Appendix F
General Fokker-Planck Collisional Operator

For Coulomb collisions, the Boltzmann collisional operator can be approximated by a general Fokker-Planck type of collisional operator,

$$C_{ab}(f_a, f_b) = - \nabla_v \cdot \left[ A_{ab} f_a - \frac{1}{2} \nabla_v \cdot (\bar{D}_{ab} f_a) \right], \qquad (F1)$$

where the higher-order derivatives in velocity space are neglected, and where A is called a dynamical friction vector and $\bar{D}$ is called a diffusion tensor. In space physics and astrophysics, various approximations for A and $\bar{D}$ are used, and if a collisional operator has the form (F1), then Equation (A1) is summarily called the Fokker-Planck equation. Summation over all of the species (including self-collisions) then defines the full operator $C(f_a) = \sum_b C_{ab}(f_a f_b)$, which can be also written as $C(f_a) = -\nabla_v \cdot [A_a f_a - (1/2) \nabla_v \cdot (\bar{D}_a f_a)]$, where one defines $A_a = \sum_b A_{ab}$ and $\bar{D}_a = \sum_b \bar{D}_{ab}$. The Fokker-Planck operators work extremely well for any collisional process where collisions with a small scattering angle dominate, and where a lot of subsequent collisions gradually yield (in the sense of a random walk) a significant deviation from a particle's original velocity direction. This is exactly the case for the scattering by the electrostatic Coulomb force, where the Rutherford scattering cross section is proportional to $1/\sin^4(\chi/2)$ and heavily dominated by events with a small scattering angle $\chi$.

For any tensor $\bar{\bar{X}}$, a general Fokker-Planck operator can be integrated according to

$$\int \bar{\bar{X}} C_{ab}(f_a, f_b) d^3 v = \int f_a A_{ab} \cdot \frac{\partial \bar{\bar{X}}}{\partial v} d^3 v + \frac{1}{2} \int f_a \bar{\bar{D}}_{ab} : \frac{\partial}{\partial v} \frac{\partial \bar{\bar{X}}}{\partial v} d^3 v, \qquad (F2)$$

and, for clarity, explicitly in the index notation

$$\int \bar{\bar{X}} C_{ab}(f_a, f_b) d^3 v = \int f_a A_i^{ab} \frac{\partial \bar{\bar{X}}}{\partial v_i} d^3 v + \frac{1}{2} \int f_a D_{ij}^{ab} \frac{\partial}{\partial v_i} \frac{\partial \bar{\bar{X}}}{\partial v_j} d^3 v. \qquad (F3)$$

The useful identities are

$$\frac{\partial |\mathbf{v}|}{\partial v_i} = \frac{v_i}{|\mathbf{v}|}; \quad \frac{\partial |\mathbf{c}|}{\partial v_i} = \frac{c_i}{|\mathbf{c}|}; \quad \frac{\partial |\mathbf{v}|^2}{\partial v_i} = 2v_i; \quad \frac{\partial |\mathbf{c}|^2}{\partial v_i} = 2c_i, \qquad (F4)$$

and the tensorial collisional contributions defined in (5) can be calculated according to

$$\mathbf{R}_{ab} = m_a \int f_a A_{ab} d^3 v; \qquad (F5)$$

$$Q_{ab} = m_a \int f_a A_{ab} \cdot c_a d^3 v + \frac{m_a}{2} \int f_a \text{Tr} \bar{\bar{D}}_{ab} d^3 v; \qquad (F6)$$

$$\bar{\bar{Q}}_{ab}^{(2)} = m_a \int f_a [A_{ab} c_a]^S d^3 v + \frac{m_a}{2} \int f_a [\bar{\bar{D}}_{ab}]^S d^3 v; \qquad (F7)$$

$$\bar{\bar{Q}}_{ab}^{(3)} = m_a \int f_a [A_{ab} c_a c_a]^S d^3 v + \frac{m_a}{2} \int f_a [\bar{\bar{D}}_{ab}^S c_a]^S d^3 v. \qquad (F8)$$

If the diffusion tensor is symmetric, then $\bar{\bar{D}}_{ab}^S = 2\bar{\bar{D}}_{ab}$ (for clarity, the symmetric operator does not act on species indices, and in general $\bar{\bar{D}}_{ab} \neq \bar{\bar{D}}_{ba}$, similar to $\nu_{ab} \neq \nu_{ba}$, the symmetric operator acts as $(\bar{\bar{D}}_{ij}^{ab})^S = \bar{\bar{D}}_{ij}^{ab} + \bar{\bar{D}}_{ji}^{ab}$). The fourth- and fifth-order collisional contributions are

$$(\bar{\bar{Q}}_{ab}^{(4)})_{ijkl} = m_a \int f_a [A_{ab} c_a c_a c_a]^S_{ijkl} d^3 v + \frac{m_a}{2} \int f_a [[\bar{\bar{D}}_{ab}^S c_a c_a]_{ijkl}^S + (\bar{\bar{D}}^{ab})^S_{ik} c_j^a c_l^a + (\bar{\bar{D}}^{ab})^S_{jl} c_i^a c_k^a] d^3 v; \qquad (F9)$$

$$(\bar{\bar{Q}}_{ab}^{(5)})_{ijklm} = m_a \int f_a [A_{ab} c_a c_a c_a c_a]_{ijklm}^S d^3 v + \frac{m_a}{2} \int f_a [[\bar{\bar{D}}_{ab}^S c_a c_a c_a]_{ijklm}^S + (\bar{\bar{D}}^{ab})^S_{ik} c_j^a c_l^a c_m^a \\ + (\bar{\bar{D}}^{ab})^S_{il} c_j^a c_k^a + (\bar{\bar{D}}^{ab})^S_{jl} c_i^a c_k^a + (\bar{\bar{D}}^{ab})^S_{jm} c_i^a c_k^a + (\bar{\bar{D}}^{ab})^S_{km} c_i^a c_j^a c_l^a] d^3 v. \qquad (F10)$$

The first integral in (F9) proportional to $A_{ab}$ contains four terms, and the second integral in (F9) proportional to $\bar{\bar{D}}_{ab}^S$ contains 12 terms. Similarly, the first integral in (F10) contains five terms, and the second integral in (F10) contains 20 terms. The second integrals in (F9)–(F10) can be written simply by picking two indices for $\bar{\bar{D}}_{ab}^S$ and giving the rest of the indices to $c_a c_a$ and $c_a c_a c_a$. The generalization to the nth-order collisional contributions defined in (A10) is done naturally by introducing a set of indices $R = \{r_1...r_n\}$, together with an ordered set $(s_1, s_2)$, and writing

$$(\bar{\bar{Q}}_{ab}^{(n)})_{r_1 r_2 \dots r_n} = m_a \int f_a [A_{r_1}^{ab} c_{r_2}^{a} \dots c_{r_n}^{a}]^S d^3 v + \frac{m_a}{2} \int f_a [D_{s_1 s_2}^{ab} c_{s_3} \dots c_{s_n}]^S d^3 v; \\ \text{where } (s_1, s_2) \in R = \{r_1 \dots r_n\}; \text{ and } s_3 \dots s_n \in R \setminus \{s_1, s_2\}, \qquad (F11)$$
so that the first integral contains (n) terms, and the second integral contains $2\binom{n}{2} = n(n-1)$ terms. Alternatively, one can replace the ordered set ($s_1, s_2$) with a nonordered set {$s_1, s_2$}, and include the symmetric operator on $\bar{\mathbf{D}}^{ab}$.

It is useful to write the collisional contributions for the contracted vectors, matrices, and scalars by assuming symmetric $\bar{\mathbf{D}}^{ab}$. We use the definitions from Section 8.3 (see Equation (188)) that were also used in Appendix D; see Equations (D4) and (D5). This yields the collisional contributions for vectors:

$$ \begin{aligned} \bar{\mathbf{Q}}_{ab}^{(2n+1)} ={}& m_a \int [(2n)(A^{ab} \cdot \mathbf{c}_a)\mathbf{c}_a |\mathbf{c}_a|^{2n-2} + A^{ab} |\mathbf{c}_a|^{2n}] f_a d^3v \\ & +m_a \int [(2n)(n-1)(\bar{\mathbf{D}}^{ab}: \mathbf{c}_a \mathbf{c}_a)\mathbf{c}_a |\mathbf{c}_a|^{2n-4} + (n)(\mathrm{Tr}\,\bar{\mathbf{D}}^{ab})\mathbf{c}_a |\mathbf{c}_a|^{2n-2} \\ & + (2n)(\mathbf{c}_a \cdot \bar{\mathbf{D}}^{ab}) |\mathbf{c}_a|^{2n-2}] f_a d^3v; \end{aligned} \quad (F12) $$

for matrices:

$$ \begin{aligned} \bar{\mathbf{Q}}_{ab}^{(2n)} ={}& m_a \int [(A^{ab} \cdot \mathbf{c}_a)^S |\mathbf{c}_a|^{2n-2} + (2n-2)(A^{ab} \cdot \mathbf{c}_a)\mathbf{c}_a \mathbf{c}_a |\mathbf{c}_a|^{2n-4}] f_a d^3v \\ & +m_a \int [\bar{\mathbf{D}}^{ab} |\mathbf{c}_a|^{2n-2} + (2n-2)(\bar{\mathbf{D}}^{ab} \cdot \mathbf{c}_a \mathbf{c}_a)^S |\mathbf{c}_a|^{2n-4} \\ & +(n-1)(\mathrm{Tr}\,\bar{\mathbf{D}}^{ab})\mathbf{c}_a \mathbf{c}_a |\mathbf{c}_a|^{2n-4} + (n-1)(2n-4)(\bar{\mathbf{D}}^{ab}: \mathbf{c}_a \mathbf{c}_a)\mathbf{c}_a \mathbf{c}_a |\mathbf{c}_a|^{2n-6}] f_a d^3v; \end{aligned} \quad (F13) $$

and for scalars:

$$ \begin{aligned} \mathcal{Q}_{ab}^{(2n)} ={}& m_a \int [(2n)(A^{ab} \cdot \mathbf{c}_a) |\mathbf{c}_a|^{2n-2} + (n)(\mathrm{Tr}\,\bar{\mathbf{D}}^{ab}) |\mathbf{c}_a|^{2n-2} \\ & + (2n)(n-1)(\bar{\mathbf{D}}^{ab}: \mathbf{c}_a \mathbf{c}_a) |\mathbf{c}_a|^{2n-4}] f_a d^3v, \end{aligned} \quad (F14) $$

all valid for $n \ge 1$. Applying a trace at (F13) recovers (F14).

## Appendix G
Landau Collisional Operator (5-moment Model)

For Coulomb collisions, a very accurate collisional operator was obtained by Landau (1936, 1937) in the following form (see, for example, Equation (1.2) in Braginskii 1958):

$$ C_{ab}(f_a, f_b) = - \frac{2\pi e^4 Z_a^2 Z_b^2 \ln \Lambda}{m_a} \frac{\partial}{\partial \mathbf{v}} \cdot \int \bar{\mathbf{V}} \cdot \left[ \frac{f_a(\mathbf{v})}{m_b} \frac{\partial f_b(\mathbf{v}')}{\partial \mathbf{v}'} - \frac{f_b(\mathbf{v}')}{m_a} \frac{\partial f_a(\mathbf{v})}{\partial \mathbf{v}} \right] d^3\mathbf{v}'; \quad (G1) $$

$$ \bar{\mathbf{V}} = \frac{\bar{\mathbf{I}}}{|\mathbf{v} - \mathbf{v}'|} - \frac{(\mathbf{v} - \mathbf{v}') \cdot (\mathbf{v} - \mathbf{v}')}{|\mathbf{v} - \mathbf{v}'|^3}. \quad (G2) $$

With this collisional operator, Equation (A1) is known as the Landau equation. The Landau collisional operator is sometimes called the Landau collisional integral, because (G1) contains an integral over $d^3\mathbf{v}'$ (i.e., it is an integro-differential operator). The operator can be rewritten into the general Fokker-Planck form (F1) by introducing Rosenbluth potentials

$$ H_b(\mathbf{v}) = \int \frac{f_b(\mathbf{v}')}{|\mathbf{v} - \mathbf{v}'|} d^3\mathbf{v}'; \quad \text{and} \quad G_b(\mathbf{v}) = \int f_b(\mathbf{v}') |\mathbf{v} - \mathbf{v}'| d^3\mathbf{v}', \quad (G3) $$

yielding (see, for example, Equations (7)–(8) of Hinton 1983)

$$ A_{ab} = 2 \frac{c_{ab}}{m_a^2} \left( 1 + \frac{m_a}{m_b} \right) \frac{\partial H_b(\mathbf{v})}{\partial \mathbf{v}}; \quad \bar{\mathbf{D}}_{ab} = 2 \frac{c_{ab}}{m_a^2} \frac{\partial^2 G_b(\mathbf{v})}{\partial \mathbf{v} \partial \mathbf{v}}; \quad c_{ab} = 2\pi e^4 Z_a^2 Z_b^2 \ln \Lambda. \quad (G4) $$

The useful identities are

$$ \frac{\partial}{\partial \mathbf{v}} \cdot \bar{\mathbf{V}} = -2 \frac{\mathbf{v} - \mathbf{v}'}{|\mathbf{v} - \mathbf{v}'|^3} = - \frac{\partial}{\partial \mathbf{v}'} \cdot \bar{\mathbf{V}}; \quad \frac{\partial^2}{\partial \mathbf{v} \partial \mathbf{v}} |\mathbf{v} - \mathbf{v}'| = \bar{\mathbf{V}}, \quad (G5) $$

and it is easy to verify that (F1), (G4) recovers the Landau operator (G1) (after one uses the Gauss-Ostrogradsky divergence theorem in velocity $d^3\mathbf{v}'$, which makes the associated integral vanish). By using Laplacian $\nabla_v^2 = \nabla_v \cdot \nabla_v$, the following identity implies

$$ \nabla_v^2 \frac{1}{|\mathbf{v} - \mathbf{v}'|} = -4\pi\delta(\mathbf{v} - \mathbf{v}'); \quad \Rightarrow \quad \nabla_v^2 H_b(\mathbf{v}) = -4\pi f_b(\mathbf{v}). \quad (G6) $$
The Rosenbluth potential $H_b(v)$ is thus completely analogous to the electrostatic potential $\Phi(x)$ (with a Poisson equation $\nabla^2\Phi(x) = -4\pi\rho_c(x)$, where $\rho_c(x)$ is the charge's spatial distribution), here just used in velocity space. Also, because of the identity $\nabla_v^2|v - v'| = 2/|v - v'|$, the Rosenbluth potentials are related by

$$H_b = \frac{1}{2} \nabla_v^2 G_b; \quad \Rightarrow \quad A_{ab} = \frac{1}{2} \left( 1 + \frac{m_a}{m_b} \right) \frac{\partial}{\partial v} \cdot \bar{D}_{ab}. \qquad (G7)$$

However, the structure of the Rosenbluth potentials implies that the Landau operator is quite complicated. Indeed, the simplest example, when prescribing Maxwellian $f_b = n_b/(\pi^{3/2}v_{\text{thb}}^3)\exp(-y^2)$ with the (vector) variable $\mathbf{y} = (\mathbf{v} - \mathbf{u}_b)/v_{\text{thb}}$ and scalar $y = |\mathbf{y}|$, yields the Rosenbluth potentials

$$H_b(\mathbf{v}) = \frac{n_b \operatorname{erf}(y)}{v_{\text{thb}} y}; \qquad (G8)$$

$$G_b(\mathbf{v}) = n_b v_{\text{thb}} \left[ \frac{1}{\sqrt{\pi}} e^{-y^2} + \left( \frac{1}{2y} + y \right) \operatorname{erf}(y) \right], \qquad (G9)$$

where the error function $\operatorname{erf}(y) = (2/\sqrt{\pi}) \int_0^y e^{-z^2} dz$ is present. These Rosenbluth potentials make the collisional contributions (F5), (F6) difficult to calculate.

For clarity on how the $H_b$ is obtained, it is useful to introduce the (vector) variable $\mathbf{x} = (\mathbf{v}' - \mathbf{v})/v_{\text{thb}}$ and the scalar $x = |\mathbf{x}|$, and change the integration into $d^3\mathbf{v}' = v_{\text{thb}}^3 d^3\mathbf{x}$, so that

$$H_b(\mathbf{v}) = \frac{n_b}{\pi^{3/2} v_{\text{thb}}^3} \int_{-\infty}^{\infty} \frac{e^{-\frac{|\mathbf{v}' - \mathbf{u}_b|^2}{v_{\text{thb}}^2}}}{|\mathbf{v}' - \mathbf{v}|} d^3\mathbf{v}' = \frac{n_b}{\pi^{3/2} v_{\text{thb}}^3} \int_{-\infty}^{\infty} \frac{e^{-|\mathbf{x} + \mathbf{y}|^2}}{x} d^3\mathbf{x}. \qquad (G10)$$

In the last integral, the variable $\mathbf{y}$ is a constant (because $\mathbf{v}$ and $\mathbf{u}_b$ are constants). One introduces spherical coordinates in the $\mathbf{x}$-space with orthogonal unit vectors $\hat{\mathbf{e}}_1$, $\hat{\mathbf{e}}_2$, $\hat{\mathbf{e}}_3$, where the direction of vector $\mathbf{y}$ forms axis $\hat{\mathbf{e}}_3 = \mathbf{y}/y$, so that the vector

$$\mathbf{x} = x \sin \theta \cos \phi \hat{\mathbf{e}}_1 + x \sin \theta \sin \phi \hat{\mathbf{e}}_2 + x \cos \theta \hat{\mathbf{e}}_3. \qquad (G11)$$

In this reference frame, $\mathbf{y} = (0, 0, y)$, and so $|\mathbf{x} + \mathbf{y}|^2 = x^2 + y^2 + 2xy \cos \theta$. Then one can calculate the integral in spherical coordinates $d^3\mathbf{x} = x^2 \sin \theta dxd\theta d\phi$, yielding

$$\begin{aligned} \int_{-\infty}^{\infty} \frac{e^{-|\mathbf{x}+\mathbf{y}|^2}}{x} d^3\mathbf{x} &= 2\pi \int_0^{\infty} \int_0^{\pi} x e^{-(x^2+y^2)} \sin\theta e^{-2xy \cos\theta} d\theta d\mathbf{x} \\ &= 2\pi \int_0^{\infty} x e^{-(x^2+y^2)} \frac{1}{2xy} (e^{+2xy} - e^{-2xy}) d\mathbf{x} = \frac{\pi}{y} \int_0^{\infty} (e^{-(x-y)^2} - e^{-(x+y)^2}) d\mathbf{x} \\ &= \frac{\pi}{y} \left( \int_{-\infty}^{\infty} e^{-z^2} dz - \int_{-\infty}^{\infty} e^{-z^2} d\mathbf{z} \right) = \frac{\pi}{y} \int_{-\infty}^{\infty} e^{-z^2} d\mathbf{z} = \frac{2\pi}{y} \int_0^{\infty} e^{-z^2} d\mathbf{z} = \frac{\pi^{3/2}}{y} \operatorname{erf}(y), \end{aligned} \qquad (G12)$$

recovering (G8). The result can be verified by calculating (G6). Similarly, the potential $G_b$ can be obtained by calculating

$$\begin{aligned} \int_{-\infty}^{\infty} x e^{-|\mathbf{x}+\mathbf{y}|^2} d^3\mathbf{x} &= \frac{\pi}{y} \int_0^{\infty} x^2 (e^{-(x-y)^2} - e^{-(x+y)^2}) d\mathbf{x} = \frac{\pi}{y} \left( \int_{-\infty}^{\infty} (z+y)^2 e^{-z^2} dz - \int_{-\infty}^{\infty} (z-y)^2 e^{-z^2} dz \right) \\ &= \frac{\pi}{y} \left( 2 \int_0^y z^2 e^{-z^2} dz + 4y \int_0^\infty z e^{-z^2} dz + 2y^2 \int_0^\infty e^{-z^2} dz \right) = \pi^{3/2} \left( y + \frac{1}{2y} \right) \operatorname{erf}(y) + \pi e^{-y^2}, \end{aligned} \qquad (G13)$$

recovering (G9), and which can be verified to satisfy (G7).

Note that because $\operatorname{erf}(0) = 0$, the error function can actually be defined as an indefinite integral:

$$\frac{2}{\sqrt{\pi}} \int e^{-x^2} dx = \operatorname{erf}(x); \qquad \frac{2}{\sqrt{\pi}} \int e^{-\frac{(x+a)^2}{b^2}} dx = \frac{\operatorname{erf}(x+a)}{b^2}.$$

The useful relations are $\operatorname{erf}(-x) = -\operatorname{erf}(x)$ and $\operatorname{erf}(\infty) = 1$. Then the calculations above can be done more elegantly, for example:

$$\int_0^\infty e^{-(x-y)^2} dx = \frac{\sqrt{\pi}}{2} \operatorname{erf}(x-y) \Big|_{x=0}^{x=\infty} = \frac{\sqrt{\pi}}{2} (1 + \operatorname{erf}(y)).$$
G.1. Momentum Exchange Rates $R_{ab}$

To obtain the momentum exchange rates $R_{ab}$, one needs to calculate

$$ 
\begin{aligned}
\mathbf{R}_{ab} &= m_a \int f_a A_{ab} d^3 v = 2 \frac{c_{ab}}{m_a} \left(1 + \frac{m_a}{m_b}\right) \int f_a \frac{\partial H_b}{\partial \mathbf{v}} d^3 v; \\
&= -2 \frac{c_{ab}}{m_a} \left(1 + \frac{m_a}{m_b}\right) \int H_b \frac{\partial f_a}{\partial \mathbf{v}} d^3 v.
\end{aligned} 
\quad (G14) 
$$

Prescribing Maxwellian $f_a = (n_a/(\pi^{3/2}v_{\text{tha}}^3))\exp(-|\mathbf{v} - \mathbf{u}_a|^2/v_{\text{tha}}^2)$ with general velocity $\mathbf{u}_a$ leads to the “runaway” effect addressed below in Appendix G.3. It is useful to first consider a simplified situation where the differences between the drift velocities $\mathbf{u}_a$ and $\mathbf{u}_b$ are small. The Maxwellian $f_a$ is rewritten with the variable $\mathbf{y}$ and the variable $\mathbf{u} = (\mathbf{u}_b - \mathbf{u}_a)/v_{\text{tha}}$, and expanded by assuming that $|\mathbf{u}| \ll 1$, so that

$$ f_a = \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-|\mathbf{y} + \mathbf{u}|^2} \approx \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-y^2 \alpha^2} (1 - 2\alpha \mathbf{y} \cdot \mathbf{u}), \quad (G15) $$

where $\alpha = v_{\text{thb}}/v_{\text{tha}}$. Then the derivative

$$ \frac{\partial f_a}{\partial \mathbf{v}} = - \frac{2n_a}{\pi^{3/2} v_{\text{tha}}^4} e^{-y^2 \alpha^2} [\mathbf{u} + \alpha \mathbf{y} - 2\alpha^2 \mathbf{y}(\mathbf{y} \cdot \mathbf{u})], \quad (G16) $$

and one needs to calculate

$$ \mathbf{R}_{ab} = \frac{4c_{ab}}{m_a} \left(1 + \frac{m_a}{m_b}\right) \frac{n_a n_b}{\pi^{3/2} v_{\text{tha}}^4 v_{\text{thb}}} \underbrace{\int_{-\infty}^{\infty} \frac{\text{erf}(y)}{y} e^{-y^2 \alpha^2} [\mathbf{u} + \alpha \mathbf{y} - 2\alpha^2 \mathbf{y}(\mathbf{y} \cdot \mathbf{u})] d^3 v}_{\textcircled{1} + \textcircled{2} + \textcircled{3}} \quad (G17) $$

where we split the integral into three parts. The integration over $d^3v$ can be changed to $v_{\text{thb}}^3 d^3y$. We will use

$$ 
\begin{aligned}
\int_0^\infty e^{-y^2 \alpha^2} y \, \text{erf}(y) \, dy &= \frac{1}{2\alpha^2 \sqrt{1 + \alpha^2}}; \\
\int_0^\infty e^{-y^2 \alpha^2} y^3 \, \text{erf}(y) \, dy &= \frac{3\alpha^2 + 2}{4\alpha^4 (1 + \alpha^2)^{3/2}}.
\end{aligned} 
\quad (G18) 
$$

The three integrals are then evaluated according to

$$ 
\begin{aligned}
\textcircled{1} &= \mathbf{u} \int_{-\infty}^{\infty} \frac{\text{erf}(y)}{y} e^{-y^2 \alpha^2} d^3 v = \mathbf{u} v_{\text{thb}}^3 4\pi \int_0^\infty y \, \text{erf}(y) e^{-y^2 \alpha^2} dy = \mathbf{u} v_{\text{thb}}^3 \frac{2\pi}{\alpha^2 \sqrt{1 + \alpha^2}}; \\
\textcircled{2} &= \alpha \int_{-\infty}^{\infty} y \frac{\text{erf}(y)}{y} e^{-y^2 \alpha^2} d^3 v = 0; \\
\textcircled{3} &= -2\alpha^2 \int_{-\infty}^{\infty} \mathbf{y}(\mathbf{y} \cdot \mathbf{u}) \frac{\text{erf}(y)}{y} e^{-y^2 \alpha^2} d^3 v = -\frac{2\alpha^2}{3} \mathbf{u} \int_{-\infty}^{\infty} y \, \text{erf}(y) e^{-y^2 \alpha^2} d^3 v \\
&= -\frac{8\pi}{3} \alpha^2 v_{\text{thb}}^3 \mathbf{u} \int_0^\infty y^3 \, \text{erf}(y) e^{-y^2 \alpha^2} dy = -\frac{8\pi}{3} \alpha^2 v_{\text{thb}}^3 \frac{3\alpha^2 + 2}{4\alpha^4 (1 + \alpha^2)^{3/2}},
\end{aligned} 
\quad (G19) $$

and so

$$ \textcircled{1} + \textcircled{3} = \mathbf{u} v_{\text{thb}}^3 \frac{2\pi}{3\alpha^2(1 + \alpha^2)^{3/2}} = \mathbf{u} \frac{2\pi}{3} \frac{v_{\text{tha}}^5 v_{\text{thb}}}{(v_{\text{tha}}^2 + v_{\text{thb}}^2)^{3/2}}. \quad (G20) $$

The entire result (G17) can then be written as (see, for example, Equations (46)–(47) of Hinton 1983)

$$ \mathbf{R}_{ab} = \rho_a \nu_{ab}(\mathbf{u}_b - \mathbf{u}_a), \quad (G21) $$

where the collisional frequency

$$ \nu_{ab} = \tau_{ab}^{-1} = \frac{16}{3} \sqrt{\pi} \frac{n_b e^4 Z_a^2 Z_b^2 \ln \Lambda}{m_a^2 (v_{\text{tha}}^2 + v_{\text{thb}}^2)^{3/2}} \left(1 + \frac{m_a}{m_b}\right), \quad (G22) $$

and the thermal speeds $v_{\text{tha}}^2 = 2T_a/m_a$. Note that $m_a n_a \nu_{ab} = m_b n_b \nu_{ba}$ holds. The collisional frequency (G22) is identical to Equation (C2) of Schunk (1977); see Equation (179).

It is useful to clarify the physical meaning of the collisional frequencies. Considering momentum equations for two species where all of the spatial gradients are neglected, so that $\partial \mathbf{u}_a/\partial t - (eZ_a/m_a)\mathbf{E} = \mathbf{R}_{ab}/\rho_a$ and $\partial \mathbf{u}_b/\partial t - (eZ_b/m_b)\mathbf{E} = \mathbf{R}_{ba}/\rho_b$, then subtracting
them and defining the difference $\delta\mathbf{u} = \mathbf{u}_b - \mathbf{u}_a$, yields an evolution equation

$$ \frac{\partial \delta \mathbf{u}}{\partial t} + \nu \delta \mathbf{u} = e\mathbf{E} \left( \frac{Z_b}{m_b} - \frac{Z_a}{m_a} \right); \quad \nu = \nu_{ab} + \nu_{ba}. \qquad (G23) $$

With no use of Maxwell's equations, and instead assuming an applied (external) constant electric field and also constant collisional frequencies, an initial velocity difference $\delta\mathbf{u}(0)$ evolves according to

$$ \delta \mathbf{u}(t) = \delta \mathbf{u}(0) e^{-\nu t} + (1 - e^{-\nu t}) \frac{e \mathbf{E}}{\nu} \left( \frac{Z_b}{m_b} - \frac{Z_a}{m_a} \right). \qquad (G24) $$

Approximately after time $\tau = 1/\nu$ (which represents many small-angle collisions), the dependence on the initial condition disappears and the difference between the velocities reaches a constant value:

$$ \mathbf{u}_b - \mathbf{u}_a = \frac{e\mathbf{E}}{\nu_{ab} + \nu_{ba}} \left( \frac{Z_b}{m_b} - \frac{Z_a}{m_a} \right) = \text{const.} \qquad (G25) $$

Provided that $Z_a/m_a \neq Z_b/m_b$, the collisional time $\tau = 1/(\nu_{ab} + \nu_{ba})$ can then be interpreted as an average time that is required for particles "a" and "b" to experience (many small-angle) collisions, so that the difference between their average fluid velocities reaches a constant value proportional to the value of the applied (external) electric field $\mathbf{E}$. For the particular case of $Z_a/m_a = Z_b/m_b$, the velocities become equal regardless of the value of applied $\mathbf{E}$.

For a particular case of a one ion-electron plasma, $\mathbf{u}_e - \mathbf{u}_i = -e\mathbf{E}/(\nu_{ei}m_e)$, which can also be directly obtained from the quasistatic electron or ion momentum equations. Prescribing the charge neutrality, $n_e = Z_i n_i$, so that the current $j = -en_e(\mathbf{u}_e - \mathbf{u}_i)$ then yields the relation $j = \sigma \mathbf{E}$ with the usual electrical conductivity $\sigma = 1/\eta = e^2 n_e/(\nu_{ei}m_e)$, where $\sigma$ does not depend on the value of current $j$ (because $j$ is assumed to be small).

## G.2. Energy Exchange Rates $Q_{ab}$

Similar calculations are used to obtain the energy exchange rates $Q_{ab}$, according to (F8). It is beneficial to notice that $\text{Tr}\tilde{D}_{ab} = (4c_{ab}/m_a^2)H_b$, and so

$$ Q_{ab} = \frac{2c_{ab}}{m_a} \left(1 + \frac{m_a}{m_b}\right) \int f_a \frac{\partial H_b}{\partial \mathbf{v}} \cdot \mathbf{c}_a d^3\mathbf{v} + \frac{2c_{ab}}{m_a} \int f_a H_b d^3\mathbf{v}; \qquad (G26) $$

$$ \frac{\partial H_b}{\partial \mathbf{v}} = \frac{n_b}{v_{\text{thb}}^2} \mathbf{y} \left( \frac{1}{y^2} \frac{2}{\sqrt{\pi}} e^{-y^2} - \frac{1}{y^3} \text{erf}(y) \right), \qquad (G27) $$

and because $\mathbf{c}_a = \mathbf{y}v_{\text{thb}} + \mathbf{u}v_{\text{tha}}$, then

$$ \frac{\partial H_b}{\partial \mathbf{v}} \cdot \mathbf{c}_a = \frac{n_b}{v_{\text{thb}}^2} (y^2 v_{\text{thb}} + (\mathbf{u} \cdot \mathbf{y}) v_{\text{tha}}) \left( \frac{1}{y^2} \frac{2}{\sqrt{\pi}} e^{-y^2} - \frac{1}{y^3} \text{erf}(y) \right). \qquad (G28) $$

Importantly, to correctly account for the $|\mathbf{u}|^2$ contributions, the $f_a$ has to be expanded further:

$$ f_a = \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-|y|} \approx \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-y^2 \alpha^2} (1 - (2\alpha y \cdot \mathbf{u} + |\mathbf{u}|^2) + 2\alpha^2 (y \cdot \mathbf{u})^2), \qquad (G29) $$

where $\alpha = v_{\text{thb}}/v_{\text{tha}}$. This distribution function yields

$$ \int f_a H_b d^3\mathbf{v} = \frac{2n_a n_b}{\sqrt{\pi} v_{\text{tha}}} \left[ \frac{1}{\sqrt{1 + \alpha^2}} - \frac{|\mathbf{u}|^2}{3(1 + \alpha^2)^{3/2}} \right]; \\ \int f_a \frac{\partial H_b}{\partial \mathbf{v}} \cdot \mathbf{c}_a d^3\mathbf{v} = \frac{2n_a n_b}{\sqrt{\pi} v_{\text{tha}}} \left[ -\frac{1}{(1 + \alpha^2)^{3/2}} + \frac{|\mathbf{u}|^2}{(1 + \alpha^2)^{5/2}} \right], \qquad (G30) $$

and the final result reads

$$ Q_{ab} = 3\rho_a \nu_{ab} \frac{T_b - T_a}{m_a + m_b} + \rho_a \nu_{ab} \frac{3}{2} \left( \frac{m_b T_a}{m_b T_a + m_a T_b} - \frac{1}{3} \frac{m_b}{m_b + m_a} \right) |\mathbf{u}_b - \mathbf{u}_a|^2, \qquad (G31) $$

or equivalently,

$$ Q_{ab} = 3\rho_a \nu_{ab} \frac{T_b - T_a}{m_a + m_b} + \rho_a \nu_{ab} \frac{m_b(3T_a m_a + 2T_a m_b - T_b m_a)}{2(T_b m_a + T_a m_b)(m_b + m_a)} |\mathbf{u}_b - \mathbf{u}_a|^2. \qquad (G32) $$

Hinton (1983) calculates only the first term, the thermal exchange rate (his Equation (52); see also Landau 1936 for an ion-electron plasma). Calculating $Q_{ab} + Q_{ba} = \rho_a \nu_{ab} |\mathbf{u}_b - \mathbf{u}_a|^2 = (\mathbf{u}_b - \mathbf{u}_a) \cdot \mathbf{R}_{ab}$ yields the energy conservation, and the result (G31) is well-
defined. (Recalculating $R_{ab}$ with the further expanded $f_0$ (G29) yields the unchanged result $R_{ab} = \rho_a \nu_{ab}(\mathbf{u}_b - \mathbf{u}_a)$). As a double check, expanding the more general expression for unrestricted drifts (G64) (by expansion $\Psi_{ab} = 1 - \epsilon^2$) yields

$$Q_{ab} = 3\rho_a \nu_{ab} \frac{T_b - T_a}{m_a + m_b} \left( 1 - \frac{|\mathbf{u}_b - \mathbf{u}_a|^2}{\frac{2T_a}{m_a} + \frac{2T_b}{m_b}} \right) + \rho_a \nu_{ab} \frac{m_b}{m_b + m_a} |\mathbf{u}_b - \mathbf{u}_a|^2. \qquad (G33)$$

Results (G33) and (G31) are equivalent, and valid for an unrestricted difference in temperature. Prescribing that the difference in temperatures is small simplifies the frictional part into

$$Q_{ab} = 3\rho_a \nu_{ab} \frac{T_b - T_a}{m_a + m_b} + \rho_a \nu_{ab} \frac{m_b}{m_a + m_b} |\mathbf{u}_b - \mathbf{u}_a|^2. \qquad (G34)$$

This frictional part is derived elegantly in the Appendix of Braginskii (1965).

