\pi\,\epsilon_0\,\mu_{ss'}\,u_{ss'}^{2}\,b},$ (3.84)

where $b$ is the impact parameter. Thus, we can write

$\displaystyle \int \frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)}=16\pi\,\ln{\mit\Lambda}_c,$ (3.85)

where the quantity

$\displaystyle \ln {\mit\Lambda}_c = \ln\left(\frac{\chi_{\rm max}}{\chi_{\rm min}}\right)=\ln\left(\frac{b_{\rm max}}{b_{\rm min}}\right)$ (3.86)

is known as the Coulomb logarithm.

It follows from the previous analysis that

$\displaystyle C_{ss'}$ $\displaystyle = \left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,\mu_{ss'}}\right)^24...
...rac{\partial}{\partial {\bf v}_{s'}}\right){\bf J}_{ss'}\right]d^3{\bf v}_{s'}.$    
    (3.87)

If we define the tensor

$\displaystyle {\bf w}_{ss'} = \frac{{\bf I}-{\bf l}{\bf l}}{u_{ss'}} = \frac{u_{ss'}^2\,{\bf I}-{\bf u}_{ss'}{\bf u}_{ss'}}{u_{ss'}^3}$ (3.88)

then it is readily seen that

$\displaystyle \left(\frac{\partial}{\partial {\bf u}_{ss'}}\cdot{\bf w}_{ss'}\right)_\alpha$ $\displaystyle \equiv \frac{\partial}{\partial u_{ss'\,\beta}}\left(\frac{\delta...
...2}}\right)=-\frac{2\,u_{ss'\,\alpha}}{(u_{ss'\,\gamma}\,u_{ss'\,\gamma})^{3/2}}$    
  $\displaystyle = - \left(\frac{2\,{\bf l}}{u_{ss'}^{2}}\right)_\alpha.$ (3.89)

Here, $\alpha$, $\beta$, and $\gamma$ run from $1$ to $3$, and correspond to Cartesian components. Moreover, we have made use of the Einstein summation convention (that repeated indices are implicitly summed from $1$ to $3$) (Riley 1974). Hence, we deduce that

$\displaystyle C_{ss'}$ $\displaystyle = \left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,\mu_{ss'}}\right)^24...
...rac{\partial}{\partial {\bf v}_{s'}}\right){\bf J}_{ss'}\right]d^3{\bf v}_{s'}.$    
    (3.90)

Integration by parts yields

$\displaystyle C_{ss'}$ $\displaystyle = \left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,\mu_{ss'}}\right)^24...
...{\bf v}_{s'}}\cdot{\bf w}_{ss'}\right)\cdot{\bf J}_{ss'}\right]d^3{\bf v}_{s'}.$    
    (3.91)

However,

$\displaystyle \frac{\partial}{\partial {\bf v}_{s'}}\cdot{\bf w}_{ss'} = - \fra...
...'}}\cdot{\bf w}_{ss'} = -\frac{\partial}{\partial {\bf v}_s}\cdot{\bf w}_{ss'},$ (3.92)

because ${\bf w}_{ss'}$ is a function of ${\bf u}_{ss'}={\bf v}_s-{\bf v}_{s'}$. Thus, we obtain the so-called Landau collision operator (Landau 1936),

$\displaystyle C_{ss'} = \frac{\gamma_{ss'}}{m_s}\,\frac{\partial}{\partial {\bf v}_s}\cdot\int {\bf w}_{ss'}\cdot{\bf J}_{ss'}\,d^3{\bf v}_{s'},$ (3.93)

where

$\displaystyle \gamma_{ss'} = \left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0}\right)^2\,2\pi\,\ln{\mit\Lambda}_c.$ (3.94)