An extended hydrodynamics model for inertial confinement fusion hohlraums

Abstract

In some inertial confinement fusion hohlraum designs, the inside plasma is not sufficiently collisional to be satisfactorily described by the Euler equations implemented in hydrodynamic simulation codes, particularly in converging regions of the expanding plasma flow. To better treat that situation, this paper presents an extended hydrodynamics model including higher moments of the particle velocity distribution function, together with physically justified closure assumptions and relaxation terms. A preliminary one-dimensional numerical implementation of the model is shown to give satisfactory results in a test case involving a high-velocity collision of two plasma flows. Paths to extend that model to three dimensions as needed for an actual hohlraum geometry are briefly discussed.

Appendices

Appendix A: Fokker–Planck relaxation times for Coulomb collisions

The expressions of the collision times used in the main part of the paper are derived here from the Fokker–Planck equation governing Coulomb collision processes, in the reference form of Rosenbluth et al. [41]. The term governing the evolution of the velocity distribution function $f_a(\mathbf {c})$ for particles of species a due to collisions on particles of species b is:

$$\begin{aligned} \left( \frac{\partial f_a}{\partial t}\right) _{a\rightarrow b} = -\frac{\partial J_{(ab)i}}{\partial c_i} \end{aligned}$$

(A.1)

where the current in velocity space ${\mathbf {J}}_{(ab)}$ is the sum of a convection term and a diffusion term:

$$\begin{aligned} J_{(ab)i} = 4\pi \varGamma _{ab}\left( -\frac{m_a}{m_b}\frac{\partial {\mathcal {S}}_b}{\partial c_i}f_a + \frac{\partial ^2{\mathcal {T}}_b}{\partial c_i\partial c_j}\frac{\partial f_a}{\partial c_j}\right) \end{aligned}$$

(A.2)

with the Rosenbluth potentials ${\mathcal {S}}_b$ and ${\mathcal {T}}_b$ defined by:

$$\begin{aligned} \varDelta _c{\mathcal {S}}_b = f_b \quad ; \quad \varDelta _c{\mathcal {T}}_b = {\mathcal {S}}_b \end{aligned}$$

(A.3)

and with

$$\begin{aligned} \varGamma _{ab} = \frac{4\pi e^4Z_a^2Z_b^2}{m_a^2}\mathrm {Log}\varLambda _{ab} \end{aligned}$$

where $Z_se$ and $m_s$ are the charge and mass of particles of species $s=a,b$ and $\mathrm {Log}\varLambda _{ab}$ is the Coulomb logarithm defined, e.g., in Ref. [71]. We use a definition of the potentials slightly different from the original $h_b$ and $g_b$ from reference [41]; the correspondence is:

$$\begin{aligned} {\mathcal {S}}_b = -\frac{m_b}{m_a+m_b}\frac{h_b}{4\pi } \quad ; \quad {\mathcal {T}}_b = -\frac{g_b}{8\pi } \end{aligned}$$

A general integral expression of the Rosenbluth potentials, which is a solution of the Poisson equations (A.3), is

$$\begin{aligned} S_b(\mathbf {c})&= \frac{-1}{4\pi }\int \frac{f_b(\mathbf {c}^\prime )}{|\mathbf {c}-\mathbf {c}^\prime |}\text{ d}^3c^\prime \end{aligned}$$

(A.4)

$$\begin{aligned} T_b(\mathbf {c})&= \frac{-1}{8\pi }\int f_b(\mathbf {c}^\prime )|\mathbf {c}-\mathbf {c}^\prime |\text{ d}^3c^\prime \end{aligned}$$

(A.5)

More useful expressions can be obtained in specific cases.

The case of a Maxwellian distribution for target particles. If $f_b$ is the Maxwellian:

$$\begin{aligned} f_b(\mathbf {c}) = n_b\left( \frac{m_b}{2\pi k_BT_b}\right) ^{3/2}e^{-u^2} \end{aligned}$$

with

$$\begin{aligned} \mathbf {u} = \left( \frac{m_b}{2k_BT_b}\right) ^{1/2}(\mathbf {c}-\mathbf {v}_b) \end{aligned}$$

where $n_b$, $\mathbf {v}_b$ and $T_b$ are the density, bulk velocity and temperature of species b, then the Rosenbluth potentials can be explicitly computed:

$$\begin{aligned} {\mathcal {S}}_b(\mathbf {c})&= -\frac{n_b}{4\pi }\left( \frac{m_b}{2k_BT_b}\right) ^{1/2}\frac{\text{ erf }(u)}{u} \\ {\mathcal {T}}_b(\mathbf {c})&= -\frac{n_b}{8\pi }\left( \frac{2k_BT_b}{m_b}\right) ^{1/2}\left( \frac{e^{-u^2}}{\sqrt{\pi }} + \left( u+\frac{1}{2u}\right) \text{ erf }(u)\right) \end{aligned}$$

For electrons, to lowest order in powers of the electron/ion mass ratio, we get:

$$\begin{aligned} \frac{\partial {\mathcal {S}}_e}{\partial c_i}&= \frac{n_e}{3\pi ^{3/2}}\left( \frac{m_e}{2k_BT_e}\right) ^{3/2}(c_i-v_{e,i}) \\ \frac{\partial ^2{\mathcal {T}}_e}{\partial c_i\partial c_j}&= -\frac{n_e}{6\pi ^{3/2}}\left( \frac{m_e}{2k_BT_e}\right) ^{1/2}\delta _{ij} \end{aligned}$$

so that the electron collision term for ions of species a finally reads:

$$\begin{aligned} \left( \frac{\partial f_a}{\partial t}\right) _{ae}&= \frac{4\sqrt{2\pi }e^4Z_a^2m_e^{1/2}n_e}{3m_a(k_BT_e)^{3/2}}\mathrm {Log}\varLambda _{ae} \\&\quad \times \frac{\partial }{\partial c_i}\left( (c_i-v_{e,i})f_a + \frac{k_BT_e}{m_a}\frac{\partial f_a}{\partial c_i}\right) \end{aligned}$$

For collisions on ions, we get:

$$\begin{aligned} \frac{\partial {\mathcal {S}}_b}{\partial c_i}&= \frac{n_b}{3\pi ^{3/2}}\left( \frac{m_b}{2k_BT_b}\right) ^{3/2}R(u)(c_i-v_{b,i})\\ \frac{\partial ^2{\mathcal {T}}_b}{\partial c_i\partial c_j}&= -\frac{n_b}{6\pi ^{3/2}}\left( \frac{m_b}{2k_BT_b}\right) ^{1/2} \nonumber \\&\quad \times \left[ \left( \delta _{ij}-\frac{u_iu_j}{u^2}\right) L(u) + \frac{u_iu_j}{u^2}R(u)\right] \nonumber \end{aligned}$$