### G.3. $R_{ab}$ and $Q_{ab}$ for Unrestricted Drifts $\mathbf{u}_b - \mathbf{u}_a$ (Runaway Effect)

Here we want to calculate $R_{ab}$ for the general Maxwellian distributions $f_a$, $f_b$, with no restriction on the value of the difference $\mathbf{u}_b - \mathbf{u}_a$. We follow Burgers (1969) and Tanenbaum (1967). Instead of using the Rosenbluth potential $H_b$ and calculating (G14), it is easier to consider

$$\mathbf{R}_{ab} = 2 \frac{c_{ab}}{m_a} \left( 1 + \frac{m_a}{m_b} \right) \iint f_a(\mathbf{v}) f_b(\mathbf{v}') \frac{\mathbf{v}' - \mathbf{v}}{|\mathbf{v}' - \mathbf{v}|^3} d^3\mathbf{v} d^3\mathbf{v}'. \qquad (G35)$$

Additionally, instead of $\mathbf{v}$ and $\mathbf{v}'$, it feels more natural to use $\mathbf{v}_a = \mathbf{v}$ and $\mathbf{v}_b = \mathbf{v}'$. It is useful to introduce the vectors $\mathbf{x} = \mathbf{v}_b - \mathbf{v}_a$ and $\mathbf{u} = \mathbf{u}_b - \mathbf{u}_a$. The integral is then calculated by introducing the “center-of-mass” velocity,

$$C = \frac{m_a \mathbf{v}_a + m_b \mathbf{v}_b}{m_a + m_b} - \frac{m_a \mathbf{u}_a + m_b \mathbf{u}_b}{m_a + m_b} + \frac{m_a m_b}{(m_a + m_b)(m_b T_a + m_a T_b)} (\mathbf{u} - \mathbf{x}), \qquad (G36)$$

which transforms

$$f_a f_b = \frac{n_a n_b}{\pi^3 v_{\text{tha}}^3 v_{\text{thb}}^3} \exp \left( - \frac{|\mathbf{v}_a - \mathbf{u}_a|^2}{v_{\text{tha}}^2} - \frac{|\mathbf{v}_b - \mathbf{u}_b|^2}{v_{\text{thb}}^2} \right), \qquad (G37)$$

into

$$f_a f_b = \frac{n_a n_b}{\pi^3 \tilde{\alpha}^3 \beta^3} \exp \left( - \frac{|\mathbf{C}|^2}{\tilde{\alpha}^2} - \frac{|\mathbf{x} - \mathbf{u}|^2}{\beta^2} \right), \qquad (G38)$$

with new thermal speeds

$$\tilde{\alpha}^2 = \frac{2T_a T_b}{m_b T_a + m_a T_b}; \qquad \beta^2 = v_{\text{tha}}^2 + v_{\text{thb}}^2. \qquad (G39)$$

Importantly, $d^3\mathbf{v}_a d^3\mathbf{v}_b = d^3\mathbf{C} d^3\mathbf{x}$ (by calculating Jacobian). For later calculations of more complicated integrals than (G35), the useful transformations are

$$\begin{aligned} c_a &= \mathbf{C} - \frac{m_b T_a}{m_b T_a + m_a T_b} (\mathbf{x} - \mathbf{u}); \\ c_b &= \mathbf{C} + \frac{m_a T_b}{m_b T_a + m_a T_b} (\mathbf{x} - \mathbf{u}). \end{aligned} \qquad (G40)$$

The integral (G35) thus transforms into

$$\mathbf{R}_{ab} = 2 \frac{c_{ab}}{m_a} \left( 1 + \frac{m_a}{m_b} \right) \frac{n_a n_b}{\pi^{3/2} \beta^3} \int \frac{\mathbf{x}}{x^3} e^{-\frac{|\mathbf{x} - \mathbf{u}|^2}{\beta^2}} d^3 \mathbf{x}, \qquad (G41)$$

where we have already integrated over $d^3\mathbf{C}$. One introduces a reference frame in the $\mathbf{x}$-space with unit vectors $\hat{\mathbf{e}}_1$, $\hat{\mathbf{e}}_2$, $\hat{\mathbf{e}}_3$, where the direction of vector $\mathbf{u}$ defines the axis $\hat{\mathbf{e}}_3 = \mathbf{u}/\mathbf{u}$, so that

$$\mathbf{x} = x \sin \theta \cos \phi \hat{\mathbf{e}}_1 + x \sin \theta \sin \phi \hat{\mathbf{e}}_2 + x \cos \theta \hat{\mathbf{e}}_3. \qquad (G42)$$

For example, the integration of (G42) over $\phi$ yields $\int_0^{2\pi} x d\phi = 2\pi x \cos \theta \hat{\mathbf{e}}_3$, i.e., the result is in the direction of $\mathbf{u}$. Furthermore, because $|\mathbf{x} - \mathbf{u}|^2 = x^2 + u^2 - 2xu \cos \theta$, the integration of (G41) over $\phi$ can be carried out, yielding

$$\int \frac{\mathbf{x}}{x^3} e^{-\frac{|\mathbf{x} - \mathbf{u}|^2}{\beta^2}} d^3 \mathbf{x} = \frac{\mathbf{u}}{u} 2\pi \int_0^\infty \int_0^\pi e^{-\frac{|\mathbf{x} - \mathbf{u}|^2}{\beta^2}} \cos \theta \sin \theta d\theta d\theta. \qquad (G43)$$
To calculate that integral, it is useful to introduce (constant) $\epsilon = u/\beta$, and change the integration into the variables

$$z = \frac{x}{\beta} - s; \quad s = \epsilon \cos \theta, \qquad (G44)$$

so that $|\mathbf{x} - \mathbf{u}|^2/\beta^2 = z^2 - s^2 + \epsilon^2$, yielding

$$\int \frac{\mathbf{x}}{x^3} e^{-\frac{|\mathbf{x}-\mathbf{u}|^2}{\beta^2}} d^3\mathbf{x} = \mathbf{u}2\pi \frac{e^{-\epsilon^2}}{\epsilon^3} \int_{-\infty}^{\infty} \int_{-\epsilon}^{\epsilon} s e^{-z^2+s^2} dz ds \\ = \mathbf{u}\pi^{3/2} \left( \frac{\operatorname{erf}(\epsilon)}{\epsilon^3} - \frac{2}{\sqrt{\pi}} \frac{e^{-\epsilon^2}}{\epsilon^2} \right). \qquad (G45)$$

In the last integral, it is necessary to first integrate over $dz$ and then over $ds$, by using

$$\int_{-\infty}^{\infty} e^{-z^2} dz = \frac{\sqrt{\pi}}{2} (1 + \operatorname{erf}(s)); \\ \int_{-\epsilon}^{\epsilon} s e^{s^2} \operatorname{erf}(s) ds = e^{\epsilon^2} \operatorname{erf}(\epsilon) - \frac{2}{\sqrt{\pi}} \epsilon. \qquad (G46)$$

The final result then reads

$$\mathbf{R}_{ab} = \rho_a \nu_{ab}(\mathbf{u}_b - \mathbf{u}_a) \Phi_{ab}; \qquad (G47)$$

$$\Phi_{ab} = \left( \frac{3}{4} \sqrt{\pi} \frac{\operatorname{erf}(\epsilon)}{\epsilon^3} - \frac{3}{2} \frac{e^{-\epsilon^2}}{\epsilon^2} \right); \quad \epsilon = \frac{|\mathbf{u}_b - \mathbf{u}_a|}{\sqrt{\nu_{\text{tha}}^2 + \nu_{\text{thb}}^2}}, \qquad (G48)$$

recovering Equation (26.4) of Burgers (1969) and Equation (25b) of Schunk (1977). For small values $\epsilon \to 0$, the contribution $\Phi \to 1$ (more precisely, $\Phi_{ab} = 1 - (3/5)\epsilon^2$) recovers the previous result (G21) with small drifts. However, for large values $\epsilon \gg 1$, the contribution $\Phi_{ab}$ decreases to zero as $\Phi_{ab} = 3\sqrt{\pi}/(4\epsilon^3)$, and thus for large differences the in drifts $|\mathbf{u}_b - \mathbf{u}_a|$, the momentum exchange rates $\mathbf{R}_{ab}$ disappear for Coulomb collisions. This phenomenon is known as the “runaway effect” (Dreicer 1959). It is also possible to write

$$\Phi_{ab} = \frac{3\sqrt{\pi}}{2\epsilon} \tilde{G}_{ab}(\epsilon); \quad \text{where} \quad \tilde{G}_{ab}(\epsilon) = \frac{\operatorname{erf}(\epsilon)}{2\epsilon^2} - \frac{e^{-\epsilon^2}}{\sqrt{\pi}\epsilon} = \frac{\operatorname{erf}(\epsilon) - \epsilon \operatorname{erf}'(\epsilon)}{2\epsilon^2}, \qquad (G49)$$

where $\tilde{G}_{ab}(\epsilon)$ is called the Chandrasekhar function (we use tilde to differentiate it from the Rosenbluth potential $G_b$), and (G47) then becomes

$$\mathbf{R}_{ab} = \frac{3}{2} \sqrt{\pi} \rho_a \nu_{ab} (v_{\text{tha}}^2 + v_{\text{thb}}^2)^{\frac{1}{2}} \frac{\mathbf{u}_b - \mathbf{u}_a}{|\mathbf{u}_b - \mathbf{u}_a|} \tilde{G}_{ab}(\epsilon). \qquad (G50)$$

In plasma books (e.g., Helander & Sigmar 2002), the Chandrasekhar function is typically introduced in velocity space as $\tilde{G}(v/v_{\text{thb}})$, i.e., without drifts and before the integration over $d^3v$. The runaway effect is then explained on a population of electron species, which gets accelerated by an applied external electric field. Because for large velocities $v$ frictional forces (collisions) decrease as $\tilde{G} \sim v_{\text{thb}}^2/(2v^2)$, the tail of the distribution function might depart and run away. In this sense, the runaway effect could be viewed as a purely kinetic effect. Nevertheless, an obviously analogous runaway effect exists in the fluid description (i.e., after the integration over $d^3v$), it is just represented through the difference in the drifts $\mathbf{u}_b - \mathbf{u}_a$ (which form a current $j$). For example, considering a one ion-electron plasma with an electric current $j = -en_e(\mathbf{u}_e - \mathbf{u}_i)$, then taking the electron momentum equation and neglecting, for simplicity, all of the terms except for the external $E$ and $\mathbf{R}_{ei}$ (including $\partial\mathbf{u}_e/\partial t$, which neglects acceleration) yields the relation

$$\mathbf{E} = \frac{\mathbf{R}_{ei}}{en_e} = \eta \mathbf{j}; \quad \eta = \frac{1}{\sigma} = \frac{\rho_e \nu_{ei}}{e^2 n_e^2} \Phi_{ei}, \qquad (G51)$$

which agrees with Equation (33.6) of Burgers (1969). The electrical resistivity $\eta$ now contains $\Phi_{ei}$, given by (G48) with $\epsilon = j/(en_e v_{\text{the}})$. For small values of current $j$, the $\eta$ is independent of $j$. The runaway effect means that with increasing current $j$, the electrical resistivity $\eta$ decreases, and for large current $j$, it becomes $\eta = (3\sqrt{\pi}/4)en_e \rho_e \nu_{ei} v_{\text{thc}}^3/j^3$. In reality, the problem is much more complex when the acceleration is considered, because, subtracting the two momentum equations, the general difference in the velocities $\delta\mathbf{u} = \mathbf{u}_b - \mathbf{u}_a$ now evolves according to a nonlinear differential equation,

$$\frac{\partial \delta \mathbf{u}}{\partial t} + \nu \Phi_{ab}(\epsilon) \delta \mathbf{u} = e \mathbf{E} \left( \frac{Z_b}{m_b} - \frac{Z_a}{m_a} \right); \quad \nu = \nu_{ab} + \nu_{ba}, \qquad (G52)$$

which does not seem to be solvable analytically. Nevertheless (after studying the solutions for some time), it is possible to conclude that there exist two distinct classes of solutions, which are typically separated by the value of the applied constant electric field $E$ with respect to a critical value $E_{\text{crit}}$, where the maximal frictional forces balance the electric forces. For $E < E_{\text{crit}}$, the solutions converge in
<figure><img src="image_9.png" /><figcaption>**Figure 6.** Left: functions $\Phi_{ab}$ (red line) and $\Psi_{ab}$ (blue line), with respect to $\epsilon$ defined in Equation (G64). Right: functions $\epsilon\Phi_{ab} \sim R_{ab}$ (red line) and $\epsilon^2\Phi_{ab} \sim Q_{ab}$ (blue line), where the temperature is fixed. Corresponding approximations for small drifts with $\Phi_{ab} = 1$ are also plotted (dotted lines). Function $\epsilon\Phi_{ab}$ reaches maximum 0.57 at $\epsilon = 0.97$, and function $\epsilon^2\Phi_{ab}$ reaches maximum 0.70 at $\epsilon = 1.51$. It is possible to conclude that the small drift approximation is reasonably accurate up to $\epsilon = 0.5$, and that very small values $\epsilon \ll 1$ are actually not required. Even though we did not calculate the runaway effect for higher-order moments, out of curiosity we include a function $\epsilon^3\Phi_{ab}$ (black dashed line) that does not decrease to zero for large drifts, but instead converges to a constant value of 1.33.</figcaption></figure>

time toward a situation where $\Phi_{ab} = 1$, and one recovers evolution Equation (G23) with static solution (G25). In contrast, for $E > E_{crit}$, the solutions evolve in time toward a situation with $\Phi_{ab} = 0$, which can be shown, for example, by considering solutions where $\Phi_{ab}(\epsilon)$ is approximated with its asymptotic expansion. For very large values of $E$, one can straightforwardly prescribe $\Phi_{ab} = 0$, yielding a (collisionless) solution:

$$ \mathbf{u}_b - \mathbf{u}_a = eE \left( \frac{Z_b}{m_b} - \frac{Z_a}{m_a} \right) t. \qquad (G53) $$

Thus, provided that $Z_a/m_a \neq Z_b/m_b$ is true, a stationary solution does not exist, and the difference in the velocities grows in time without bounds, before the beam/stream plasma instabilities with the associated development of turbulence (and, in extreme cases, eventually relativistic effects) restrict its further growth. For the particular case $Z_a/m_a = Z_b/m_b$, the runaway effect does not exist, and the difference in the velocities will converge to zero according to (G52). The frictional forces $\epsilon\Phi_{ab}(\epsilon)$ are plotted as a red curve in the right-hand panel of Figure 6. They reach its maximum value $[\epsilon\Phi_{ab}(\epsilon)]_{max} = 0.57$ at $\epsilon = 0.97$ (often rounded as $\epsilon = 1$). The critical electric field $E_{crit}$ is determined by making the maximum frictional forces equal to the electric forces, so that (G52) becomes $\partial\delta\mathbf{u}/\partial t = 0$, yielding

$$ E_{\text{crit}} = \frac{[\epsilon \Phi_{ab}(\epsilon)]_{\text{max}} \sqrt{v_{\text{tha}}^2 + v_{\text{thb}}^2}}{0.57} \frac{(\nu_{ab} + \nu_{ba})}{e} \frac{m_a m_b}{(Z_b m_a - Z_a m_b)} \frac{\mathbf{u}_b - \mathbf{u}_a}{|\mathbf{u}_b - \mathbf{u}_a|} \qquad (G54) $$

Alternatively, one might use the Chandrasekhar function, where $[\epsilon\Phi_{ab}]_{max} = (3/2)\sqrt{\pi}[\tilde{G}_{ab}]_{max}$, and $[\tilde{G}_{ab}]_{max} = 0.214$. The runaway effect thus exists for

$$ E > E_{\text{crit}} = \frac{[\tilde{G}_{ab}(\epsilon)]_{\text{max}} \hat{E}_D}{0.214} \qquad (G55) $$

$$ \hat{E}_D = \frac{3\sqrt{\pi}}{2} \sqrt{v_{\text{tha}}^2 + v_{\text{thb}}^2} \frac{(\nu_{ab} + \nu_{ba})}{e} \frac{m_a m_b}{|Z_b m_a - Z_a m_b|}, \qquad (G56) $$

where $\hat{E}_D$ can be viewed as a generalized Dreicer electric field for two species with arbitrary masses, charges, and temperatures. By further substituting for the collisional frequencies (we take $\ln \lambda$ to be constant),

$$ \hat{E}_D = 8\pi \frac{(m_a + m_b)}{m_a m_b} \frac{(\rho_a + \rho_b)}{|Z_b m_a - Z_a m_b|} \frac{e^3 Z_a^2 Z_b^2 \ln \lambda}{(v_{\text{tha}}^2 + v_{\text{thb}}^2)}, \qquad (G57) $$

which for an ion-electron plasma yields the usual Dreicer electric field

$$ E_D = \frac{4\pi n_i e^3 Z_i^2 \ln \lambda}{T_e}. \qquad (G58) $$

In the paper of Dreicer (1959), his reference field is defined as $E_c = E_D/2$, so in his notation the runaway effect exists for $E > 0.43E_c$ instead of $E > 0.214E_D$. In most of the recent literature, definition (G58) is used. It is sometimes incorrectly stated that the runaway effect exists for $E$ exceeding $E_D$, whereas the correct value, as calculated by Dreicer, is almost five times smaller. Note the
dependence of (G58) on $T_e$, meaning that for any given value of electric field, the runaway effect will appear if the temperatures are sufficiently high. For $Z_a/m_a = Z_b/m_b$, the $\hat{E}_D$ becomes infinitely large, and the runaway effect between these species is not present. For an ion-electron plasma, the Dreicer electric field is also discussed, for example by Tanenbaum (1967, p. 258) and Balescu (1988, p. 775). We found it useful to consider the situation for two arbitrary (charged) species.

Similar to $R_{ab}$, the $Q_{ab}$ is obtained by calculating two integrals in (G26), and the first integral yields

$$ 
\begin{aligned}
\int f_a \frac{\partial H_b}{\partial \mathbf{v}_a} \cdot \mathbf{c}_a d^3 v_a &= \int \int f_a f_b \frac{\mathbf{v}_b - \mathbf{v}_a}{|\mathbf{v}_b - \mathbf{v}_a|^3} \cdot \mathbf{c}_a d^3 v_a d^3 v_b \\
&= - \frac{n_a n_b}{\pi^{3/2} \beta^3} \frac{m_b T_a}{m_b T_a + m_a T_b} \int \frac{\mathbf{x}}{x^3} \cdot (\mathbf{x} - \mathbf{u}) e^{-\frac{|\mathbf{x}-\mathbf{u}|^2}{\beta^2}} d^3 x \\
&= - \frac{2 n_a n_b}{\sqrt{\pi} \beta^3} v_{\text{tha}}^2 e^{-\epsilon^2},
\end{aligned}
\quad (G59) $$

where we have used

$$ \int \frac{1}{x} e^{-\frac{|\mathbf{x}-\mathbf{u}|^2}{\beta^2}} d^3 x = \pi^{3/2} \beta^2 \frac{\operatorname{erf}(\epsilon)}{\epsilon}; \quad (G60) $$

$$ \int \frac{\mathbf{x}}{x} \cdot (\mathbf{x} - \mathbf{u}) e^{-\frac{|\mathbf{x}-\mathbf{u}|^2}{\beta^2}} d^3 x = 2\pi\beta^2 e^{-\epsilon^2}. \quad (G61) $$

The second integral in (G26) yields

$$ \int f_a H_b d^3 v_a = \int \int f_a f_b \frac{1}{|v_b - v_a|} d^3 v_a d^3 v_b = \frac{n_a n_b}{\beta} \frac{\operatorname{erf}(\epsilon)}{\epsilon} = \frac{n_a n_b}{\beta^3} \frac{\operatorname{erf}(\epsilon)}{\epsilon^3} |\mathbf{u}|^2. \quad (G62) $$

The entire Equation (G26) then becomes

$$ Q_{ab} = \rho_a v_{ab} \left[ -\frac{3T_a}{m_a} e^{-\epsilon^2} + \frac{3}{4} \sqrt{\pi} \frac{\operatorname{erf}(\epsilon)}{\epsilon^3} \frac{m_b}{m_a + m_b} |\mathbf{u}_b - \mathbf{u}_a|^2 \right], \quad (G63) $$

and the difference in the temperatures $T_b - T_a$ is not directly visible. Nevertheless, the solution can be rewritten into

$$ 
\begin{aligned}
Q_{ab} &= \rho_a v_{ab} \left[ 3 \frac{T_b - T_a}{m_a + m_b} \Psi_{ab} + \frac{m_b}{m_a + m_b} |\mathbf{u}_b - \mathbf{u}_a|^2 \Phi_{ab} \right]; \\
\Psi_{ab} &= e^{-\epsilon^2}; \quad \Phi_{ab} = \left( \frac{3}{4} \sqrt{\pi} \frac{\operatorname{erf}(\epsilon)}{\epsilon^3} - \frac{3}{2} \frac{e^{-\epsilon^2}}{\epsilon^2} \right); \quad \epsilon = \frac{|\mathbf{u}_b - \mathbf{u}_a|}{\sqrt{v_{\text{tha}}^2 + v_{\text{thb}}^2}},
\end{aligned}
\quad (G64) $$

recovering Equation (26.8) of Burgers (1969) and Equation (25c) of Schunk (1977). Similar to $R_{ab}$, for large differences in drifts, the $Q_{ab}$ disappears.

It is of interest to explore the validity of the results with small drifts, obtained in Appendices G.1 and G.2. The functions $\Phi_{ab}$ and $\Psi_{ab}$ are plotted in the left-hand panel of Figure 6. Both functions are decreasing, and thus in fluid models with the small drift approximation, the effects of the collisions are overestimated. We fix the temperature (so that $v_{ab} = \text{const.}$), and in the right-hand panel of Figure 6 we plot the function $\epsilon\Phi_{ab}$, which corresponds to $R_{ab}$ (red line), and function $\epsilon^2\Phi_{ab}$, which corresponds to $Q_{ab}$ (blue line). For large drifts $\epsilon \gg 1$, functions $\epsilon\Phi_{ab} \sim 3\sqrt{\pi}/(4\epsilon^2)$ and $\epsilon^2\Phi_{ab} \sim 3\sqrt{\pi}/(4\epsilon)$.

### G.4. Difficulties with Rosenbluth Potentials

It is interesting to analyze why it seems impossible to calculate the runaway effect for $R_{ab}$ through the Rosenbluth potentials, and why one needs to use the “center-of-mass” transformation instead. An attempt to calculate the runaway effect yields

$$ 
\begin{aligned}
\mathbf{R}_{ab} &= m_a \int f_a \mathbf{A}_{ab} d^3 v \\
&= -4 \frac{c_{ab}}{m_a} \left(1 + \frac{m_a}{m_b}\right) \frac{n_a n_b v_{\text{thb}}}{\pi^{3/2} v_{\text{tha}}^3} \int e^{-|\alpha y + u|^2} \frac{\mathbf{y}}{y} \left( \frac{\operatorname{erf}(y)}{2y^2} - \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y} \right) d^3 y,
\end{aligned}
\quad (G65) $$
where $\alpha = v_{\text{thb}}/v_{\text{tha}}$ and $\mathbf{u} = (\mathbf{u}_b - \mathbf{u}_a)/v_{\text{tha}}$, and we have also identified the Chandrasekhar function. First, integrating over $d\phi$, where the direction of $\mathbf{u}$ forms the axis $\hat{\mathbf{e}}_3 = \mathbf{u}/u$, yields

$$ \int e^{-|\alpha y + u|^2} \frac{\mathbf{y}}{y} \left( \frac{\text{erf}(y)}{2y^2} - \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y} \right) d^3 y \\ = 2\pi \frac{\mathbf{u}}{u} e^{-u^2} \int_0^\infty \int_0^\pi e^{-\alpha^2 y^2} e^{-2\alpha y u \cos \theta} \cos \theta \sin \theta \left( \frac{\text{erf}(y)}{2} - \frac{y}{\sqrt{\pi}} e^{-y^2} \right) dy d\theta. \qquad (G66) $$

Then one can perform the integration over $d\theta$; however, the subsequent integration over $dy$ does not seem possible. Or, one can first attempt the integration over $dy$, using the substitutions $s = u \cos \theta$, $z = \alpha y + s$, so that $|\alpha \mathbf{y} + \mathbf{u}|^2 = z^2 - s^2 + u^2$ yields

$$ (G66) = 2\pi \frac{\mathbf{u}}{u^3} e^{-u^2} \int_s^\infty \int_{-u}^u se^{+s^2} e^{-z^2} \left[ \frac{1}{2} \text{erf}\left(\frac{z-s}{\alpha}\right) - \frac{z-s}{\alpha\sqrt{\pi}} e^{-\frac{(z-s)^2}{\alpha^2}} \right] dz ds, \qquad (G67) $$

but the one-dimensional integrals over $dz$ again appear impossible to calculate. The problem is the “drift” “s,” and also the constants $\alpha$. For example, the following indefinite integral is easily calculated by parts:

$$ \int e^{-(az+b)^2} \text{erf}(az+b) dz = \frac{\sqrt{\pi}}{4a} \text{erf}^2(az+b), \qquad (G68) $$

but the result is not useful. Obviously, a different approach has to be used to integrate over the Chandrasekhar function if $f_a^{(0)}$ is a Maxwellian with unrestricted drifts.

Importantly, from Appendix G.3, where the “center-of-mass” transformation is used, we know that the correct answer has to be

$$ \int f_a^{(0)} \frac{\mathbf{y}}{y} \tilde{G}_{ab}(y) d^3 v \stackrel{!}{=} - n_a \frac{v_{\text{thb}}^2}{v_{\text{tha}}^2 + v_{\text{thb}}^2} \frac{\mathbf{u}_b - \mathbf{u}_a}{|\mathbf{u}_b - \mathbf{u}_a|} \tilde{G}_{ab}(\epsilon), \qquad (G69) $$

where $\mathbf{y} = (\mathbf{v} - \mathbf{u}_b)/v_{\text{thb}}$, $\epsilon = |\mathbf{u}_b - \mathbf{u}_a|/\sqrt{v_{\text{tha}}^2 + v_{\text{thb}}^2}$, and $d^3v = v_{\text{thb}}^3 d^3y$; or, written in a full form,

$$ \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} \int e^{-\frac{|\mathbf{v}-\mathbf{u}_a|^2}{v_{\text{tha}}^2}} \frac{\mathbf{y}}{y} \left( \frac{\text{erf}(y)}{2y^2} - \frac{e^{-y^2}}{\sqrt{\pi} y} \right) d^3 v \\ \stackrel{!}{=} - n_a \frac{v_{\text{thb}}^2}{v_{\text{tha}}^2 + v_{\text{thb}}^2} \frac{\mathbf{u}_b - \mathbf{u}_a}{|\mathbf{u}_b - \mathbf{u}_a|} \left( \frac{\text{erf}(\epsilon)}{2\epsilon^2} - \frac{e^{-\epsilon^2}}{\sqrt{\pi} \epsilon} \right). \qquad (G70) $$

Finally, written in perhaps the prettiest form, when not referring to any physical quantities (i.e., a form suitable for integral tables),

$$ \int_{-\infty}^{\infty} e^{-|\alpha y + u|^2} \frac{\mathbf{y}}{y} \left( \frac{\text{erf}(y)}{2y^2} - \frac{e^{-y^2}}{\sqrt{\pi} y} \right) d^3 y \\ \stackrel{!}{=} - \frac{\pi^{3/2}}{\alpha(1+\alpha^2)} \frac{\mathbf{u}}{u} \left( \frac{\text{erf}(\epsilon)}{2\epsilon^2} - \frac{e^{-\epsilon^2}}{\sqrt{\pi} \epsilon} \right); \quad \text{where} \quad \epsilon = \frac{u}{\sqrt{1+\alpha^2}}; \quad \alpha > 0. \qquad (G71) $$

It is remarkable that the integral has such a striking symmetry, even though the integral seems impossible to calculate directly, i.e., the integral “transfers” a Chandrasekhar function in $y$-variable to a Chandrasekhar function in $\epsilon$-variable. The result seems well-defined, even for $\alpha < 0$, so the restriction is $\alpha \neq 0$ and real (the integral is divergent for $\alpha = 0$). The limit $u \to 0$ yields zero. The “proof” of (G71) can be viewed as analogous when evaluating the one-dimensional Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2} dx$ through $\iint e^{-(x^2+y^2)} dxdy$ in polar coordinates, where, instead of integrating over $d^3v$, a trick is used to integrate over $d^3vd^3v'$.

# Appendix H
## 8-moment Model (Heat Flux and Thermal Force)

To obtain the collisional contributions with the heat flux, one uses the following 8-moment distribution function of Grad:

$$ f_b(\mathbf{v}') = \frac{n_b}{\pi^{3/2} v_{\text{thb}}^3} e^{-|c_b|^2} \left[ 1 - \frac{m_b}{T_b p_b} \left( 1 - \frac{m_b |c_b|^2}{5 T_b} \right) q_b \cdot c_b \right]. \qquad (H1) $$

The calculations done by Burgers (1969), Schunk (1977), and Killie et al. (2004) were performed using the “center-of-mass” transformation, described in Appendix G.3. Here, to do something slightly different, we verify the calculations by using the Rosenbluth potentials. The route through the Rosenbluth potentials has a great disadvantage, as error functions are encountered even if we are interested only in expressions with small drift velocities (with respect to thermal velocities). This is because the Rosenbluth potentials have to be derived with the exact (H1), and not expanded for small drifts from the beginning. Nevertheless, the route has an
advantage, as it is possible to do a double check in the middle of the calculations, because there are identities that the Rosenbluth potentials must satisfy.

