A Conservative, Entropic Multispecies BGK Model

Abstract

We derive a conservative multispecies BGK model that follows the spirit of the original, single species BGK model by making the specific choice to conserve species masses, total momentum, and total kinetic energy and to satisfy Boltzmann’s $\mathcal {H}$-Theorem. The derivation emphasizes the connection to the Boltzmann operator which allows for direct inclusion of information from higher-fidelity collision physics models. We also develop a complete hydrodynamic closure via the Chapman-Enskog expansion, including a general procedure to generate symmetric diffusion coefficients based on this model. We numerically investigate velocity and temperature relaxation in dense plasmas and compare the model with previous multispecies BGK models and discuss the trade-offs that are made in defining and using them. In particular, we demonstrate that the BGK model in the NRL plasma formulary does not conserve momentum or energy in general.

Appendices

Positivity of the Temperature $T_{ij}$

In this section, we show the temperature $T_{ij}$ defined in (36) is non-negative.

Proposition 2

Suppose that $T_i$ and $T_j$ are non-negative and that $\rho _i \nu _{ij} + {\rho _j \nu _{ji}} > 0$. Then the temperature $T_{ij}$ defined in (36) is also non-negative.

Proof

A quick inspection of (36) shows that it sufficient to show that

$$\begin{aligned} \rho _i \nu _{ij} \bigg (v_i^2 - v_{ij}^2\bigg ) + \rho _j \nu _{ji} \bigg (v_i^2 - v_{ij}^2\bigg ) \end{aligned}$$

(112)

is non-negative. We write the formula for $\mathbf {v}_{ij}$ in (34) as

$$\begin{aligned} \mathbf {v}_{ij} = \lambda _{ij} \mathbf {v}_i + \lambda _{ji} \mathbf {v}_j, \end{aligned}$$

(113)

where the coefficients

$$\begin{aligned} \lambda _{ij} = \frac{\rho _i \nu _{ij}}{\rho _i \nu _{ij} + {\rho _j \nu _{ji}}} \quad \text {and} \quad \lambda _{ij} = \frac{\rho _j \nu _{ji}}{\rho _i \nu _{ij} + {\rho _j \nu _{ji}}} \end{aligned}$$

(114)

satisfy $\lambda _{ij} + \lambda _{ji} = 1$.

We divide (112) by $\rho _i \nu _{ij} + {\rho _j \nu _{ji}}$ and, using the fact that

$$\begin{aligned} v_{ij}^2 = \lambda _{ij}^2 v_i^2 + \lambda _{ji}^2 v_j^2 + 2\lambda _{ij}\lambda _{ji} \mathbf {v}_i \cdot \mathbf {v}_j \end{aligned}$$

(115)

compute

$$\begin{aligned} \lambda _{ij} (v_i^2 - v_{ij}^2) + \lambda _{ji} (v_j^2 - v_{ij}^2)&= \lambda _{ij} v_i^2 + \lambda _{ji} v_j^2 - v_{ij}^2, \nonumber \\&= \lambda _{ij}(1 - \lambda _{ij})v_i^2 + \lambda _{ji}(1 - \lambda _{ji}) v_j^2 - 2\lambda _{ij}\lambda _{ji} \mathbf {v}_i \cdot \mathbf {v}_j, \nonumber \\&= \lambda _{ij}\lambda _{ji} v_i^2 + \lambda _{ji}\lambda _{ij} v_j^2 - 2\lambda _{ij}\lambda _{ji} \mathbf {v}_i \cdot \mathbf {v}_j, \nonumber \\&= \lambda _{ij}\lambda _{ji}(\mathbf {v}_i - \mathbf {v}_j)^2, \end{aligned}$$

(116)

which is non-negative. Hence the expression in (112) is non-negative, and this completes the proof. $\square$

Momentum and Temperature Transfer Rates for the Boltzmann Equation

In this section, we reproduce the binary mixture results of Morse [35] in more detail in order to clarify the approximations made in deriving the collision rates in Sect. 4. Morse provides formulas for the momentum and energy relaxation rates for some specific cross sections. Here we provide a more general formula for these quantities that takes a general momentum transfer cross section as an input.