(A.6)

where the following functions have been defined:

$$\begin{aligned} R(u)&= \frac{3}{2u^2}\left( \frac{\sqrt{\pi }\mathrm {erf}(u)}{2u}-e^{-u^2}\right) \\&\mathrel {\mathop {\sim }\limits _{0}} 1-\frac{3u^2}{5} \dots \\&\mathrel {\mathop {\sim }\limits _{\infty }} \frac{3\sqrt{\pi }}{4u^3} \\ L(u)&= \frac{3}{4u^2}\left( e^{-u^2}+\left( 2u-\frac{1}{u}\right) \frac{\sqrt{\pi }}{2}\mathrm {erf}(u)\right) \\&\mathrel {\mathop {\sim }\limits _{0}} 1-\frac{u^2}{5} \dots \\&\mathrel {\mathop {\sim }\limits _{\infty }} \frac{3\sqrt{\pi }}{4u} \end{aligned}$$

To get synthetic formulas for the orders of magnitude of collision times, we will use a more practical approximation of R(u), which reproduces those two limits:

$$\begin{aligned} R(u) \rightarrow \left( 1+\left( \frac{2}{9\pi }\right) ^{1/3}2u^2\right) ^{-3/2} \end{aligned}$$

Definition of a slowing-down time From the expression (A.2) of the current in velocity space, the slowing-down rate of distribution a by target particles b is obtained:

$$\begin{aligned} \frac{{\mathrm{d}}v_{a,i}}{{\mathrm{d}}t}&= \frac{1}{n_a}\int c_i \left( \frac{\partial f_a}{\partial t}\right) _{a\rightarrow b}d^3c \nonumber \\&= -\frac{4\pi \varGamma _{ab}}{n_a}\left( 1+\frac{m_a}{m_b}\right) \int \frac{\partial {\mathcal {S}}_b}{\partial c_i}f_a({\mathbf {c}})d^3c \end{aligned}$$

(A.7)

Among the two terms in the latter expression, the second one comes directly from the “slowing-down” term in the Fokker–Planck equation, and the first one comes from the variation of the diffusion tensor inside the region where $f_a$ takes on non-negligible values. Let us notice that so far, no approximation was made, and that expression is exact; in particular it conserves momentum in $a\leftrightarrow b$ collisions, since it is in the form $1/n_am_a$ $\times$ a factor which is symmetric in the exchange $a\leftrightarrow b$ $\times$ the integral which is antisymmetric in the exchange $a\leftrightarrow b$, as can be seen after three integrations by parts, taking into account that $\varDelta _cS_a = f_a$. We thus obtain

$$\begin{aligned} n_am_a\frac{{\mathrm{d}}v_{a,i}}{{\mathrm{d}}t} + n_bm_b\frac{{\mathrm{d}}v_{b,i}}{{\mathrm{d}}t} = 0 \end{aligned}$$

When the distribution $f_a$ is very localized (very cold), in the integral (A.7) we can factor out $\partial {\mathcal {S}}_b/\partial c_i$. But then, the symmetry which leads to the explicit momentum conservation is broken, because a further assumption was made about $f_a$ with respect to $f_b$. If we further assume that $f_b$ is Maxwellian, inserting Eq. (A.6), we obtain:

$$\begin{aligned} \frac{{\mathrm{d}}v_{a,i}}{{\mathrm{d}}t}&= -4\pi \varGamma _{ab}\left( 1+\frac{m_a}{m_b}\right) \frac{n_b}{3}\left( \frac{m_b}{2\pi k_BT_b}\right) ^{3/2} \nonumber \\&\times R(u)(v_{a,i}-v_{b,i}) \end{aligned}$$

(A.8)

where we recall that:

$$\begin{aligned} u = \left( \frac{m_b}{2k_BT_b}\right) ^{1/2}|\mathbf {v}_a-\mathbf {v}_b| \end{aligned}$$

The exact expression of R(u) was derived above in the case of a Maxwellian $f_b$. If in addition $f_b$ is also localized (with a thermal velocity $k_BT_b/m_b$ much smaller than the relative velocity $\mathbf {v}_a-\mathbf {v}_b$), then the lost symmetry is recovered, since we know (see above) that in that case

$$\begin{aligned} R(u) \rightarrow \frac{3\sqrt{\pi }}{4u^3} \end{aligned}$$

so that the slowing-down rate reads:

$$\begin{aligned} \frac{{\mathrm{d}}v_{a,i}}{{\mathrm{d}}t} = -n_b\varGamma _{ab}\left( 1+\frac{m_a}{m_b}\right) \frac{v_{a,i}-v_{b,i}}{|\mathbf {v}_a-\mathbf {v}_b|^3} \end{aligned}$$

We can check that the above expression is in the form $1/n_am_a$ $\times$ a term which is antisymmetric in the exchange $a\leftrightarrow b$. To recover the symmetry needed for momentum conservation outside of the cold distribution limit we can fix the “faulty” part in expression (A.8):

$$\begin{aligned} \frac{4\pi }{3} \left( \frac{m_b}{2\pi k_BT_b}\right) ^{3/2}R(u) \end{aligned}$$

by replacing throughout the quadratic thermal velocity of target particles with an expression which is symmetric in the exchange $a\leftrightarrow b$, for example the mean:

$$\begin{aligned} \frac{k_BT_b}{m_b} \rightarrow \left( \frac{k_BT_b}{m_b}\right) ^* = \frac{1}{n_a+n_b}\left( n_a\frac{k_BT_a}{m_a}+n_b\frac{k_BT_b}{m_b}\right) \end{aligned}$$

which is an easily accessible quantity in practice since it is the ratio of total pressure to total density, or the sum:

$$\begin{aligned} \frac{k_BT_b}{m_b} \rightarrow \left( \frac{k_BT_b}{m_b}\right) ^* = \frac{k_BT_a}{m_a}+\frac{k_BT_b}{m_b} \end{aligned}$$