## H.1. Rosenbluth Potentials

Using the same variables $x = (v' - v)/v_{\text{thb}}$ and $y = (v - u_b)/v_{\text{thb}}$ as before, so that $c_b = (x + y)v_{\text{thb}}$, we need to obtain the Rosenbluth potentials

$$H_b(v) = \int \frac{f_b(v')}{|v' - v|} d^3v' \\ = \frac{n_b}{\pi^{3/2} v_{\text{thb}}} \int \frac{e^{-|x+y|^2}}{x} \left[ 1 - \frac{m_b v_{\text{thb}}}{T_b p_b} q_b \cdot (x + y) \left( 1 - \frac{2}{5} |x + y|^2 \right) \right] d^3x; \qquad (H2)$$

$$G_b(v) = \int |v' - v| f_b(v') d^3v' \\ = \frac{n_b v_{\text{thb}}}{\pi^{3/2}} \int x e^{-|x+y|^2} \left[ 1 - \frac{m_b v_{\text{thb}}}{T_b p_b} q_b \cdot (x + y) \left( 1 - \frac{2}{5} |x + y|^2 \right) \right] d^3x. \qquad (H3)$$

It is possible to calculate the following integrals (directly obtainable with Maple in spherical geometry, after the vector integrals containing $x$ are first integrated by hand over $d\phi$):

$$\int \frac{1}{x} e^{-|x+y|^2} d^3x = \pi^{3/2} \frac{\operatorname{erf}(y)}{y}; \qquad (H4)$$

$$\int \frac{1}{x} e^{-|x+y|^2} \left(1 - \frac{2}{5}|\mathbf{x} + \mathbf{y}|^2\right) d^3x = \frac{2}{5} \pi^{3/2} \frac{\operatorname{erf}(y)}{y} + \frac{2}{5} \pi e^{-y^2}; \qquad (H5)$$

$$\int \frac{\mathbf{x}}{x} e^{-|\mathbf{x}+\mathbf{y}|^2} \left(1 - \frac{2}{5}|\mathbf{x} + \mathbf{y}|^2\right) d^3x = -\frac{2}{5} \pi^{3/2} \mathbf{y} \frac{\operatorname{erf}(y)}{y}, \qquad (H6)$$

and similarly:

$$\int x e^{-|x+y|^2} d^3x = \pi^{3/2} \left(y + \frac{1}{2y}\right) \operatorname{erf}(y) + \pi e^{-y^2}; \qquad (H7)$$

$$\int x e^{-|x+y|^2} \left(1 - \frac{2}{5}|\mathbf{x} + \mathbf{y}|^2\right) d^3x = \frac{2}{5} \pi^{3/2} y \operatorname{erf}(y) + \frac{2}{5} \pi e^{-y^2}; \qquad (H8)$$

$$\int x \mathbf{x} e^{-|x+y|^2} \left(1 - \frac{2}{5}|\mathbf{x} + \mathbf{y}|^2\right) d^3x = -\frac{2}{5} \mathbf{y} \left[ \pi^{3/2} \left(y + \frac{1}{4y^3}\right) \operatorname{erf}(y) + \pi \left(1 - \frac{1}{2y^2}\right) e^{-y^2} \right]. \qquad (H9)$$

This yields the final Rosenbluth potentials for the 8-moment model, in the following form:

$$H_b(v) = \frac{n_b}{v_{\text{thb}}} \left[ \frac{\operatorname{erf}(y)}{y} - \frac{2 m_b v_{\text{thb}}}{5 T_b p_b} (q_b \cdot y) \frac{1}{\sqrt{\pi}} e^{-y^2} \right]; \qquad (H10)$$

$$G_b(v) = n_b v_{\text{thb}} \left[ \left( y + \frac{1}{2y} \right) \operatorname{erf}(y) + \frac{1}{\sqrt{\pi}} e^{-y^2} \right] \\ + \frac{2 n_b}{5 p_b} (q_b \cdot y) \left[ \frac{\operatorname{erf}(y)}{2y^3} - \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} \right]. \qquad (H11)$$

We will need a vector

$$\frac{\partial H_b}{\partial \mathbf{v}} = \frac{n_b}{v_{\text{thb}}^2} \left[ \mathbf{y} \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\operatorname{erf}(y)}{y^3} \right) - \frac{2 m_b v_{\text{thb}}}{5 T_b p_b} (\mathbf{q}_b - 2\mathbf{y}(\mathbf{q}_b \cdot \mathbf{y})) \frac{1}{\sqrt{\pi}} e^{-y^2} \right], \qquad (H12)$$
and a matrix

$$ \frac{\partial^2 G_b}{\partial \mathbf{v} \partial \mathbf{v}} = \frac{n_b}{v_{\text{thb}}} \left( \bar{\mathbf{I}} - \frac{\mathbf{y}\mathbf{y}}{y^2} \right) \left[ \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} + \left( \frac{1}{y} - \frac{1}{2y^3} \right) \text{erf}(y) \right] + \frac{n_b}{v_{\text{thb}}} \frac{\mathbf{y}\mathbf{y}}{y^2} \left[ \frac{\text{erf}(y)}{y^3} - \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} \right] \\ + \frac{n_b m_b}{5T_b p_b} \left\{ \left[ \mathbf{q}_b \mathbf{y} + \mathbf{y} \mathbf{q}_b + (\mathbf{q}_b \cdot \mathbf{y}) \left( \bar{\mathbf{I}} - \frac{\mathbf{y}\mathbf{y}}{y^2} \right) \right] \left[ -\frac{3}{2} \frac{\text{erf}(y)}{y^5} + \frac{1}{\sqrt{\pi}} \left( \frac{2}{y^2} + \frac{3}{y^4} \right) e^{-y^2} \right] \right. \\ \left. + (\mathbf{q}_b \cdot \mathbf{y}) \frac{\mathbf{y}\mathbf{y}}{y^2} \left[ \frac{6}{y^5} \text{erf}(y) - \frac{4}{\sqrt{\pi}} \left( 1 + \frac{2}{y^2} + \frac{3}{y^4} \right) e^{-y^2} \right] \right\}. \qquad (H13) $$

As a double check, applying $(\partial/\partial\mathbf{v}) \cdot$ on (H12) recovers $-4\pi f_b$, and applying $(1/2)\text{Tr}$ on (H13) recovers $H_b$. The dynamical friction vector then reads

$$ A_{ab} = 2 \frac{c_{ab}}{m_a^2} \left( 1 + \frac{m_a}{m_b} \right) \frac{n_b}{v_{\text{thb}}^2} \left[ \mathbf{y} \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\text{erf}(y)}{y^3} \right) - \frac{2}{5} \frac{m_b v_{\text{thb}}}{T_b p_b} (\mathbf{q}_b - 2\mathbf{y}(\mathbf{q}_b \cdot \mathbf{y})) \frac{1}{\sqrt{\pi}} e^{-y^2} \right], \qquad (H14) $$

and after slight rearrangement, the diffusion tensor becomes

$$ \bar{\mathbf{D}}_{ab} = 2 \frac{c_{ab}}{m_a^2} \left\{ \frac{n_b}{v_{\text{thb}}} \bar{\mathbf{I}} \left[ \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} + \left( \frac{1}{y} - \frac{1}{2y^3} \right) \text{erf}(y) \right] + \frac{n_b}{v_{\text{thb}}} \frac{\mathbf{y}\mathbf{y}}{y^2} \left[ \left( \frac{3}{2y^3} - \frac{1}{y} \right) \text{erf}(y) - \frac{3}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} \right] \right. \\ + \frac{n_b m_b}{5T_b p_b} [\mathbf{q}_b \mathbf{y} + \mathbf{y} \mathbf{q}_b + (\mathbf{q}_b \cdot \mathbf{y}) \bar{\mathbf{I}}] \left[ -\frac{3}{2} \frac{\text{erf}(y)}{y^5} + \frac{1}{\sqrt{\pi}} \left( \frac{2}{y^2} + \frac{3}{y^4} \right) e^{-y^2} \right] \\ \left. + \frac{n_b m_b}{5T_b p_b} (\mathbf{q}_b \cdot \mathbf{y}) \frac{\mathbf{y}\mathbf{y}}{y^2} \left[ \frac{15}{2} \frac{\text{erf}(y)}{y^5} - \frac{1}{\sqrt{\pi}} \left( 4 + \frac{10}{y^2} + \frac{15}{y^4} \right) e^{-y^2} \right] \right\}. \qquad (H15) $$

## H.2. Momentum Exchange Rates $R_{ab}$

Then, similar to $f_b$ according to (H1), one prescribes for species “a”:

$$ f_a(\mathbf{v}) = \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-\frac{|c_a|^2}{v_{\text{tha}}^2}} \left[ 1 - \frac{m_a}{T_a p_a} \left( 1 - \frac{m_a |c_a|^2}{5T_a} \right) \mathbf{q}_a \cdot \mathbf{c}_a \right], \qquad (H16) $$

and introduces the variable $\mathbf{u} = (\mathbf{u}_b - \mathbf{u}_a)/v_{\text{tha}}$, so that $\mathbf{c}_a = \mathbf{y}v_{\text{thb}} + \mathbf{u}v_{\text{tha}}$. However, the resulting integrals would yield the runaway effect, and were never evaluated. It is necessary to get rid of the runaway effect, and approximate the $f_a$ with small drifts $u \ll 1$, and in the first step

$$ f_a(\mathbf{v}) \approx \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-\alpha^2 y^2} \left[ 1 - 2\alpha(\mathbf{y} \cdot \mathbf{u}) - u^2 + 2\alpha^2(\mathbf{y} \cdot \mathbf{u})^2 \right] \left[ 1 - \frac{m_a}{T_a p_a} \mathbf{q}_a \cdot (\mathbf{y}v_{\text{thb}} + \mathbf{u}v_{\text{tha}}) \right. \\ \left. + \frac{m_a^2}{5T_a^2 p_a} \mathbf{q}_a \cdot (\mathbf{y}v_{\text{thb}} + \mathbf{u}v_{\text{tha}})(\mathbf{y}^2 v_{\text{thb}}^2 + 2\mathbf{y} \cdot \mathbf{u}v_{\text{tha}} v_{\text{thb}} + u^2 v_{\text{tha}}^2) \right], \qquad (H17) $$

where $\alpha = v_{\text{thb}}/v_{\text{tha}}$. The distribution function (H17) needs to be further reduced to the “semilinear approximation,” where the difference in the temperatures is not restricted, but one keeps only the precision $o(\mathbf{u})$ and also neglects all of the cross terms such as $\mathbf{q}_a \cdot \mathbf{u}$, keeping only

$$ f_a(\mathbf{v}) \approx \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-\alpha^2 y^2} \left[ 1 - 2\alpha(\mathbf{y} \cdot \mathbf{u}) - \frac{m_a}{T_a p_a} (\mathbf{q}_a \cdot \mathbf{y}) v_{\text{thb}} \left( 1 - \frac{2}{5}\alpha^2 y^2 \right) \right]. \qquad (H18) $$

We want to obtain

$$ \mathbf{R}_{ab} = 2 \frac{c_{ab}}{m_a} \left( 1 + \frac{m_a}{m_b} \right) \iint f_a \frac{\partial H_b}{\partial \mathbf{v}} d^3 \mathbf{v}, \qquad (H19) $$
and we split the calculation into two integrals of (H12). The first integral ~y calculates

$$ 
\begin{aligned}
& \int \mathbf{y} \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\operatorname{erf}(y)}{y^3} \right) f_a \, d^3\mathbf{v} \\
& = \frac{n_a v_{\text{thb}}^3}{\pi^{3/2} v_{\text{tha}}^3} 4\pi \left[ \frac{\alpha}{3\alpha^2(1 + \alpha^2)^{3/2}} \mathbf{u} + \frac{m_a}{T_a p_a} \mathbf{q}_a v_{\text{thb}} \frac{1}{10\alpha^2(1 + \alpha^2)^{5/2}} \right],
\end{aligned}
\quad (H20) $$

where we have used

$$ \int_0^\infty e^{-\alpha^2 y^2} y^4 \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\operatorname{erf}(y)}{y^3} \right) dy = - \frac{1}{2\alpha^2(1 + \alpha^2)^{3/2}}; \quad (H21) $$

$$ \int_0^\infty e^{-\alpha^2 y^2} y^4 \left( 1 + \frac{2}{5}\alpha^2 y^2 \right) \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\operatorname{erf}(y)}{y^3} \right) dy = - \frac{3}{10\alpha^2(1 + \alpha^2)^{5/2}}, \quad (H22) $$

and the second part of (H12) calculates

$$ \frac{2}{\sqrt{\pi}} \int e^{-y^2} (\mathbf{q}_b - 2\mathbf{y}(\mathbf{q}_b \cdot \mathbf{y})) f_a \, d^3\mathbf{v} = \frac{n_a v_{\text{thb}}^3}{\pi^{3/2} v_{\text{tha}}^3} \mathbf{q}_b 2\pi \frac{\alpha^2}{(1 + \alpha^2)^{5/2}}. \quad (H23) $$

For a quick conversion to the collisional frequencies, it is useful to write

$$ \nu_{ab} = \frac{8}{3\sqrt{\pi}} \frac{n_b}{v_{\text{tha}}^3 (1 + \alpha^2)^{3/2}} \frac{c_{ab}}{m_a^2} \left( 1 + \frac{m_a}{m_b} \right). \quad (H24) $$

Putting the results together yields the final result:

$$ \mathbf{R}_{ab} = \rho_a \nu_{ab} (\mathbf{u}_b - \mathbf{u}_a) + \nu_{ab} \frac{3}{5} \frac{\mu_{ab}}{T_{ab}} \left( \mathbf{q}_a - \frac{\rho_a}{\rho_b} \mathbf{q}_b \right), \quad (H25) $$

recovering Equation (41b) of Schunk (1977; derived before by Burgers 1969). Alternatively, $\mu_{ab}/T_{ab} = 2/(v_{\text{tha}}^2 + v_{\text{thb}}^2)$. As a double check, $\mathbf{R}_{ab} = -\mathbf{R}_{ba}$, and for self-collisions, $\mathbf{R}_{aa} = 0$, as it should be. The contribution coming from the heat flux is known as the thermal force.

### H.3. Heat Flux Exchange Rates

To calculate the heat flux contributions, one needs to calculate

$$ \mathbf{Q}_{ab}^{(3) \prime} = \frac{\delta \mathbf{q}_{ab}^{\prime}}{\delta t} = \frac{1}{2} \mathrm{Tr} \bar{\mathbf{Q}}_{ab}^{(3)} - \frac{5}{2} \frac{p_a}{\rho_a} \mathbf{R}_{ab} - \frac{1}{\rho_a} \mathbf{R}_{ab} \cdot \bar{\mathbf{\Pi}}_{a}^{(2)}, \quad (H26) $$

where $\bar{\mathbf{\Pi}}_{a}^{(2)} = 0$ for the 8-moment model (the cross term $\mathbf{R}_{ab} \cdot \bar{\mathbf{\Pi}}_{a}^{(2)}$ would be neglected anyway) and where

$$ \frac{1}{2} \mathrm{Tr} \bar{\mathbf{Q}}_{ab}^{(3)} = m_a \int f_a \left[ (\mathbf{A}_{ab} \cdot \mathbf{c}_a) \mathbf{c}_a + \frac{1}{2} \mathbf{A}_{ab} |\mathbf{c}_a|^2 \right] d^3\mathbf{v} \\ + m_a \int f_a \left[ \frac{1}{2} (\mathrm{Tr} \bar{\mathbf{D}}_{ab}) \mathbf{c}_a + \bar{\mathbf{D}}_{ab} \cdot \mathbf{c}_a \right] d^3\mathbf{v}. \quad (H27) $$

We have used $\mathrm{Tr}[\mathbf{A}\mathbf{c}]^S = 2(\mathbf{A} \cdot \mathbf{c})\mathbf{c} + \mathbf{A}|\mathbf{c}|^2$, and because the diffusion tensor is symmetric, $\bar{\mathbf{D}}^S = 2\bar{\mathbf{D}}$ and $\mathrm{Tr}[\bar{\mathbf{D}}^S\mathbf{c}]^S = 2(\mathrm{Tr}\bar{\mathbf{D}})\mathbf{c} + 4\bar{\mathbf{D}} \cdot \mathbf{c}$. By assuming no restriction on the temperature difference, we have verified (with the great help of Maple) that “semilinear” heat flux contributions (45)–(49) of Schunk (1977; derived before by Burgers 1969) are indeed correct for Coulomb collisions (with $z_{\text{st}} = 3/5$, $z_{\text{st}'} = 13/10$, $z_{\text{st}''} = 2$, and also $z_{\text{st}'''} = 4$). For Coulomb collisions, the final result (after the subtraction of $\frac{5}{2} \frac{p_a}{\rho_a} \mathbf{R}_a$) is written in a compact form in Section 2.3; see Equation (32).

In the “linear approximation,” where the temperature differences are small, the result simplifies into

$$ \frac{1}{2} \mathrm{Tr} \bar{\mathbf{Q}}_{ab}^{(3)} = \frac{\delta \mathbf{q}_{ab}}{\delta t} = \nu_{ab} \left[ -\mathbf{q}_a D_{ab(1)} + \mathbf{q}_b \frac{\rho_a}{\rho_b} D_{ab(4)} + p_a (\mathbf{u}_b - \mathbf{u}_a) \frac{m_b + \frac{5}{2} m_a}{m_a + m_b} \right], \quad (H28) $$
where the introduced constants are defined in (H30), (H31). Alternatively, by summing over all of the “b” species and separating the self-collisions,

$$ \frac{1}{2} \mathrm{Tr} \bar{\mathcal{Q}}_a^{(3)} = \frac{\delta \mathbf{q}_a}{\delta t} = -\frac{4}{5} \nu_{aa} \mathbf{q}_a - \sum_{b \neq a} \nu_{ab} \left[ D_{ab(1)} \mathbf{q}_a - D_{ab(4)} \frac{\rho_a}{\rho_b} \mathbf{q}_b - p_a (\mathbf{u}_b - \mathbf{u}_a) \frac{m_b + \frac{5}{2} m_a}{m_a + m_b} \right]; \quad (H29) $$

$$ D_{ab(1)} = \frac{1}{(m_a + m_b)^2} \left( 3m_a^2 + \frac{1}{10}m_a m_b - \frac{1}{5}m_b^2 \right); \quad (H30) $$

$$ D_{ab(4)} = \frac{1}{(m_a + m_b)^2} \left( \frac{6}{5}m_b^2 - \frac{3}{2}m_a m_b \right), \quad (H31) $$

recovering Equations (41e)-(43) of Schunk (1977); see also Equations (34)-(36) of Killie et al. (2004). The entire heat flux contributions are thus

$$ \begin{aligned} \mathcal{Q}_a^{(3) \prime} &= \frac{\delta \mathbf{q}_a}{\delta t} = \frac{1}{2} \mathrm{Tr} \bar{\mathcal{Q}}_a^{(3)} - \frac{5}{2} \frac{p_a}{\rho_a} \mathbf{R}_a \\ &= -\mathbf{q}_a \left[ \frac{4}{5} \nu_{aa} + \sum_{b \neq a} \nu_{ab} \left( D_{ab(1)} + \frac{3}{2} \frac{p_a}{\rho_a} \frac{\mu_{ab}}{T_{ab}} \right) \right] + \sum_{b \neq a} \mathbf{q}_b \nu_{ab} \frac{\rho_a}{\rho_b} \left( D_{ab(4)} + \frac{3}{2} \frac{p_a}{\rho_a} \frac{\mu_{ab}}{T_{ab}} \right) \\ &\quad - \frac{3}{2} p_a \sum_{b \neq a} \nu_{ab} \frac{m_b}{m_a + m_b} (\mathbf{u}_b - \mathbf{u}_a), \end{aligned} \quad (H32) $$

and enter the right-hand side of the evolution equation for the heat flux vector, for example, in its simplest form:

$$ \frac{d_a \mathbf{q}_a}{dt} + \Omega_a \hat{\mathbf{b}} \times \mathbf{q}_a + \frac{5}{2} \frac{p_a}{m_a} \nabla T_a = \mathcal{Q}_a^{(3) \prime}. \quad (H33) $$

Importantly, in comparison to the BGK operator, the right-hand side also contains all of the heat fluxes $\mathbf{q}_b$. Formally, it is still possible to obtain a result for $\mathbf{q}_a$ in a quasistatic approximation, as a solution of the equation

$$ \hat{\mathbf{b}} \times \mathbf{q}_a + \frac{\bar{\nu}_a}{\Omega_a} \mathbf{q}_a = - \frac{\mathbf{a}_a}{\Omega_a}, \quad (H34) $$

where we define

$$ \bar{\nu}_a = \frac{4}{5} \nu_{aa} + \sum_{b \neq a} \nu_{ab} \left( D_{ab(1)} + \frac{3}{2} \frac{p_a}{\rho_a} \frac{\mu_{ab}}{T_{ab}} \right); \quad (H35) $$

$$ \begin{aligned} \mathbf{a}_a = \frac{5}{2} \frac{p_a}{m_a} \nabla T_a & - \sum_{b \neq a} \mathbf{q}_b \nu_{ab} \frac{\rho_a}{\rho_b} \left( D_{ab(4)} + \frac{3}{2} \frac{p_a}{\rho_a} \frac{\mu_{ab}}{T_{ab}} \right) \\ & + \frac{3}{2} p_a \sum_{b \neq a} \nu_{ab} \frac{m_b}{m_a + m_b} (\mathbf{u}_b - \mathbf{u}_a), \end{aligned} \quad (H36) $$

which has the following exact solution:

$$ \mathbf{q}_a = - \frac{1}{\bar{\nu}_a} (\mathbf{a}_a \cdot \hat{\mathbf{b}}) \hat{\mathbf{b}} - \frac{\bar{\nu}_a}{\Omega_a^2 + \bar{\nu}_a^2} \mathbf{a}_a \perp + \frac{\Omega_a}{\Omega_a^2 + \bar{\nu}_a^2} \hat{\mathbf{b}} \times \mathbf{a}_a. \quad (H37) $$

Nevertheless, the heat fluxes of various species are coupled.

### H.4. One Ion–Electron Plasma

Considering a one ion–electron plasma (so $n_e = Z_i n_i$) with small differences in temperature, and neglecting the ratios $m_e/m_i$, the ion and electron heat fluxes decouple. For the electron species $D_{ei(1)} = -1/5$, $D_{ei(4)} = 6/5$, and $\mu_{ei} = m_e$, by using abbreviation $\delta \mathbf{u} = \mathbf{u}_e - \mathbf{u}_i$,

$$ \mathbf{R}_e = - \rho_e \nu_{ei} \delta \mathbf{u} + \nu_{ei} \frac{\rho_e}{p_e} \frac{3}{5} \mathbf{q}_e; \quad (H38) $$

$$ \frac{\delta \mathbf{q}_e}{\delta t} = - \mathbf{q}_e \left( \frac{4}{5} \nu_{ee} - \frac{1}{5} \nu_{ei} \right) - \nu_{ei} p_e \delta \mathbf{u}. \quad (H39) $$
The entire heat flux contributions are

$$Q_e^{(3)} = - \bar{\nu}_e q_e + \frac{3}{2} \nu_{ei} p_e \delta \mathbf{u}; \qquad (H40)$$

$$\bar{\nu}_e = \frac{4}{5} \nu_{ee} + \frac{13}{10} \nu_{ei}; \qquad (H41)$$

$$\mathbf{a}_e = \frac{5}{2} \frac{p_e}{m_e} \nabla T_e - \frac{3}{2} \nu_{ei} p_e \delta \mathbf{u}, \qquad (H42)$$

yielding a solution for the electron heat flux (split into a thermal part and a frictional part):

$$\mathbf{q}_e^T = - \kappa_{\parallel}^e \nabla_{\parallel} T_e - \kappa_{\perp}^e \nabla_{\perp} T_e + \kappa_{\times}^e \hat{\mathbf{b}} \times \nabla T_e; \qquad (H43)$$

$$\mathbf{q}_e^u = + \frac{3}{2} \frac{\nu_{ei}}{\bar{\nu}_e} p_e \delta \mathbf{u}_{\parallel} + \frac{3}{2} \frac{\bar{\nu}_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} p_e \delta \mathbf{u}_{\perp} - \frac{3}{2} \frac{\Omega_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} p_e \hat{\mathbf{b}} \times \delta \mathbf{u}, \qquad (H44)$$

with the thermal conductivities

$$\kappa_{\parallel}^e = \frac{5}{2} \frac{p_e}{\bar{\nu}_e m_e}; \qquad \kappa_{\perp}^e = \frac{5}{2} \frac{p_e}{m_e} \frac{\bar{\nu}_e}{(\Omega_e^2 + \bar{\nu}_e^2)}; \qquad \kappa_{\times}^e = \frac{5}{2} \frac{p_e}{m_e} \frac{\Omega_e}{(\Omega_e^2 + \bar{\nu}_e^2)}. \qquad (H45)$$

The thermal conductivities have the same form as the BGK conductivities. The only difference is that while $\bar{\nu}_e = \nu_{ee} + \nu_{ei}$ for the BGK operator, now we have to use (H41). By using $\nu_{ee} = \nu_{ei}/(Z_i \sqrt{2})$ from Equation (182),

$$\bar{\nu}_e = \left( \frac{1}{Z_i \sqrt{2}} \frac{4}{5} + \frac{13}{10} \right) \nu_{ei}; \quad \text{for } Z_i = 1: \quad \bar{\nu}_e = 1.866 \nu_{ei}. \qquad (H46)$$

The momentum exchange rates are also split into a frictional part and a thermal part:

$$\mathbf{R}_e^u = - \rho_e \nu_{ei} \left[ \left(1 - \frac{9}{10} \frac{\nu_{ei}}{\bar{\nu}_e}\right) \delta \mathbf{u}_{\parallel} + \left(1 - \frac{9}{10} \frac{\bar{\nu}_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2}\right) \delta \mathbf{u}_{\perp} + \frac{9}{10} \frac{\Omega_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} \hat{\mathbf{b}} \times \delta \mathbf{u} \right]; \qquad (H47)$$

$$\mathbf{R}_e^T = - \frac{3}{2} \frac{\nu_{ei}}{\bar{\nu}_e} n_e \nabla_{\parallel} T_e - \frac{3}{2} \frac{\bar{\nu}_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} n_e \nabla_{\perp} T_e + \frac{3}{2} \frac{\Omega_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} n_e \hat{\mathbf{b}} \times \nabla T_e. \qquad (H48)$$

In comparison, the Braginskii (1965) result for $Z_i = 1$ reads

$$\mathbf{R}_e^u = - \rho_e \nu_{ei} \left[ 0.51 \delta \mathbf{u}_{\parallel} + \left(1 - \frac{6.42x^2 + 1.84}{x^4 + 14.79x^2 + 3.77}\right) \delta \mathbf{u}_{\perp} + \frac{x(1.70x^2 + 0.78)}{x^4 + 14.79x^2 + 3.77} \hat{\mathbf{b}} \times \delta \mathbf{u} \right]; \\ \mathbf{R}_e^T = -0.71 n_e \nabla_{\parallel} T_e - \frac{5.10x^2 + 2.68}{x^4 + 14.79x^2 + 3.77} n_e \nabla_{\perp} T_e + \frac{x((3/2)x^2 + 3.05)}{x^4 + 14.79x^2 + 3.77} n_e \hat{\mathbf{b}} \times \nabla T_e, \qquad (H49)$$

where $x = \Omega_e/\nu_{ei}$. The heat flux and associated thermal force of Burgers and Schunk therefore finally explains the entire mathematical structure of the Braginskii equations, i.e., all of the terms are finally present, only the numerical values are different.

Examining the obtained numerical values in the limit of a strong magnetic field with $Z_i = 1$ (where, for simplicity, we neglect all of the ratios $\nu_{ei}/\Omega_e$), yields

$$\mathbf{R}_e = - \rho_e \nu_{ei} (0.518 \delta \mathbf{u}_{\parallel} + \delta \mathbf{u}_{\perp}) - 0.80 n_e \nabla_{\parallel} T_e; \\ \mathbf{q}_e^u = +0.80 p_e \delta \mathbf{u}_{\parallel}, \qquad (H50)$$

which is very close to the Braginskii values

$$\mathbf{R}_e = - \rho_e \nu_{ei} (0.513 \delta \mathbf{u}_{\parallel} + \delta \mathbf{u}_{\perp}) - 0.71 n_e \nabla_{\parallel} T_e; \\ \mathbf{q}_e^u = +0.71 p_e \delta \mathbf{u}_{\parallel}. \qquad (H51)$$

Note that both results (H50), (H51) contain the same symmetrical constants 0.8 and 0.71 in the frictional heat flux $\mathbf{q}_e^u$ and the thermal force $\mathbf{R}_e^T$. This is known as the Onsager symmetry, and it is also valid for a general magnetic field strength and a general charge, as can be seen by comparing (H44) and (H48).

Continuing with the strong magnetic field and examining the perpendicular heat conductivities yields ($Z_i = 1$ for $\kappa_{\perp}^e$)

$$\kappa_{\perp}^e = 4.66 \frac{p_e \nu_{ei}}{m_e \Omega_e^2}; \qquad \kappa_{\times}^e = \frac{5}{2} \frac{p_e}{m_e \Omega_e}, \qquad (H52)$$
and both match Braginskii exactly. Nevertheless, the parallel heat conductivity (which is independent of magnetic field strength; $Z_i = 1$)

$$ \kappa_{\parallel}^{e} = 1.34 \frac{p_e}{\nu_{ei} m_e}, \qquad (H53) $$

which is quite low in comparison to the Braginskii value of 3.16.

### H.5. Ion Species

For ion species $D_{ie(1)} = 3$, $D_{ie(4)} = -3m_e/(2m_i)$, and identical proton and electron temperatures, the momentum exchange rates (H25) yield

$$ \mathbf{R}_i = \rho_i \nu_{ie} \delta \mathbf{u} - \nu_{ie} \frac{\rho_i}{p_e} \frac{3}{5} \mathbf{q}_e = - \mathbf{R}_e, \qquad (H54) $$

and $R_e$ has already been calculated. Furthermore, the collisional heat flux contributions (H32)–(H37) simplify into

$$ Q_i^{(3)} = - \bar{\nu}_i q_i; \qquad (H55) $$

$$ \bar{\nu}_i = \frac{4}{5} \nu_{ii} + 3 \nu_{ie}; \qquad (H56) $$

$$ \mathbf{a}_i = \frac{5}{2} \frac{p_i}{m_i} \nabla T_i, \qquad (H57) $$

where the electron heat flux $q_e$ notably cancels out exactly for equal temperatures. Ion frequencies should thus be added according to

$$ \begin{aligned} \bar{\nu}_i &= \left( \frac{4}{5} + 3 \frac{\sqrt{2}}{Z_i} \sqrt{\frac{m_e}{m_i}} \right) \nu_{ii}; && \text{for } Z_i = 1: \quad \bar{\nu}_i = 0.899 \nu_{ii}; \\ &= \frac{4}{5} \left( 1 + \frac{15}{2Z_i} \sqrt{\frac{m_e}{2m_i}} \right) \nu_{ii}. && \end{aligned} \qquad (H58) $$

The model of Burgers–Schunk yields the ion heat flux

$$ \mathbf{q}_i = - \kappa_{\parallel}^i \nabla_{\parallel} T_i - \kappa_{\perp}^i \nabla_{\perp} T_i + \kappa_{\times}^i \hat{\mathbf{b}} \times \nabla T_i, \qquad (H59) $$

with the ion thermal conductivities

$$ \kappa_{\parallel}^{i} = \frac{5}{2} \frac{p_i}{\bar{\nu}_i m_i}; \qquad \kappa_{\perp}^{i} = \frac{5}{2} \frac{p_i}{m_i} \frac{\bar{\nu}_i}{(\Omega_i^2 + \bar{\nu}_i^2)}; \qquad \kappa_{\times}^{i} = \frac{5}{2} \frac{p_i}{m_i} \frac{\Omega_i}{(\Omega_i^2 + \bar{\nu}_i^2)}, \qquad (H60) $$

where frequencies are added according to (H58). Importantly, the ion–electron contributions are not completely negligible, and without them $\bar{\nu}_i = (4/5)\nu_{ii} = 0.8\nu_{ii}$.

However, in the work of Braginskii (1965), the ion–electron collisions are neglected for the ion heat fluxes and viscosities, and only ion self-collisions are accounted for. This can be seen from his ion coefficients that do not depend on $Z_i$. Neglecting the ion–electron collisions, the model of Burgers–Schunk yields

$$ \kappa_{\parallel}^{i} = \frac{25}{8} \frac{p_i}{\nu_{ii} m_i}; \qquad \kappa_{\perp}^{i} = 2 \frac{p_i}{m_i} \frac{\nu_{ii}}{\Omega_i^2 + (4/5)^2 \nu_{ii}^2}; \qquad \kappa_{\times}^{i} = \frac{5}{2} \frac{p_i}{m_i} \frac{\Omega_i}{\Omega_i^2 + (4/5)^2 \nu_{ii}^2}. \qquad (H61) $$

For the parallel conductivity, $\kappa_{\parallel}^i \sim 25/8 = 3.125$, in comparison to Braginskii’s 3.906. In the strong magnetic field limit,

$$ \kappa_{\perp}^{i} = 2 \frac{p_i}{m_i} \frac{\nu_{ii}}{\Omega_i^2}; \qquad \kappa_{\times}^{i} = \frac{5}{2} \frac{p_i}{m_i \Omega_i}, \qquad (H62) $$

and both match Braginskii exactly (!). If ion–electron collisions are taken into account, these Burger–Schunk coefficients change into (for $Z_i = 1$) $\kappa_{\parallel}^i \sim 2.78$, $\kappa_{\perp}^i \sim 2.24$, and $\kappa_{\times}^i \sim 5/2$, and the perpendicular $\kappa_{\perp}^i$ would suddenly not match Braginskii. It would not make sense for the electron $\kappa_{\perp}^e$ to match Braginskii exactly (for a strong B-field) and the ion $\kappa_{\perp}^i$ not to, which is a definitive indication that ion–electron collisions are neglected in Braginskii.

Including the ion–electron collisions, the $\kappa_{\perp}^i$ in the strong B-limit reads

$$ \kappa_{\perp}^{i} = \frac{p_i \nu_{ii}}{m_i \Omega_i^2} \left( 2 + \frac{15}{Z_i} \sqrt{\frac{m_e}{2m_i}} \right). \qquad (H63) $$

Neglecting ion–electron collisions with respect to ion–ion (self) collisions is analogous to neglecting 0.1 with respect to 0.8—the contribution is not tiny.
<table><caption>**Table 8**
Parallel Friction Force $\mathbf{R}_e^u = - \alpha_0 \rho_e \nu_{ei} \delta\mathbf{u}_\parallel$, Coefficient $\alpha_0$ is Plotted, or Parallel Electrical Resistivity $\eta_\parallel = 1/\sigma_\parallel = \alpha_0 m_e \nu_{ei} / (e^2 n_e)$</caption><thead><tr><th>∥ Friction Force <b>R</b><sub>e</sub><sup>u</sup></th><th>Z<sub>i</sub> = 1</th><th>Z<sub>i</sub> = 2</th><th>Z<sub>i</sub> = 3</th><th>Z<sub>i</sub> = 4</th><th>Z<sub>i</sub> = 16</th><th>Z<sub>i</sub> = ∞</th></tr></thead><tbody><tr><td>Burgers–Schunk (N = 1)</td><td>0.518</td><td>0.431</td><td>0.395</td><td>0.376</td><td>0.326</td><td>0.308</td></tr><tr><td>Killie et al.</td><td>0.597</td><td>0.460</td><td>0.391</td><td>0.349</td><td>0.231</td><td>0.182</td></tr><tr><td>Braginskii (N = 2)</td><td>0.513</td><td><b>0.431</b></td><td>0.395</td><td>0.375</td><td>0.319</td><td>0.2949</td></tr><tr><td>Landshoff (N = 4)</td><td>0.508</td><td>0.430</td><td>0.395</td><td></td><td></td><td>0.29455</td></tr><tr><td>Spitzer–Härm (N = ∞)</td><td>0.506</td><td>0.431</td><td></td><td>0.375</td><td>0.319</td><td>0.2945</td></tr></tbody></table>

**Note.** The model of Burgers–Schunk is more precise than that of Killie et al. The model of Landshoff for $N=1$ matches Burgers–Schunk, and for $N=2$ it matches Braginskii. For $Z_i=1$, the value of Landshoff ($N=4$) is slightly corrected ($0.509 \to 0.508$, emphasized with bold font) from the more precise work of Kaneko (1960), and the values of Landshoff for other $Z_i$ might be slightly incorrect. The values of Braginskii for $Z_i=2, 3$ in his Table II are slightly incorrect, and we have used the values from analytic expression (56), which now also match Landshoff ($N=2$). The Braginskii value for $Z_i=16$ is also from (56). From Kaneko & Taguchi (1978), Kaneko & Yamao (1980), and Ji & Held (2013), the “final” value for $Z_i=1$ is $\alpha_0 = 0.50612$, and the result of Spitzer–Härm is correct. Note that by keeping $n_e$ and $T_e$ constant in the definition of $\nu_{ei}$, the friction force $\sim \alpha_0 \nu_{ei}$ actually increases with increasing $Z_i$ (and the electrical conductivity decreases).

# Appendix I
## Comparison of Various Models with Braginskii (Electrons)

Focusing on the parallel direction, the momentum exchange rates $\mathbf{R}_{e\parallel}$ and electron heat flux $\mathbf{q}_{e\parallel}$ can be written in a general form:

$$ 
\begin{aligned}
\mathbf{R}_{e\parallel} &= -\alpha_0 \rho_e \nu_{ei} \delta\mathbf{u}_\parallel - \beta_0 n_e \nabla_\parallel T_e; \\
\mathbf{q}_{e\parallel} &= +\beta_0^* p_e \delta\mathbf{u}_\parallel - \gamma_0 \frac{p_e}{m_e \nu_{ei}} \nabla_\parallel T_e.
\end{aligned} 
\quad (11) 
$$

The Braginskii (1965) values of $\alpha_0$; $\beta_0 = \beta_0^*$ and $\gamma_0$ are given in his Table 2 (p. 25). The model of Burgers (1969)–Schunk (1977) is given by

$$ \alpha_0 = 1 - \frac{9}{10} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \beta_0 = \beta_0^* = \frac{3}{2} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \gamma_0 = \frac{5}{2} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \bar{\nu}_e = \left( \frac{1}{Z_i \sqrt{2}} \frac{4}{5} + \frac{13}{10} \right) \nu_{ei}, \quad (12) $$

or equivalently,

$$ \alpha_0 = \frac{\sqrt{2} + Z_i}{\sqrt{2} + (13/4)Z_i}; \quad \beta_0 = \beta_0^* = \frac{15Z_i}{4\sqrt{2} + 13Z_i}; \quad \gamma_0 = \frac{25Z_i}{4\sqrt{2} + 13Z_i}. \quad (13) $$

The model of Killie et al. (2004), discussed in Appendix I.2, yields

$$ \alpha_0 = 1 - \frac{9}{35} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \beta_0 = \frac{3}{7} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \beta_0^* = \frac{3}{2} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \gamma_0 = \frac{5}{2} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \bar{\nu}_e = \left( \frac{1}{Z_i \sqrt{2}} \frac{16}{35} + \frac{11}{35} \right) \nu_{ei}. \quad (14) $$

The other included models are described below.