To derive the relaxation rates, we consider a spatially homogeneous gas mixture in which each species has a Maxwellian distribution with number density $n_i$, bulk velocity $\mathbf {v}_i$, and temperature $T_i$ that are slightly out of equilibrium with each other. Taking moments of the Boltzmann equation, we find the formulas for momentum and energy relaxation for species i due to collisions with species j:

$$\begin{aligned} \frac{\partial (\rho _i \mathbf {v}_i)}{\partial t}&= \int m_i \mathbf {c}(\mathcal {M}_i(\mathbf {c}') \mathcal {M}_j(\mathbf {c}_*') - \mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*)) g \sigma _{ij}\, d\varvec{\Omega }d\mathbf {c}_*d\mathbf {c}, \nonumber \\&= \int m_i (\mathbf {c}' - \mathbf {c}) \mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*)g \sigma _{ij} \, d\varvec{\Omega }d\mathbf {c}_*d\mathbf {c}, \end{aligned}$$

(117)

$$\begin{aligned} \frac{\partial (n_i E_i)}{\partial t}&= \int \frac{m_i c ^2}{2} (\mathcal {M}_i(\mathbf {c}') \mathcal {M}_j(\mathbf {c}_*') - \mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*)) g \sigma _{ij}\, d\varvec{\Omega }d\mathbf {c}_*d\mathbf {c}, \nonumber \\&= \int \frac{m_i}{2} ((c')^2 - c^2) \mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*)g \sigma _{ij}\, d\varvec{\Omega }d\mathbf {c}_*d\mathbf {c}. \end{aligned}$$

(118)

Here $n_i E_i = \int \frac{m_i}{2} c^2 f_i d\mathbf {c}$ is the total kinetic energy of species i.

To compute the above integrals, we make a change of variables from $\mathbf {c}$ and $\mathbf {c}_*$ to the relative velocity $\mathbf {g}$ and the center of mass velocity $\mathbf {G}$:

$$\begin{aligned} \mathbf {g}= \mathbf {c}- \mathbf {c}_*\quad \text {and} \quad \mathbf {G}= \frac{m_i \mathbf {c}+ m_j \mathbf {c}_*}{m_i + m_j}. \end{aligned}$$

(119)

In terms of $\mathbf {g}$ and $\mathbf {G}$, (117) and (118) take the form

$$\begin{aligned} \frac{\partial (\rho _i \mathbf {v}_i)}{\partial t}&= -\mu _{ij} \int \mathbf {g}\mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*) g \sigma _{ij}^{(1)} (g)\, d\mathbf {g}d\mathbf {G}, \end{aligned}$$

(120)

$$\begin{aligned} \frac{\partial (n_i E_i)}{\partial t}&= -\mu _{ij} \int \mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*) (\mathbf {G}\cdot \mathbf {g})\, g \sigma _{ij}^{(1)}(g) d\mathbf {g}d\mathbf {G}, \end{aligned}$$

(121)

where $\sigma _{ij}^{(1)}$ is the momentum transfer cross section defined by

$$\begin{aligned} \sigma _{ij}^{(1)} = 2\pi \int (1 - \cos \theta ) \sigma (g,\cos \theta ) \sin \theta d\theta . \end{aligned}$$

(122)

This formula for $\sigma _{ij}$ arises when choosing $\mathbf {g}$ as the polar direction for $\varvec{\Omega }$ in the angular integrals in (117) and (118); that is

$$\begin{aligned} \int (\mathbf {g}- g \mathbf{\Omega } ) \sigma _{ij}(g,\varvec{\Omega }) d\varvec{\Omega }&= 2\pi \mathbf {g}\int (1- \cos \theta ) \sin \theta \sigma _{ij}(g,\cos \theta ) d\theta . \end{aligned}$$

(123)