(A.9)

which is more satisfactory from a physical point of view because it is supposed to be the squared average relative velocity in the collision of a particle a on a particle b. Using the synthetic expression for R(u) given above, the slowing-down rate is finally put in a form with the requested symmetry:

$$\begin{aligned}&\frac{{\mathrm{d}}v_{a,i}}{{\mathrm{d}}t} = -\frac{n_bm_b}{n_am_a+n_bm_b}\frac{v_{a,i}-v_{b,i}}{\tau _R} \\&\frac{{\mathrm{d}}}{{\mathrm{d}}t}(v_{a,i}-v_{b,i}) = -\frac{v_{a,i}-v_{b,i}}{\tau _R} \end{aligned}$$

where the slowing-down time $\tau _R$ of ions by target particles reads, in the case of ions (labelled by subscript b):

$$\begin{aligned} \tau _{Rab}&= \frac{m_a^2m_b^2}{4\pi e^4Z_a^2Z_b^2 (m_a+m_b)(n_am_a+n_bm_b)\mathrm {Log}\varLambda _{ab}} \nonumber \\&\quad \times \left( |\mathbf {v}_a-\mathbf {v}_b|^2 + \left( \frac{9\pi }{2}\right) ^{1/3}\left( \frac{k_BT_b}{m_b}\right) ^* \right) ^{3/2} \end{aligned}$$

(A.10)

It may be questionable to use, in the limit of a vanishing relative velocity, an expression of the slowing-down rate which is strictly valid for a single particle a colliding on target particles b. In the limit $\mathbf {v}_a-\mathbf {v}_b\rightarrow 0$ the slowing-down rate can be computed exactly if the two distributions are assumed to remain Maxwellian (although this is questionable when $T_a\ne T_b$).

Thus if $f_a$ is the Maxwellian for particles of mass $m_a$ with parameters $n_a$, $\mathbf {v}_a$ and $T_a$, we can write, to first order in $|\mathbf {v}_a-\mathbf {v}_b|$:

$$\begin{aligned} f_a({\mathbf {c}})&\sim n_a\left( \frac{m_a}{2\pi k_BT_a}\right) ^{3/2}e^{-m_a|\mathbf {c}-\mathbf {v}_b|^2/2k_BT_a} \\&\quad \times \left( 1 + \frac{m_a}{k_BT_a}(v_{a,i}-v_{b,i})(c_i-v_{b,i})\right) \end{aligned}$$

Using expression (A.6) we can then explicitly compute the slowing-down rate:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(v_{a,i}-v_{b,i}) \sim -\frac{v_{a,i}-v_{b,i}}{\tau _{RM}} \end{aligned}$$

with the slowing-down time

$$\begin{aligned} \tau _{RM} = \frac{3m_a^2m_b^2\left( \frac{k_BT_a}{m_a}+\frac{k_BT_b}{m_b}\right) ^{3/2}}{4\sqrt{2\pi } e^4Z_a^2Z_b^2 (m_a+m_b)(n_am_a+n_bm_b) \mathrm {Log}\varLambda _{ab}} \end{aligned}$$

The vanishing-velocity limit of the approximate expression (A.10) is recovered provided the second definition of the average temperature (A.9) (the sum of the quadratic mean velocities) is used.

In the case of electrons, we only write the part corresponding to the time derivative on the ion velocity:

$$\begin{aligned} \frac{{\mathrm{d}}v_{a,i}}{{\mathrm{d}}t} = -\frac{v_{a,i}}{\tau _{Rae}} \end{aligned}$$

since momentum conservation in collisions involves all ion species simultaneously, and the electron bulk velocity relaxes very quickly to a value determined by that set of species. We then get, to lowest order in powers of the electron/ion mass ratio:

$$\begin{aligned} \tau _{Rae} \sim \frac{3m_a(k_BT_e)^{3/2}}{4\sqrt{2\pi }e^4Z_a^2m_e^{1/2}n_e\mathrm {Log}\varLambda _{ae}} \end{aligned}$$

Definition of a diffusion time From the form of the second Fokker–Planck collision term, a velocity diffusion (or thermalization) time $\tau _D$ of ions by target particles can be defined in the following way:

$$\begin{aligned} \frac{\partial f}{\partial t} = \frac{\partial }{\partial c_i}\left( D_{ij}\frac{\partial f}{\partial c_j}\right) \end{aligned}$$

with

$$\begin{aligned} \text{ Tr }({\mathbf {D}}) = \frac{1}{2}\frac{{\mathrm{d}}\langle c^2\rangle }{{\mathrm{d}}t} = \frac{\langle c^2\rangle }{\tau _D} \end{aligned}$$

where $\langle c^2\rangle$ is the mean quadratic width of the distribution f, which reads for a Maxwellian:

$$\begin{aligned} \langle c^2\rangle = 3\frac{k_BT}{m} \end{aligned}$$

In the case of electrons this is:

$$\begin{aligned} \tau _{\mathrm{Dae}} = \frac{T_a}{T_e} \tau _{\mathrm{Rae}} \end{aligned}$$

and for ions (labelled by subscript b):

$$\begin{aligned} \tau _{\mathrm{Dab}} = \frac{T_a}{T_b} \tau _{\mathrm{Rab}} \frac{3R(u)}{R(u)+2L(u)} \end{aligned}$$

where

$$\begin{aligned} u = \left( \frac{m_b}{2k_BT_b}\right) ^{1/2}v \end{aligned}$$

This also reads

$$\begin{aligned} \tau _{\mathrm{Dab}} = \frac{3m_ak_BT_a}{4\pi e^4Z_a^2Z_b^2n_b\mathrm {Log}\varLambda _{ab}}\frac{v}{\mathrm {erf}\left( \left( \frac{m_b}{2k_BT_b}\right) ^{1/2}v\right) } \end{aligned}$$

A practical approximate formula for the ion-ion diffusion time, with the correct limits for $u\rightarrow 0$ and $u\rightarrow \infty$, can be designed in the same way as for the slowing-down time:

$$\begin{aligned} \tau _{Dab} = \frac{3m_ak_BT_a}{4\pi e^4Z_a^2Z_b^2n_b\mathrm {Log}\varLambda _{ab}}\left( v^2+\frac{\pi }{2}\frac{k_BT_b}{m_b}\right) ^{1/2} \end{aligned}$$

(A.11)

Close to thermal equilibrium, for all particle species a Fokker–Planck term is recovered, which takes on the form:

$$\begin{aligned} \frac{\partial f_a}{\partial t} = \frac{1}{\tau _{\mathrm{Rab}}}\frac{\partial }{\partial c_i}\left( c_if_a + \frac{k_BT_b}{m_a}\frac{\partial f_a}{\partial c_i}\right) \end{aligned}$$

with a characteristic time which is the slowing-down time $\tau _{Rab}$ of particles a by the distribution of target particles b.