In Table 8, we compare the parallel friction force; in Table 9, the parallel thermal force; in Table 10, the parallel thermal heat flux (thermal conductivity $\kappa_\parallel^e$); and in Table 11, the parallel frictional heat flux. Furthermore, in Table 12, we compare $\kappa_\perp^e$ in the strong magnetic field limit.

We include the numerical model of Spitzer & Härm (1953; see also Spitzer 1962), with their notation being discussed in Appendix I.1, which reads:

$$ \alpha_0 = \frac{3\pi}{32\gamma_E}; \quad \beta_0 = \frac{3}{2}\frac{\gamma_T}{\gamma_E}; \quad \beta_0^* = 4\frac{\delta_E}{\gamma_E} - \frac{5}{2}; \quad \gamma_0 = \epsilon\delta_T\frac{320}{3\pi}, \quad (15) $$

with the numerical values of $\gamma_E$, $\gamma_T$, $\delta_E$, $\delta_T$, and $\epsilon$ given by Table III in Spitzer & Härm (1953). For the Lorentzian plasma ($Z_i = \infty$), the coefficients are $\gamma_E = \gamma_T = \delta_E = \delta_T = 1$ and $\epsilon = 2/5$. We also include the model of Landshoff (1949, 1951), who calculated several transport coefficients (with the inclusion of a magnetic field) before Spitzer and Braginskii, and studied convergence with increasing Laguerre polynomials from $N=1$ to $N=4$ (in his work, $i=N+1$). The model is interesting because for $N=1$, it matches the values of Burgers–Schunk, and for $N=2$, it matches Braginskii. His model can be figured out to be

$$ \alpha_0 = \frac{1}{Z_i}(\Delta_{00}/\Delta)^{-1}; \quad \beta_0 = \beta_0^* = \frac{5}{2} \frac{\Delta_{01}/\Delta}{\Delta_{00}/\Delta}; \quad \gamma_0 = \frac{25}{4} Z_i \left( \frac{\Delta_{11}}{\Delta} - \frac{(\Delta_{01}/\Delta)^2}{\Delta_{00}/\Delta} \right), \quad (16) $$
<table><caption>**Table 9**
Parallel Thermal Force $R_e^T = -\beta_0 n_e \nabla_{\parallel} T_e$, Coefficient $\beta_0$ is Plotted</caption><thead><tr><th>‖ Thermal Force R<sub>e</sub><sup>T</sup></th><th>Z<sub>i</sub> = 1</th><th>Z<sub>i</sub> = 2</th><th>Z<sub>i</sub> = 3</th><th>Z<sub>i</sub> = 4</th><th>Z<sub>i</sub> = 16</th><th>Z<sub>i</sub> = ∞</th></tr></thead><tbody><tr><td>Burgers–Schunk</td><td>0.804</td><td>0.948</td><td>1.008</td><td>1.041</td><td>1.123</td><td>1.154</td></tr><tr><td>Killie et al.</td><td>0.672</td><td>0.901</td><td>1.015</td><td>1.085</td><td>1.281</td><td>1.364</td></tr><tr><td>Braginskii</td><td>0.711</td><td>0.905</td><td>1.016</td><td>1.090</td><td>1.362</td><td>1.521</td></tr><tr><td>Landshoff (N = 4)</td><td><b>0.709</b></td><td>0.904</td><td>1.016</td><td></td><td></td><td>1.5005</td></tr><tr><td>Spitzer–Härm</td><td>0.703</td><td>0.908</td><td></td><td>1.092</td><td>1.346</td><td>3/2</td></tr></tbody></table>

**Note.** The model of Killie et al. is more precise than that of Burgers–Schunk. The model of Landshoff for $N = 1$ matches Burgers–Schunk, and for $N = 2$ it matches Braginskii. For $Z_i = 1$, the Landshoff ($N = 4$) value is slightly corrected ($0.710 \rightarrow 0.709$) from Kaneko. The final value for $Z_i = 1$ from Kaneko et al. and Ji & Held reads $\beta_0 = 0.70287$, and the Spitzer–Härm result is correct.

<table><caption>**Table 10**
Parallel Electron Heat Conductivity $\kappa_{\parallel}^e = \gamma_0 p_e / (m_e \nu_{ei})$ (Thermal Heat Flux $q_e^T = -\kappa_{\parallel}^e \nabla_{\parallel} T_e$), Coefficient $\gamma_0$ is Plotted</caption><thead><tr><th>‖ Heat Conductivity κ<sub>‖</sub><sup>e</sup></th><th>Z<sub>i</sub> = 1</th><th>Z<sub>i</sub> = 2</th><th>Z<sub>i</sub> = 3</th><th>Z<sub>i</sub> = 4</th><th>Z<sub>i</sub> = 16</th><th>Z<sub>i</sub> = ∞</th></tr></thead><tbody><tr><td>Burgers–Schunk</td><td>1.34</td><td>1.58</td><td>1.68</td><td>1.73</td><td>1.87</td><td>1.92</td></tr><tr><td>Killie et al.</td><td>3.92</td><td>5.25</td><td>5.92</td><td>6.33</td><td>7.47</td><td>7.95</td></tr><tr><td>Braginskii</td><td>3.1616</td><td>4.890</td><td>6.064</td><td>6.920</td><td>10.334</td><td>12.471</td></tr><tr><td>Landshoff (N = 4)</td><td><b>3.178</b></td><td>4.902</td><td>6.069</td><td></td><td></td><td>13.572</td></tr><tr><td>Spitzer–Härm</td><td>3.203</td><td>4.960</td><td></td><td>6.983</td><td>10.629</td><td>13.581</td></tr></tbody></table>

**Note.** The model of Killie et al. is a significant improvement over that of Burgers–Schunk. The model of Landshoff for $N = 1$ matches Burgers–Schunk, and for $N = 2$ it approximately matches Braginskii. For $Z_i = 1$, the Landshoff ($N = 4$) value is slightly corrected ($3.175 \rightarrow 3.178$) from Kaneko. The final value for $Z_i = 1$ from Kaneko et al. and Ji & Held reads $\gamma_0 = 3.2031$, and the Spitzer–Härm result is correct. Note that by keeping $n_e$ and $T_e$ constant in the definition of $\nu_{ei}$, the heat conductivity $\gamma_0/\nu_{ei}$ actually decreases with increasing $Z_i$.

<table><caption>**Table 11**
Parallel Electron Frictional Heat Flux $q_e^u = \beta_0^* p_e \delta u_\parallel$</caption><thead><tr><th>‖ Frictional Heat Flux q<sub>e</sub><sup>u</sup></th><th>Z<sub>i</sub> = 1</th><th>Z<sub>i</sub> = 2</th><th>Z<sub>i</sub> = 3</th><th>Z<sub>i</sub> = 4</th><th>Z<sub>i</sub> = 16</th><th>Z<sub>i</sub> = ∞</th></tr></thead><tbody><tr><td>Killie et al.</td><td>2.35</td><td>3.15</td><td>3.55</td><td>3.80</td><td>4.48</td><td>4.77</td></tr><tr><td>Spitzer–Härm</td><td>0.699</td><td>0.888</td><td></td><td>1.089</td><td>1.346</td><td>3/2</td></tr></tbody></table>

**Note.** For the models of Burgers–Schunk, Braginskii, and Landshoff, the Onsager symmetry $\beta_0^* = \beta_0$ holds exactly with the values given in Table 9. For the model of Spitzer–Härm, the Onsager symmetry is satisfied only approximately, with the largest discrepancy for $Z_i = 2$, of around 2%. For the model of Killie et al., the Onsager symmetry is broken, and the frictional heat flux values are quite large.

<table><caption>**Table 12**
Perpendicular Electron Heat Conductivity $\kappa_{\perp}^e = \gamma_1' p_e \nu_{ei} / (m_e \Omega_e^2)$, in the Limit of a Strong Magnetic Field, Coefficient $\gamma_1'$ is Plotted</caption><thead><tr><th>⊥ Heat Conductivity κ<sub>⊥</sub><sup>e</sup></th><th>Z<sub>i</sub> = 1</th><th>Z<sub>i</sub> = 2</th><th>Z<sub>i</sub> = 3</th><th>Z<sub>i</sub> = 4</th><th>Z<sub>i</sub> = ∞</th></tr></thead><tbody><tr><td>Burgers–Schunk</td><td>4.664</td><td>3.957</td><td>3.721</td><td>3.604</td><td>3.25</td></tr><tr><td>Killie et al.</td><td>1.59</td><td>1.19</td><td>1.06</td><td>0.99</td><td>0.79</td></tr><tr><td>Braginskii</td><td>4.664</td><td>3.957</td><td>3.721</td><td>3.604</td><td>3.25</td></tr></tbody></table>

**Note.** The Braginskii values are from his Table II. Interestingly, the Burgers–Schunk model matches the Braginskii values exactly. In fact, both models yield the same analytic expression $\gamma_1' = (\sqrt{2}/Z_i) + 13/4$, see (60), so the numerical comparison between Burgers–Schunk and Braginskii is a bit meaningless (and the reason why the $Z_i = 16$ value was omitted from our table). The table shows that the model of Killie et al. is imprecise.

with the coefficients from Table I of Landshoff (1951). We plot his highest-order model for $N = 4$. The models of Landshoff were calculated with higher numerical precision in the work of Kaneko (1960), where the following conversion has to be used:

$$ \alpha_0 = \frac{1}{e^{I(0)}}, \quad \beta_0 = \beta_0^* = - \frac{5}{2} \frac{b^{I(0)}}{e^{I(0)}}, \quad \gamma_0 = \frac{25}{4} \left[ b^{I(-1)} - \frac{(b^{I(0)})^2}{e^{I(0)}} \right], \qquad (17) $$
with the values in his Tables I, II, and III. In his work, $M=N+1$, and values for the models from $N=1$ to $N=5$ are given, although only for $Z_i=1$. The model is easily comparable with that of Landshoff (1951) because the same coefficients are given. In our comparison tables, we slightly correct these $Z_i=1$ values of Landshoff ($N=4$) with the more precise ones of Kaneko. In the later works of Kaneko & Taguchi (1978) and Kaneko & Yamao (1980), calculations with up to $M=50$ were made, and the notation is changed into $b^{I(0)} \rightarrow b_1^{I(0)}$, $b^{I(-1)} \rightarrow b_1^{I(1)}$. From their work and the recent work of Ji & Held (2013), who used up to 160 Laguerre polynomials, the correct values for charge $Z_i=1$ read $\alpha_0=0.50612$, $\beta_0=0.70287$, and $\gamma_0=3.2031$.

For the work of Balescu (1988), who was the first to recover Braginskii with the moment approach of Grad, the following conversion has to be used:

$$ \alpha_0 = \frac{1}{\tilde{\sigma}_{\parallel}}; \quad \beta_0 = \beta_0^* = -\sqrt{\frac{5}{2} \frac{\tilde{\alpha}_{\parallel}}{\tilde{\sigma}_{\parallel}}}; \quad \gamma_0 = \frac{5}{2} \left( \tilde{\kappa}_{\parallel}^e - \frac{\tilde{\alpha}_{\parallel}^2}{\tilde{\sigma}_{\parallel}} \right), \qquad (18) $$

with the numerical values for $Z_i=1$ given in his Table 4.1 (p. 239). For his 13-moment model ($N=1$), the results are equal to Burgers–Schunk, and for his 21-moment model ($N=2$), the results are equal to Braginskii. However, for his 29-moment model ($N=3$), the coefficients of Balescu were shown to be imprecise by Ji & Held (2013), see their Table I, who were able to pinpoint exactly the analytic errors in the collisional matrices of Balescu. That the Balescu $N=3$ values are indeed incorrect can be quickly double-checked by comparison with the $M=4$ model of Kaneko (1960), from where the Balescu parameters should be $\tilde{\sigma}_{\parallel} = e^{I(0)} = 1.964$, $\tilde{\alpha}_{\parallel} = \sqrt{5/2} b^{I(0)} = -0.887$, and $\tilde{\kappa}_{\parallel}^e = (5/2)b^{I(-1)} = 1.666$, agreeing with the modern calculations of Ji & Held (2013).

## 1.1. Notation of Spitzer–Härm (1953)

The exact values of the parallel transport coefficients (with the exception of parallel viscosity) were first numerically obtained by Spitzer & Härm (1953). Essentially, the perturbation $\phi_e$ (or $f_e^{(1)}$) around a Maxwellian $f_e = f_e^{(0)}(1-\phi_e)$ that satisfies the Fokker-Planck equation was found numerically, and the obtained result was used to calculate the transport coefficients. No magnetic field is present in their work, and the results can be interpreted as applying to unmagnetized plasmas, or to magnetized plasmas in the direction parallel to magnetic field lines. Similar to Braginskii (Chapters 2 and 4), the paper treats a one ion–electron plasma (with $n_e = Z_i n_i$).

The notation of Spitzer & Härm (1953) can be very confusing. The results are given in a form

$$ j = \sigma E + \alpha \nabla T_e; \qquad (19) $$

$$ \mathbf{q}_e^{\text{Spitzer}} = -\beta E - K \nabla T_e, \qquad (110) $$

with the coefficients $\sigma$, $\alpha$, $\beta$, and $K$ given by their Equations (33)–(36). These coefficients contain a quantity $C^2$. This quantity is only defined by a sentence following Equation (16) in their previous paper by Cohen et al. (1950), which reads “$C^2$ is the mean square electron velocity,” meaning $C = \sqrt{3T_e/m_e}$ with the important factor of 3 present (we use the same notation as Braginskii, with the Boltzmann constant equal to one). Rewriting their coefficients in (19), (110) to our notation yields

$$ \begin{aligned} \sigma &= \frac{32}{3\pi} \frac{e^2 n_e}{m_e \nu_{ei}} \gamma_E; & \alpha &= \frac{16}{\pi} \frac{e n_e}{m_e \nu_{ei}} \gamma_T; \\ \beta &= \frac{128}{3\pi} \frac{e p_e}{m_e \nu_{ei}} \delta_E; & K &= \frac{320}{3\pi} \frac{p_e}{m_e \nu_{ei}} \delta_T, \end{aligned} \qquad (111) $$

where the numerical values of $\gamma_E$, $\gamma_T$, $\delta_E$, and $\delta_T$ are given in Table III of Spitzer & Härm (1953). The coefficients (111) are essentially normalized with respect to a Lorentzian plasma $Z_i = \infty$ (meaning when electron–electron collisions are negligible), in which case $\gamma_E = \gamma_T = \delta_E = \delta_T = 1$.

Unfortunately, Spitzer & Härm (1953) do not define their $\mathbf{q}_e^{\text{Spitzer}}$, and only describe it as a “the rate of flow of heat.” The heat flux is also not defined in the book of Spitzer (1962), but he notes (Equation (5.45)) that, from the thermodynamics of irreversible processes, the model closely satisfies

$$ \beta = \alpha T_e + \frac{5 T_e}{2 e} \sigma. \qquad (112) $$

Equation (112) should be the Onsager symmetry. In the historical literature, there are three other major possibilities regarding how to define the heat flux. The first two choices are:

$$ \mathbf{q}_a^{***} = \frac{m_a}{2} \int \mathbf{v} |\mathbf{v}|^2 f_a d^3 \mathbf{v} = \mathbf{q}_a + \frac{5}{2} p_a \mathbf{u}_a + \mathbf{u}_a \cdot \bar{\mathbf{\Pi}}_a^{(2)} + \frac{\rho_a}{2} |\mathbf{u}_a|^2 \mathbf{u}_a; \qquad (113) $$

$$ \mathbf{q}_a^{**} = \int \mathbf{v} \left( \frac{m_a \nu^2}{2} - \frac{5}{2} T_a \right) f_a d^3 \mathbf{v} = \mathbf{q}_a + \mathbf{u}_a \cdot \bar{\mathbf{\Pi}}_a^{(2)} + \frac{\rho_a}{2} |\mathbf{u}_a|^2 \mathbf{u}_a. \qquad (114) $$
The nonlinear terms can be neglected. Spitzer is not using the second choice, and the first choice is almost correct, except that for the electron heat flux, only (5/2)p_e u_e would be created, and not the whole current u_e - u_i. The third choice is the definition of Chapman & Cowling (1939), where the heat flux is defined with respect to the *average* velocity of all of the species 〈u〉 = (Σ_a ρ_a u_a) / (Σ_a ρ_a), according to

$$ q_a^* = \frac{m_a}{2} \int (v - \langle u \rangle) |v - \langle u \rangle|^2 f_a d^3 v = q_a + \frac{5}{2} p_a w_a + \frac{\rho_a}{2} |w_a|^2 w_a + w_a \cdot \bar{\Pi}_a^{(2)}, \quad (115) $$

where w_a = u_a - 〈u〉. For an ion-electron plasma 〈u〉 = u_i and w_e = u_e - u_i. Thus, to satisfy (112), the correct interpretation seems to be

$$ q_e^{\text{Spitzer}} = q_e^* = q_e + \frac{5}{2} p_e \delta u = q_e - \frac{5 T_e}{2 e} j, \quad (116) $$

where j = -en_e δu and δu = u_e - u_i.

Result (19) should be viewed as part of the evolution equation for ∂u_e/∂t (here written in a steady state with all other terms neglected), and substituting the electric field into (110) then yields

$$ \begin{gathered} en_e E = R_e = \frac{en_e}{\sigma} j - en_e \frac{\alpha}{\sigma} \nabla T_e; \\ q_e = -\left(\frac{\beta}{\sigma} - \frac{5 T_e}{2 e}\right) j - \epsilon K \nabla T_e; \quad \text{where} \quad \epsilon = 1 - \frac{\alpha \beta}{\sigma K} = 1 - \frac{3}{5} \frac{\delta_E \gamma_T}{\delta_T \gamma_E}. \end{gathered} \quad (117) $$

The numerical coefficient ϵ is given in Table III of Spitzer & Härm (1953) as well. Or, equivalently, by using (111),

$$ \begin{gathered} R_e = -\frac{3\pi}{32\gamma_E} \rho_e \nu_{ei} \delta u - \frac{3}{2} \frac{\gamma_T}{\gamma_E} n_e \nabla T_e; \\ q_e = +\left(4\frac{\delta_E}{\gamma_E} - \frac{5}{2}\right) p_e \delta u - \epsilon \delta_T \frac{320}{3\pi} \frac{p_e}{m_e \nu_{ei}} \nabla T_e. \end{gathered} \quad (118) $$

In this form, the results can be directly compared to Braginskii, with the relations

$$ \alpha_0 = \frac{3\pi}{32\gamma_E}; \quad \beta_0 = \frac{3}{2} \frac{\gamma_T}{\gamma_E}; \quad \beta_0^* = 4 \frac{\delta_E}{\gamma_E} - \frac{5}{2}; \quad \gamma_0 = \epsilon \delta_T \frac{320}{3\pi}. $$

The Onsager symmetry then reads

$$ \frac{3}{2} \gamma_T = 4 \delta_E - \frac{5}{2} \gamma_E, \quad (119) $$

which the model satisfies approximately, and for the Lorentz case exactly. The largest difference appears for Z_i = 2, where the left-hand side of (119) is 0.621 and the right-hand side is 0.607, so Spitzer's claim that Equation (112) is satisfied to about one part in a thousand seems a bit exaggerated, or we are interpreting his results incorrectly. The model of Spitzer & Härm (1953) and Spitzer (1962) is criticized in the monograph of Balescu (1988, Part 1, p. 266). Nevertheless, the coefficients α_0, β_0, and γ_0 in the model of Spitzer & Härm (1953) are the correct answer, and in comparison with Kaneko & Taguchi (1978), Kaneko & Yamao (1980), or Ji & Held (2013), these coefficients are valid for three decimal digits. For numerical simulations that employ the heat flux of Spitzer & Härm (1953), it seems logical to simply ignore the imprecise β_0^* values, and enforce the Onsager symmetry β_0^* = β_0 in their model by hand.

## 1.2. Model of Killie et al. (2004)

Instead of the 8-moment distribution function of Grad (H1) used in the model of Burgers-Schunk, Killie et al. (2004) argued that it is better to use

$$ f_a = f_a^{(0)} \left[ 1 - \frac{m_a^2 |c_a|^2}{5 T_a^2 p_a} \left( 1 - \frac{m_a |c_a|^2}{7 T_a} \right) q_a \cdot c_a \right], \quad (120) $$

yielding the collisional contributions (which we did not verify) for small temperature differences

$$ \mathbf{R}_a = \rho_a \sum_b \nu_{ab} (\mathbf{u}_b - \mathbf{u}_a) + \sum_b \nu_{ab} \frac{3}{5} \frac{\mu_{ab}}{T_{ab}} \left[ \mathbf{q}_a \left( 1 - \frac{5}{7} \frac{m_b}{m_a + m_b} \right) - \mathbf{q}_b \frac{\rho_a}{\rho_b} \left( 1 - \frac{5}{7} \frac{m_a}{m_a + m_b} \right) \right], \quad (121) $$
and

$$ \frac{1}{2} \mathrm{Tr} \bar{\mathcal{Q}}_a^{(3)} = \frac{\delta \mathbf{q}_a}{\delta t} = - \frac{16}{35} \nu_{aa} \mathbf{q}_a - \sum_{b \neq a} \nu_{ab} \left[ D_{ab}^{(1)} \mathbf{q}_a - D_{ab}^{(4)} \frac{\rho_a}{\rho_b} \mathbf{q}_b - p_a (\mathbf{u}_b - \mathbf{u}_a) \frac{m_b + \frac{5}{2} m_a}{m_a + m_b} \right]; \quad (122) $$

$$ D_{ab}^{(1)} = \frac{1}{(m_a + m_b)^3} \left( 3m_a^3 - \frac{1}{2}m_a^2 m_b - \frac{2}{5}m_a m_b^2 - \frac{4}{35}m_b^3 \right); \quad (123) $$

$$ D_{ab}^{(4)} = \frac{1}{(m_a + m_b)^3} \left( \frac{6}{5}m_b^3 - \frac{171}{70}m_b^2 m_a - \frac{3}{7}m_b m_a^2 \right). \quad (124) $$

Similar to Burgers–Schunk, they also provide equations for unrestricted temperature differences. Considering an ion–electron plasma yields $D_{ei}^{(1)} = -4/35$, $D_{ei}^{(4)} = 6/5$, and

$$ \mathbf{R}_e = - \rho_e \nu_{ei} \delta \mathbf{u} + \nu_{ei} \frac{\rho_e}{p_e} \frac{6}{35} \mathbf{q}_e; \quad (125) $$

$$ \frac{1}{2} \mathrm{Tr} \bar{\mathcal{Q}}_e^{(3)} = \frac{\delta \mathbf{q}_e}{\delta t} = - \mathbf{q}_e \left( \frac{16}{35} \nu_{ee} - \frac{4}{35} \nu_{ei} \right) - \nu_{ei} p_e \delta \mathbf{u}, \quad (126) $$

with total collisional contributions

$$ \mathbf{Q}_e^{(3)\prime} = - \bar{\nu}_e \mathbf{q}_e + \frac{3}{2} \nu_{ei} p_e \delta \mathbf{u}; \quad (127) $$

$$ \bar{\nu}_e = \frac{16}{35} \nu_{ee} + \frac{11}{35} \nu_{ei}; \quad (128) $$

$$ \mathbf{a}_e = \frac{5}{2} \frac{p_e}{m_e} \nabla T_e - \frac{3}{2} \nu_{ei} p_e \delta \mathbf{u}. \quad (129) $$

This yields the heat flux solution equivalent to Equations (H43)–(H45), with the only difference being that the frequencies are now added according to

$$ \bar{\nu}_e = \left( \frac{1}{Z_i \sqrt{2}} \frac{16}{35} + \frac{11}{35} \right) \nu_{ei}; \quad \text{for } Z_i = 1: \quad \bar{\nu}_e = 0.6375 \nu_{ei}. \quad (130) $$

The momentum exchange rates then read

$$ \mathbf{R}_e^u = - \rho_e \nu_{ei} \left[ \left(1 - \frac{9}{35} \frac{\nu_{ei}}{\bar{\nu}_e}\right) \delta \mathbf{u}_\parallel + \left(1 - \frac{9}{35} \frac{\bar{\nu}_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2}\right) \delta \mathbf{u}_\perp + \frac{9}{35} \frac{\Omega_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} \hat{\mathbf{b}} \times \delta \mathbf{u} \right]; \quad (131) $$

$$ \mathbf{R}_e^T = - \frac{3}{7} \frac{\nu_{ei}}{\bar{\nu}_e} n_e \nabla_{\parallel} T_e - \frac{3}{7} \frac{\bar{\nu}_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} n_e \nabla_{\perp} T_e + \frac{3}{7} \frac{\Omega_e \nu_{ei}}{\Omega_e^2 + \bar{\nu}_e^2} n_e \hat{\mathbf{b}} \times \nabla T_e, \quad (132) $$

and a direct comparison with Braginskii is done according to

$$ \alpha_0 = 1 - \frac{9}{35} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \beta_0 = \frac{3}{7} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \beta_0^* = \frac{3}{2} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \gamma_0 = \frac{5}{2} \frac{\nu_{ei}}{\bar{\nu}_e}; \quad \bar{\nu}_e = \left( \frac{1}{Z_i \sqrt{2}} \frac{16}{35} + \frac{11}{35} \right) \nu_{ei}. \quad (133) $$

Examining the numerical values for $Z_i = 1$, for example, the parallel heat conductivity reads $\kappa_\parallel^e = 3.92 p_e / (\nu_{ei} m_e)$. This is a big improvement in the model of Killie et al. (2004): the conductivity is almost three times larger than the 1.34 value of Burgers–Schunk, and is much closer to the correct value of 3.20. The other results are (strong B-field, $Z_i = 1$)

$$ \begin{aligned} \mathbf{R}_e &= -\rho_e \nu_{ei} (0.60 \delta \mathbf{u}_\parallel + \delta \mathbf{u}_\perp) - 0.67 n_e \nabla_{\parallel} T_e; \\ \mathbf{q}_e^u &= 2.35 p_e \delta \mathbf{u}_\parallel, \end{aligned} \quad (134) $$

and the thermal force value of 0.67 is now closer to the correct value of 0.70 as well. However, the frictional heat flux $\mathbf{q}_e^u$ is quite large (over three times larger than it should be, 2.35 versus 0.70). Importantly, the Onsager symmetry between $\mathbf{q}_e^u$ and $\mathbf{R}_e^T$ is broken, which can also be seen from the general results (I32), (H48). Nevertheless, the model indeed improves the parallel thermal heat flux and the parallel thermal force of Burgers–Schunk.
## Appendix J
10-moment Model (Viscosity)

To calculate the collisional contributions for the stress tensor with the Landau operator, one uses the following 10-moment distribution function of Grad:

$$f_b(\mathbf{v}') = \frac{n_b}{\pi^{3/2} v_{\text{thb}}^3} e^{-\frac{|c_b|^2}{v_{\text{thb}}^2}} \left[ 1 + \frac{m_b}{2T_b p_b} \bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{c}_b \mathbf{c}_b \right]. \qquad (J1)$$

As a reminder, $\bar{\mathbf{\Pi}}_b^{(2)}: \bar{\mathbf{I}} = 0$. By using symmetries and a Gaussian integration, it is possible to show that

$$\int \mathbf{c}_b \mathbf{c}_b e^{-\frac{|c_b|^2}{v_{\text{thb}}^2} d^3\mathbf{v}'} = \frac{\pi^{3/2}}{2} v_{\text{thb}}^5 \bar{\mathbf{I}}; \qquad \bar{\mathbf{\Pi}}_b^{(2)}: \int \mathbf{c}_b \mathbf{c}_b \mathbf{v}' e^{-\frac{|c_b|^2}{v_{\text{thb}}^2} d^3\mathbf{v}'} = 0; \\ \bar{\mathbf{\Pi}}_b^{(2)}: \int \mathbf{c}_b \mathbf{c}_b \mathbf{c}_b e^{-\frac{|c_b|^2}{v_{\text{thb}}^2} d^3\mathbf{v}'} = \frac{\pi^{3/2}}{2} v_{\text{thb}}^7 \bar{\mathbf{\Pi}}_b^{(2)}. \qquad (J2)$$

The last integral is a special case of (J48). Thus, the distribution function (J1) correctly reproduces the density, fluid velocity, and full pressure tensor $m_b \int \mathbf{c}_b \mathbf{c}_b f_b d^3\mathbf{c}_b = p_b \bar{\mathbf{I}} + \bar{\mathbf{\Pi}}_b^{(2)}$, so the distribution function is well defined.

### J.1. Rosenbluth Potentials

Using the variables $\mathbf{x} = (\mathbf{v}' - \mathbf{v})/v_{\text{thb}}$ and $\mathbf{y} = (\mathbf{v} - \mathbf{u}_b)/v_{\text{thb}}$ with $\mathbf{c}_b = (\mathbf{x} + \mathbf{y})v_{\text{thb}}$, we need to calculate the Rosenbluth potentials

$$H_b(\mathbf{v}) = \int \frac{f_b(\mathbf{v}')}{|\mathbf{v}' - \mathbf{v}|} d^3\mathbf{v}' \\ = \frac{n_b}{\pi^{3/2} v_{\text{thb}}} \int \frac{e^{-|\mathbf{x} + \mathbf{y}|^2}}{x} \left[ 1 + \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} : (\mathbf{x} + \mathbf{y})(\mathbf{x} + \mathbf{y}) \right] d^3\mathbf{x}; \qquad (J3)$$

$$G_b(\mathbf{v}) = \int |\mathbf{v}' - \mathbf{v}| f_b(\mathbf{v}') d^3\mathbf{v}' \\ = \frac{n_b v_{\text{thb}}}{\pi^{3/2}} \int x e^{-|\mathbf{x} + \mathbf{y}|^2} \left[ 1 + \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} : (\mathbf{x} + \mathbf{y})(\mathbf{x} + \mathbf{y}) \right] d^3\mathbf{x}. \qquad (J4)$$

Using the integrals (J40) and (J44), the final results for the Rosenbluth potentials are

$$H_b = \frac{n_b}{v_{\text{thb}}} \left\{ \frac{\text{erf}(y)}{y} + \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} : \mathbf{yy} \left[ \text{erf}(y) \frac{3}{4y^5} - \frac{e^{-y^2}}{\sqrt{\pi}} \left( \frac{1}{y^2} + \frac{3}{2y^4} \right) \right] \right\}; \\ G_b = n_b v_{\text{thb}} \left\{ \frac{e^{-y^2}}{\sqrt{\pi}} + \left( y + \frac{1}{2y} \right) \text{erf}(y) + \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} : \mathbf{yy} \left[ -\frac{3}{4\sqrt{\pi}} \frac{e^{-y^2}}{y^4} + \left( -\frac{1}{4y^3} + \frac{3}{8y^5} \right) \text{erf}(y) \right] \right\}. \qquad (J5)$$

We will need the derivative

$$\frac{\partial H_b}{\partial \mathbf{v}} = \frac{n_b}{v_{\text{thb}}^2} \mathbf{y} \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\text{erf}(y)}{y^3} \right) \\ + \frac{n_b}{v_{\text{thb}}^2} \frac{2}{p_b} (\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) \left[ \text{erf}(y) \frac{3}{4y^5} - \frac{e^{-y^2}}{\sqrt{\pi}} \left( \frac{1}{y^2} + \frac{3}{2y^4} \right) \right] \\ + \frac{n_b}{v_{\text{thb}}^2} (\bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy}) \mathbf{y} \left[ \frac{2}{\sqrt{\pi}} \left( \frac{1}{y^2} + \frac{5}{2y^4} + \frac{15}{4y^6} \right) e^{-y^2} - \frac{15}{4} \frac{\text{erf}(y)}{y^7} \right]. \qquad (J6)$$
As a double check, applying $\partial/\partial \mathbf{v}$ at the last expression recovers $-4\pi f_b(\mathbf{v})$, where, for example,

$$ \frac{\partial}{\partial \mathbf{v}} \cdot (\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) = 0; \qquad \frac{\partial}{\partial \mathbf{v}} \cdot [(\bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy})\mathbf{y}] = \frac{5}{v_{\text{thb}}} \bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy}. \qquad (J7) $$

The entire dynamical friction vector for the 10-moment model then becomes

$$ A_{ab} = 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b}{v_{\text{thb}}^2} \left\{ \begin{aligned} & \mathbf{y} \left( \frac{2}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} - \frac{\text{erf}(y)}{y^3} \right) \\ & + \frac{2}{p_b} (\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) \left[ \text{erf}(y) \frac{3}{4y^5} - \frac{e^{-y^2}}{\sqrt{\pi}} \left( \frac{1}{y^2} + \frac{3}{2y^4} \right) \right] \\ & + \frac{1}{p_b} (\bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy}) \mathbf{y} \left[ \frac{2}{\sqrt{\pi}} \left( \frac{1}{y^2} + \frac{5}{2y^4} + \frac{15}{4y^6} \right) e^{-y^2} - \frac{15}{4} \frac{\text{erf}(y)}{y^7} \right] \end{aligned} \right\}. \qquad (J8) $$

For the diffusion tensor, to perform the subsequent analytic calculations in a clear way, it is useful to write the second Rosenbluth potential $G_b$ by introducing $A_1, A_2$:

$$ G_b = n_b v_{\text{thb}} \left[ A_1 + \frac{1}{p_b} (\bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy}) A_2 \right], \qquad (J9) $$

where

$$ \begin{aligned} A_1 &= \frac{e^{-y^2}}{\sqrt{\pi}} + \left(y + \frac{1}{2y}\right) \text{erf}(y); \\ A_2 &= -\frac{3}{4\sqrt{\pi}} \frac{e^{-y^2}}{y^4} + \left(-\frac{1}{4y^3} + \frac{3}{8y^5}\right) \text{erf}(y). \end{aligned} \qquad (J10) $$

The required derivatives are then

$$ \frac{\partial G_b}{\partial \mathbf{v}} = n_b \left[ \frac{\mathbf{y}}{y} A_1' + \frac{2}{p_b} (\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) A_2 + \left( \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} : \mathbf{yy} \right) \frac{\mathbf{y}}{y} A_2' \right], \qquad (J11) $$

and

$$ \begin{aligned} \frac{\partial G_b}{\partial \mathbf{v} \partial \mathbf{v}} = & \frac{n_b}{v_{\text{thb}}} \left\{ \left( \frac{\bar{\mathbf{I}}}{y} - \frac{\mathbf{y}\mathbf{y}}{y^3} \right) A_1' + \frac{\mathbf{y}\mathbf{y}}{y^2} A_1'' \right. \\ & + \frac{2}{p_b} \left[ \frac{\mathbf{y}}{y} (\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) + (\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) \frac{\mathbf{y}}{y} \right] A_2' + \frac{2}{p_b} \bar{\mathbf{\Pi}}_b^{(2)} A_2 \\ & \left. + \left( \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} : \mathbf{yy} \right) \left[ \left( \frac{\bar{\mathbf{I}}}{y} - \frac{\mathbf{y}\mathbf{y}}{y^3} \right) A_2' + \frac{\mathbf{y}\mathbf{y}}{y^2} A_2'' \right] \right\}. \end{aligned} \qquad (J12) $$

As a double check, applying $(1/2)\text{Tr}$ at the last expression recovers $H_b$.

