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Next: Coulomb Logarithm Up: Collisions Previous: Rutherford Scattering Cross-Section

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 (3.23) and (3.88), the Boltzmann collision operator for two-body Coulomb collisions between particles of type $ 1$ (with mass $ m_1$ and charge $ e_1$ ) and particles of type $ 2$ (with mass $ m_2$ and charge $ e_2$ ) can be written

$\displaystyle C_{12}= \int\!\!\int\!\!\int u\,\frac{d\sigma}{d{\mit\Omega}}\,(f_1'\,f_2'-f_1\,f_2)\, d^3{\bf v}_2\,d{\mit\Omega},$ (3.89)


$\displaystyle \frac{d\sigma}{d{\mit\Omega}} = \frac{1}{4}\left(\frac{e_1\,e_2}{4\pi\,\epsilon_0\,\mu_{12}\,u^2}\right)^2 \frac{1}{\sin^4(\chi/2)}.$ (3.90)

Here, $ {\bf u}$ 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_1'$ , $ f_2'$ , $ f_1$ , and $ f_2$ are short-hand for $ f_1({\bf r}, {\bf v}_1',t)$ , $ f_2({\bf r}, {\bf v}_2',t)$ , $ f_1({\bf r}, {\bf v}_1,t)$ , and $ f_2({\bf r}, {\bf v}_2,t)$ , respectively. Finally, $ \mu_{12}=(1/m_1+1/m_2)^{-1}$ .

The type $ 1$ and type $ 2$ particle velocities prior to the collision are $ {\bf v}_1$ and $ {\bf v}_2$ , respectively, so that $ {\bf u}={\bf v_1}-{\bf v}_2$ . Let us write the corresponding velocities after the collision as (see Section 3.3)

$\displaystyle {\bf v}_1'$ $\displaystyle = {\bf v}_1 + \frac{\mu_{12}}{m_1}\,{\bf g},$ (3.91)
$\displaystyle {\bf v}_2'$ $\displaystyle = {\bf v}_2 -\frac{\mu_{12}}{m_2}\,{\bf g}.$ (3.92)

Here, $ {\bf g}= {\bf u}'-{\bf u}$ is assumed to be small, which implies that the angle of deflection is also small. Expanding $ f_1'\equiv f_1({\bf r},{\bf v}_1', t)$ to second order in $ {\bf g}$ , we obtain

$\displaystyle f_1({\bf v}_1') \simeq f_1({\bf v}_1) + \frac{\mu_{12}}{m_1}\,{\b...
...{\bf g}: \frac{\partial^2 f_1({\bf v}_1)}{\partial {\bf v}_1\partial{\bf v}_1}.$ (3.93)

Likewise, expanding $ f_2'\equiv f_2({\bf r},{\bf v}_2', t)$ , we get

$\displaystyle f_2({\bf v}_2') \simeq f_2({\bf v}_1) - \frac{\mu_{12}}{m_2}\,{\b...
...{\bf g}: \frac{\partial^2 f_2({\bf v}_2)}{\partial {\bf v}_2\partial{\bf v}_2}.$ (3.94)

Note that, in writing the previous two equations, we have neglected the $ {\bf r}$ and $ t$ dependence of $ f_1({\bf r}, {\bf v}_1',t)$ , et cetera, for ease of notation. Hence,

$\displaystyle f_1'\,f_2' - f_1\,f_2$ $\displaystyle \simeq \mu_{12}\,{\bf g}\cdot\left( \frac{\partial f_1}{\partial ...
...frac{f_2}{m_1} -\frac{f_1}{m_2}\,\frac{\partial f_2}{\partial {\bf v}_2}\right)$    
  $\displaystyle \phantom{=}+\frac{1}{2}\,\mu_{12}^{\,2}\,{\bf g}{\bf g}: \left( \...
...partial f_1}{\partial{\bf v}_1}\,\frac{\partial f_2}{\partial{\bf v}_2}\right).$ (3.95)

It follows that

$\displaystyle C_{12}$ $\displaystyle \simeq \frac{1}{4}\left(\frac{e_1\,e_2}{4\pi\,\epsilon_0\,\mu_{12...
...2}\right){\bf J}\right]\frac{d^3{\bf v}_2\,d{\mit\Omega}}{u^3\,\sin^4(\chi/2)},$    


$\displaystyle {\bf J} = \frac{\partial f_1}{\partial {\bf v}_1}\,\frac{f_2}{m_1} -\frac{f_1}{m_2}\,\frac{\partial f_2}{\partial {\bf v}_2}.$ (3.97)

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

$\displaystyle {\bf g} \simeq u\left[(\cos\chi-1)\,{\bf l} + \sin\chi\,\cos\phi\,{\bf m} +\sin\chi\,\sin\phi\,{\bf n}\right].$ (3.98)


$\displaystyle \int \frac{{\bf g}\,d{\mit\Omega}}{\sin^4(\chi/2)}$ $\displaystyle =\int_0^\pi\oint \frac{{\bf g}\,\sin\chi\,d\chi\,d\phi}{\sin^4(\chi/2)}= {\bf u} \int\frac{(\cos\chi-1)\,d{\mit\Omega}}{\sin^4(\chi/2)}$    
  $\displaystyle \simeq -{\bf u} \int\frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)},$ (3.99)


