Rosenbluth Potentials

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

$$C_{ss'} = -\frac{1}{m_s}\,\frac{\partial}{\partial {\bf v}_s}\cdot {\bf A}_{ss'},$$ (3.95)

where

$${\bf A}_{ss'} = {\bf B}_{ss'}\,f_s-{\bf D}_{ss'}\cdot \frac{\partial f_s}{\partial {\bf v}_s},$$ (3.96)

and

$${\bf B}_{ss'} =\frac{\gamma_{ss'}}{m_{s'}}\int {\bf w}_{ss'}\cdot\frac{\partial f_{s'}}{\partial {\bf v}_{s'}}\,d^3{\bf v}_{s'},$$ (3.97)

$${\bf D}_{ss'} =\frac{\gamma_{ss'}}{m_s}\int {\bf w}_{ss'}\,f_{s'}\,d^3{\bf v}_{s'}.$$ (3.98)

Let

$$G_{s'}({\bf v}_s) = \int u_{ss'}\,f_{s'}\,d^3{\bf v}_{s'},$$ (3.99)

$$H_{s'}({\bf v}_s) = \int u_{ss'}^{-1}\,f_{s'}\,d^3{\bf v}_{s'}.$$ (3.100)

Now, from Equation (3.88),

$$w_{ss'\,\alpha\beta} = \frac{\delta_{\alpha\beta}}{u_{ss'}} - \frac{u_{ss'\,\alpha}\,u_{ss'\,\beta}}{u_{ss'}^{3}}.$$ (3.101)

Moreover,

$$\frac{\partial u_{ss'}}{\partial u_{ss'\,\alpha}} = \frac{u_{ss'\,\alpha}}{u_{ss'}},$$ (3.102)

$$\frac{\partial u_{ss'\,\alpha}}{\partial u_{ss'\,\beta}} = \delta_{\alpha\beta}.$$ (3.103)

Hence, it is easily demonstrated that

$$w_{ss'\,\alpha\beta} = \frac{\partial^2 u_{ss'}}{\partial u_{ss'\,\alpha}\,\partial u_{ss'\,\beta}},$$ (3.104)

$$\frac{\partial w_{ss'\,\alpha\beta}}{\partial u_{ss'\,\beta}} = \frac{\partial w_{ss'\,\beta\beta}}{\partial u_{ss'\,\alpha}} = 2\,\frac{\partial}{\partial u_{ss' \,\alpha}}\left(\frac{1}{u_{ss'}}\right).$$ (3.105)

According to Equations (3.97) and (3.98),

$${\bf B}_{ss'} =\frac{\gamma_{ss'}}{m_{s'}}\int {\bf w}_{ss'}\cdot\frac{\partial... ...{\partial}{\partial {\bf u}_{ss'}}\cdot {\bf w}_{ss'}\,f_{s'}\,d^3{\bf v}_{s'},$$ (3.106)

$${\bf D}_{ss'} =\frac{\gamma_{ss'}}{m_s}\int {\bf w}_{ss'}\,f_{s'}\,d^3{\bf v}_{s'},$$ (3.107)

where we have integrated the first equation by parts, making use of Equation (3.92). Thus, we deduce from Equations (3.104) and (3.105) that

$${\bf B}_{ss'} =\frac{2\,\gamma_{ss'}}{m_{s'}}\,\frac{\partial H_{s'}}{\partial {\bf v}_s},$$ (3.108)

$${\bf D}_{ss'} = \frac{\gamma_{ss'}}{m_s}\,\frac{\partial^2 G_{s'}}{\partial {\bf v}_s\partial{\bf v}_s}.$$ (3.109)

The quantities $H_{s'}({\bf v})$ and $G_{s'}({\bf v})$ are known as Rosenbluth potentials (Rosenbluth, MacDonald, and Judd 1957), and can easily be seen to satisfy

$$\nabla^{2}_v H_{s'} = -4\pi\,f_{s'}({\bf v}),$$ (3.110)

$$\nabla^{2}_v G_{s'} = 2\,H_{s'}({\bf v}),$$ (3.111)

where $\nabla^{2}_v$ denotes a velocity-space Laplacian operator. The former result follows because $\nabla^{2}_v(1/v)=-4\pi\,\delta ({\bf v})$, and the latter because $\nabla^{2}_v(v)=2/v$.

When expressed in terms of the Rosenbluth potentials, the Landau collision operator, (3.93), takes the form

$$C_{ss'} = -{\mit\Gamma}_{ss'}\,\frac{\partial}{\partial v_\alpha}\!\left(... ...l v_{\alpha}\,\partial v_{\beta}}\,\frac{\partial f_s}{\partial v_\beta}\right)$$

$$=-{\mit\Gamma}_{ss'}\left[\left(1+\frac{m_s}{m_{s'}}\right)\frac{... ...partial^2\,G_{s'}}{\partial v_{\alpha}\,\partial v_{\beta}}\,f_s\right)\right],$$ (3.112)

where

$${\mit\Gamma}_{ss'} = \frac{2\,\gamma_{ss'}}{m_s^{2}}= \left(\frac{e_s\,e_{s'}}{4\pi\,\epsilon_0\,m_s}\right)^2\,4\pi\,\ln{\mit\Lambda}_c.$$ (3.113)