After a slight rearrangement suitable for the calculations, the entire diffusion tensor then becomes

$$ \begin{aligned} D_{ab} = 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{v_{\text{thb}}} \left\{ \begin{aligned} & \bar{\mathbf{I}} \frac{A_1'}{y} + \frac{\mathbf{y}\mathbf{y}}{y^2} \left( A_1'' - \frac{A_1'}{y} \right) \\ & + \frac{1}{p_b} \left[ 2\mathbf{y}(\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y}) + 2(\bar{\mathbf{\Pi}}_b^{(2)} \cdot \mathbf{y})\mathbf{y} + (\bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy})\bar{\mathbf{I}} \right] \frac{A_2'}{y} \\ & + \frac{2}{p_b} \bar{\mathbf{\Pi}}_b^{(2)} A_2 + \frac{1}{p_b} (\bar{\mathbf{\Pi}}_b^{(2)} : \mathbf{yy}) \frac{\mathbf{y}\mathbf{y}}{y^2} \left( A_2'' - \frac{A_2'}{y} \right) \end{aligned} \right\}. \end{aligned} \qquad (J13) $$
with the “coefficients”

$$
\begin{align*}
\frac{A'_1}{y} &= \left(\frac{1}{y} - \frac{1}{2y^3}\right) \operatorname{erf}(y) + \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2}; \\
A''_1 &= \left(-\frac{1}{y} + \frac{3}{2y^3}\right) \operatorname{erf}(y) - \frac{3}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2}; \\
A'_2 &= \left(\frac{3}{4y^4} - \frac{15}{8y^6}\right) \operatorname{erf}(y) + \frac{e^{-y^2}}{\sqrt{\pi}} \left(\frac{1}{y^3} + \frac{15}{4y^5}\right); \\
A''_2 &= \left(\frac{3}{4y^5} - \frac{15}{8y^7}\right) \operatorname{erf}(y) + \frac{e^{-y^2}}{\sqrt{\pi}} \left(\frac{1}{y^4} + \frac{15}{4y^6}\right); \\
A''_2 - \frac{A'_2}{y} &= \left(-\frac{15}{4y^5} + \frac{105}{8y^7}\right) \operatorname{erf}(y) - \frac{e^{-y^2}}{\sqrt{\pi}} \left(\frac{2}{y^2} + \frac{10}{y^4} + \frac{105}{4y^6}\right). \tag{J14}
\end{align*}
$$

Or, explicitly, in its entire form:

$$
\begin{equation}
\begin{aligned}
D_{ab} = 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{v_{\text{thb}}} & \left\{ \bar{\bar{I}} \left[ \left( \frac{1}{y} - \frac{1}{2y^3} \right) \text{erf}(y) + \frac{1}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} \right] + \frac{yy}{y^2} \left[ \left( -\frac{1}{y} + \frac{3}{2y^3} \right) \text{erf}(y) - \frac{3}{\sqrt{\pi}} \frac{e^{-y^2}}{y^2} \right] \right. \\
& + \frac{1}{p_b} \left[ 2y \left( \bar{\bar{\Pi}}_b^{(2)} \cdot y \right) + 2 \left( \bar{\bar{\Pi}}_b^{(2)} \cdot y \right) y + \left( \bar{\bar{\Pi}}_b^{(2)} : yy \right) \bar{\bar{I}} \right] \left[ \left( \frac{3}{4y^5} - \frac{15}{8y^7} \right) \text{erf}(y) + \frac{e^{-y^2}}{\sqrt{\pi}} \left( \frac{1}{y^4} + \frac{15}{4y^6} \right) \right] \\
& \left. + \frac{2}{p_b} \bar{\bar{\Pi}}_b^{(2)} \left[ -\frac{3}{4\sqrt{\pi}} \frac{e^{-y^2}}{y^4} + \left( -\frac{1}{4y^3} + \frac{3}{8y^5} \right) \text{erf}(y) \right] \right. \\
& \left. + \left( \frac{\bar{\bar{\Pi}}_b^{(2)} : yy}{p_b} \right) \frac{yy}{y^2} \left[ \left( -\frac{15}{4y^5} + \frac{105}{8y^7} \right) \text{erf}(y) - \frac{e^{-y^2}}{\sqrt{\pi}} \left( \frac{2}{y^2} + \frac{10}{y^4} + \frac{105}{4y^6} \right) \right] \right\}.
\end{aligned}
\tag{J15}
\end{equation}
$$

J.2. Viscosity Calculation

For species “a,” the distribution function in a semilinear approximation reads

$$
f_a(\mathbf{v}) = \frac{n_a}{\pi^{3/2} v_{\text{tha}}^3} e^{-\alpha^2 y^2} \left[ 1 - 2\alpha(\mathbf{y} \cdot \mathbf{u}) + \frac{\alpha^2 \bar{\bar{\Pi}}_a^{(2)} : \mathbf{y}\mathbf{y}}{p_a} \right]. \quad (J16)
$$

It can be seen that, at the semilinear level, there is no new contribution to the momentum equation. For the pressure tensor equation,
we need to calculate the following collisional contributions:

$$
\bar{\bar{Q}}_{ab}^{(2)} = m_a \int f_a [A_{ab} c_a]^S d^3 v + m_a \int f_a \bar{\bar{D}}_{ab} d^3 v, \quad (J17)
$$

where the diffusion tensor is symmetric. Starting with the second term, and using the derived formulas (J45)–(J48), the integration
over the diffusion tensor then yields

$$
\begin{equation}
\begin{split}
& m_a \int f_a \bar{\bar{D}}_{ab} d^3 v = 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{v_{\text{thb}}} \frac{\rho_a}{\pi^{3/2}} \alpha^3 \\
& \qquad \left\{ + \bar{\bar{I}} \frac{4\pi}{3} \int_0^\infty (2A_1' y + A_1'' y^2) e^{-\alpha^2 y^2} dy \right. \\
& \qquad \left. + \frac{\bar{\bar{\Pi}}_b^{(2)}}{p_b} 8\pi \int_0^\infty \left[ \frac{3}{5} y^3 A_2' + y^2 A_2 + \frac{1}{15} y^4 A_2'' \right] e^{-\alpha^2 y^2} dy \right. \\
& \qquad \left. + \frac{\bar{\bar{\Pi}}_a^{(2)}}{p_a} \frac{8\pi}{15} \alpha^2 \int_0^\infty y^4 \left( A_1'' - \frac{A_1'}{y} \right) e^{-\alpha^2 y^2} dy \right\},
\end{split}
\tag{J18}
\end{equation}
$$
and the further one-dimensional integration brings the following result:

$$ m_a \int f_a \bar{\mathbf{D}}_{ab} d^3 v = 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{v_{\text{thb}}} \frac{\rho_a}{\pi^{3/2}} \alpha^3 \left\{ + \bar{\mathbf{I}} \frac{4\pi}{3} \frac{1}{\alpha^2 \sqrt{1 + \alpha^2}} - \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} \frac{4\pi}{15} \frac{1}{(1 + \alpha^2)^{3/2}} - \frac{\bar{\mathbf{\Pi}}_a^{(2)}}{p_a} \frac{4\pi}{15} \frac{1}{\alpha^2 (1 + \alpha^2)^{3/2}} \right\} \\ = \rho_a \nu_{ab} \frac{m_b}{m_a + m_b} \left[ \bar{\mathbf{I}} (v_{\text{tha}}^2 + v_{\text{thb}}^2) - \frac{\bar{\mathbf{\Pi}}_b^{(2)} v_{\text{thb}}^2}{p_b} - \frac{\bar{\mathbf{\Pi}}_a^{(2)} v_{\text{tha}}^2}{p_a} \right]. \qquad (J19) $$

Similarly, the first term in (J17) calculates

$$ m_a \int f_a [\mathbf{A}_{ab} \mathbf{c}_a]^S d^3 v = \rho_a \nu_{ab} \left[ - \bar{\mathbf{I}} v_{\text{tha}}^2 + \frac{\bar{\mathbf{\Pi}}_b^{(2)}}{p_b} \frac{3}{5} \frac{v_{\text{thb}}^2 v_{\text{tha}}^2}{(v_{\text{tha}}^2 + v_{\text{thb}}^2)} - \frac{\bar{\mathbf{\Pi}}_a^{(2)}}{p_a} \frac{5 v_{\text{thb}}^2 + 2 v_{\text{tha}}^2}{5 (v_{\text{tha}}^2 + v_{\text{thb}}^2)} v_{\text{tha}}^2 \right]. \qquad (J20) $$

Adding (J19)+(J20) yields the final collisional contributions for the right-hand side of the pressure tensor equation, which can be written in the convenient following form:

$$ \bar{\mathbf{Q}}_{ab}^{(2)} = 2 \frac{\rho_a \nu_{ab}}{m_a + m_b} (T_b - T_a) \bar{\mathbf{I}} - 2 \frac{m_a \nu_{ab}}{m_a + m_b} \frac{T_b}{T_{ab}} \left( \bar{\mathbf{\Pi}}_a^{(2)} - \frac{T_a n_a}{T_b n_b} \bar{\mathbf{\Pi}}_b^{(2)} \right) \\ - \frac{\nu_{ab}}{m_a + m_b} \left[ \frac{6}{5} m_b - \frac{4}{5} \mu_{ab} \frac{T_b - T_a}{T_{ab}} \right] \left( \bar{\mathbf{\Pi}}_a^{(2)} + \frac{\rho_a}{\rho_b} \bar{\mathbf{\Pi}}_b^{(2)} \right), \qquad (J21) $$

with reduced mass and reduced temperature

$$ \mu_{ab} = \frac{m_a m_b}{m_a + m_b}; \qquad T_{ab} = \frac{m_a T_b + m_b T_a}{m_a + m_b}. $$

Introducing $\Sigma_b$ over all of the species, result (J21) identifies with Equation (44) of Schunk (1977; derived before by Burgers). It is valid in the semilinear approximation, for unrestricted temperature differences. For Coulomb collisions, the viscosity calculated through the Rosenbluth potentials (for the Landau collisional operator) thus yields the same result as the Boltzmann collisional operator. By explicitly separating the self-collisions,

$$ \bar{\mathbf{Q}}_{a}^{(2)} = \sum_{b} \bar{\mathbf{Q}}_{ab}^{(2)} = \frac{\delta \bar{\mathbf{p}}_{a}}{\delta t} \\ = -\frac{6}{5} \nu_{aa} \bar{\mathbf{\Pi}}_{a}^{(2)} + \sum_{b \neq a} \left[ 2 \frac{\rho_{a} \nu_{ab}}{m_{a} + m_{b}} (T_{b} - T_{a}) \bar{\mathbf{I}} - 2 \frac{m_{a} \nu_{ab}}{m_{a} + m_{b}} \frac{T_{b}}{T_{ab}} \left( \bar{\mathbf{\Pi}}_{a}^{(2)} - \frac{T_{a} n_{a}}{T_{b} n_{b}} \bar{\mathbf{\Pi}}_{b}^{(2)} \right) \right] \\ - \sum_{b \neq a} \left[ \frac{\nu_{ab}}{m_{a} + m_{b}} \left( \frac{6}{5} m_{b} - \frac{4}{5} \mu_{ab} \frac{T_{b} - T_{a}}{T_{ab}} \right) \left( \bar{\mathbf{\Pi}}_{a}^{(2)} + \frac{\rho_{a}}{\rho_{b}} \bar{\mathbf{\Pi}}_{b}^{(2)} \right) \right], \qquad (J22) $$

where the “famous” 6/5 constant is present. As a double check, calculating the energy exchange rates yields

$$ Q_{ab} = \frac{1}{2} \mathrm{Tr} \bar{\mathbf{Q}}_{ab}^{(2)} = 3 \frac{\rho_a \nu_{ab}}{m_a + m_b} (T_b - T_a), \qquad (J23) $$

as it should be.

The collisional contributions for the stress tensor are thus

$$ \begin{aligned} \bar{\mathbf{Q}}_{a}^{(2)}' &= \frac{\delta \bar{\mathbf{\Pi}}_{a}^{(2)}}{\delta t} = \bar{\mathbf{Q}}_{a}^{(2)} - \frac{\bar{\mathbf{I}}}{3} \mathrm{Tr} \bar{\mathbf{Q}}_{a}^{(2)} \\ &= -\frac{6}{5} \nu_{aa} \bar{\mathbf{\Pi}}_{a}^{(2)} - \sum_{b \neq a} \left[ 2 \frac{m_a \nu_{ab}}{m_a + m_b} \frac{T_b}{T_{ab}} \left( \bar{\mathbf{\Pi}}_{a}^{(2)} - \frac{T_a n_a}{T_b n_b} \bar{\mathbf{\Pi}}_{b}^{(2)} \right) \right] \\ &\quad - \sum_{b \neq a} \left[ \frac{\nu_{ab}}{m_a + m_b} \left( \frac{6}{5} m_b - \frac{4}{5} \mu_{ab} \frac{T_b - T_a}{T_{ab}} \right) \left( \bar{\mathbf{\Pi}}_{a}^{(2)} + \frac{\rho_a}{\rho_b} \bar{\mathbf{\Pi}}_{b}^{(2)} \right) \right], \end{aligned} \qquad (J24) $$
and enter the right-hand side of its evolution equation, for example, written in its simplest form:

$$ \frac{d_a \bar{\Pi}_a^{(2)}}{dt} + \Omega_a (\hat{\mathbf{b}} \times \bar{\Pi}_a^{(2)})^S + p_a \bar{W}_a = \frac{\delta \bar{\Pi}_a^{(2)}}{\delta t}. \qquad (J25) $$

Importantly, in the collisionless regime, the right-hand side of (J25) simply goes to zero. It is possible to write a general solution in a quasistatic approximation, but the stress tensors of various species are coupled.

### J.3. Small Temperature Differences

For a particular case of small temperature differences between species,

$$ \bar{\mathcal{Q}}_a^{(2)} = \frac{\delta \bar{\mathbf{p}}_a}{\delta t} = - \frac{6}{5} \nu_{aa} \bar{\Pi}_a^{(2)} - 2 \sum_{b \neq a} \frac{m_a \nu_{ab}}{m_a + m_b} \left[ \bar{\mathbf{p}}_a - \frac{n_a}{n_b} \bar{\mathbf{p}}_b + \frac{3}{5} \frac{m_b}{m_a} \left( \bar{\Pi}_a^{(2)} + \frac{\rho_a}{\rho_b} \bar{\Pi}_b^{(2)} \right) \right], \qquad (J26) $$

where one uses $\bar{\mathbf{p}} = p\bar{\mathbf{I}} + \bar{\Pi}^{(2)}$, recovering Equation (41d) of Schunk (1977). Finally, for the stress tensor,

$$ \bar{\mathcal{Q}}_a^{(2)} = \frac{\delta \bar{\Pi}_a^{(2)}}{\delta t} = - \frac{6}{5} \nu_{aa} \bar{\Pi}_a^{(2)} - 2 \sum_{b \neq a} \frac{m_a \nu_{ab}}{m_a + m_b} \left[ \left( 1 + \frac{3}{5} \frac{m_b}{m_a} \right) \bar{\Pi}_a^{(2)} - \frac{2}{5} \frac{n_a}{n_b} \bar{\Pi}_b^{(2)} \right]. \qquad (J27) $$

### J.4. One Ion–Electron Plasma

For a plasma consisting of one ion species and electrons, in the first step

$$ \bar{\mathcal{Q}}_i^{(2)} = \frac{\delta \bar{\Pi}_i^{(2)}}{\delta t} = - \left( \frac{6}{5} \nu_{ii} + 2 \nu_{ie} \right) \bar{\Pi}_i^{(2)} + \frac{4}{5} \nu_{ie} \frac{n_i}{n_e} \bar{\Pi}_e^{(2)}; \qquad (J28) $$

$$ \bar{\mathcal{Q}}_e^{(2)} = \frac{\delta \bar{\Pi}_e^{(2)}}{\delta t} = - \frac{6}{5} (\nu_{ee} + \nu_{ei}) \bar{\Pi}_e^{(2)} + \frac{4}{5} \nu_{ei} \frac{\rho_e}{\rho_i} \bar{\Pi}_i^{(2)}. \qquad (J29) $$

Nevertheless, because for example for the parallel viscosity the ion $\bar{\Pi}_i^{(2)}$ is larger than the electron $\bar{\Pi}_e^{(2)}$ by a factor of $\sqrt{m_i/m_e}$, the coupling is only weak and the last terms in the above expressions can be neglected for simplicity. Then,

$$ \bar{\mathcal{Q}}_i^{(2)} = \frac{\delta \bar{\Pi}_i^{(2)}}{\delta t} = - \bar{\nu}_i \bar{\Pi}_i^{(2)}; \qquad \bar{\nu}_i = \frac{6}{5} \nu_{ii} + 2 \nu_{ie} = \frac{6}{5} \left( 1 + \frac{5}{3} \frac{\sqrt{2}}{Z_i} \sqrt{\frac{m_e}{m_i}} \right) \nu_{ii}; \qquad (J30) $$

$$ \bar{\mathcal{Q}}_e^{(2)} = \frac{\delta \bar{\Pi}_e^{(2)}}{\delta t} = - \bar{\nu}_e \bar{\Pi}_e^{(2)}; \qquad \bar{\nu}_e = \frac{6}{5} (\nu_{ee} + \nu_{ei}) = \frac{6}{5} \left( 1 + \frac{1}{Z_i \sqrt{2}} \right) \nu_{ei}. \qquad (J31) $$

In a quasistatic approximation, one derives the following viscosity coefficients:

$$ \eta_0^a = \frac{p_a}{\bar{\nu}_a}; \quad \eta_1^a = \frac{p_a \bar{\nu}_a}{4\Omega_a^2 + \bar{\nu}_a^2}; \quad \eta_2^a = \frac{p_a \bar{\nu}_a}{\Omega_a^2 + \bar{\nu}_a^2}; \quad \eta_3^a = \frac{2p_a \Omega_a}{4\Omega_a^2 + \bar{\nu}_a^2}; \quad \eta_4^a = \frac{p_a \Omega_a}{\Omega_a^2 + \bar{\nu}_a^2}, \qquad (J32) $$

which have the same form as the BGK viscosities. The difference is that while $\bar{\nu}_i = \nu_{ii} + \nu_{ie}$ and $\bar{\nu}_e = \nu_{ee} + \nu_{ei}$ for the BGK operator, here the frequencies have to be added according to (J30), (J31).

Importantly, because Braginskii (1965) neglected the ion–electron collisions for ion viscosities, direct comparison with Braginskii has to done with $\bar{\nu}_i = (6/5)\nu_{ii}$. Using this approximation, the parallel viscosities of the Burger–Schunk model are

$$ \eta_0^i = \frac{5}{6} \frac{p_i}{\nu_{ii}}; \qquad \eta_0^e = \frac{5}{6} \frac{Z_i \sqrt{2}}{(1 + Z_i \sqrt{2})} \frac{p_e}{\nu_{ei}}, \qquad (J33) $$

where $5/6 = 0.83$, contrasting with the Braginskii ion value of 0.96. Considering a specific case $Z_i = 1$, the electron viscosity $\eta_0^e = 0.49p_e/\nu_{ei}$, contrasting with Braginskii’s value of 0.73.

### J.5. Strong Magnetic Field Limit

Examining the strong magnetic field limit, the viscosities for ions become

$$ \eta_1^i = \frac{3}{10} \frac{p_i \nu_{ii}}{\Omega_i^2}; \quad \eta_2^i = \frac{6}{5} \frac{p_i \nu_{ii}}{\Omega_i^2}; \quad \eta_3^i = \frac{1}{2} \frac{p_i}{\Omega_i}; \quad \eta_4^i = \frac{p_i}{\Omega_i}, \qquad (J34) $$
(with relations $\eta_1^a = 4\eta_1^a$, $\eta_4^a = 2\eta_3^a$ valid for both electrons and ions). All four viscosities match Braginskii exactly! Similarly, for electrons in the strong magnetic field limit, the Burgers–Schunk model yields

$$ \eta_1^e = \frac{3}{10} \left( 1 + \frac{1}{Z_i \sqrt{2}} \right) \frac{p_e \nu_{ei}}{\Omega_e^2}; \quad \eta_2^e = \frac{6}{5} \left( 1 + \frac{1}{Z_i \sqrt{2}} \right) \frac{p_e \nu_{ei}}{\Omega_e^2}; \quad \eta_3^e = \frac{1}{2} \frac{p_e}{\Omega_e}; \quad \eta_4^e = \frac{p_e}{\Omega_e}. \qquad (J35) $$

Evaluation for $Z_i = 1$ yields $\eta_1^e = 0.51 p_e \nu_{ei} / \Omega_e^2$, and again all match Braginskii exactly. If Braginskii had provided electron viscosities for different $Z_i$ values, all four viscosity coefficients (except for parallel $\eta_0$) would match his results exactly.

If ion–electron collisions are considered, the gyroviscosities $\eta_3^i$, $\eta_4^i$ given by (J34) remain unchanged, and the perpendicular viscosities become

$$ \eta_1^i = \frac{p_i \nu_{ii}}{\Omega_i^2} \frac{3}{10} \left( 1 + \frac{5}{3} \frac{\sqrt{2}}{Z_i} \sqrt{\frac{m_e}{m_i}} \right); \quad \eta_2^i = \frac{p_i \nu_{ii}}{\Omega_i^2} \frac{6}{5} \left( 1 + \frac{5}{3} \frac{\sqrt{2}}{Z_i} \sqrt{\frac{m_e}{m_i}} \right), \qquad (J36) $$

where again $\eta_2^i = 4\eta_1^i$ holds. That the result (J36) is indeed correct, and can be checked against the 2-Laguerre Equation (89b) of Ji & Held (2013) when written in a strong B-limit. (Use $\zeta = (1/Z_i)\sqrt{m_e/m_i}$, $r_i = \Omega_i \hat{\tau}_{ii}$, and $\eta_2^i = \hat{\eta}_2^i p_i \hat{\tau}_{ii}$, with conversion $\hat{\tau}_{ii} = \tau_{ii} / \sqrt{2}$, because we use Braginskii’s definition of $\tau_{ii}$; see Section 8.2). Interestingly, the result does not change in their 3-Laguerre model (or higher-order models). The same is true for the perpendicular heat conductivities $\kappa_\perp^a$.

### J.6. Table of Integrals

To calculate the first Rosenbluth potential $H_b$, we use the following integrals:

$$ \mathcal{Y} \int \frac{1}{x} e^{-|x+y|^2} d^3x = \mathcal{Y} \mathcal{Y} \pi^{3/2} \frac{\operatorname{erf}(y)}{y}; \qquad (J37) $$

$$ \int \frac{\mathbf{x}}{x} e^{-|\mathbf{x}+\mathbf{y}|^2} d^3\mathbf{x} = - \mathbf{y} \pi \left[ \frac{e^{-y^2}}{y^2} + \sqrt{\pi} \left( \frac{1}{y} - \frac{1}{2y^3} \right) \operatorname{erf}(y) \right]; \qquad (J38) $$

$$ \int \frac{\mathbf{x}\mathbf{x}}{x} e^{-|\mathbf{x}+\mathbf{y}|^2} d^3\mathbf{x} = \bar{\mathcal{I}}\pi \left[ \frac{e^{-y^2}}{2y^2} + \frac{\sqrt{\pi}}{2} \operatorname{erf}(y) \left( \frac{1}{y} - \frac{1}{2y^3} \right) \right] \\ + \mathcal{Y}\mathcal{Y}\pi \left[ e^{-y^2} \left( \frac{1}{y^2} - \frac{3}{2y^4} \right) + \sqrt{\pi} \operatorname{erf}(y) \left( \frac{1}{y} - \frac{1}{y^3} + \frac{3}{4y^5} \right) \right], \qquad (J39) $$

and so

$$ \int \frac{(\mathbf{x} + \mathbf{y})(\mathbf{x} + \mathbf{y})}{x} e^{-|\mathbf{x}+\mathbf{y}|^2} d^3\mathbf{x} = \bar{\mathcal{I}}\pi \left[ \frac{e^{-y^2}}{2y^2} + \frac{\sqrt{\pi}}{2} \operatorname{erf}(y) \left( \frac{1}{y} - \frac{1}{2y^3} \right) \right] \\ + \mathcal{Y}\mathcal{Y}\pi \left[ \sqrt{\pi} \operatorname{erf}(y) \frac{3}{4y^5} - e^{-y^2} \left( \frac{1}{y^2} + \frac{3}{2y^4} \right) \right]. \qquad (J40) $$

To calculate the second Rosenbluth $G_b$, we use

$$ \mathcal{Y} \int x e^{-|x+y|^2} d^3x = \mathcal{Y} \mathcal{Y} \pi \left[ e^{-y^2} + \sqrt{\pi} \left( y + \frac{1}{2y} \right) \operatorname{erf}(y) \right]; \qquad (J41) $$

$$ \int \mathbf{x}\mathbf{x} e^{-|\mathbf{x}+\mathbf{y}|^2} d^3\mathbf{x} = - \mathbf{y}\pi \left[ \left( 1 + \frac{1}{2y^2} \right) e^{-y^2} + \sqrt{\pi} \left( y + \frac{1}{y} - \frac{1}{4y^3} \right) \operatorname{erf}(y) \right]; \qquad (J42) $$

$$ \int \mathbf{x}\mathbf{x}\mathbf{x} e^{-|\mathbf{x}+\mathbf{y}|^2} d^3\mathbf{x} = \bar{\mathcal{I}}\pi \left[ \left( \frac{1}{2} + \frac{1}{4y^2} \right) e^{-y^2} + \sqrt{\pi} \left( \frac{y}{2} - \frac{1}{8y^3} + \frac{1}{2y} \right) \operatorname{erf}(y) \right] \\ + \mathcal{Y}\mathcal{Y}\mathcal{Y}\pi \left[ \left( 1 + \frac{1}{y^2} - \frac{3}{4y^4} \right) e^{-y^2} + \sqrt{\pi} \left( y + \frac{3}{2y} - \frac{3}{4y^3} + \frac{3}{8y^5} \right) \operatorname{erf}(y) \right], \qquad (J43) $$
and so

$$ \int x(\mathbf{x} + \mathbf{y})(\mathbf{x} + \mathbf{y})e^{-|\mathbf{x}+\mathbf{y}|^2}d^3\mathbf{x} = \bar{\bar{I}}\pi \left[ \left(\frac{1}{2} + \frac{1}{4y^2}\right)e^{-y^2} + \sqrt{\pi}\left(\frac{y}{2} - \frac{1}{8y^3} + \frac{1}{2y}\right)\text{erf}(y) \right] \\ +yy\pi \left[ -\frac{3}{4y^4}e^{-y^2} + \sqrt{\pi}\left(-\frac{1}{4y^3} + \frac{3}{8y^5}\right)\text{erf}(y) \right]. \quad (J44) $$

To calculate the viscosity, the Rosenbluth potentials are integrated by the following scheme:

$$ \int yyf(y)e^{-\alpha^2 y^2} d^3 y = \frac{\bar{\bar{I}}}{3} \int y^2 f(y)e^{-\alpha^2 y^2} d^3 y = \bar{\bar{I}} \frac{4\pi}{3} \int_0^\infty y^4 f(y)e^{-\alpha^2 y^2} dy; \quad (J45) $$

$$ \int y(\bar{\bar{\Pi}}_b^{(2)} \cdot \mathbf{y})f(y)e^{-\alpha^2 y^2} d^3 y = \bar{\bar{\Pi}}_b^{(2)} \frac{4\pi}{3} \int_0^\infty y^4 f(y)e^{-\alpha^2 y^2} dy; \quad (J46) $$

$$ \bar{\bar{\Pi}}_b^{(2)}: \int yyf(y)e^{-\alpha^2 y^2} d^3 y = 0, \quad (J47) $$

where, in our case, functions $f(y)$ are well behaved, so these integrals hold. Additionally, for any symmetric $(3 \times 3)$ matrix $\bar{\bar{A}}$,

$$ \bar{\bar{A}}: \int yyyf(y)e^{-\alpha^2 y^2} d^3 y = \left[ \bar{\bar{A}} + (\text{Tr}\bar{\bar{A}})\frac{\bar{\bar{I}}}{2} \right] \frac{8\pi}{15} \int_0^\infty y^6 f(y)e^{-\alpha^2 y^2} dy, \quad (J48) $$

and for the stress tensor, $\text{Tr} \bar{\bar{\Pi}}_b^{(2)} = 0$ (the integral can be calculated by splitting $\bar{\bar{A}}: yy$ explicitly into components, and then by using symmetries, for example).

### J0.6.1. Spherical Integration

To obtain the integrals (J39), for example, one introduces an orthogonal reference frame in the x-space with unit vectors $\hat{e}_1$, $\hat{e}_2$, $\hat{e}_3$, where the direction of y forms axis $\hat{e}_3 = \mathbf{y}/y$, so that

$$ \mathbf{x} = x \sin \theta \cos \phi \hat{e}_1 + x \sin \theta \sin \phi \hat{e}_2 + x \cos \theta \hat{e}_3, \quad (J49) $$

which then allows one to first perform the integral over $d\phi$,

$$ \int_0^{2\pi} x d\phi = 2\pi x \cos \theta \hat{e}_3; \quad (J50) $$

$$ \int_0^{2\pi} x x d\phi = \pi x^2 \sin^2 \theta \bar{\bar{I}} + \pi x^2 (3 \cos^2 \theta - 1) \hat{e}_3 \hat{e}_3, \quad (J51) $$

and then over $d\theta dx$.

## Appendix K
### Braginskii Heat Flux (11-moment Model)

We use the usual *reducible* Hermite polynomials with a perturbation of the distribution function $f_b = f_b^{(0)}(1 + \chi_b)$ (see the details in Appendix B):

$$ \chi_b = \frac{1}{10} \tilde{h}_i^{b(3)} \tilde{H}_i^{b(3)} + \frac{1}{280} \tilde{h}_i^{b(5)} \tilde{H}_i^{b(5)}, \quad (K1) $$

where

$$ \begin{aligned} \tilde{H}_i^{b(3)} &= \delta_{jk} \tilde{H}_{ijk}^{b(3)} = \tilde{c}_i^b (\tilde{c}_b^2 - 5); \\ \tilde{H}_i^{b(5)} &= \delta_{jk} \delta_{lm} \tilde{H}_{ijklm}^{b(5)} = \tilde{c}_i^b (\tilde{c}_b^4 - 14\tilde{c}_b^2 + 35). \end{aligned} \quad (K2) $$

For the clarity of the calculations, here we only consider the heat flux part of $\chi_b$ (i.e., the 11-moment model), but the full 21-moment model can be implicitly assumed for the final collisional contributions at the semilinear level. The orthogonality relations are (species indices are dropped)

$$ \int \tilde{H}_i^{(3)} \tilde{H}_j^{(3)} \phi^{(0)} d^3 \tilde{c} = 10 \delta_{ij}; \quad \int \tilde{H}_i^{(5)} \tilde{H}_j^{(5)} \phi^{(0)} d^3 \tilde{c} = 280 \delta_{ij}, \quad (K3) $$
yielding (K1). By using this perturbation $\chi_b$, one can directly calculate the heat flux vector and the fifth-order moment vector

$$ \vec{q}_i^b = \frac{m_b}{2} \int f_b c_i c^2 d^3c = \frac{p_b}{2} \sqrt{\frac{T_b}{m_b}} \tilde{h}_i^{b(3)}; $$

$$ X_i^{b(5)} = m_b \int f_b c_i c^4 d^3c = p_b \frac{T_b}{m_b} \sqrt{\frac{T_b}{m_b}} (\tilde{h}_i^{b(5)} + 14\tilde{h}^{b(3)}), \qquad (K4) $$

or one can directly calculate the Hermite moments

$$ \begin{aligned} \tilde{h}_i^{b(3)} &= \frac{2}{p_b} \sqrt{\frac{m_b}{T_b}} \vec{q}_i^b; \\ \tilde{h}_i^{b(5)} &= \frac{1}{p_b} \sqrt{\frac{m_b}{T_b}} \left( \frac{m_b}{T_b} X_i^{b(5)} - 28 \vec{q}_i^b \right). \end{aligned} \qquad (K5) $$

Note that we have chosen to define all of the vectors and tensors (including $X_i^{b(5)}$, $\tilde{H}_i^{(3)}$, $\tilde{H}_i^{(5)}$, etc.) without any additional normalization factors, so they are directly obtained from higher-order tensors by just applying contractions. The sole exception is the heat flux vector, which contains a factor of 1/2, to match its usual definition. As also noted after Equation (B41), the reminder of this exception in the index notation is the arrow on the heat flux vector components $q_i$. We will again use the Rosenbluth potentials, and not the center-of-mass transformation. However, this time we will keep working with the Hermite fluid moments, which has a nice advantage in that the expressions can be kept in a partially dimensionless form.