Next we rewrite the Maxwellians that appear in (120) and (121) in terms of the variables $\mathbf {G},\mathbf {g}$ and the velocity difference $\Delta \mathbf {v}= \mathbf {v}_j - \mathbf {v}_i$:

$$\begin{aligned} \mathcal {M}_i(\mathbf {c}) \mathcal {M}_j(\mathbf {c}_*)&= \frac{n_in_j (m_im_j)^{3/2}}{(2\pi T_i)^{3/2}(2\pi T_j)^{3/2}}&e^{-\frac{m_j T_i + m_i T_j}{2 T_i T_j}\left[ \mathbf {G}+ \frac{T_j(\mu _{ij}\mathbf {g}- m_i \mathbf {v}_i) - T_i(\mu _{ij}\mathbf {g}+ m_j \mathbf {v}_j)}{m_iT_j + m_jT_i}\right] ^2} \nonumber \\&\ \quad \times \; e^{-\frac{m_im_j}{2(m_iT_j + m_jT_i)}(\mathbf {g}+ \Delta \mathbf {v})^2}. \end{aligned}$$

(124)

Performing the $\mathbf {G}$ integration in (120) and (121), we obtain

$$\begin{aligned} \frac{\partial (\rho _i \mathbf {v}_i)}{\partial t}&= -\frac{n_i n_j (m_i m_j)^{3/2} \mu _{ij}}{(2\pi )^{3/2} (m_i T_j + m_j T_i)^{3/2}} \int \mathbf {g}g \sigma _{ij}^{(1)}(g) e^{-\frac{m_im_j}{2(m_iT_j + m_jT_i)}(\mathbf {g}+ \Delta \mathbf {v})^2} d\mathbf {g},\end{aligned}$$

(125)

$$\begin{aligned} \frac{\partial n_i E_i}{\partial t}&= -\frac{n_i n_j (m_i m_j)^{3/2} \mu _{ij}}{(2\pi )^{3/2} (m_i T_j + m_j T_i)^{5/2}} \nonumber \\&\quad \times \int \left[ \mu _{ij}(T_i - T_j)g^2 + \mathbf {g}\cdot (m_i T_j \mathbf {v}_i + m_j T_i \mathbf {v}_j)\right] g \sigma _{ij}^{(1)}(g) e^{-\frac{m_im_j}{2(m_iT_j + m_jT_i)}(\mathbf {g}+ \Delta \mathbf {v})^2} d\mathbf {g}. \end{aligned}$$

(126)

Next we recast the energy transfer Eq. (126) as temperature transfer via

$$\begin{aligned} \frac{\partial n_i E_i}{\partial t}&= \frac{3}{2} n_i \frac{\partial T_i}{\partial t} + \frac{\partial \rho _i \mathbf {v}_i}{\partial t} \cdot \mathbf {v}_i. \end{aligned}$$

(127)

Thus, using (125), we have

$$\begin{aligned} \frac{3}{2} n_i \frac{\partial T_i}{\partial t}&= -\frac{n_i n_j (m_i m_j)^{3/2} \mu _{ij}}{(2\pi )^{3/2} (m_i T_j + m_j T_i)^{5/2}} \nonumber \\&\quad \times \; \int \left[ \mu _{ij}(T_i - T_j)g^2 + m_j T_i \mathbf {g}\cdot \Delta \mathbf {v}\right] g \sigma _{ij}^{(1)}(g) e^{-\frac{m_im_j}{2(m_iT_j + m_jT_i)}(\mathbf {g}+ \Delta \mathbf {v})^2} d\mathbf {g}. \end{aligned}$$

(128)