A global relaxation time The collision times estimated in the preceding sections pertain to the evolution of localized parts of the test-particle distribution in velocity space. Those times will acquire a global meaning for the whole distribution if they can be defined so as to keep, at least approximately, the same value over the region where the test distribution function is not negligible. This clearly applies to the relaxation of an ion distribution on electrons, thanks to the large difference in characteristic velocities, or for ion-ion collisions in the case of a plasma interpenetration with a relative velocity larger than the ion thermal velocity. Moreover, in those two cases, the self-collisions of the ion distribution draw it back to the Maxwellian, which strengthens the global character of the interaction with target particles.

In the case of ion-ion collisions with a relative velocity comparable with the thermal velocity, as already mentioned by Kogan at the end of his paper [69], it is more difficult to define a global relaxation coefficient, even though it can be explicitly calculated in the case of two Maxwellians. The result given by Kogan for temperature relaxation with a vanishing relative velocity (once corrected for a missing factor with respect to the Rosenbluth collision term [41]), is:

$$\begin{aligned} \tau _{ab} = \frac{3(m_bk_BT_a+m_ak_BT_b)^{3/2}}{8\sqrt{2\pi m_am_b}e^4Z_a^2Z_b^2n_b\mathrm {Log}\varLambda _{ab}} \end{aligned}$$

(A.12)

which is the characteristic time to use in the temperature relaxation equation:

$$\begin{aligned} \frac{{\mathrm{d}}T_a}{{\mathrm{d}}t} = \frac{T_b-T_a}{\tau _{ab}} \end{aligned}$$

which leads to the symmetric rate

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(T_a-T_b) = \frac{T_b-T_a}{\tau _{TM}} \end{aligned}$$

where

$$\begin{aligned} \tau _{TM}&= \left( \frac{1}{\tau _{ab}}+\frac{1}{\tau _{ba}}\right) ^{-1} \\&= \frac{3m_am_b\left( \frac{k_BT_a}{m_a}+\frac{k_BT_b}{m_b}\right) ^{3/2}}{8\sqrt{2\pi } e^4Z_a^2Z_b^2 (n_a+n_b) \mathrm {Log}\varLambda _{ab}} \end{aligned}$$

This time is very similar to the limit slowing-down time $\tau _{RM}$, and is actually the same in the case of equal-mass particles $m_a=m_b$. This seems to make expression (A.10) a decent candidate for a global relaxation time taking into account plasma interpenetration and/or pressure anisotropy.

But actually the faster particles in the distribution will relax more slowly, so that the distribution will be distorted away from the Maxwellian. In particular when the temperatures are very different ($T_a\gg T_b$) we know [79] that the test-particle distribution will acquire a colder component in the target particle region, while the rest of the distribution will slow down with an almost vanishing-divergence current in velocity space, which can actually be used to model the slowing-down of the fast $\alpha$ particles from fusion reactions in ICF [28, 80]. Even when the temperatures are the same, as discussed in the text, in actual interpenetration calculations performed with a kinetic code, it is found (see the animation provided as supplementary material: animation1.gif) that at the end of the relaxation the distribution shifts from the two-beam to a bi-Maxwellian shape, and accordingly the limit relaxation rate near isotropy is closer to the analytic value found in the latter case (see Appendix B). Thus obviously, instead of looking for a single analytic formula valid for all cases, we have to design a heuristic relaxation rate accounting for the actual behaviour of the plasma, supposedly found in kinetic calculations.

Of course we should not expect a relaxation rate, however cleverly designed, to account for the diversity of kinetic effects. It will only be used as a reasonable order of magnitude in the situations expected to occur in hohlraum plasmas, and specifically as an important input in their modeling through extended hydrodynamics.

Appendix B: Relaxation of the anisotropy of a bi-Maxwellian

Kogan [69] has given an analytic expression of the rate of self-collision relaxation of a bi-Maxwellian, i.e. a distribution reading:

$$\begin{aligned} f(c_x,c_\bot )&= N\left( \frac{m}{2\pi k_BT_\parallel }\right) ^{1/2}\frac{m}{2\pi k_BT_\bot } \nonumber \\&\quad \times \mathrm {exp}\left( \frac{-m}{2k_B}\left( \frac{c_x^2}{T_\parallel }+\frac{c_\bot ^2}{T_\bot }\right) \right) \end{aligned}$$

(B.13)

The general result, valid for all values of the degree of anisotropy, is the following (a correction factor 2 was included, bringing Kogan’s expression of the collision term in agreement with that of Rosenbluth et al. [41]):

$$\begin{aligned} \frac{{\mathrm{d}}T_\parallel }{{\mathrm{d}}t}&= \frac{8Z^4e^4N\text{ Log }\varLambda }{5}\left( \frac{\pi }{m (k_BT)^3}\right) ^{1/2} \nonumber \\&\quad \times (T-T_\parallel )F(\frac{T_\parallel -T}{T}) \end{aligned}$$

(B.14)

where the function F(x) reads:

$$\begin{aligned} F(x)&= -\frac{5\sqrt{1+x}}{x^2} \left[ 1 + \frac{1}{\sqrt{6}}\left( \frac{\sqrt{x}}{2\sqrt{1+x}}\right. \right. \nonumber \\&\quad \left. \left. - \frac{\sqrt{1+x}}{\sqrt{x}}\right) \text{ Log }\frac{\sqrt{1+x}+\sqrt{\frac{3}{2}x}}{\sqrt{1+x}-\sqrt{\frac{3}{2}x}}\right] \end{aligned}$$

(B.15)

That expression is valid without restrictions for $x>0$ ($T_\parallel >T_\bot$), and in the reverse case its analytic extension in the complex plane of values of $\sqrt{x}$ must be used, noticing that for all complex values of z

$$\begin{aligned} \text{ Log }\frac{1+iz}{1-iz} = 2i\text{ Arctg }z \end{aligned}$$

so that for $x<0$ :

$$\begin{aligned} F(x)&= -\frac{5\sqrt{1+x}}{x^2} \left[ 1 - \frac{1}{\sqrt{6}}\left( \frac{\sqrt{-x}}{\sqrt{1+x}}\right. \right. \\&\quad \left. \left. +2\frac{\sqrt{1+x}}{\sqrt{-x}}\right) \text{ Arctg }\frac{\sqrt{-\frac{3}{2}x}}{\sqrt{1+x}}\right] \end{aligned}$$

F(x) is plotted as the dashed green curve on Fig. 5. For a small anisotropy $F(x)\mathrel {\mathop {\rightarrow }\limits _{x\rightarrow 0}}1$, leading to the following definition of the ion-ion collision time for a near-Maxwellian distribution:

$$\begin{aligned} \frac{{\mathrm{d}}T_\parallel }{{\mathrm{d}}t} = \frac{T-T_\parallel }{\tau _{\mathrm{Max}}} \end{aligned}$$

with

$$\begin{aligned} \tau _{\mathrm{Max}} = \frac{5m_i^{1/2}(k_BT_i)^{3/2}}{8\sqrt{\pi }Z^4e^4N_i\text{ Log }\varLambda _{ii}} \end{aligned}$$

(B.16)

The particular numerical factor in the above expression of the collision time arises from the expansion of the relaxation rate about isotropy ($x\sim 0$), as can be cross-checked through a direct calculation from the Rosenbluth potentials, given in the next paragraph. It is specific to the relaxation of the anisotropy of a bi-Maxwellian distribution, and numerically different from those found in other collisional relaxation processes near isotropy, even though its order of magnitude and functional dependencies on mass, density and temperature are the same.