### K.1. Rosenbluth Potentials

By introducing

$$ \tilde{c}_b = \sqrt{\frac{m_b}{T_b}} (\mathbf{v}' - \mathbf{u}_b); \quad |\mathbf{v}' - \mathbf{v}| = \sqrt{\frac{T_b}{m_b}} |\tilde{c}_b - \tilde{\mathbf{y}}|; \quad \tilde{\mathbf{y}} = \sqrt{\frac{m_b}{T_b}} (\mathbf{v} - \mathbf{u}_b), \qquad (K6) $$

so that our previously used $\mathbf{y} = \tilde{\mathbf{y}}/\sqrt{2}$, the Rosenbluth potentials read

$$ \begin{aligned} H_b(\mathbf{v}) &= \int \frac{f_b(\mathbf{v}')}{|\mathbf{v}' - \mathbf{v}|} d^3\mathbf{v}' = n_b \sqrt{\frac{m_b}{T_b}} \int \frac{\phi_b^{(0)}}{|\tilde{c}_b - \tilde{\mathbf{y}}|} (1 + \chi_b) d^3\tilde{c}_b; \\ G_b(\mathbf{v}) &= \int |\mathbf{v}' - \mathbf{v}| f_b(\mathbf{v}') d^3\mathbf{v}' = n_b \sqrt{\frac{T_b}{m_b}} \int |\tilde{c}_b - \tilde{\mathbf{y}}| \phi_b^{(0)} (1 + \chi_b) d^3\tilde{c}_b, \end{aligned} \qquad (K7) $$

and calculate

$$ \begin{aligned} H_b(\mathbf{v}) &= n_b \sqrt{\frac{m_b}{T_b}} \left[ \frac{1}{\tilde{\mathbf{y}}} \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{10} \left(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)} + (\tilde{\mathbf{y}}^2 - 5)\frac{\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)}}{28}\right) \right]; \\ G_b(\mathbf{v}) &= n_b \sqrt{\frac{T_b}{m_b}} \left[ \begin{aligned} &\sqrt{\frac{2}{\pi}} e^{-\tilde{\mathbf{y}}^2/2} + \left(\tilde{\mathbf{y}} + \frac{1}{\tilde{\mathbf{y}}}\right) \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) \\ &+ \left(\frac{\operatorname{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{5\tilde{\mathbf{y}}^3} - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{5\tilde{\mathbf{y}}^2}\right) \tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)} - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{140} \tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)} \end{aligned} \right]. \end{aligned} \qquad (K8) $$

The derivatives calculate using $\partial/\partial v_i = \sqrt{m_b/T_b} \partial/\partial \tilde{y}_i$ and

$$ \frac{\partial H_b}{\partial \mathbf{v}} = \frac{n_b m_b}{T_b} \left[ \tilde{\mathbf{y}} \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{\tilde{\mathbf{y}}^2} - \frac{\operatorname{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{\tilde{\mathbf{y}}^3} \right) - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{10} \left( \tilde{\mathbf{h}}^{b(3)} - \tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)}) + (\tilde{\mathbf{y}}^2 - 5) \frac{\tilde{\mathbf{h}}^{b(5)}}{28} - (\tilde{\mathbf{y}}^2 - 7) \frac{\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)})}{28} \right) \right], \qquad (K9) $$

and further applying $(\partial/\partial \mathbf{v}) \cdot$ recovers $-4\pi f_b$. It is useful to write the second Rosenbluth potential as

$$ G_b(\mathbf{v}) = n_b \sqrt{\frac{T_b}{m_b}} [\tilde{A}_1 + \tilde{A}_3 \tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)} + \tilde{A}_5 \tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)}], \qquad (K10) $$
where

$$
\begin{align}
\tilde{A}_1 &= \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \left(\tilde{y} + \frac{1}{\tilde{y}}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \nonumber \\
\tilde{A}_3 &= \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{5\tilde{y}^3} - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{5\tilde{y}^2}; \nonumber \\
\tilde{A}_5 &= -\sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{140}, \tag{K11}
\end{align}
$$

so that the second derivative calculates easily:

$$
\begin{equation}
\begin{split}
\frac{\partial}{\partial \mathbf{v}} \frac{\partial G_b}{\partial \mathbf{v}} = n_b \sqrt{\frac{m_b}{T_b}} \left\{
\begin{aligned}[t]
& \bar{\mathcal{I}} \frac{\tilde{A}'_1}{\tilde{y}} + \frac{\tilde{y} \tilde{y}}{\tilde{y}^2} (\tilde{A}''_1 - \frac{\tilde{A}'_1}{\tilde{y}}) \\
& + (\tilde{y} \tilde{\mathcal{H}}^{b(3)} + \tilde{\mathcal{H}}^{b(3)} \tilde{y} + \bar{\mathcal{I}} (\tilde{y} \cdot \tilde{\mathcal{H}}^{b(3)})) \frac{\tilde{A}'_3}{\tilde{y}} + \frac{\tilde{y} \tilde{y}}{\tilde{y}^2} (\tilde{y} \cdot \tilde{\mathcal{H}}^{b(3)}) (\tilde{A}''_3 - \frac{\tilde{A}'_3}{\tilde{y}}) \\
& + (\tilde{y} \tilde{\mathcal{H}}^{b(5)} + \tilde{\mathcal{H}}^{b(5)} \tilde{y} + \bar{\mathcal{I}} (\tilde{y} \cdot \tilde{\mathcal{H}}^{b(5)})) \frac{\tilde{A}'_5}{\tilde{y}} + \frac{\tilde{y} \tilde{y}}{\tilde{y}^2} (\tilde{y} \cdot \tilde{\mathcal{H}}^{b(5)}) (\tilde{A}''_5 - \frac{\tilde{A}'_5}{\tilde{y}})
\end{aligned}
\right\},
\end{split}
\tag{K12}
\end{equation}
$$

and applying (1/2)Tr recovers $H_b$. The coefficients are

$$
\begin{align*}
\tilde{A}'_1 &= \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}} + \left(1 - \frac{1}{\tilde{y}^2}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}'_3 &= \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}} + \frac{3}{\tilde{y}^3}\right) \frac{e^{-\tilde{y}^2/2}}{5} - \frac{3}{5\tilde{y}^4} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}'_5 &= \sqrt{\frac{2}{\pi}} \frac{\tilde{y} e^{-\tilde{y}^2/2}}{140}; \\
\tilde{A}''_1 &= -\sqrt{\frac{2}{\pi}} \frac{2}{\tilde{y}^2} e^{-\tilde{y}^2/2} + \frac{2}{\tilde{y}^3} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}''_3 &= -\sqrt{\frac{2}{\pi}} \left(1 + \frac{4}{\tilde{y}^2} + \frac{12}{\tilde{y}^4}\right) \frac{e^{-\tilde{y}^2/2}}{5} + \frac{12}{5\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}''_5 &= -\sqrt{\frac{2}{\pi}} (\tilde{y}^2 - 1) \frac{e^{-\tilde{y}^2/2}}{140}, \tag{K13}
\end{align*}
$$

and so

$$
\begin{align}
\tilde{A}_1'' - \frac{\tilde{A}_1'}{\tilde{y}} &= -\sqrt{\frac{2}{\pi}} \frac{3}{\tilde{y}^2} e^{-\tilde{y}^2/2} - \left(\frac{1}{\tilde{y}} - \frac{3}{\tilde{y}^3}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \nonumber \\
\tilde{A}_3'' - \frac{\tilde{A}_3'}{\tilde{y}} &= -\sqrt{\frac{2}{\pi}} \left(1 + \frac{5}{\tilde{y}^2} + \frac{15}{\tilde{y}^4}\right) \frac{e^{-\tilde{y}^2/2}}{5} + \frac{3}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \nonumber \\
\tilde{A}_5'' - \frac{\tilde{A}_5'}{\tilde{y}} &= -\sqrt{\frac{2}{\pi}} \frac{\tilde{y}^2 e^{-\tilde{y}^2/2}}{140}. \tag{K14}
\end{align}
$$

K.2. Dynamical Friction Vector and Diffusion Tensor

The dynamical friction vector thus reads

$$
\begin{equation}
\begin{split}
\boldsymbol{A}^{ab} = {}& 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \left[ \tilde{\mathbf{y}} \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}^3} \right) \right. \\
& \left. - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{10} \left( \tilde{\mathbf{h}}^{b(3)} - \tilde{\mathbf{y}} (\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)}) + (\tilde{y}^2 - 5) \frac{\tilde{\mathbf{h}}^{b(5)}}{28} - (\tilde{y}^2 - 7) \frac{\tilde{\mathbf{y}} (\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)})}{28} \right) \right],
\end{split}
\tag{K15}
\end{equation}
$$
and the diffusion tensor

$$ \bar{D}^{ab} = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \left\{ \bar{\mathcal{I}} \left( \frac{\tilde{A}_1'}{\tilde{y}} + \frac{\tilde{y} \tilde{\mathcal{I}}}{\tilde{y}^2} \left( \tilde{A}_1'' - \frac{\tilde{A}_1'}{\tilde{y}} \right) \right) \right. \\ \left. + (\tilde{y} \tilde{\mathcal{H}}^{b(3)} + \tilde{\mathcal{H}}^{b(3)} \tilde{\mathcal{I}} + \bar{\mathcal{I}} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(3)})) \frac{\tilde{A}_3'}{\tilde{y}} + \frac{\tilde{y} \tilde{\mathcal{I}}}{\tilde{y}^2} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(3)}) \left( \tilde{A}_3'' - \frac{\tilde{A}_3'}{\tilde{y}} \right) \right. \\ \left. + (\tilde{y} \tilde{\mathcal{H}}^{b(5)} + \tilde{\mathcal{H}}^{b(5)} \tilde{\mathcal{I}} + \bar{\mathcal{I}} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(5)})) \frac{\tilde{A}_5'}{\tilde{y}} + \frac{\tilde{y} \tilde{\mathcal{I}}}{\tilde{y}^2} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(5)}) \left( \tilde{A}_5'' - \frac{\tilde{A}_5'}{\tilde{y}} \right) \right\}, \quad (K16) $$

or in its entire beauty:

$$ \bar{D}^{ab} = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \left\{ \bar{\mathcal{I}} \left[ \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} + \left( \frac{1}{\tilde{y}} - \frac{1}{\tilde{y}^3} \right) \operatorname{erf} \left( \frac{\tilde{y}}{\sqrt{2}} \right) \right] \right. \\ + \frac{\tilde{y} \tilde{\mathcal{I}}}{\tilde{y}^2} \left[ - \sqrt{\frac{2}{\pi}} \frac{3}{\tilde{y}^2} e^{-\tilde{y}^2/2} - \left( \frac{1}{\tilde{y}} - \frac{3}{\tilde{y}^3} \right) \operatorname{erf} \left( \frac{\tilde{y}}{\sqrt{2}} \right) \right] \\ + (\tilde{y} \tilde{\mathcal{H}}^{b(3)} + \tilde{\mathcal{H}}^{b(3)} \tilde{\mathcal{I}} + \bar{\mathcal{I}} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(3)})) \left[ \sqrt{\frac{2}{\pi}} \left( \frac{1}{\tilde{y}^2} + \frac{3}{\tilde{y}^4} \right) \frac{e^{-\tilde{y}^2/2}}{5} - \frac{3}{5\tilde{y}^5} \operatorname{erf} \left( \frac{\tilde{y}}{\sqrt{2}} \right) \right] \\ + \frac{\tilde{y} \tilde{\mathcal{I}}}{\tilde{y}^2} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(3)}) \left[ - \sqrt{\frac{2}{\pi}} \left( 1 + \frac{5}{\tilde{y}^2} + \frac{15}{\tilde{y}^4} \right) \frac{e^{-\tilde{y}^2/2}}{5} + \frac{3}{\tilde{y}^5} \operatorname{erf} \left( \frac{\tilde{y}}{\sqrt{2}} \right) \right] \\ + (\tilde{y} \tilde{\mathcal{H}}^{b(5)} + \tilde{\mathcal{H}}^{b(5)} \tilde{\mathcal{I}} + \bar{\mathcal{I}} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(5)})) \left[ \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{140} \right] \\ \left. + \frac{\tilde{y} \tilde{\mathcal{I}}}{\tilde{y}^2} (\tilde{\mathcal{I}} \cdot \tilde{\mathcal{H}}^{b(5)}) \left[ - \sqrt{\frac{2}{\pi}} \frac{\tilde{y}^2 e^{-\tilde{y}^2/2}}{140} \right] \right\}. \quad (K17) $$

As a reminder,

$$ \tilde{\mathbf{y}} = \sqrt{\frac{m_b}{T_b}} (\mathbf{v} - \mathbf{u}_b); \quad c_{ab} = 2\pi e^4 Z_a^2 Z_b^2 \ln \Lambda. \quad (K18) $$

### K.3. Distribution Function for Species “a”

The general distribution function for species “a” reads

$$ f_a = f_a^{(0)} (1 + \chi_a) = n_a \left( \frac{m_a}{T_a} \right)^{3/2} \phi_a^{(0)} (1 + \chi_a); \quad \phi_a^{(0)} = \frac{e^{-\tilde{c}_a^2/2}}{(2\pi)^{3/2}}; \quad \tilde{c}_a = \sqrt{\frac{m_a}{T_a}} (\mathbf{v} - \mathbf{u}_a), \quad (K19) $$

where the perturbation

$$ \chi_a = \frac{1}{10} \tilde{h}_i^{a(3)} \tilde{H}_i^{a(3)} (\tilde{c}_a) + \frac{1}{280} \tilde{h}_i^{a(5)} \tilde{H}_i^{a(5)} (\tilde{c}_a). \quad (K20) $$

To avoid the complicated runaway effect, the weight has to be expanded with small drifts, for example, by defining

$$ \tilde{\mathbf{u}} = (\mathbf{u}_b - \mathbf{u}_a) \sqrt{\frac{m_a}{T_a}}; \quad \alpha = \frac{\sqrt{T_b/m_b}}{\sqrt{T_a/m_a}}; \quad \tilde{c}_a = \alpha \tilde{\mathbf{y}} + \tilde{\mathbf{u}}, \quad (K21) $$

so that the expansion for small drifts

$$ e^{-\tilde{c}_a^2/2} = e^{-|\alpha \tilde{\mathbf{y}} + \tilde{\mathbf{u}}|^2/2} \approx e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} (1 - \alpha \tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}}). \quad (K22) $$

In comparison to our previously used normalization, $\tilde{\mathbf{y}} = \mathbf{y}\sqrt{2}$ and $\tilde{\mathbf{u}} = \mathbf{u}\sqrt{2}$ and $\tilde{c}_a = c_a\sqrt{2}/v_{\text{tha}}$. The perturbation $\chi_a$ contains Hermite polynomials, and these also have to be expanded in the semilinear approximation. Importantly, after contraction with the
Hermite (fluid) moments,

$$
\begin{align}
\tilde{h}_i^{a(3)} \tilde{H}_i^{a(3)} (\tilde{c}_a) &\approx \tilde{h}_i^{a(3)} \alpha \tilde{y}_i (\alpha^2 \tilde{y}^2 - 5) = \tilde{h}_i^{a(3)} \tilde{H}_i^{a(3)} (\alpha \tilde{y}); \nonumber \\
\tilde{h}_i^{a(5)} \tilde{H}_i^{a(5)} (\tilde{c}_a) &\approx \tilde{h}_i^{a(5)} \alpha \tilde{y}_i (\alpha^4 \tilde{y}^4 - 14 \alpha^2 \tilde{y}^2 + 35) = \tilde{h}_i^{a(5)} \tilde{H}_i^{a(5)} (\alpha \tilde{y}), \tag{K23}
\end{align}
$$

where all of the drift $\tilde{u}$ contributions such as $\tilde{h}_i^{a(3)} \tilde{u}_i$ are neglected in the semilinear approximation. The expanded distribution function thus reads

$$
f_a = n_a \left(\frac{m_a}{T_a}\right)^{3/2} \frac{e^{-\alpha^2 \tilde{y}^2/2}}{(2\pi)^{3/2}} (1 - \alpha \tilde{y} \cdot \tilde{u} + \chi_a), \quad (K24)
$$

with the perturbation

$$
\chi_a = \frac{1}{10} \tilde{h}_i^{a(3)} \tilde{H}_i^{a(3)}(\alpha \tilde{y}) + \frac{1}{280} \tilde{h}_i^{a(5)} \tilde{H}_i^{a(5)}(\alpha \tilde{y}). \quad (K25)
$$

The integrals are evaluated with $d^3v = (T_b/m_b)^{3/2}d^3\tilde{y}$, so a useful shortcut is

$$
\int f_a d^3 v = n_a \alpha^3 \int \frac{e^{-\alpha^2 \tilde{y}^2/2}}{(2\pi)^{3/2}} (1 - \alpha \tilde{y} \cdot \tilde{u} + \chi_a) d^3 \tilde{y}. \quad (K26)
$$

Also, it is useful to express $c_{ab}$ directly through the collisional frequencies $\nu_{ab}$, according to

$$
2 \frac{c_{ab} n_b}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) = 3 \nu_{ab} \sqrt{\frac{\pi}{2}} (1 + \alpha^2)^{3/2} \left(\frac{T_a}{m_a}\right)^{3/2}. \quad (K27)
$$

K.4. Momentum Exchange Rates R<sub>ab</sub>

The momentum exchange rates calculate

$$
\begin{align}
\mathbf{R}_{ab} &= m_a \int f_a A^{ab} d^3 v \nonumber \\
&= \nu_{ab} \rho_a \sqrt{\frac{T_a}{m_a}} \tilde{u} + \frac{3}{5} \nu_{ab} \frac{\mu_{ab}}{T_{ab}} \left[ \frac{p_a}{2} \sqrt{\frac{T_a}{m_a}} \tilde{h}^{a(3)} - \frac{\rho_a}{\rho_b} \frac{p_b}{2} \sqrt{\frac{T_b}{m_b}} \tilde{h}^{b(3)} \right] \nonumber \\
&\quad - \frac{3}{56} \nu_{ab} \left( \frac{\mu_{ab}}{T_{ab}} \right)^2 \left[ p_a \left( \frac{T_a}{m_a} \right)^{3/2} \tilde{h}^{a(5)} - \frac{\rho_a}{\rho_b} p_b \left( \frac{T_b}{m_b} \right)^{3/2} \tilde{h}^{b(5)} \right], \tag{K28}
\end{align}
$$

or expressed through the usual fluid variables,

$$
\begin{equation}
\begin{split}
\mathbf{R}_{ab} = \nu_{ab} \rho_a (\mathbf{u}_b - \mathbf{u}_a) &+ \frac{3}{5} \nu_{ab} \frac{\mu_{ab}}{T_{ab}} \left[ \mathbf{q}^a - \frac{\rho_a}{\rho_b} \mathbf{q}^b \right] \\
& - \frac{3}{56} \nu_{ab} \left( \frac{\mu_{ab}}{T_{ab}} \right)^2 \left[ \left( \mathbf{X}^{a(5)} - 28 \frac{T_a}{m_a} \mathbf{q}^a \right) - \frac{\rho_a}{\rho_b} \left( \mathbf{X}^{b(5)} - 28 \frac{T_b}{m_b} \mathbf{q}^b \right) \right].
\end{split}
\tag{K29}
\end{equation}
$$

Note that $\mathbf{R}_{ab} = -\mathbf{R}_{ba}$. An alternative form reads

$$
\begin{equation}
\begin{split}
\mathbf{R}_{ab} = \nu_{ab} \rho_a (\mathbf{u}_b - \mathbf{u}_a) &+ \nu_{ab} \frac{\mu_{ab}}{T_{ab}} \left[ \mathbf{q}^a \left( \frac{3}{5} + \frac{3}{2} \frac{\mu_{ab}}{m_a} \frac{T_a}{T_{ab}} \right) - \frac{\rho_a}{\rho_b} \mathbf{q}^b \left( \frac{3}{5} + \frac{3}{2} \frac{\mu_{ab}}{m_b} \frac{T_b}{T_{ab}} \right) \right] \\
& - \frac{3}{56} \nu_{ab} \left( \frac{\mu_{ab}}{T_{ab}} \right)^2 \left[ \mathbf{X}^{a(5)} - \frac{\rho_a}{\rho_b} \mathbf{X}^{b(5)} \right],
\end{split}
\tag{K30}
\end{equation}
$$

or yet another one:

$$
\begin{equation}
\begin{split}
\mathbf{R}_{ab} = \nu_{ab} \rho_a (\mathbf{u}_b - \mathbf{u}_a) &+ \nu_{ab} \frac{\mu_{ab}}{T_{ab}} \left[ \frac{\mathbf{q}^a \frac{21}{10} T_a m_b + \frac{3}{5} T_b m_a}{T_a m_b + T_b m_a} - \frac{\rho_a}{\rho_b} \frac{\mathbf{q}^b \frac{3}{5} T_a m_b + \frac{21}{10} T_b m_a}{T_a m_b + T_b m_a} \right] \\
& - \frac{3}{56} \nu_{ab} \left( \frac{\mu_{ab}}{T_{ab}} \right)^2 \left[ \mathbf{X}^{a(5)} - \frac{\rho_a}{\rho_b} \mathbf{X}^{b(5)} \right].
\end{split}
\tag{K31}
\end{equation}
$$
### K.5. Heat Flux Exchange Rates

We need to calculate the collisional contributions for the heat flux

$$ \frac{1}{2} \mathrm{Tr} \bar{\mathbf{Q}}_{ab}^{(3)} = \frac{\delta \mathbf{q}_{ab}}{\delta t} = m_a \int f_a \left[ (\mathbf{A}_{ab} \cdot \mathbf{c}_a) \mathbf{c}_a + \frac{1}{2} \mathbf{A}_{ab} |\mathbf{c}_a|^2 \right] d^3v \\ + m_a \int f_a \left[ \frac{1}{2} (\mathrm{Tr} \bar{\mathbf{D}}_{ab}) \mathbf{c}_a + \bar{\mathbf{D}}_{ab} \cdot \mathbf{c}_a \right] d^3v, \qquad (K32) $$

where the velocity

$$ \mathbf{c}_a = \sqrt{\frac{T_a}{m_a}} \tilde{\mathbf{c}}_a = \sqrt{\frac{T_a}{m_a}} (\alpha \tilde{\mathbf{y}} + \tilde{\mathbf{u}}). \qquad (K33) $$

Before attempting the integration of (K32), it is useful to apply the semilinear approximation, which yields step by step

$$ \mathbf{A}^{ab} \cdot \mathbf{c}_a \approx 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \left[ \begin{aligned} & \left(\alpha \tilde{\mathbf{y}}^2 + \tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}}\right) \left(\sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{\tilde{\mathbf{y}}^2} - \frac{\mathrm{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{\tilde{\mathbf{y}}^3}\right) \\ & - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{10} \alpha \left( (\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)})(1-\tilde{\mathbf{y}}^2) + \frac{(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)})(8\tilde{\mathbf{y}}^2 - \tilde{\mathbf{y}}^4 - 5)}{28} \right) \end{aligned} \right]; \qquad (K34) $$

$$ (\mathbf{A}^{ab} \cdot \mathbf{c}_a) \mathbf{c}_a \approx 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \frac{T_a}{m_a} \left[ \begin{aligned} & + \alpha(\alpha \tilde{\mathbf{y}}^2 \tilde{\mathbf{y}} + \tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}}) + \tilde{\mathbf{y}}^2 \tilde{\mathbf{u}}) \left(\sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{\tilde{\mathbf{y}}^2} - \frac{\mathrm{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{\tilde{\mathbf{y}}^3}\right) \\ & - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{10} \alpha^2 \left(\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)})(1-\tilde{\mathbf{y}}^2) + \frac{\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)})(8\tilde{\mathbf{y}}^2 - \tilde{\mathbf{y}}^4 - 5)}{28}\right) \end{aligned} \right]. \qquad (K35) $$

Furthermore, in the semilinear approximation,

$$ |\mathbf{c}_a|^2 \approx \frac{T_a}{m_a}(\alpha^2 \tilde{\mathbf{y}}^2 + 2\alpha \tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}}), \qquad (K36) $$

and thus

$$ \mathbf{A}^{ab} |\mathbf{c}_a|^2 \approx 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \frac{T_a}{m_a} \left[ \begin{aligned} & + \alpha(\alpha \tilde{\mathbf{y}}^2 \tilde{\mathbf{y}} + 2\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}})) \left(\sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{\tilde{\mathbf{y}}^2} - \frac{\mathrm{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{\tilde{\mathbf{y}}^3}\right) \\ & - \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{10} \alpha^2 \tilde{\mathbf{y}}^2 \left(\tilde{\mathbf{h}}^{b(3)} - \tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)} + (\tilde{\mathbf{y}}^2 - 5)\frac{\tilde{\mathbf{h}}^{b(5)}}{28} - (\tilde{\mathbf{y}}^2 - 7)\frac{\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)})}{28}\right) \end{aligned} \right]. \qquad (K37) $$

For the diffusion tensor,

$$ \mathrm{Tr} \bar{\mathbf{D}}^{ab} = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \left\{ \left( 2 \frac{\tilde{A}_1'}{\tilde{y}} + \tilde{A}_1'' \right) \right. \\ \left. + (\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(3)}) \left( 4 \frac{\tilde{A}_3'}{\tilde{y}} + \tilde{A}_3'' \right) + (\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{b(5)}) \left( 4 \frac{\tilde{A}_5'}{\tilde{y}} + \tilde{A}_5'' \right) \right\}, \qquad (K38) $$
and in the semilinear approximation:

$$ \frac{1}{2}(\text{Tr}\bar{\mathbf{D}}^{ab})\mathbf{c}_a \approx 2\frac{c_{ab}}{m_a^2}\frac{n_b}{\alpha}\left\{ (\alpha\tilde{\mathbf{y}} + \tilde{\mathbf{u}})\left(\frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} + \frac{\tilde{\mathbf{A}}_1''}{2}\right) \right. \\ \left. + \alpha\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(3)})\left(2\frac{\tilde{\mathbf{A}}_3'}{\tilde{\mathbf{y}}} + \frac{\tilde{\mathbf{A}}_3''}{2}\right) + \alpha\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(5)})\left(2\frac{\tilde{\mathbf{A}}_5'}{\tilde{\mathbf{y}}} + \frac{\tilde{\mathbf{A}}_5''}{2}\right) \right\}; \quad (K39) $$

$$ \bar{\mathbf{D}}^{ab} \cdot \mathbf{c}_a \approx 2\frac{c_{ab}}{m_a^2}\frac{n_b}{\alpha}\left\{ \alpha\tilde{\mathbf{y}}\tilde{\mathbf{A}}_1'' + \tilde{\mathbf{u}}\frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} + \frac{\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}})}{\tilde{\mathbf{y}}^2}\left(\tilde{\mathbf{A}}_1'' - \frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}}\right) \right. \\ \left. + \alpha\tilde{\mathbf{h}}^{(3)}\tilde{\mathbf{y}}\tilde{\mathbf{A}}_3' + \alpha\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(3)})\left(\frac{\tilde{\mathbf{A}}_3'}{\tilde{\mathbf{y}}} + \tilde{\mathbf{A}}_3''\right) \right. \\ \left. + \alpha\tilde{\mathbf{h}}^{(5)}\tilde{\mathbf{y}}\tilde{\mathbf{A}}_5' + \alpha\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(5)})\left(\frac{\tilde{\mathbf{A}}_5'}{\tilde{\mathbf{y}}} + \tilde{\mathbf{A}}_5''\right) \right\}. \quad (K40) $$

Collecting all of the results together, the first part of (K32) becomes

$$ (A^{ab} \cdot c_a)c_a + \frac{1}{2}A^{ab}|c_a|^2 \approx 2\frac{c_{ab}}{m_a^2}\left(1 + \frac{m_a}{m_b}\right)\frac{n_b}{\alpha^2}\left\{ \alpha\left(\frac{3}{2}\alpha\tilde{\mathbf{y}}^2\tilde{\mathbf{y}} + 2\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}}) + \tilde{\mathbf{y}}^2\tilde{\mathbf{u}}\right)\left(\sqrt{\frac{2}{\pi}}\frac{e^{-\tilde{\mathbf{y}}^2/2}}{\tilde{\mathbf{y}}^2} - \frac{\text{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{\tilde{\mathbf{y}}^3}\right) \right. \\ \left. - \sqrt{\frac{2}{\pi}}\frac{e^{-\tilde{\mathbf{y}}^2/2}}{10}\alpha^2\left(\frac{1}{2}\tilde{\mathbf{y}}^2\tilde{\mathbf{h}}^{(3)} + \tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(3)})(1 - \frac{3}{2}\tilde{\mathbf{y}}^2) \right. \right. \\ \left. + \frac{1}{2}\tilde{\mathbf{y}}^2(\tilde{\mathbf{y}}^2 - 5)\frac{\tilde{\mathbf{h}}^{(5)}}{28} + \frac{\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(5)})}{28}\left(\frac{23}{2}\tilde{\mathbf{y}}^2 - \frac{3}{2}\tilde{\mathbf{y}}^4 - 5\right) \right\}; \quad (K41) $$

and the second part of (K32) becomes

$$ \frac{1}{2}(\text{Tr}\bar{\mathbf{D}}^{ab})\mathbf{c}_a + \bar{\mathbf{D}}^{ab} \cdot \mathbf{c}_a \approx 2\frac{c_{ab}}{m_a^2}\frac{n_b}{\alpha}\left\{ \alpha\tilde{\mathbf{y}}\left(\frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} + \frac{3\tilde{\mathbf{A}}_1''}{2}\right) + \tilde{\mathbf{u}}\left(2\frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} + \frac{\tilde{\mathbf{A}}_1''}{2}\right) + \frac{\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}})}{\tilde{\mathbf{y}}^2}\left(\tilde{\mathbf{A}}_1'' - \frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}}\right) \right. \\ \left. + \alpha\tilde{\mathbf{h}}^{(3)}\tilde{\mathbf{y}}\tilde{\mathbf{A}}_3' + \alpha\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(3)})\left(3\frac{\tilde{\mathbf{A}}_3'}{\tilde{\mathbf{y}}} + \frac{3}{2}\tilde{\mathbf{A}}_3''\right) \right. \\ \left. + \alpha\tilde{\mathbf{h}}^{(5)}\tilde{\mathbf{y}}\tilde{\mathbf{A}}_5' + \alpha\tilde{\mathbf{y}}(\tilde{\mathbf{y}} \cdot \tilde{\mathbf{h}}^{(5)})\left(3\frac{\tilde{\mathbf{A}}_5'}{\tilde{\mathbf{y}}} + \frac{3}{2}\tilde{\mathbf{A}}_5''\right) \right\}. \quad (K42) $$

Now (K32) can be directly integrated, again by applying a semilinear approximation during the integration. By using (K26) and (K27), the entire collisional integral (K32) can be written in a symbolic form:

$$ \frac{1}{2} \text{Tr} \bar{\mathbf{Q}}_{ab}^{(3)} = m_a n_a 3 \alpha^3 \nu_{ab} \sqrt{\frac{\pi}{2}} (1 + \alpha^2)^{3/2} \left(\frac{T_a}{m_a}\right)^{3/2} \\ \times \int \frac{e^{-\alpha^2 \tilde{\mathbf{y}}^2/2}}{(2\pi)^{3/2}} (1 - \alpha \tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}} + \chi_a) \left[ \frac{1}{\alpha^2} \{\text{K41}\} + \frac{1}{(1 + \frac{m_a}{m_b})\alpha} \{\text{K42}\} \right] d^3 \tilde{\mathbf{y}}, \quad (K43) $$

where {K41} and {K42} represent only parts of the corresponding equations that are inside the curly brackets. The final result of the integration reads

$$ \frac{1}{2} \text{Tr} \bar{\mathbf{Q}}_{ab}^{(3)} = \frac{\delta \mathbf{q}_{ab}}{\delta t} = - \nu_{ab} p_a (\mathbf{u}_b - \mathbf{u}_a) U_{ab(1)} \\ - \nu_{ab} D_{ab(1)} \frac{p_a}{2} \sqrt{\frac{T_a}{m_a}} \tilde{\mathbf{h}}^{a(3)} + \nu_{ab} D_{ab(2)} \frac{\rho_a}{\rho_b} \frac{p_b}{2} \sqrt{\frac{T_b}{m_b}} \tilde{\mathbf{h}}^{b(3)} \\ + \nu_{ab} E_{ab(1)} p_a \sqrt{\frac{T_a}{m_a}} \tilde{\mathbf{h}}^{a(5)} + \nu_{ab} E_{ab(2)} \frac{\rho_a}{\rho_b} p_b \sqrt{\frac{T_b}{m_b}} \tilde{\mathbf{h}}^{b(5)}, \quad (K44) $$
with the mass-ratio coefficients

$$U_{ab(1)} = \frac{(4T_a - 11T_b)m_a m_b - 2T_a m_b^2 - 5T_b m_a^2}{2(T_a m_b + T_b m_a)(m_b + m_a)};$$

$$D_{ab(1)} = - \frac{6T_a^2 m_a m_b^2 + 2T_a^2 m_b^3 + 21T_a T_b m_a^2 m_b - 5T_a T_b m_a m_b^2 - 30T_b^2 m_a^3 - 52T_b^2 m_a^2 m_b}{10(T_a m_b + T_b m_a)^2(m_b + m_a)};$$

$$D_{ab(2)} = \frac{3m_b T_a [(10T_a - 11T_b)m_a m_b + 4T_a m_b^2 - 5T_b m_a^2]}{10(T_a m_b + T_b m_a)^2(m_b + m_a)};$$

$$E_{ab(1)} = - \frac{3T_a m_b [6T_a^2 m_a m_b^2 + 2T_a^2 m_b^3 + 27T_a T_b m_a^2 m_b - 11T_a T_b m_a m_b^2 - 84T_b^2 m_a^3 - 118T_b^2 m_a^2 m_b]}{560(T_a m_b + T_b m_a)^3(m_b + m_a)};$$

$$E_{ab(2)} = - \frac{3m_a m_b T_a T_b [16T_a m_a m_b + 10T_a m_b^2 - 5T_b m_a^2 - 11T_b m_a m_b]}{112(T_a m_b + T_b m_a)^3(m_b + m_a)}. \qquad (K45)$$

As a double check, we have verified that neglecting the fifth-order Hermite moments $\tilde{h}^{(5)}$ in (K44) yields a model that matches Burgers–Schunk; see Equations (45)–(49) in Schunk (1977; after there prescribing Coulomb collisions). For small temperature differences, the mass-ratio coefficients simplify into

$$U_{ab(1)} = - \frac{(5/2)m_a + m_b}{m_a + m_b};$$

$$D_{ab(1)} = \frac{3m_a^2 + \frac{1}{10}m_a m_b - \frac{1}{5}m_b^2}{(m_a + m_b)^2}; \qquad D_{ab(2)} = \frac{\frac{6}{5}m_b^2 - \frac{3}{2}m_a m_b}{(m_a + m_b)^2};$$

$$E_{ab(1)} = \frac{3 m_b (84m_a^2 + 7m_a m_b - 2m_b^2)}{560 (m_a + m_b)^3}; \qquad E_{ab(2)} = \frac{15 m_a m_b (m_a - 2m_b)}{112 (m_a + m_b)^3}. \qquad (K46)$$