$\displaystyle \int \frac{{\bf g}{\bf g}\,d{\mit\Omega}}{\sin^4(\chi/2)}$ $\displaystyle \simeq \frac{u^2}{2}\,({\bf m}{\bf m}+{\bf n}{\bf n})\int\frac{\s...
...simeq u^2\,({\bf I}-{\bf l}{\bf l})\int\frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)},$ (3.100)

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


$\displaystyle \int \frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)}=\int\frac{4\pi\,\sin...
...)}{\sin(\chi/2)}= 16\pi\,\ln\left(\frac{\chi_{\rm max}}{\chi_{\rm min}}\right),$ (3.101)

where $ \chi_{\rm max}$ and $ \chi_{\rm min}$ are the maximum and minimum angles of deflection, respectively. However, according to Equation (3.82), small-angle two-body Coulomb collisions are characterized by

$\displaystyle \chi\simeq \frac{e_1\,e_2}{2\pi\,\epsilon_0\,\mu_{12}\,u^2\,b},$ (3.102)

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

$\displaystyle \int \frac{2\,d{\mit\Omega}}{\sin^2(\chi/2)}=16\pi\,\ln{\mit\Lambda}_c,$ (3.103)

where the quantity

$\displaystyle \ln {\mit\Lambda}_c = \ln\left(\frac{\chi_{\rm max}}{\chi_{\rm min}}\right)=\ln\left(\frac{b_{\rm max}}{b_{\rm min}}\right),$ (3.104)

is known as the Coulomb logarithm.

It follows from the previous analysis that

$\displaystyle C_{12}$ $\displaystyle = \left(\frac{e_1\,e_2}{4\pi\,\epsilon_0\,\mu_{12}}\right)^24\pi\...
...{1}{m_2}\,\frac{\partial}{\partial {\bf v}_2}\right){\bf J}\right]d^3{\bf v}_2.$    

If we define the tensor

$\displaystyle {\bf w} = \frac{{\bf I}-{\bf l}{\bf l}}{u} = \frac{u^2\,{\bf I}-{\bf u}{\bf u}}{u^3}$ (3.106)

then it is readily seen that

$\displaystyle \left(\frac{\partial}{\partial {\bf u}}\cdot{\bf w}\right)_i \equ...
...ght)=-\frac{2\,u_i}{(u_k\,u_k)^{3/2}}= - \left(\frac{2\,{\bf l}}{u^2}\right)_i.$ (3.107)

Here, $ i$ , $ j$ , et cetera, 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

$\displaystyle C_{12}$ $\displaystyle = \left(\frac{e_1\,e_2}{4\pi\,\epsilon_0\,\mu_{12}}\right)^24\pi\...
...{1}{m_2}\,\frac{\partial}{\partial {\bf v}_2}\right){\bf J}\right]d^3{\bf v}_2.$    

Integration by parts yields

$\displaystyle C_{12}$ $\displaystyle = \left(\frac{e_1\,e_2}{4\pi\,\epsilon_0\,\mu_{12}}\right)^24\pi\...
...partial}{\partial {\bf v}_2}\cdot{\bf w}\right)\cdot{\bf J}\right]d^3{\bf v}_2.$    


$\displaystyle \frac{\partial}{\partial {\bf v}_2}\cdot{\bf w} = - \frac{\partia...
...artial {\bf u}}\cdot{\bf w} = -\frac{\partial}{\partial {\bf v}_1}\cdot{\bf w},$ (3.110)

because $ {\bf w}$ is a function of $ {\bf u}= {\bf v}_1-{\bf v}_2$ . Thus, we obtain the so-called Landau collision operator (Laudau 1936),

$\displaystyle C_{12} = \frac{\gamma_{12}}{m_1}\,\frac{\partial}{\partial {\bf v}_1}\cdot\int {\bf w}\cdot{\bf J}\,d^3{\bf v}_2,$ (3.111)


$\displaystyle \gamma_{12} = \left(\frac{e_1\,e_2}{4\pi\,\epsilon_0}\right)^2\,2\pi\,\ln{\mit\Lambda}_c.$ (3.112)

It is sometimes convenient to write the Landau collision operator in the form

$\displaystyle C_{12} = -\frac{1}{m_1}\,\frac{\partial}{\partial {\bf v}_1}\cdot {\bf A}_{12},$ (3.113)


$\displaystyle {\bf A}_{12} = {\bf B}_{12}\,f_1-{\bf D}_{12}\cdot \frac{\partial f_1}{\partial {\bf v}_1},$ (3.114)


$\displaystyle {\bf B}_{12}$ $\displaystyle =\frac{\gamma_{12}}{m_2}\int {\bf w}\cdot\frac{\partial f_2}{\partial {\bf v}_2}\,d^3{\bf v}_2,$ (3.115)
$\displaystyle {\bf D}_{12}$ $\displaystyle =\frac{\gamma_{12}}{m_1}\int {\bf w}\,f_2\,d^3{\bf v}_2.$ (3.116)

next up previous
Next: Coulomb Logarithm Up: Collisions Previous: Rutherford Scattering Cross-Section
Richard Fitzpatrick 2016-01-23