Direct calculation from the Rosenbluth potentials The evolution of the second-order moment of the distribution due to collisions reads:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(Nk_BT_\parallel )&= \int mc_x^2\left( \frac{\partial f}{\partial t}\right) _c(\mathbf {c}){\mathrm{d}}^3c \\&= 8\pi m\varGamma \int \left( {\mathcal {S}}\frac{\partial }{\partial c_x}(2c_xf) + \frac{\partial {\mathcal {T}}}{\partial c_x}\frac{\partial f}{\partial c_x}\right) {\mathrm{d}}^3c \end{aligned}$$

where expressions (A.1)–(A.3) were inserted, dropping all species-specific subscripts since a single species is involved. Using the integral expressions of the potentials (A.4) and (A.5), this reads:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(Nk_BT_\parallel )&= -m\varGamma \int \int \frac{f(\mathbf {c}^\prime )}{|\mathbf {c}-\mathbf {c}^\prime |} \left( 4f(\mathbf {c})\right. \\&\quad \left. +(5c_x-c_{x}^\prime )\frac{\partial f}{\partial c_x}(\mathbf {c})\right) d^3cd^3c^\prime \end{aligned}$$

Using Kogan’s change of variables (with unit Jacobian):

$$\begin{aligned} (\mathbf {c},\mathbf {c}^\prime ) \rightarrow \left( \mathbf {u} = \mathbf {c}-\mathbf {c}^\prime ,\mathbf {t} = \frac{\mathbf {c}+\mathbf {c}^\prime }{2}\right) \end{aligned}$$

and taking into account that f is the bi-Maxwellian (B.13) to write its $c_x$-derivative, we obtain:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(Nk_BT_\parallel )&= -4m\varGamma \int \int \left( 1 - \frac{m}{k_BT_\parallel }\left( t_x+\frac{3}{4}u_x\right) \right. \\&\left. \times \!\left( t_x\!+\!\frac{u_x}{2}\right) \right) f\left( {\mathbf {t}}\!-\!\frac{{\mathbf {u}}}{2}\right) f\left( {\mathbf {t}}\!+\!\frac{{\mathbf {u}}}{2}\right) \frac{{\mathrm{d}}^3u}{|\mathbf {u}|}{\mathrm{d}}^3t \end{aligned}$$

or, inserting the expression of the distribution function:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(k_BT_\parallel )&= -4m\varGamma N\left( \frac{m}{2\pi k_BT_\bot }\right) ^2\frac{m}{2\pi k_BT_\parallel } \\&\quad \times \int \int \left( 1 \!-\! \frac{m}{k_BT_\parallel }\left( t_x\!+\!\frac{3}{4}u_x\right) \left( t_x\!+\!\frac{u_x}{2}\right) \right) \\&\quad \times \mathrm {exp}\left( \frac{-m}{k_B}\left( \frac{t_x^2}{T_\parallel }+\frac{t_\bot ^2}{T_\bot }\right) \right) \\&\quad \times \mathrm {exp}\left( \frac{-m}{4k_B}\left( \frac{u_x^2}{T_\parallel }+\frac{u_\bot ^2}{T_\bot }\right) \right) \frac{{\mathrm{d}}^3u}{|\mathbf {u}|}d^3t \end{aligned}$$

Integrations over $\mathbf {t}$ are straightforward, and we are left with the following integral over $\mathbf {u}$:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(k_BT_\parallel )&= -2m\varGamma N\frac{m}{4\pi k_BT_\bot } \left( \frac{m}{4\pi k_BT_\parallel }\right) ^{1/2} \\&\quad \times \int \left( 1 - \frac{3mu_x^2}{4k_BT_\parallel }\right) \mathrm {exp}\\&\quad \left( \frac{-m}{4k_B}\left( \frac{u_x^2}{T_\parallel }+\frac{u_\bot ^2}{T_\bot }\right) \right) \frac{{\mathrm{d}}^3u}{|\mathbf {u}|} \end{aligned}$$

We now define:

$$\begin{aligned} \left( \frac{m}{4k_BT_\parallel }\right) ^{1/2}u_x = r\cos \theta \end{aligned}$$

and

$$\begin{aligned} \left( \frac{m}{4k_BT_\bot }\right) ^{1/2}u_\bot = r\sin \theta \end{aligned}$$

which splits the integral into an angular part and a radial part which can be integrated in a straightforward way, finally leading to:

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(k_BT_\parallel )&= \!-\!\frac{m\varGamma N}{\sqrt{\pi }}\int _0^\pi \frac{(1\!-\!3\cos ^2\theta )\sin \theta d\theta }{\left( \frac{k_BT_\parallel }{m}\cos ^2\theta \!+\!\frac{k_BT_\bot }{m}\sin ^2\theta \right) ^{1/2}} \\&= \frac{8\sqrt{\pi }Z^4e^4N\text{ Log }\varLambda }{\sqrt{mk_BT}}\frac{\sqrt{1+x}}{x} \\&\quad \times \left[ 1 - \frac{1}{\sqrt{6}}\left( \sqrt{\frac{1+x}{x}}-\frac{1}{2}\sqrt{\frac{x}{1+x}}\right) \right. \\&\quad \left. \mathrm {Log}\frac{\sqrt{1+x}+\sqrt{\frac{3x}{2}}}{\sqrt{1+x}-\sqrt{\frac{3x}{2}}}\right] \end{aligned}$$

where we inserted $x=(T_\parallel -T)/T$. It can be checked that the expression (B.14)–(B.15) obtained by Kogan is recovered (including the previously mentioned correction factor). Expanding the above expression about isotropy ($x\sim 0$) we find

$$\begin{aligned} \frac{{\mathrm{d}}}{{\mathrm{d}}t}(k_BT_\parallel )&\mathrel {\mathop {\sim }\limits _{x\sim 0}} -\frac{8\sqrt{\pi }Z^4e^4N\text{ Log }\varLambda }{5\sqrt{mk_BT}}x \\&\mathrel {\mathop {\sim }\limits _{x\sim 0}} \frac{8\sqrt{\pi }Z^4e^4N\text{ Log }\varLambda }{5\sqrt{m(k_BT)^3}}(k_BT-k_BT_\parallel ) \end{aligned}$$