The model is easily changed from Hermite moments to fluid moments by

$$\begin{align*} \boldsymbol{q}_a &= \frac{p_a}{2} \sqrt{\frac{T_a}{m_a}} \tilde{\boldsymbol{h}}^{a(3)}; & p_a \sqrt{\frac{T_a}{m_a}} \tilde{\boldsymbol{h}}^{a(5)} &= \frac{m_a}{T_a} \boldsymbol{X}_a^{(5)} - 28 \boldsymbol{q}_a; \\ \boldsymbol{q}_b &= \frac{p_b}{2} \sqrt{\frac{T_b}{m_b}} \tilde{\boldsymbol{h}}^{b(3)}; & p_b \sqrt{\frac{T_b}{m_b}} \tilde{\boldsymbol{h}}^{b(5)} &= \frac{m_b}{T_b} \boldsymbol{X}_b^{(5)} - 28 \boldsymbol{q}_b. \end{align*} \qquad (K47)$$

The heat flux exchange rates become

$$\begin{align} \boldsymbol{Q}_{ab}^{(3) \prime} &= \frac{\delta \boldsymbol{q}_{ab}^{\prime}}{\delta t} = \frac{1}{2} \mathrm{Tr} \bar{\boldsymbol{Q}}_{ab}^{(3)} - \frac{5}{2} \frac{p_a}{\rho_a} \boldsymbol{R}_{ab} \nonumber \\ &= -\nu_{ab} p_a (\boldsymbol{u}_b - \boldsymbol{u}_a) \hat{U}_{ab(1)} - \nu_{ab} \hat{D}_{ab(1)} \boldsymbol{q}_a + \nu_{ab} \hat{D}_{ab(2)} \frac{\rho_a}{\rho_b} \boldsymbol{q}_b \nonumber \\ &\quad + \nu_{ab} \hat{E}_{ab(1)} \frac{\rho_a}{p_a} \boldsymbol{X}_a^{(5)} - \nu_{ab} \hat{E}_{ab(2)} \frac{\rho_a}{\rho_b} \frac{\rho_b}{p_b} \boldsymbol{X}_b^{(5)}, \tag{K48} \end{align}$$

with mass-ratio coefficients (introducing hat)

$$\begin{align*} \hat{U}_{ab(1)} &= U_{ab(1)} + \frac{5}{2}; \\ \hat{D}_{ab(1)} &= D_{ab(1)} + 28E_{ab(1)} + \frac{3}{2} \frac{T_a}{m_a} \frac{\mu_{ab}}{T_{ab}} + \frac{15}{4} \frac{T_a^2}{m_a^2} \frac{\mu_{ab}^2}{T_{ab}^2}; \\ \hat{D}_{ab(2)} &= D_{ab(2)} - 28E_{ab(2)} + \frac{3}{2} \frac{T_a}{m_a} \frac{\mu_{ab}}{T_{ab}} + \frac{15}{4} \frac{T_a T_b}{m_a m_b} \frac{\mu_{ab}^2}{T_{ab}^2}; \\ \hat{E}_{ab(1)} &= E_{ab(1)} + \frac{15}{112} \frac{T_a^2}{m_a^2} \frac{\mu_{ab}^2}{T_{ab}^2}; \\ \hat{E}_{ab(2)} &= - \left( E_{ab(2)} - \frac{15}{112} \frac{T_a T_b}{m_a m_b} \frac{\mu_{ab}^2}{T_{ab}^2} \right). \end{align*} \qquad (K49)$$

By introducing summation over all of the “b” species and separating the self-collisions, the final results are given by (18), (19).
### K.6. Fifth-order Moment Exchange Rates

We need to calculate the collisional contributions for the right-hand side of the evolution equation for vector $X_i^{a(5)}$, which is obtained by calculating

$$ 
\begin{aligned}
\mathcal{Q}_{ab}^{(5)} &= \text{TrTr}\bar{\mathcal{Q}}_{ab}^{(5)} = \frac{\delta X_{ab}^{(5)}}{\delta t} = m_a \int \mathbf{c}_a |\mathbf{c}_a|^4 C_{ab}(f_a) d^3v \\
&= m_a \int f_a (A^{ab} |\mathbf{c}_a|^4 + 4(A^{ab} \cdot \mathbf{c}_a) |\mathbf{c}_a|^2 \mathbf{c}_a) d^3v \\
&\quad + m_a \int f_a (4(\bar{\mathcal{D}}^{ab} \cdot \mathbf{c}_a) |\mathbf{c}_a|^2 + 4(\bar{\mathcal{D}}^{ab}: \mathbf{c}_a \mathbf{c}_a) \mathbf{c}_a + 2(\text{Tr}\bar{\mathcal{D}}^{ab}) |\mathbf{c}_a|^2 \mathbf{c}_a) d^3v.
\end{aligned}
\quad (K50) $$

Again, before the integration of (K50), it is useful to apply the semilinear approximation, which calculates step by step

$$ A^{ab} |\mathbf{c}_a|^4 \approx 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \left(\frac{T_a}{m_a}\right)^2 \left[ \begin{aligned} &+ \alpha(\alpha^3 \tilde{y}^4 \tilde{y} + 4\alpha^2 \tilde{y}^2 \tilde{y} (\tilde{y} \cdot \tilde{u})) \left(\sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} - \frac{\text{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}^3}\right) \\ &- \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{10} \alpha^4 \tilde{y}^4 \left(\tilde{h}^{b(3)} - \tilde{y}(\tilde{y} \cdot \tilde{h}^{b(3)}) + (\tilde{y}^2 - 5)\frac{\tilde{h}^{b(5)}}{28} - (\tilde{y}^2 - 7)\frac{\tilde{y}(\tilde{y} \cdot \tilde{h}^{b(5)})}{28}\right) \end{aligned} \right]; \quad (K51) $$

$$ (A^{ab} \cdot \mathbf{c}_a) \mathbf{c}_a |\mathbf{c}_a|^2 \approx 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \left(\frac{T_a}{m_a}\right)^2 \left[ \begin{aligned} &+ \alpha(\alpha^3 \tilde{y}^4 \tilde{y} + \alpha^2 \tilde{y}^4 \tilde{u} + 3\alpha^2 \tilde{y}^2 \tilde{y} (\tilde{y} \cdot \tilde{u})) \left(\sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} - \frac{\text{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}^3}\right) \\ &- \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{10} \alpha^4 \tilde{y}^2 \left(\tilde{y}(\tilde{y} \cdot \tilde{h}^{b(3)})(1-\tilde{y}^2) + \frac{\tilde{y}(\tilde{y} \cdot \tilde{h}^{b(5)})(8\tilde{y}^2 - \tilde{y}^4 - 5)}{28}\right) \end{aligned} \right]; \quad (K52) $$

$$ \bar{\mathcal{D}}^{ab}: \mathbf{c}_a \mathbf{c}_a \approx 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \frac{T_a}{m_a} \left\{ \begin{aligned} & (\alpha^2 \tilde{y}^2 + 2\alpha(\tilde{y} \cdot \tilde{u})) A_1'' \\ & + \alpha^2 \tilde{y}^2 (\tilde{y} \cdot \tilde{h}^{b(3)}) (\tilde{A}_3'' + 2\frac{\tilde{A}_3'}{\tilde{y}}) + \alpha^2 \tilde{y}^2 (\tilde{y} \cdot \tilde{h}^{b(5)}) (\tilde{A}_5'' + 2\frac{\tilde{A}_5'}{\tilde{y}}) \end{aligned} \right\}; \quad (K53) $$

$$ \bar{\mathcal{D}}^{ab} \cdot \mathbf{c}_a |\mathbf{c}_a|^2 \approx 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{\alpha} \frac{T_a}{m_a} \left\{ \begin{aligned} & \alpha^3 \tilde{y}^2 \tilde{y} \tilde{A}_1'' + \alpha^2 \tilde{u} \tilde{y} \tilde{A}_1' + \alpha^2 \tilde{y} (\tilde{y} \cdot \tilde{u}) (3 \tilde{A}_1'' - \frac{\tilde{A}_1'}{\tilde{y}}) \\ & + \alpha^3 \tilde{y}^2 \left[ \tilde{h}^{b(3)} \tilde{y} \tilde{A}_3' + \tilde{y} (\tilde{y} \cdot \tilde{h}^{b(3)}) \left( \frac{\tilde{A}_3'}{\tilde{y}} + \tilde{A}_3'' \right) \right] \\ & + \alpha^3 \tilde{y}^2 \left[ \tilde{h}^{b(5)} \tilde{y} \tilde{A}_5' + \tilde{y} (\tilde{y} \cdot \tilde{h}^{b(5)}) \left( \frac{\tilde{A}_5'}{\tilde{y}} + \tilde{A}_5'' \right) \right] \end{aligned} \right\}; \quad (K54) $$

$$ (\bar{\mathcal{D}}^{ab}: \mathbf{c}_a \mathbf{c}_a) \mathbf{c}_a \approx 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{\alpha} \frac{T_a}{m_a} \left\{ \begin{aligned} & (\alpha^3 \tilde{y}^2 \tilde{y} + \alpha^2 \tilde{y}^2 \tilde{u} + 2\alpha^2 \tilde{y} (\tilde{y} \cdot \tilde{u})) A_1'' \\ & + \alpha^3 \tilde{y}^2 \tilde{y} (\tilde{y} \cdot \tilde{h}^{b(3)}) \left( 2\frac{\tilde{A}_3'}{\tilde{y}} + \tilde{A}_3'' \right) + \alpha^3 \tilde{y}^2 \tilde{y} (\tilde{y} \cdot \tilde{h}^{b(5)}) \left( 2\frac{\tilde{A}_5'}{\tilde{y}} + \tilde{A}_5'' \right) \end{aligned} \right\}; \quad (K55) $$

$$ 2(\text{Tr}\bar{\mathcal{D}}^{ab}) \mathbf{c}_a |\mathbf{c}_a|^2 \approx 2 \frac{c_{ab}}{m_a^2} \frac{n_b}{\alpha} \frac{T_a}{m_a} \left\{ \begin{aligned} & (\alpha^3 \tilde{y}^2 \tilde{y} + \alpha^2 \tilde{y}^2 \tilde{u} + 2\alpha^2 \tilde{y} (\tilde{y} \cdot \tilde{u})) \left( 4\frac{\tilde{A}_1'}{\tilde{y}} + 2\tilde{A}_1'' \right) \\ & + \alpha^3 \tilde{y}^2 \tilde{y} (\tilde{y} \cdot \tilde{h}^{b(3)}) \left( 8\frac{\tilde{A}_3'}{\tilde{y}} + 2\tilde{A}_3'' \right) + \alpha^3 \tilde{y}^2 \tilde{y} (\tilde{y} \cdot \tilde{h}^{b(5)}) \left( 8\frac{\tilde{A}_5'}{\tilde{y}} + 2\tilde{A}_5'' \right) \end{aligned} \right\}. \quad (K56) $$
Collecting the results together, the first part of (K50) becomes

$$
\begin{equation}
\begin{aligned}
A^{ab} |c_a|^4 + 4(A^{ab} \cdot c_a)c_a |c_a|^2 &\approx 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b T_a}{\alpha^2 m_a} \left\{ \right. \\
& \qquad \left. + \alpha (5\alpha^3 \bar{y}^4 \bar{\mathbf{y}} + 4\alpha^2 \bar{y}^4 \bar{\mathbf{u}} + 16\alpha^2 \bar{y}^2 \bar{\mathbf{y}} (\bar{\mathbf{y}} \cdot \bar{\mathbf{u}})) \right. \\
& \qquad \left. \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\bar{y}^2/2}}{10} \alpha^4 \bar{y}^2 \left( \bar{y}^2 \bar{\mathbf{h}}^{(3)} + \bar{\mathbf{y}} (\bar{\mathbf{y}} \cdot \bar{\mathbf{h}}^{(3)})(4 - 5\bar{y}^2) + \bar{y}^2 (\bar{y}^2 - 5) \frac{\bar{\mathbf{h}}^{(5)}}{28} \right. \right. \\
& \qquad \left. \left. + \frac{\bar{\mathbf{y}} (\bar{\mathbf{y}} \cdot \bar{\mathbf{h}}^{(5)})}{28} (39\bar{y}^2 - 5\bar{y}^4 - 20) \right) \right\},
\end{aligned}
\tag{K57}
\end{equation}
$$

and the second part of (K50) reads

$$
\begin{equation}
\begin{split}
& 4(\bar{\mathbf{D}}^{ab} \cdot c_a)|c_a|^2 + 4(\bar{\mathbf{D}}^{ab}: c_a c_a)c_a + 2(\mathrm{Tr}\,\bar{\mathbf{D}}^{ab})|c_a|^2 c_a \\
& \approx 2 \frac{c_{ab} n_b T_a}{m_a^2 \alpha m_a} \\
& \times \left\{ \begin{aligned}
& \alpha^3 \bar{y}^2 \bar{\mathbf{y}} \left( 4 \frac{\bar{A}_1'}{\bar{y}} + 10 \bar{A}_1'' \right) + \alpha^2 \bar{y}^2 \bar{\mathbf{u}} \left( 8 \frac{\bar{A}_1'}{\bar{y}} + 6 \bar{A}_1'' \right) + \alpha^2 \bar{\mathbf{y}} (\bar{\mathbf{y}} \cdot \bar{\mathbf{u}}) \left( 4 \frac{\bar{A}_1'}{\bar{y}} + 24 \bar{A}_1'' \right) \\
& + 4 \alpha^3 \bar{y}^3 \bar{\mathbf{h}}^{(3)} \bar{A}_3' + \alpha^3 \bar{y}^2 \bar{\mathbf{y}} (\bar{\mathbf{y}} \cdot \bar{\mathbf{h}}^{(3)}) \left( 20 \frac{\bar{A}_3'}{\bar{y}} + 10 \bar{A}_3'' \right) \\
& + 4 \alpha^3 \bar{y}^3 \bar{\mathbf{h}}^{(5)} \bar{A}_5' + \alpha^3 \bar{y}^2 \bar{\mathbf{y}} (\bar{\mathbf{y}} \cdot \bar{\mathbf{h}}^{(5)}) \left( 20 \frac{\bar{A}_5'}{\bar{y}} + 10 \bar{A}_5'' \right)
\end{aligned} \right\}.
\end{split}
\tag{K58}
\end{equation}
$$

Now (K50) can be integrated, and the entire collisional integral can be written in a symbolic form:

$$
\begin{equation}
\begin{split}
\boldsymbol{Q}_{ab}^{(5)} ={}& m_a n_a 3 \alpha^3 \nu_{ab} \sqrt{\frac{\pi}{2}} (1 + \alpha^2)^{3/2} \left(\frac{T_a}{m_a}\right)^{5/2} \\
& \times \int \frac{e^{-\alpha^2 \bar{y}^2/2}}{(2\pi)^{3/2}} (1 - \alpha \bar{\mathbf{y}} \cdot \bar{\mathbf{u}} + \chi_a) \left[ \frac{1}{\alpha^2} \{\text{K57}\} + \frac{1}{(1 + \frac{m_a}{m_b})\alpha} \{\text{K58}\} \right] d^3\bar{\mathbf{y}},
\end{split}
\tag{K59}
\end{equation}
$$

where {K57} and {K58} represent only parts of the corresponding equations that are inside the curly brackets. The integration yields

$$
\begin{equation}
\begin{split}
\boldsymbol{Q}_{ab}^{(5)} = \operatorname{Tr}\operatorname{Tr}\bar{\boldsymbol{Q}}_{ab}^{(5)} = \frac{\delta\boldsymbol{X}_{ab}^{(5)}}{\delta t} = \nu_{ab} \frac{T_a}{m_a} & \left\{ + p_a (\boldsymbol{u}_b - \boldsymbol{u}_a) U_{ab(2)} \right. \\
& \left. - F_{ab(1)} \frac{P_a}{2} \sqrt{\frac{T_a}{m_a}} \bar{\boldsymbol{h}}^{(3)} - F_{ab(2)} \frac{\rho_a}{\rho_b} \frac{P_b}{2} \sqrt{\frac{T_b}{m_b}} \bar{\boldsymbol{h}}^{(3)} \right. \\
& \left. - G_{ab(1)} P_a \sqrt{\frac{T_a}{m_a}} \bar{\boldsymbol{h}}^{(5)} + G_{ab(2)} \frac{\rho_a}{\rho_b} P_b \sqrt{\frac{T_b}{m_b}} \bar{\boldsymbol{h}}^{(5)} \right\},
\end{split}
\tag{K60}
\end{equation}
$$

with mass-ratio coefficients

$$
\begin{align*}
U_{ab(2)} &= - \frac{16 T_a^2 m_a m_b^2 - 8 T_a^2 m_b^3 + 56 T_a T_b m_a^2 m_b - 52 T_a T_b m_a m_b^2 - 35 T_b^2 m_a^3 - 119 T_b^2 m_a^2 m_b}{(T_a m_b + T_b m_a)^2 (m_b + m_a)}; \\
F_{ab(1)} &= \{40 T_a^4 m_a m_b^3 + 8 T_a^4 m_b^4 + 180 T_a^3 T_b m_a^2 m_b^2 + 68 T_a^3 T_b m_a m_b^3 + 315 T_a^2 T_b^2 m_a^3 m_b + 207 T_a^2 T_b^2 m_a^2 m_b^2 \\
&\quad + 700 T_a^3 m_a^4 + 392 T_a T_b^3 m_a^3 m_b - 280 T_b^4 m_a^4\} [5(T_a m_b + T_b m_a)^3 (m_b + m_a) T_a]^{-1}; \\
F_{ab(2)} &= - \frac{3 T_a m_b [16 T_a^2 m_b^3 + 140 T_a T_b m_a^2 m_b + 72 T_a T_b m_a m_b^2 - 35 T_b^2 m_a^3 - 119 T_b^2 m_a^2 m_b]}{5(T_a m_b + T_b m_a)^3 (m_b + m_a)}; \\
G_{ab(1)} &= - \{40 T_a^4 m_a m_b^4 + 8 T_a^4 m_b^5 + 220 T_a^3 T_b m_a^2 m_b^3 + 140 T_a^3 T_b m_a m_b^4 + 495 T_a^2 T_b^2 m_a^3 m_b^2 \\
&\quad + 627 T_a^2 T_b^2 m_a^2 m_b^3 + 3640 T_a T_b^3 m_a^4 m_b + 1916 T_a T_b^3 m_a^3 m_b^2 - 1400 T_b^4 m_a^5 \\
&\quad - 3304 T_b^4 m_a^4 m_b\} [280(T_a m_b + T_b m_a)^4 (m_a + m_b)]^{-1}; \\
G_{ab(2)} &= \frac{3 T_a T_b m_a^2 m_b [8 T_a^2 m_b^2 - 32 T_a T_b m_a m_b - 28 T_a T_b m_b^2 + 5 T_b^2 m_a^2 + 17 T_b^2 m_a m_b]}{8(T_a m_b + T_b m_a)^4 (m_a + m_b)}. \tag{K61}
\end{align*}
$$
For small temperature differences, the mass-ratio coefficients simplify into

$$
\begin{align*}
U_{ab(2)} &= \frac{35m_a^2 + 28m_a m_b + 8m_b^2}{(m_a + m_b)^2}; \\
F_{ab(1)} &= \frac{420m_a^3 + 287m_a^2 m_b + 100m_a m_b^2 + 8m_b^3}{5(m_a + m_b)^3}; \\
F_{ab(2)} &= \frac{3}{5} \frac{m_b(35m_a^2 - 56m_a m_b - 16m_b^2)}{(m_a + m_b)^3}; \\
G_{ab(1)} &= \frac{1400m_a^4 - 1736m_a^3 m_b - 675m_a^2 m_b^2 - 172m_a m_b^3 - 8m_b^4}{280(m_a + m_b)^4}; \\
G_{ab(2)} &= \frac{15}{8} \frac{m_a^2 m_b(m_a - 4m_b)}{(m_a + m_b)^4}. \tag{K62}
\end{align*}
$$

Rewritten with fluid moments, the exchange rates for the fifth-order moment become

$$
\begin{equation}
\begin{split}
\boldsymbol{Q}_{ab}^{(5)\prime} = \boldsymbol{Q}_{ab}^{(5)} - 35 \frac{p_a^2}{\rho_a^2} \boldsymbol{R}_{ab} = \frac{\delta \boldsymbol{X}_{ab}^{(5)\prime}}{\delta t} ={}& \nu_{ab} \left\{ -\frac{p_a^2}{\rho_a} (\boldsymbol{u}_b - \boldsymbol{u}_a) \hat{U}_{ab(2)} \right. \\
& \left. - \hat{F}_{ab(1)} \frac{P_a}{\rho_a} \boldsymbol{q}_a + \hat{F}_{ab(2)} \frac{P_a}{\rho_a} \frac{\rho_a}{\rho_b} \boldsymbol{q}_b - \hat{G}_{ab(1)} \boldsymbol{X}_a^{(5)} - \hat{G}_{ab(2)} \frac{P_a}{P_b} \boldsymbol{X}_b^{(5)} \right\},
\end{split}
\tag{K63}
\end{equation}
$$

with mass-ratio coefficients (introducing hat)

$$
\begin{align*}
\hat{U}_{ab(2)} &= -(U_{ab(2)} - 35); \\
\hat{F}_{ab(1)} &= F_{ab(1)} - 28G_{ab(1)} + 35 \frac{T_a}{m_a} \frac{\mu_{ab}}{T_{ab}} V_{ab(1)}; \\
\hat{F}_{ab(2)} &= -\left( F_{ab(2)} + 28G_{ab(2)} - 35 \frac{T_a}{m_a} \frac{\mu_{ab}}{T_{ab}} V_{ab(2)} \right); \\
\hat{G}_{ab(1)} &= G_{ab(1)} - \frac{15}{8} \frac{T_a^2}{m_a^2} \frac{\mu_{ab}^2}{T_{ab}^2}; \\
\hat{G}_{ab(2)} &= -\left( G_{ab(2)} - \frac{15}{8} \frac{T_a T_b}{m_a m_b} \frac{\mu_{ab}^2}{T_{ab}^2} \right). \tag{K64}
\end{align*}
$$

The final results are given by (20), (21).

Appendix L
Braginskii Viscosity (15-moment Model)

We use polynomials derived from the *reducible* Hermite polynomials (see the details in Appendix B), with the perturbation of the distribution function $f_b(v') = f_b^{(0)}(1 + \chi_b)$

$$
\chi_b = \frac{1}{2} \tilde{h}_{ij}^{b(2)} \tilde{H}_{ij}^{b(2)} + \frac{1}{28} \hat{h}_{ij}^{b(4)} \hat{H}_{ij}^{b(4)}. \qquad (L1)
$$

For the clarity of the calculations, we only consider the viscous part of $\chi_b$ (i.e., the 15-moment model) here, but the full 22-moment model can be implicitly assumed for the final collisional contributions at the semilinear level. The Hermite polynomials are (dropping species index “b” for the polynomials and velocities $\tilde{c}$)

$$
\begin{align*}
\tilde{H}_{ij}^{(2)} &= \tilde{c}_i \tilde{c}_j - \delta_{ij}; \\
\tilde{H}_{ij}^{(4)} &= \tilde{H}_{ijk}^{(4)} = \left( \tilde{c}_i \tilde{c}_j - \frac{\delta_{ij}}{3} \tilde{c}^2 \right) (\tilde{c}^2 - 7) + \frac{\delta_{ij}}{3} \tilde{H}^{(4)}; \\
\tilde{H}^{(4)} &= \tilde{H}_{iik}^{(4)} = \tilde{c}^4 - 10 \tilde{c}^2 + 15; \\
\hat{H}_{ij}^{(4)} &= \tilde{H}_{ij}^{(4)} - \frac{\delta_{ij}}{3} \tilde{H}^{(4)} = \left( \tilde{c}_i \tilde{c}_j - \frac{\delta_{ij}}{3} \tilde{c}^2 \right) (\tilde{c}^2 - 7). \tag{L2}
\end{align*}
$$
The irreducible polynomials yield the same perturbation $\chi_b$. By using the perturbation (L1), one can calculate the fluid moments $\Pi_{ij}^{b(2)}$, $\Pi_{ij}^{b(4)}$, or one can directly calculate the Hermite moments

$$ \tilde{h}_{ij}^{b(2)} = \frac{1}{p_b} \Pi_{ij}^{b(2)}; \quad \hat{h}_{ij}^{b(4)} = \frac{\rho_b}{p_b^2} \Pi_{ij}^{b(4)} - \frac{7}{p_b} \Pi_{ij}^{b(2)}, \qquad (L3) $$

yielding the same relations. Both $\tilde{h}_{ij}^{(2)}$ and $\hat{h}_{ij}^{(4)}$ are traceless (and $\tilde{h}_{ij}^{(2)} = \hat{h}_{ij}^{(2)}$).

### L.1. Rosenbluth Potentials

The notation reads

$$ \tilde{c}_b = \sqrt{\frac{m_b}{T_b}} (\mathbf{v}' - \mathbf{u}_b); \quad |\mathbf{v}' - \mathbf{v}| = \sqrt{\frac{T_b}{m_b}} |\tilde{c}_b - \tilde{\mathbf{y}}|; \quad \tilde{\mathbf{y}} = \sqrt{\frac{m_b}{T_b}} (\mathbf{v} - \mathbf{u}_b), $$

the Rosenbluth potentials are

$$ H_b(\mathbf{v}) = \int \frac{f_b(\mathbf{v}')}{|\mathbf{v}' - \mathbf{v}|} d^3\mathbf{v}' = n_b \sqrt{\frac{m_b}{T_b}} \int \frac{\phi_b^{(0)}}{|\tilde{c}_b - \tilde{\mathbf{y}}|} (1 + \chi_b) d^3\tilde{c}_b; $$

$$ G_b(\mathbf{v}) = \int |\mathbf{v}' - \mathbf{v}| f_b(\mathbf{v}') d^3\mathbf{v}' = n_b \sqrt{\frac{T_b}{m_b}} \int |\tilde{c}_b - \tilde{\mathbf{y}}| \phi_b^{(0)} (1 + \chi_b) d^3\tilde{c}_b, $$

and further calculate

$$ H_b(\mathbf{v}) = n_b \sqrt{\frac{m_b}{T_b}} \left\{ \frac{1}{\tilde{\mathbf{y}}} \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) + \frac{1}{2} (\tilde{\mathbf{h}}^{b(2)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) \left[ \frac{3}{\tilde{\mathbf{y}}^5} \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{\mathbf{y}}^2} + \frac{3}{\tilde{\mathbf{y}}^4}\right) e^{-\tilde{\mathbf{y}}^2/2} \right] - \frac{1}{28} (\tilde{\mathbf{h}}^{b(4)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) \sqrt{\frac{2}{\pi}} e^{-\tilde{\mathbf{y}}^2/2} \right\}; \qquad (L4) $$

$$ G_b(\mathbf{v}) = n_b \sqrt{\frac{T_b}{m_b}} \left\{ \begin{aligned} & \sqrt{\frac{2}{\pi}} e^{-\tilde{\mathbf{y}}^2/2} + \left(\tilde{\mathbf{y}} + \frac{1}{\tilde{\mathbf{y}}}\right) \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) \\ & -\frac{1}{2} (\tilde{\mathbf{h}}^{b(2)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) \left[ \frac{3}{\tilde{\mathbf{y}}^4} \sqrt{\frac{2}{\pi}} e^{-\tilde{\mathbf{y}}^2/2} + \left(\frac{1}{\tilde{\mathbf{y}}^3} - \frac{3}{\tilde{\mathbf{y}}^5}\right) \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) \right] \\ & -\frac{1}{14} (\tilde{\mathbf{h}}^{b(4)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) \left[ \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{\mathbf{y}}^2} + \frac{3}{\tilde{\mathbf{y}}^4}\right) e^{-\tilde{\mathbf{y}}^2/2} - \frac{3}{\tilde{\mathbf{y}}^5} \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) \right] \end{aligned} \right\}. \qquad (L5) $$

The derivative of the first Rosenbluth potential becomes

$$ \begin{aligned} \frac{\partial H_b}{\partial \mathbf{v}} = & \frac{n_b m_b}{T_b} \left\{ \tilde{\mathbf{y}} \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{\mathbf{y}}^2/2}}{\tilde{\mathbf{y}}^2} - \frac{\operatorname{erf}(\tilde{\mathbf{y}}/\sqrt{2})}{\tilde{\mathbf{y}}^3} \right) \right. \\ & + (\tilde{\mathbf{h}}^{b(2)} : \tilde{\mathbf{y}}) \left[ \frac{3}{\tilde{\mathbf{y}}^5} \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{\mathbf{y}}^2} + \frac{3}{\tilde{\mathbf{y}}^4}\right) e^{-\tilde{\mathbf{y}}^2/2} \right] \\ & + \frac{1}{2} (\tilde{\mathbf{h}}^{b(2)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) \tilde{\mathbf{y}} \left[ -\frac{15}{\tilde{\mathbf{y}}^7} \operatorname{erf}\left(\frac{\tilde{\mathbf{y}}}{\sqrt{2}}\right) + \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{\mathbf{y}}^2} + \frac{5}{\tilde{\mathbf{y}}^4} + \frac{15}{\tilde{\mathbf{y}}^6}\right) e^{-\tilde{\mathbf{y}}^2/2} \right] \\ & \left. -\frac{1}{14} (\tilde{\mathbf{h}}^{b(4)} : \tilde{\mathbf{y}}) \sqrt{\frac{2}{\pi}} e^{-\tilde{\mathbf{y}}^2/2} + \frac{1}{28} (\tilde{\mathbf{h}}^{b(4)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) \tilde{\mathbf{y}} \sqrt{\frac{2}{\pi}} e^{-\tilde{\mathbf{y}}^2/2} \right\}. \qquad (L6) \end{aligned} $$

For the second Rosenbluth potential, it is useful to use the form

$$ G_b(\mathbf{v}) = n_b \sqrt{\frac{T_b}{m_b}} [\tilde{A}_1 + \tilde{A}_2 (\tilde{\mathbf{h}}^{b(2)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}}) + \tilde{A}_4 (\tilde{\mathbf{h}}^{b(4)} : \tilde{\mathbf{y}}\tilde{\mathbf{y}})], \qquad (L7) $$
where

$$
\begin{align*}
\tilde{A}_1 &= \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \left(\tilde{y} + \frac{1}{\tilde{y}}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}_2 &= -\frac{1}{2} \left[ \frac{3}{\tilde{y}^4} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \left(\frac{1}{\tilde{y}^3} - \frac{3}{\tilde{y}^5}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) \right]; \\
\tilde{A}_4 &= -\frac{1}{14} \left[ \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{3}{\tilde{y}^4}\right) e^{-\tilde{y}^2/2} - \frac{3}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) \right]. \tag{L8}
\end{align*}
$$

Its second derivative then calculates

$$
\begin{equation}
\begin{split}
\frac{\partial}{\partial \mathbf{v}} \frac{\partial G_b}{\partial \mathbf{v}} = n_b \sqrt{\frac{m_b}{T_b}} \left\{ \bar{\mathcal{I}} \frac{\tilde{A}_1'}{\tilde{y}} + \frac{\tilde{y} \tilde{y}}{\tilde{y}^2} \left( \tilde{A}_1'' - \frac{\tilde{A}_1'}{\tilde{y}} \right) \right. \\
\left. \qquad + \left[ 2\tilde{y} \left( \tilde{\mathcal{h}}^{b(2)} \cdot \tilde{y} \right) + 2 \left( \tilde{\mathcal{h}}^{b(2)} \cdot \tilde{y} \right) \tilde{y} + \bar{\mathcal{I}} \left( \tilde{\mathcal{h}}^{b(2)} : \tilde{y} \tilde{y} \right) \right] \frac{\tilde{A}_2'}{\tilde{y}} \right. \\
\left. \qquad + 2 \tilde{A}_2 \tilde{\mathcal{h}}^{b(2)} + \frac{\tilde{y} \tilde{y}}{\tilde{y}^2} \left( \tilde{\mathcal{h}}^{b(2)} : \tilde{y} \tilde{y} \right) \left( \tilde{A}_2'' - \frac{\tilde{A}_2'}{\tilde{y}} \right) \right. \\
\left. \qquad + \left[ 2\tilde{y} \left( \hat{\mathcal{h}}^{b(4)} \cdot \tilde{y} \right) + 2 \left( \hat{\mathcal{h}}^{b(4)} \cdot \tilde{y} \right) \tilde{y} + \bar{\mathcal{I}} \left( \hat{\mathcal{h}}^{b(4)} : \tilde{y} \tilde{y} \right) \right] \frac{\tilde{A}_4'}{\tilde{y}} \right. \\
\left. \qquad + 2 \tilde{A}_4 \hat{\mathcal{h}}^{b(4)} + \frac{\tilde{y} \tilde{y}}{\tilde{y}^2} \left( \hat{\mathcal{h}}^{b(4)} : \tilde{y} \tilde{y} \right) \left( \tilde{A}_4'' - \frac{\tilde{A}_4'}{\tilde{y}} \right) \right\},
\end{split}
\tag{L9}
\end{equation}
$$

with coefficients

$$
\begin{align*}
\tilde{A}_1' &= \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}} + \left(1 - \frac{1}{\tilde{y}^2}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}_2' &= \left(\frac{1}{\tilde{y}^3} + \frac{15}{2\tilde{y}^5}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{3}{2} \left(\frac{1}{\tilde{y}^4} - \frac{5}{\tilde{y}^6}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}_4' &= \frac{1}{14} \left(\frac{1}{\tilde{y}} + \frac{5}{\tilde{y}^3} + \frac{15}{\tilde{y}^5}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \frac{15}{14\tilde{y}^6} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}_1'' &= -\frac{2}{\tilde{y}^2} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{2}{\tilde{y}^3} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}_2'' &= -\left(\frac{1}{\tilde{y}^2} + \frac{9}{\tilde{y}^4} + \frac{45}{\tilde{y}^6}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \left(-\frac{6}{\tilde{y}^5} + \frac{45}{\tilde{y}^7}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \\
\tilde{A}_4'' &= -\frac{1}{14} \left(1 + \frac{6}{\tilde{y}^2} + \frac{30}{\tilde{y}^4} + \frac{90}{\tilde{y}^6}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{45}{7\tilde{y}^7} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right), \tag{L10}
\end{align*}
$$

and

$$
\begin{align}
\tilde{A}_1'' - \frac{\tilde{A}_1'}{\tilde{y}} &= -\frac{3}{\tilde{y}^2} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \left(\frac{1}{\tilde{y}} - \frac{3}{\tilde{y}^3}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \nonumber \\
\tilde{A}_2'' - \frac{\tilde{A}_2'}{\tilde{y}} &= -\left(\frac{1}{\tilde{y}^2} + \frac{10}{\tilde{y}^4} + \frac{105}{2\tilde{y}^6}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \frac{15}{2}\left(\frac{1}{\tilde{y}^5} - \frac{7}{\tilde{y}^7}\right) \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right); \nonumber \\
\tilde{A}_4'' - \frac{\tilde{A}_4'}{\tilde{y}} &= -\frac{1}{14}\left(1 + \frac{7}{\tilde{y}^2} + \frac{35}{\tilde{y}^4} + \frac{105}{\tilde{y}^6}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{15}{2\tilde{y}^7} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right). \tag{L11}
\end{align}
$$
As a double check, applying (1/2)Tr on (L9) yields

$$ \frac{1}{2} \mathrm{Tr} \frac{\partial}{\partial \mathbf{v}} \frac{\partial \mathbf{G}_b}{\partial \mathbf{v}} = n_b \sqrt{\frac{m_b}{T_b}} \frac{1}{2} \left\{ \tilde{A}_1'' + 2 \frac{\tilde{A}_1'}{\tilde{y}} + (\tilde{\tilde{h}}^{b(2)} : \tilde{y}\tilde{y}) \left( \tilde{A}_2'' + 6 \frac{\tilde{A}_2'}{\tilde{y}} \right) \right. \\ \left. + (\hat{\tilde{h}}^{b(4)} : \tilde{y}\tilde{y}) \left( \tilde{A}_4'' + 6 \frac{\tilde{A}_4'}{\tilde{y}} \right) \right\} = H_b, \qquad (L12) $$

recovering the first Rosenbluth potential (L4). Similarly, applying $(\partial/\partial\mathbf{v})$ on (L6) recovers $-4\pi f_b(\mathbf{v})$. Both Rosenbluth potentials seem to be calculated correctly.