Appendix C: Moments of a two-component distribution with azimuthal symmetry

The velocity distribution function is assumed to be in the form $f(\mathbf {c}) = f^{(1)}(\mathbf {c}) + f^{(2)}(\mathbf {c})$ with

$$\begin{aligned} f^{(n)}(\mathbf {c}) = \frac{\rho _n}{m}f_\parallel ^{(n)}(\mathbf {c})f_\bot ^{(n)}(\mathbf {c}) \end{aligned}$$

and the factors are defined by

$$\begin{aligned} f_\parallel ^{(n)}(\mathbf {c})= & {} \left( \frac{m}{2k_BT_{\parallel n}}\right) ^{1/2} \nonumber \\&\times F_\parallel \left( \left( \frac{m}{2k_BT_{\parallel n}}\right) ^{1/2}\varOmega _i(c_i-v_{ni})\right) \nonumber \\ \end{aligned}$$

(C.17)

and

$$\begin{aligned} f_\bot ^{(n)}(\mathbf {c})= & {} \frac{m}{2k_BT_{\bot n}}F_\bot \left( (c_i-v_{ni})\right. \nonumber \\&\times \left. \left[ \frac{m}{2k_BT_{\bot n}}(\delta _{ij}-\varOmega _i\varOmega _j)\right] (c_j-v_{nj})\right) \nonumber \\ \end{aligned}$$

(C.18)

$T_{\parallel n}$ and $T_{\bot n}$ are the parallel and perpendicular temperatures of beam number n, $\mathbf {\Omega }=(\mathbf {v}_2-\mathbf {v}_1)/|\mathbf {v}_2-\mathbf {v}_1|$, and the functions $F_\parallel$ and $F_\bot$ are normalized as follows:

$$\begin{aligned} \int _{-\infty }^\infty F_\parallel (x){\mathrm{d}}x = 1 \quad ; \quad \int _0^\infty F_\bot (x){\mathrm{d}}x = \frac{1}{\pi } \end{aligned}$$

and for higher moments:

$$\begin{aligned}&\int _{-\infty }^\infty xF_\parallel (x){\mathrm{d}}x = 0 \quad ; \quad \int _{-\infty }^\infty x^2F_\parallel (x){\mathrm{d}}x = \frac{1}{2} \\&\int _0^\infty xF_\bot (x){\mathrm{d}}x = \frac{1}{\pi } \end{aligned}$$

In the specific case of a bi-Maxwellian distribution, we have:

$$\begin{aligned} F_\parallel (x) = \frac{1}{\sqrt{\pi }}\mathrm {e}^{-x^2} \quad \mathrm {and} \quad F_\bot (x) = \frac{1}{\pi }\mathrm {e}^{-x} \end{aligned}$$

The first two moments are obviously

$$\begin{aligned} \rho&= m\int \left( f^{(1)}(\mathbf {c}) + f^{(2)}(\mathbf {c})\right) {\mathrm{d}}^3c = \rho _1+\rho _2 \\ \rho \mathbf {v}&= m\int \mathbf {c}\left( f^{(1)}(\mathbf {c}) + f^{(2)}(\mathbf {c})\right) {\mathrm{d}}^3c = \rho _1\mathbf {v}_1+\rho _2\mathbf {v}_2 \end{aligned}$$

The bulk velocity is the barycentre of the distribution:

$$\begin{aligned} \mathbf {v} = \frac{\rho _1\mathbf {v}_1+\rho _2\mathbf {v}_2}{\rho _1+\rho _2} \end{aligned}$$

The moment of order 2 reads, using notations from [81]:

$$\begin{aligned} mM_{ij}^2 = m\int c_ic_jf(\mathbf {c}){\mathrm{d}}^3c = \rho v_iv_j+P_{ij} \end{aligned}$$

where $P_{ij}$ is the pressure tensor:

$$\begin{aligned} P_{ij}&= m\int (c_i-v_i)(c_j-v_j)\left( f^{(1)}(\mathbf {c})+f^{(2)}(\mathbf {c})\right) {\mathrm{d}}^3c \nonumber \\&= P_{ij}^{(1)}+P_{ij}^{(2)} + \frac{\rho _1\rho _2}{\rho _1+\rho _2}|\mathbf {v}_2-\mathbf {v}_1|^2\varOmega _i\varOmega _j \end{aligned}$$

(C.19)

where

$$\begin{aligned} P_{ij}^{(n)} = m\int (c_i-v_{ni})(c_j-v_{nj})f^{(n)}(\mathbf {c}){\mathrm{d}}^3c \end{aligned}$$

The vector $\mathbf {\Omega }$ being given, an orthonormal basis $(\mathbf {\Omega },\mathbf {U},\mathbf {V})$ can be defined, in which the vector $\mathbf {x}=\mathbf {c}-\mathbf {v}_n$ has components $(x_\parallel ,x_U,x_V)$. The pressure tensor of component n then reads

$$\begin{aligned} P_{ij}^{(n)}= & {} \int (x_\parallel ^2\varOmega _i\varOmega _j+x_U^2U_iU_j+x_V^2V_iV_j + \cdots ) \\&\times \frac{\rho _nm^{3/2}}{(2k_BT_{\parallel n})^{1/2}2k_BT_{\bot n}} F_\parallel \left( \left( \frac{m}{2k_BT_{\parallel n}}\right) ^{1/2}x_\parallel \right) \\&\times F_\bot \left( -\left[ \frac{m(x_U^2+x_V^2)}{2k_BT_{\bot n}}\right] \right) {\mathrm{d}}^3x \end{aligned}$$

where the ellipsis stands for crossed terms such as $x_\parallel x_U\varOmega _iU_j$ whose integral against $F_\bot$ vanishes. The sum of non-vanishing terms is

$$\begin{aligned} P_{ij}^{(n)}&= \rho _n\frac{k_BT_{\parallel n}}{m}\varOmega _i\varOmega _j + \rho _n\frac{k_BT_{\bot n}}{m}(U_iU_j+V_iV_j) \\&= \rho _n\frac{k_BT_{\parallel n}}{m}\varOmega _i\varOmega _j + \rho _n\frac{k_BT_{\bot n}}{m}(\delta _{ij}-\varOmega _i\varOmega _j) \end{aligned}$$

Substituting that expression into (C.19) we finally get the pressure tensor for a two-component distribution with azimuthal symmetry around the axis $\mathbf {\Omega }$:

$$\begin{aligned} P_{ij} = P_\parallel \varOmega _i\varOmega _j + P_\bot (\delta _{ij}-\varOmega _i\varOmega _j) \end{aligned}$$

with

$$\begin{aligned} P_\parallel&= \rho _1\frac{k_BT_{\parallel 1}}{m}+\rho _2\frac{k_BT_{\parallel 2}}{m} +\frac{\rho _1\rho _2}{\rho _1+\rho _2}|\mathbf {v}_2-\mathbf {v}_1|^2 \\ P_\bot&= \rho _1\frac{k_BT_{\bot 1}}{m}+\rho _2\frac{k_BT_{\bot 2}}{m} \end{aligned}$$

The moment of order 3 reads, using notations from [81]:

$$\begin{aligned} mM_{ijk}^3&= m\int c_ic_jc_kf(\mathbf {c}){\mathrm{d}}^3c \\&= \rho v_iv_jv_k+v_iP_{jk}+v_jP_{ik}+v_kP_{ij}+Q_{ijk} \end{aligned}$$

where $Q_{ijk}$ is twice the heat flux tensor:

$$\begin{aligned} Q_{ijk}&= m\int (c_i-v_i)(c_j-v_j)(c_k-v_k) \\&\quad \times \left( f^{(1)}(\mathbf {c})+f^{(2)}(\mathbf {c})\right) {\mathrm{d}}^3c = Q_{ijk}^{(1)}+Q_{ijk}^{(2)} \\&\quad +\! (v_{1i}\!-\!v_i)P_{jk}^{(1)}\!+\!(v_{1j}\!-\!v_j)P_{ik}^{(1)}\!+\!(v_{1k}\!-\!v_k)P_{ij}^{(1)}\\&\quad +(v_{2i}-v_i)P_{jk}^{(2)} +(v_{2j}-v_j)P_{ik}^{(2)}\\&\quad +(v_{2k}-v_k)P_{ij}^{(2)} +\frac{\rho _1\rho _2(\rho _1-\rho _2)}{(\rho _1+\rho _2)^2}\\&\quad |\mathbf {v}_2-\mathbf {v}_1|^3\varOmega _i\varOmega _j\varOmega _k \end{aligned}$$

$Q_{ijk}^{(n)}$ is the moment of order 3 restricted to component n and computed in the reference frame centered on its bulk velocity $\mathbf {v}_{n}$:

$$\begin{aligned} Q_{ijk}^{(n)} = m\int (c_i-v_{ni})(c_j-v_{nj})(c_k-v_{nk})f^{(n)}(\mathbf {c}){\mathrm{d}}^3c \end{aligned}$$

Assuming $\int _{-\infty }^\infty x^3F_\parallel (x){\mathrm{d}}x=0$, $Q_{ijk}^{(n)}$ can be computed in the same way as $P_{ij}^{(n)}$ above. As expected, the result vanishes since the integral contains only terms in which at least one of the factors $x_p$ enters to an odd power. Inserting values already obtained for $v_{ni}-v_i$ and $P_{ij}^{(n)}$ and considering the dependence of the result on degrees of freedom parallel and perpendicular to $\mathbf {\Omega }$, we finally get

$$\begin{aligned} Q_{ijk}= & {} Q_\parallel \varOmega _i\varOmega _j\varOmega _k + Q_\bot [\varOmega _i(\delta _{jk}-\varOmega _j\varOmega _k) \\&+\varOmega _j(\delta _{ik}-\varOmega _i\varOmega _k)+\varOmega _k(\delta _{ij}-\varOmega _i\varOmega _j)] \end{aligned}$$

with

$$\begin{aligned} Q_\parallel&= \frac{\rho _1\rho _2}{\rho _1+\rho _2}|\mathbf {v}_2-\mathbf {v}_1|\left[ 3\left( \frac{k_BT_{\parallel 2}}{m}-\frac{k_BT_{\parallel 1}}{m}\right) \right. \\&\quad \left. +\frac{\rho _1-\rho _2}{\rho _1+\rho _2}|\mathbf {v}_2-\mathbf {v}_1|^2\right] \\ Q_\bot&= \frac{\rho _1\rho _2}{\rho _1+\rho _2}|\mathbf {v}_2-\mathbf {v}_1|\left( \frac{k_BT_{\bot 2}}{m}-\frac{k_BT_{\bot 1}}{m}\right) \end{aligned}$$

The heat flux vector is

$$\begin{aligned} q_i=\frac{1}{2}Q_{ijj} = \frac{1}{2}(Q_\parallel +2Q_\bot )\varOmega _i \end{aligned}$$

The intrinsic moment of order 4 (computed in the reference frame centered on the global bulk velocity $\mathbf {v}$) reads

$$\begin{aligned} R_{ijkm}&= m\int (c_i-v_i)(\ldots )\left( f^{(1)}(\mathbf {c})+f^{(2)}(\mathbf {c})\right) {\mathrm{d}}^3c \\&= R_{ijkm}^{(1)} +\left[ (v_{1i}-v_i)(v_{1j}-v_j)P_{km}^{(1)} + \cdots \right] \\&\quad +\rho _1(v_{1i}-v_i)(v_{1j}-v_j)(v_{1k}-v_k)(v_{1m}-v_m) \\&\quad +R_{ijkm}^{(2)} +\left[ (v_{2i}-v_i)(v_{2j}-v_j)P_{km}^{(2)} + \cdots \right] \\&\quad +\rho _2(v_{2i}-v_i)(v_{2j}-v_j)(v_{2k}-v_k)(v_{2m}-v_m) \end{aligned}$$

where the ellipsis stands for permutations making the preceding expression symmetric, and $R_{ijkm}^{(n)}$ is the moment of order 4 restricted to component n and computed in the reference frame centered on its bulk velocity $\mathbf {v}_{n}$:

$$\begin{aligned} R_{ijkm}^{(n)} = m\int x_ix_jx_kx_mf^{(n)}(\mathbf {x}){\mathrm{d}}^3x \end{aligned}$$

where the same definition as above $x_i=c_i-v_{ni}=x_\parallel \varOmega _i+x_UU_i+x_VV_i$ is used. Hence,

$$\begin{aligned} R_{ijkm}^{(n)}&= m\int x_\parallel ^4f^{(n)}(\mathbf {x}){\mathrm{d}}^3x \varOmega _i\varOmega _j\varOmega _k\varOmega _m \\&\quad + m\int x_\parallel ^2x_U^2f^{(n)}(\mathbf {x}){\mathrm{d}}^3x \left[ \varOmega _i\varOmega _j\left( \delta _{km}-\varOmega _k\varOmega _m\right) \right. \\&\quad \left. +\! \cdots \right] \!+\! m\int x_U^2x_V^2f^{(n)}(\mathbf {x})d^3x \left[ U_iU_jV_kV_m \!+\!\!\cdots \right] \\&\quad +\! m\int x_U^4f^{(n)}(\mathbf {x})d^3x \left( U_iU_jU_kU_m \!+\! V_iV_jV_kV_m\right) \end{aligned}$$