## L.2. Dynamical Friction Vector and Diffusion Tensor

The dynamical friction vector becomes

$$ \begin{aligned} A^{ab} = 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \left\{ \tilde{y} \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}^3} \right) \right. \\ \left. + (\tilde{\tilde{h}}^{b(2)} \cdot \tilde{y}) \left[ \frac{3}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{3}{\tilde{y}^4}\right) e^{-\tilde{y}^2/2} \right] \right. \\ \left. + \frac{1}{2} (\tilde{\tilde{h}}^{b(2)} : \tilde{y}\tilde{y}) \tilde{y} \left[ -\frac{15}{\tilde{y}^7} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{5}{\tilde{y}^4} + \frac{15}{\tilde{y}^6}\right) e^{-\tilde{y}^2/2} \right] \right. \\ \left. - \frac{1}{14} (\hat{\tilde{h}}^{b(4)} \cdot \tilde{y}) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{1}{28} (\hat{\hat{h}}^{b(4)} : \tilde{y}\tilde{y}) \tilde{y} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}, \end{aligned} \qquad (L13) $$

and the diffusion tensor

$$ \begin{aligned} \bar{\bar{D}}^{ab} = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \left\{ \bar{\bar{I}} \frac{\tilde{A}_1'}{\tilde{y}} + \frac{\tilde{y}\tilde{y}}{\tilde{y}^2} \left( \tilde{A}_1'' - \frac{\tilde{A}_1'}{\tilde{y}} \right) \right. \\ \left. + [2\tilde{y}(\tilde{\tilde{h}}^{b(2)} \cdot \tilde{y}) + 2(\tilde{\tilde{h}}^{b(2)} \cdot \tilde{y})\tilde{y} + \bar{\bar{I}}(\tilde{\tilde{h}}^{b(2)} : \tilde{y}\tilde{y})] \frac{\tilde{A}_2'}{\tilde{y}} \right. \\ \left. + 2\tilde{A}_2\tilde{\tilde{h}}^{b(2)} + \frac{\tilde{y}\tilde{y}}{\tilde{y}^2}(\tilde{\tilde{h}}^{b(2)} : \tilde{y}\tilde{y}) \left( \tilde{A}_2'' - \frac{\tilde{A}_2'}{\tilde{y}} \right) \right. \\ \left. + [2\tilde{y}(\hat{\tilde{h}}^{b(4)} \cdot \tilde{y}) + 2(\hat{\tilde{h}}^{b(4)} \cdot \tilde{y})\tilde{y} + \bar{\bar{I}}(\hat{\tilde{h}}^{b(4)} : \tilde{y}\tilde{y})] \frac{\tilde{A}_4'}{\tilde{y}} \right. \\ \left. + 2\tilde{A}_4\hat{\hat{h}}^{b(4)} + \frac{\tilde{y}\tilde{y}}{\tilde{y}^2}(\hat{\hat{h}}^{b(4)} : \tilde{y}\tilde{y}) \left( \tilde{A}_4'' - \frac{\tilde{A}_4'}{\tilde{y}} \right) \right\}, \end{aligned} \qquad (L14) $$

where $c_{ab} = 2\pi e^4 Z_a^2 Z_b^2 \ln \Lambda$.

## L.3. Distribution Function for Species “a”

To avoid the complicated runaway effect, the distribution function $f_a(\mathbf{v}) = f_a^{(0)}(1 + \chi_a)$ has to be expanded for small drifts, in the semilinear approximation. Following the derivation and notation introduced in Appendix C.3, the expanded distribution function becomes

$$ f_a = n_a \left( \frac{m_a}{T_a} \right)^{3/2} \frac{e^{-\alpha^2 \tilde{y}^2 / 2}}{(2\pi)^{3/2}} (1 - \alpha \tilde{y} \cdot \tilde{u} + \chi_a), \qquad (L15) $$

now with the perturbation

$$ \chi_a = \frac{1}{2} \tilde{h}_{ij}^{a(2)} \tilde{H}_{ij}^{a(2)} (\alpha \tilde{y}) + \frac{1}{28} \hat{h}_{ij}^{a(4)} \hat{H}_{ij}^{a(4)} (\alpha \tilde{y}), \qquad (L16) $$
where

$$ \begin{aligned} \tilde{h}_{ij}^{a(2)} \tilde{H}_{ij}^{a(2)} (\alpha \tilde{y}) &= \tilde{h}_{ij}^{a(2)} \alpha^2 \tilde{y}_i \tilde{y}_j; \\ \hat{h}_{ij}^{a(4)} \hat{H}_{ij}^{a(4)} (\alpha \tilde{y}) &= \hat{h}_{ij}^{a(4)} \alpha^2 \tilde{y}_i \tilde{y}_j (\alpha^2 \tilde{y}^2 - 7), \end{aligned} \qquad (L17) $$

so the perturbation reads

$$ \chi_a = \frac{\alpha^2}{2} (\tilde{\tilde{h}}^{a(2)} : \tilde{\tilde{y}}\tilde{\tilde{y}}) + \frac{\alpha^2}{28} (\hat{\tilde{h}}^{a(4)} : \tilde{\tilde{y}}\tilde{\tilde{y}}) (\alpha^2 \tilde{y}^2 - 7). \qquad (L18) $$

As a reminder,

$$ \tilde{u} = (u_b - u_a) \sqrt{\frac{m_a}{T_a}}; \qquad \alpha = \frac{\sqrt{T_b/m_b}}{\sqrt{T_a/m_a}}. $$

## L.4. Pressure Tensor Exchange Rates

We need to calculate the collisional contributions for the right-hand side of the pressure tensor equation, and these contributions read

$$ \bar{Q}_{ab}^{(2)} = m_a \int f_a [A_{ab} c_a]^S d^3 v + m_b \int f_b \bar{D}_{ab} d^3 v. \qquad (L19) $$

By implying

$$ c_a = \sqrt{\frac{T_a}{m_a}} (\alpha \tilde{y} + \tilde{u}), $$

in the semilinear approximation,

$$ \begin{aligned} A^{ab} c_a \approx & 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \left\{ (\alpha \tilde{y} \tilde{y} + \tilde{y} \tilde{u}) \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}^3} \right) \right. \\ & + \alpha (\tilde{\tilde{h}}^{b(2)} \cdot \tilde{y}) \tilde{y} \left[ \frac{3}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{3}{\tilde{y}^4}\right) e^{-\tilde{y}^2/2} \right] \\ & + \frac{\alpha}{2} (\tilde{\tilde{h}}^{b(2)} : \tilde{y} \tilde{y}) \tilde{y} \tilde{y} \left[ -\frac{15}{\tilde{y}^7} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{5}{\tilde{y}^4} + \frac{15}{\tilde{y}^6}\right) e^{-\tilde{y}^2/2} \right] \\ & \left. - \frac{\alpha}{14} (\hat{\tilde{h}}^{b(4)} \cdot \tilde{y}) \tilde{y} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{\alpha}{28} (\hat{\tilde{h}}^{b(4)} : \tilde{y} \tilde{y}) \tilde{y} \tilde{y} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}, \end{aligned} \qquad (L20) $$

and

$$ \begin{aligned} [A^{ab} c_a]^S \approx & 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \left\{ (2\alpha \tilde{y} \tilde{y} + \tilde{y} \tilde{u} + \tilde{u} \tilde{y}) \left( \sqrt{\frac{2}{\pi}} \frac{e^{-\tilde{y}^2/2}}{\tilde{y}^2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}^3} \right) \right. \\ & + \alpha ((\tilde{\tilde{h}}^{b(2)} \cdot \tilde{y}) \tilde{y} + \tilde{y} (\tilde{\tilde{h}}^{b(2)} \cdot \tilde{y})) \left[ \frac{3}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{3}{\tilde{y}^4}\right) e^{-\tilde{y}^2/2} \right] \\ & + \alpha (\tilde{\tilde{h}}^{b(2)} : \tilde{y} \tilde{y}) \tilde{y} \tilde{y} \left[ -\frac{15}{\tilde{y}^7} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \sqrt{\frac{2}{\pi}} \left(\frac{1}{\tilde{y}^2} + \frac{5}{\tilde{y}^4} + \frac{15}{\tilde{y}^6}\right) e^{-\tilde{y}^2/2} \right] \\ & \left. - \frac{\alpha}{14} ((\hat{\tilde{h}}^{b(4)} \cdot \tilde{y}) \tilde{y} + \tilde{y} (\hat{\tilde{h}}^{b(4)} \cdot \tilde{y})) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{\alpha}{14} (\hat{\tilde{h}}^{b(4)} : \tilde{y} \tilde{y}) \tilde{y} \tilde{y} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}. \end{aligned} \qquad (L21) $$
The first term of (L19) is rewritten as

$$m_a \int f_a [\mathbf{A}_{ab} \mathbf{c}_a]^S d^3\mathbf{y} = m_a n_a \alpha^3 \int \frac{e^{-\alpha^2 \tilde{\mathbf{y}}^2/2}}{(2\pi)^{3/2}} (1 - \alpha \tilde{\mathbf{y}} \cdot \tilde{\mathbf{u}} + \chi_a) [\mathbf{A}_{ab} \mathbf{c}_a]^S d^3\tilde{\mathbf{y}}, \quad (L22)$$

and by using the following integrals:

$$\begin{align*}
\int \tilde{\mathbf{y}} \tilde{\mathbf{y}} f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d^3\mathbf{y} &= \bar{\mathbf{I}} \frac{4\pi}{3} \int_0^\infty \tilde{\mathbf{y}}^4 f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d\tilde{\mathbf{y}}; \\
\int \tilde{\mathbf{y}} (\tilde{\mathbf{h}}^{\tilde{\mathbf{y}}})^2 \cdot \tilde{\mathbf{y}}) f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d^3\tilde{\mathbf{y}} &= \bar{\mathbf{h}}^{\tilde{\mathbf{y}}} \frac{4\pi}{3} \int_0^\infty \tilde{\mathbf{y}}^4 f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d\tilde{\mathbf{y}}; \\
\tilde{\mathbf{h}}^{\tilde{\mathbf{y}}} : \int \tilde{\mathbf{y}} \tilde{\mathbf{y}} f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d^3\tilde{\mathbf{y}} &= 0; \\
\tilde{\mathbf{h}}^{\tilde{\mathbf{y}}} : \int \tilde{\mathbf{y}} \tilde{\mathbf{y}} \tilde{\mathbf{y}} \tilde{\mathbf{y}} f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d^3\tilde{\mathbf{y}} &= \bar{\mathbf{h}}^{\tilde{\mathbf{y}}} \frac{8\pi}{15} \int_0^\infty \tilde{\mathbf{y}}^6 f(\tilde{\mathbf{y}}) e^{-\alpha^2 \tilde{\mathbf{y}}^2/2} d\tilde{\mathbf{y}}, \tag{L23}
\end{align*}$$

and by further applying the semilinear approximation, it integrates

$$m_a \int f_a [\mathbf{A}_{ab} \mathbf{c}_a]^S d^3\mathbf{v} = \rho_a \nu_{ab} \left[ -2 \frac{T_a}{m_a} \bar{\mathbf{I}} + \frac{6 \tilde{\mathbf{h}}^{\tilde{\mathbf{y}}} (2)}{5} \frac{T_a T_b}{T_a m_b + T_b m_a} - \frac{3}{7} \frac{\hat{\mathbf{h}}^{\tilde{\mathbf{y}}} (4)}{\mathbf{h}^{\tilde{\mathbf{y}}}} \frac{m_a T_a T_b^2}{(T_a m_b + T_b m_a)^2} \right. \\ \left. - \frac{2}{5} \frac{T_a}{m_a} \frac{\tilde{\mathbf{h}}^{\tilde{\mathbf{y}}} (2)}{T_a m_b + T_b m_a} + \frac{3}{35} \frac{T_a}{m_a} \frac{\hat{\mathbf{h}}^{\tilde{\mathbf{y}}} (4)}{\mathbf{h}^{\tilde{\mathbf{y}}}} \frac{T_a m_b (2 T_a m_b + 7 T_b m_a)}{(T_a m_b + T_b m_a)^2} \right]. \quad (L24)$$

Similarly, the second term of (L19) integrates

$$m_a \int f_a \bar{\mathbf{D}}_{ab} d^3\mathbf{v} = \frac{\rho_a \nu_{ab}}{m_a + m_b} \left[ \bar{\mathbf{I}} \frac{2}{m_a} (T_a m_b + T_b m_a) - \frac{2}{5} T_b \tilde{\mathbf{h}}^{\tilde{\mathbf{y}}} (2) + \frac{3}{35} \frac{m_a T_b^2}{T_a m_b + T_b m_a} \hat{\mathbf{h}}^{\tilde{\mathbf{y}}} (4) \right. \\ \left. - \frac{2}{5} \frac{T_a m_b}{m_a} \tilde{\mathbf{h}}^{\tilde{\mathbf{y}}} (2) + \frac{3}{35} \frac{m_b^2 T_a^2}{m_a (T_a m_b + T_b m_a)} \hat{\mathbf{h}}^{\tilde{\mathbf{y}}} (4) \right]. \quad (L25)$$

Adding the last two equations together finally yields

$$\bar{\mathbf{Q}}_{ab}^{(2)} = \frac{\rho_a \nu_{ab}}{m_a + m_b} \left[ +2(T_b - T_a)\bar{\mathbf{I}} - K_{ab(1)} T_a \tilde{\mathbf{h}}^{\tilde{\mathbf{y}}(2)} + K_{ab(2)} T_b \tilde{\mathbf{h}}^{\tilde{\mathbf{y}}(2)} \right. \\ \left. + L_{ab(1)} T_a \hat{\mathbf{h}}^{\tilde{\mathbf{y}}(4)} - L_{ab(2)} T_b \hat{\mathbf{h}}^{\tilde{\mathbf{y}}(4)} \right], \quad (L26)$$

with mass-ratio coefficients

$$\begin{align*}
K_{ab(1)} &= \frac{2(2T_a m_a m_b + 3T_a m_b^2 + 5T_b m_a^2 + 6T_b m_a m_b)}{5(T_a m_b + T_b m_a)m_a}; \\
K_{ab(2)} &= \frac{2(3T_a m_a + 2T_a m_b - T_b m_a)}{5(T_a m_b + T_b m_a)}; \\
L_{ab(1)} &= \frac{3T_a m_b (2T_a m_a m_b + 3T_a m_b^2 + 7T_b m_a^2 + 8T_b m_a m_b)}{35(T_a m_b + T_b m_a)^2 m_a}; \\
L_{ab(2)} &= \frac{3m_a T_b (5T_a m_a + 4T_a m_b - T_b m_a)}{35(T_a m_b + T_b m_a)^2}. \tag{L27}
\end{align*}$$

As a partial double check of the entire formulation, by neglecting the fourth-order Hermite moments $\hat{\mathbf{h}}^{\tilde{\mathbf{y}}(4)}$ in (L26), it can be verified that the model is then equivalent to Burgers–Schunk; see Equation (44) in Schunk (1977), or our previous Equation (J21). For a
particular case of small temperature differences, the mass-ratio coefficients simplify into

$$ 
\begin{aligned}
K_{ab(1)} &= \frac{2(5m_a + 3m_b)}{5m_a}; & K_{ab(2)} &= \frac{4}{5}; \\
L_{ab(1)} &= \frac{3(7m_a + 3m_b)m_b}{35m_a(m_b + m_a)}; & L_{ab(2)} &= \frac{12m_a}{35(m_a + m_b)},
\end{aligned} 
\quad (L28) $$

and for self-collisions, $K_{aa(1)} = 16/5$, $K_{aa(2)} = 4/5$, $L_{aa(1)} = 3/7$, and $L_{aa(2)} = 6/35$.

### L.5. Viscosity Tensor Exchange Rates

The collisional contributions for the viscosity tensor $\bar{\bar{\Pi}}_a^{(2)}$ become

$$ 
\begin{aligned}
\bar{\bar{Q}}_{ab}^{(2)'} &= \frac{\delta \bar{\bar{\Pi}}_{ab}^{(2)}}{\delta t} = \bar{\bar{Q}}_{ab}^{(2)} - \frac{\bar{\bar{I}}}{3} \mathrm{Tr} \bar{\bar{Q}}_{ab}^{(2)} \\
&= \frac{\rho_a \nu_{ab}}{m_a + m_b} \left[ -K_{ab(1)} T_a \bar{\bar{h}}^{a(2)} + K_{ab(2)} T_b \bar{\bar{h}}^{b(2)} \right. \\
&\qquad \left. + L_{ab(1)} T_a \bar{\bar{h}}^{a(4)} - L_{ab(2)} T_b \bar{\bar{h}}^{b(4)} \right],
\end{aligned} 
\quad (L29) $$

and introducing summation over all of the “b” species, and rewritten with fluid moments,

$$ 
\begin{aligned}
\bar{\bar{Q}}_a^{(2)'} = & -\frac{21}{10}\nu_{aa}\bar{\bar{\Pi}}_a^{(2)} + \frac{9}{70}\nu_{aa}\frac{\rho_a}{p_a}\bar{\bar{\Pi}}_a^{(4)} \\
& + \sum_{b \neq a} \frac{\rho_a \nu_{ab}}{m_a + m_b} \left[ -(K_{ab(1)} + 7L_{ab(1)}) \frac{1}{n_a}\bar{\bar{\Pi}}_a^{(2)} + (K_{ab(2)} + 7L_{ab(2)}) \frac{1}{n_b}\bar{\bar{\Pi}}_b^{(2)} \right. \\
& \left. + L_{ab(1)} \frac{\rho_a}{n_a p_a}\bar{\bar{\Pi}}_a^{(4)} - L_{ab(2)} \frac{\rho_b}{n_b p_b}\bar{\bar{\Pi}}_b^{(4)} \right].
\end{aligned} 
\quad (L30) $$

It is useful to define (introducing hat)

$$ \hat{K}_{ab(1)} = K_{ab(1)} + 7L_{ab(1)}; \quad \hat{K}_{ab(2)} = K_{ab(2)} + 7L_{ab(2)}, \quad (L31) $$

and the final mass-ratio coefficients are given by (23).

### L.6. Fourth-order Moment Exchange Rates

We need to calculate the collisional contributions

$$ 
\begin{aligned}
\mathrm{Tr} \bar{\bar{Q}}_{ab}^{(4)} &= \frac{\delta \mathrm{Tr} \bar{\bar{r}}_{ab}}{\delta t} = m_a \int \mathbf{c}_a \mathbf{c}_a |\mathbf{c}_a|^2 C_{ab}(f_a) d^3 v \\
&= m_a \int f_a [(\mathbf{A}^{ab} \mathbf{c}_a)^S |\mathbf{c}_a|^2 + 2 (\mathbf{A}^{ab} \cdot \mathbf{c}_a) \mathbf{c}_a \mathbf{c}_a] d^3 v \\
&\quad + m_a \int f_a [(\mathrm{Tr} \bar{\bar{D}}^{ab}) \mathbf{c}_a \mathbf{c}_a + \bar{\bar{D}}^{ab} |\mathbf{c}_a|^2 + 2 ((\bar{\bar{D}}^{ab} \cdot \mathbf{c}_a) \mathbf{c}_a)^S] d^3 v.
\end{aligned} 
\quad (L32) $$
There will be no $\tilde{u}$ contributions at the end, and it is simpler to suppress these from the beginning ($\tilde{u} = 0$), and just use $\mathbf{c}_a = \sqrt{T_a/m_a} \alpha \tilde{y}$. Then one evaluates step by step:

$$ \mathbf{A}^{ab} \cdot \mathbf{c}_a = 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \alpha \left\{ \left(\sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}}\right) \right. \\ \left. + (\tilde{\mathbf{h}}^{b(2)} : \tilde{y}\tilde{y}) \left[ -\frac{9}{2\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \frac{1}{2}\left(1 + \frac{3}{\tilde{y}^2} + \frac{9}{\tilde{y}^4}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right] \right. \\ \left. + \frac{1}{28}(\tilde{\mathbf{h}}^{b(4)} : \tilde{y}\tilde{y})(-2 + \tilde{y}^2) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}; \qquad (L33) $$

$$ 2(\mathbf{A}^{ab} \cdot \mathbf{c}_a)\mathbf{c}_a\mathbf{c}_a = 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \frac{T_a}{m_a} \alpha^3 \left\{ 2\tilde{y}\tilde{y} \left(\sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}}\right) \right. \\ \left. + (\tilde{\mathbf{h}}^{b(2)} : \tilde{y}\tilde{y})\tilde{y}\tilde{y} \left[ -\frac{9}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \left(1 + \frac{3}{\tilde{y}^2} + \frac{9}{\tilde{y}^4}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right] \right. \\ \left. + \frac{1}{14}(\tilde{\mathbf{h}}^{b(4)} : \tilde{y}\tilde{y})\tilde{y}\tilde{y}(-2 + \tilde{y}^2) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}; \qquad (L34) $$

$$ (\mathbf{A}^{ab} \mathbf{c}_a)^S |\mathbf{c}_a|^2 = 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \frac{T_a}{m_a} \alpha^3 \left\{ 2\tilde{y}\tilde{y} \left(\sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}}\right) \right. \\ \left. + ((\tilde{\mathbf{h}}^{b(2)} \cdot \tilde{y})\tilde{y} + \tilde{y}(\tilde{\mathbf{h}}^{b(2)} \cdot \tilde{y})) \left[ \frac{3}{\tilde{y}^3} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) - \left(1 + \frac{3}{\tilde{y}^2}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right] \right. \\ \left. + (\tilde{\mathbf{h}}^{b(2)} : \tilde{y}\tilde{y})\tilde{y}\tilde{y} \left[ -\frac{15}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \left(1 + \frac{5}{\tilde{y}^2} + \frac{15}{\tilde{y}^4}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right] \right. \\ \left. - \frac{1}{14}((\tilde{\mathbf{h}}^{b(4)} \cdot \tilde{y})\tilde{y} + \tilde{y}(\tilde{\mathbf{h}}^{b(4)} \cdot \tilde{y}))\tilde{y}^2 \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{1}{14}(\tilde{\mathbf{h}}^{b(4)} : \tilde{y}\tilde{y})\tilde{y}\tilde{y} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}, \qquad (L35) $$

and adding the last two results together,

$$ (\mathbf{A}^{ab} \mathbf{c}_a)^S |\mathbf{c}_a|^2 + 2(\mathbf{A}^{ab} \cdot \mathbf{c}_a)\mathbf{c}_a\mathbf{c}_a \\ = 2 \frac{c_{ab}}{m_a^2} \left(1 + \frac{m_a}{m_b}\right) \frac{n_b m_b}{T_b} \sqrt{\frac{T_a}{m_a}} \frac{T_a}{m_a} \alpha^3 \left\{ 4\tilde{y}\tilde{y} \left(\sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} - \frac{\operatorname{erf}(\tilde{y}/\sqrt{2})}{\tilde{y}}\right) \right. \\ \left. + ((\tilde{\mathbf{h}}^{b(2)} \cdot \tilde{y})\tilde{y} + \tilde{y}(\tilde{\mathbf{h}}^{b(2)} \cdot \tilde{y})) \left[ \frac{3}{\tilde{y}^3} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) - \sqrt{\frac{2}{\pi}} \left(1 + \frac{3}{\tilde{y}^2}\right) e^{-\tilde{y}^2/2} \right] \right. \\ \left. + (\tilde{\mathbf{h}}^{b(2)} : \tilde{y}\tilde{y})\tilde{y}\tilde{y} \left[ -\frac{24}{\tilde{y}^5} \operatorname{erf}\left(\frac{\tilde{y}}{\sqrt{2}}\right) + \left(2 + \frac{8}{\tilde{y}^2} + \frac{24}{\tilde{y}^4}\right) \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right] \right. \\ \left. - \frac{1}{14}((\tilde{\mathbf{h}}^{b(4)} \cdot \tilde{y})\tilde{y} + \tilde{y}(\tilde{\mathbf{h}}^{b(4)} \cdot \tilde{y}))\tilde{y}^2 \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} + \frac{1}{14}(\tilde{\mathbf{h}}^{b(4)} : \tilde{y}\tilde{y})\tilde{y}\tilde{y} \sqrt{\frac{2}{\pi}} e^{-\tilde{y}^2/2} \right\}. \qquad (L36) $$
Similarly, for the diffusion tensor, calculating step by step,

$$ \bar{\mathbf{D}}^{ab} \cdot \mathbf{c}_a = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \sqrt{\frac{T_a}{m_a}} \alpha \left\{ \begin{aligned} & \tilde{\mathbf{y}} \tilde{\mathbf{A}}_1'' + (\hat{\tilde{\mathbf{h}}}^{b(2)} \cdot \tilde{\mathbf{y}})(2\tilde{\mathbf{y}}\tilde{\mathbf{A}}_2' + 2\tilde{\mathbf{A}}_2) + \tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})(\tilde{\mathbf{A}}_2'' + 2\frac{\tilde{\mathbf{A}}_2'}{\tilde{\mathbf{y}}}) \\ & + (\hat{\tilde{\mathbf{h}}}^{b(4)} \cdot \tilde{\mathbf{y}})(2\tilde{\mathbf{y}}\tilde{\mathbf{A}}_4' + 2\tilde{\mathbf{A}}_4) + \tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})(\tilde{\mathbf{A}}_4'' + 2\frac{\tilde{\mathbf{A}}_4'}{\tilde{\mathbf{y}}}) \end{aligned} \right\}; \quad (L37) $$

$$ \mathrm{Tr} \bar{\mathbf{D}}^{ab} = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \left\{ \begin{aligned} & \tilde{\mathbf{A}}_1'' + 2 \frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} \\ & + (\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})(\tilde{\mathbf{A}}_2'' + 6 \frac{\tilde{\mathbf{A}}_2'}{\tilde{\mathbf{y}}}) + (\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})(\tilde{\mathbf{A}}_4'' + 6 \frac{\tilde{\mathbf{A}}_4'}{\tilde{\mathbf{y}}}) \end{aligned} \right\}; \quad (L38) $$

$$ \mathrm{Tr} \bar{\mathbf{D}}^{ab} \mathbf{c}_a \mathbf{c}_a = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \frac{T_a}{m_a} \alpha^2 \left\{ \begin{aligned} & \tilde{\mathbf{y}}\tilde{\mathbf{y}}(\tilde{\mathbf{A}}_1'' + 2\frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}}) \\ & + (\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\tilde{\mathbf{y}}\tilde{\mathbf{y}}(\tilde{\mathbf{A}}_2'' + 6\frac{\tilde{\mathbf{A}}_2'}{\tilde{\mathbf{y}}}) + (\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\tilde{\mathbf{y}}\tilde{\mathbf{y}}(\tilde{\mathbf{A}}_4'' + 6\frac{\tilde{\mathbf{A}}_4'}{\tilde{\mathbf{y}}}) \end{aligned} \right\}; \quad (L39) $$

$$ \begin{aligned} \bar{\mathbf{D}}^{ab} |\mathbf{c}_a|^2 = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \frac{T_a}{m_a} \alpha^2 & \left\{ \begin{aligned} & \bar{\mathbf{I}} \tilde{\mathbf{y}} \tilde{\mathbf{A}}_1' + \tilde{\mathbf{y}} \tilde{\mathbf{y}} \left( \tilde{\mathbf{A}}_1'' - \frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} \right) \\ & + [2\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(2)} \cdot \tilde{\mathbf{y}}) + 2(\hat{\tilde{\mathbf{h}}}^{b(2)} \cdot \tilde{\mathbf{y}})\tilde{\mathbf{y}} + \bar{\mathbf{I}}(\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})]\tilde{\mathbf{y}}\tilde{\mathbf{A}}_2' \\ & + 2\tilde{\mathbf{y}}^2\tilde{\mathbf{A}}_2\hat{\tilde{\mathbf{h}}}^{b(2)} + \tilde{\mathbf{y}}\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\left(\tilde{\mathbf{A}}_2'' - \frac{\tilde{\mathbf{A}}_2'}{\tilde{\mathbf{y}}}\right) \\ & + [2\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(4)} \cdot \tilde{\mathbf{y}}) + 2(\hat{\tilde{\mathbf{h}}}^{b(4)} \cdot \tilde{\mathbf{y}})\tilde{\mathbf{y}} + \bar{\mathbf{I}}(\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})]\tilde{\mathbf{y}}\tilde{\mathbf{A}}_4' \\ & + 2\tilde{\mathbf{y}}^2\tilde{\mathbf{A}}_4\hat{\tilde{\mathbf{h}}}^{b(4)} + \tilde{\mathbf{y}}\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\left(\tilde{\mathbf{A}}_4'' - \frac{\tilde{\mathbf{A}}_4'}{\tilde{\mathbf{y}}}\right) \end{aligned} \right\}; \quad (L40) $$

$$ \begin{aligned} 2[(\bar{\mathbf{D}}^{ab} \cdot \mathbf{c}_a)\mathbf{c}_a]^S = 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \frac{T_a}{m_a} \alpha^2 & \left\{ \begin{aligned} & 4\tilde{\mathbf{y}}\tilde{\mathbf{y}}\tilde{\mathbf{A}}_1'' \\ & + 2[(\hat{\tilde{\mathbf{h}}}^{b(2)} \cdot \tilde{\mathbf{y}})\tilde{\mathbf{y}}]^S (2\tilde{\mathbf{y}}\tilde{\mathbf{A}}_2' + 2\tilde{\mathbf{A}}_2) + 4\tilde{\mathbf{y}}\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\left(\tilde{\mathbf{A}}_2'' + 2\frac{\tilde{\mathbf{A}}_2'}{\tilde{\mathbf{y}}}\right) \\ & + 2[(\hat{\tilde{\mathbf{h}}}^{b(4)} \cdot \tilde{\mathbf{y}})\tilde{\mathbf{y}}]^S (2\tilde{\mathbf{y}}\tilde{\mathbf{A}}_4' + 2\tilde{\mathbf{A}}_4) + 4\tilde{\mathbf{y}}\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\left(\tilde{\mathbf{A}}_4'' + 2\frac{\tilde{\mathbf{A}}_4'}{\tilde{\mathbf{y}}}\right) \end{aligned} \right\}, \quad (L41) $$

and adding the last three results together,

$$ \begin{aligned} & (\mathrm{Tr} \bar{\mathbf{D}}^{ab}) \mathbf{c}_a \mathbf{c}_a + \bar{\mathbf{D}}^{ab} |\mathbf{c}_a|^2 + 2[(\bar{\mathbf{D}}^{ab} \cdot \mathbf{c}_a)\mathbf{c}_a]^S \\ = & 2 \frac{c_{ab}}{m_a^2} n_b \sqrt{\frac{m_b}{T_b}} \frac{T_a}{m_a} \alpha^2 \left\{ \begin{aligned} & \bar{\mathbf{I}} \tilde{\mathbf{y}} \tilde{\mathbf{A}}_1' + \tilde{\mathbf{y}} \tilde{\mathbf{y}} \left( 6\tilde{\mathbf{A}}_1'' + \frac{\tilde{\mathbf{A}}_1'}{\tilde{\mathbf{y}}} \right) \\ & + 2[(\hat{\tilde{\mathbf{h}}}^{b(2)} \cdot \tilde{\mathbf{y}})\tilde{\mathbf{y}}]^S (3\tilde{\mathbf{y}}\tilde{\mathbf{A}}_2' + 2\tilde{\mathbf{A}}_2) + \bar{\mathbf{I}}(\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\tilde{\mathbf{y}}\tilde{\mathbf{A}}_2' \\ & + 2\tilde{\mathbf{y}}^2\tilde{\mathbf{A}}_2\hat{\tilde{\mathbf{h}}}^{b(2)} + \tilde{\mathbf{y}}\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(2)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\left(6\tilde{\mathbf{A}}_2'' + 13\frac{\tilde{\mathbf{A}}_2'}{\tilde{\mathbf{y}}}\right) \\ & + 2[(\hat{\tilde{\mathbf{h}}}^{b(4)} \cdot \tilde{\mathbf{y}})\tilde{\mathbf{y}}]^S (3\tilde{\mathbf{y}}\tilde{\mathbf{A}}_4' + 2\tilde{\mathbf{A}}_4) + \bar{\mathbf{I}}(\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\tilde{\mathbf{y}}\tilde{\mathbf{A}}_4' \\ & + 2\tilde{\mathbf{y}}^2\tilde{\mathbf{A}}_4\hat{\tilde{\mathbf{h}}}^{b(4)} + \tilde{\mathbf{y}}\tilde{\mathbf{y}}(\hat{\tilde{\mathbf{h}}}^{b(4)}: \tilde{\mathbf{y}}\tilde{\mathbf{y}})\left(6\tilde{\mathbf{A}}_4'' + 13\frac{\tilde{\mathbf{A}}_4'}{\tilde{\mathbf{y}}}\right) \end{aligned} \right\}. \quad (L42) $$