In the latter, it was noticed that

$$\begin{aligned} \int x_U^{2p}f^{(n)}(\mathbf {x}){\mathrm{d}}^3x = \int x_V^{2p}f^{(n)}(\mathbf {x}){\mathrm{d}}^3x \end{aligned}$$

since the distribution is assumed isotropic in the transverse velocity plane. After some manipulations we get

$$\begin{aligned} \left[ U_iU_jV_kV_m +\cdots \right]&= \left( \delta _{ij}-\varOmega _i\varOmega _j\right) \left( \delta _{km}-\varOmega _k\varOmega _m\right) \\&\quad + \left( \delta _{ik}-\varOmega _i\varOmega _k\right) \left( \delta _{jm}-\varOmega _j\varOmega _m\right) \\&\quad + \left( \delta _{im}-\varOmega _i\varOmega _m\right) \left( \delta _{jk}-\varOmega _j\varOmega _k\right) \\&\quad -3\left( U_iU_jU_kU_m + V_iV_jV_kV_m\right) \end{aligned}$$

and since, once again due to isotropy in the transverse velocity plane,

$$\begin{aligned} \int x_U^4 f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x = 3\int x_U^2 x_V^2 f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x \end{aligned}$$

the expression for $R_{ijkm}^{(n)}$ can be simplified somewhat:

$$\begin{aligned} R_{ijkm}^{(n)} =&\,\, m\int x_\parallel ^4f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x\ \varOmega _i\varOmega _j\varOmega _k\varOmega _m \\&\quad + m\int x_\parallel ^2x_U^2f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x\\&\quad \left[ \varOmega _i\varOmega _j\left( \delta _{km}-\varOmega _k\varOmega _m\right) + \cdots \right] \\&\quad + m\int x_U^4f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x \ \\&\quad \times \frac{1}{3}\left[ \left( \delta _{ij}-\varOmega _i\varOmega _j\right) \left( \delta _{km}-\varOmega _k\varOmega _m\right) \right. \\&\quad + \left( \delta _{ik}-\varOmega _i\varOmega _k\right) \left( \delta _{jm}-\varOmega _j\varOmega _m\right) \\&\quad \left. + \left( \delta _{im}-\varOmega _i\varOmega _m\right) \left( \delta _{jk}-\varOmega _j \varOmega _k\right) \right] \end{aligned}$$

Inserting the form chosen for the distribution components, we get

$$\begin{aligned} m\int x_\parallel ^4f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x&= R_{\parallel \parallel }^{(n)} \\ m\int x_\parallel ^2x_U^2f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x&= \rho _n\frac{k_BT_{\parallel n}}{m}\frac{k_BT_{\bot n}}{m} \\ m\int x_U^4f^{(n)}({\mathbf {x}}){\mathrm{d}}^3x&= 3\rho _n\left( \frac{k_BT_{\bot n}}{m}\right) ^2 \end{aligned}$$

where, in the specific cases of a bi-Maxwellian or a “waterbag” (flat-top) distribution, respectively:

$$\begin{aligned} R_{\parallel \parallel }^{(n)} = 3\rho _n\left( \frac{k_BT_{\parallel n}}{m}\right) ^2 \quad \mathrm {or}\quad \frac{9}{5}\rho _n\left( \frac{k_BT_{\parallel n}}{m}\right) ^2 \end{aligned}$$

The final expression for the tensor of order 4 is thus

$$\begin{aligned} R_{ijkl}= & {} R_{\parallel \parallel } \varOmega _i\varOmega _j\varOmega _k\varOmega _l + R_{\parallel \bot }[\varOmega _i\varOmega _j(\delta _{kl}-\varOmega _k\varOmega _l) + \cdots ] \\&+R_{\bot \bot } [(\delta _{ij}-\varOmega _i\varOmega _j)(\delta _{kl}-\varOmega _k\varOmega _l) +\cdots ] \end{aligned}$$

where the first symmetrized bracket contains 6 terms, and the second one 3 terms. The components are

$$\begin{aligned} R_{\parallel \parallel }= & {} R_{\parallel \parallel }^{(1)} + \rho _1\left[ 6\frac{k_BT_{\parallel 1}}{m}\frac{\rho _2^2|\mathbf {v}_2-\mathbf {v}_1|^2}{(\rho _1+\rho _2)^2} + \frac{\rho _2^4|\mathbf {v}_2-\mathbf {v}_1|^4}{(\rho _1+\rho _2)^4}\right] \nonumber \\&+R_{\parallel \parallel }^{(2)} + \rho _2\left[ 6\frac{k_BT_{\parallel 2}}{m}\frac{\rho _1^2|\mathbf {v}_2-\mathbf {v}_1|^2}{(\rho _1+\rho _2)^2} + \frac{\rho _1^4|\mathbf {v}_2-\mathbf {v}_1|^4}{(\rho _1+\rho _2)^4}\right] \nonumber \\\end{aligned}$$

(C.20)

$$\begin{aligned} R_{\parallel \bot }= & {} \rho _1\frac{k_BT_{\bot 1}}{m}\left( \frac{k_BT_{\parallel 1}}{m}+\frac{\rho _2^2|\mathbf {v}_2-\mathbf {v}_1|^2}{(\rho _1+\rho _2)^2}\right) \nonumber \\&+ \rho _2\frac{k_BT_{\bot 2}}{m}\left( \frac{k_BT_{\parallel 2}}{m}+\frac{\rho _1^2|\mathbf {v}_2-\mathbf {v}_1|^2}{(\rho _1+\rho _2)^2}\right) \end{aligned}$$

(C.21)

$$\begin{aligned} R_{\bot \bot }= & {} \rho _1\left( \frac{k_BT_{\bot 1}}{m}\right) ^2+\rho _2\left( \frac{k_BT_{\bot 2}}{m}\right) ^2 \end{aligned}$$

(C.22)