To perform the integration in (128), we rewrite $\mathbf {g}$ using spherical coordinates with $\Delta \mathbf {v}$ as the polar direction and obtain

$$\begin{aligned} \frac{3}{2} n_i \frac{\partial T_i}{\partial t}&= -\frac{ n_i n_j (m_i m_j)^{3/2} \mu _{ij}}{(2\pi )^{3/2} (m_i T_j + m_j T_i)^{5/2}} \\&\quad \times \; \Bigg [ 4\pi \mu _{ij}(T_i - T_j) \int _0^\infty g^5 \sigma _{ij}^{(1)}(g) e^{-K(g^2 + \Delta v^2)} \frac{\sinh (2K g \Delta v) }{2Kg\Delta v} dg \nonumber \\&- 4\pi m_j T_i (\Delta v)^2 \int _0^\infty g^4 \sigma _{ij}^{(1)}(g) e^{-K(g^2 + \Delta v^2)} \frac{2Kg\Delta v \cosh (2Kg\Delta v) - \sinh (2Kg\Delta v)}{4 K^2 g^2 (\Delta v)^3} dg\Bigg ], \nonumber \end{aligned}$$

(129)

where $K = \frac{m_i m_j}{2(m_iT_j + m_j T_i)}$. Next, we make the approximations that $K (\Delta v)^2$ and $(T_j -T_i)$ are small. Taylor expanding the hyperbolic functions and making the change of variable $K g^2 = w^2$, the result is

$$\begin{aligned} \frac{3}{2} n_i \frac{\partial T_i}{\partial t}&= - \frac{ n_i n_j (m_i m_j)^{3/2} \mu _{ij}}{(2\pi )^{3/2} (m_i T_j + m_j T_i)^{5/2}} \nonumber \\ \times&\Bigg [ \frac{4\pi }{K^3} \mu _{ij}(T_i - T_j) \int _0^\infty w^5 \sigma _{ij}^{(1)}(w) e^{-w^2} e^{-K\Delta v^2} \left( 1 + \frac{2}{3} w^2 K (\Delta v)^2 + O(K^2(\Delta v)^4)\right) dw \nonumber \\&- \frac{4\pi m_j T_i}{K^3} K(\Delta v)^2 \int _0^\infty w^5 \sigma _{ij}^{(1)}(w) e^{-w^2} e^{-K\Delta v^2} \left( \frac{2}{3} + \frac{16}{75}w^2 K (\Delta v)^2 + O(K^2 (\Delta v)^4) \right) dw\Bigg ]. \end{aligned}$$

(130)

Neglecting terms of $O(K^2 (\Delta v)^4)$ and higher in the above equation, we obtain

$$\begin{aligned} \frac{3}{2} n_i \frac{\partial T_i}{\partial t}&= - \frac{ 4\pi n_i n_j (m_i m_j)^{3/2} \mu _{ij}}{(2\pi )^{3/2} K^3 (m_i T_j + m_j T_i)^{5/2}} \nonumber \\&\quad \times \; e^{-K\Delta v^2} \left( \mu _{ij}I_5^{ij} \left( (T_i - T_j) - \frac{1}{3} m_j (\Delta v)^2\right) + \frac{2}{3} \mu _{ij}(I_7^{ij} - I_5^{ij})(T_i - T_j) K (\Delta v)^2 \right) , \end{aligned}$$

(131)

where

$$\begin{aligned} I_n^{ij} = \int _0^\infty w^n \sigma ^{(1)}_{ij}(w) e^{-w^2} dw \quad \text {and} \quad w^2 = K g^2. \end{aligned}$$

(132)

Neglecting terms of order $(T_j - T_i) K (\Delta v)^2$ in (131) we obtain the general form¹¹

$$\begin{aligned} \frac{3}{2} n_i \frac{\partial T_i}{\partial t}&= \frac{ 8 \pi n_i n_j (m_i m_j)^{3/2} \mu _{ij}^2}{3(2\pi )^{3/2} K^3 (m_i T_j + m_j T_i)^{5/2}} I_5^{ij} \left( \frac{3}{2} (T_j - T_i) + \frac{1}{2} m_j (\Delta v)^2\right) , \nonumber \\&= \frac{ 64 \pi n_i n_j (m_i T_j + m_j T_i)^{1/2} \mu _{ij}^2}{3(2\pi )^{3/2} (m_i m_j)^{3/2} } I_5^{ij} \left( \frac{3}{2} (T_j - T_i) + \frac{1}{2} m_j (\Delta v)^2\right) , \nonumber \\&= \alpha _{ij} \left( \frac{3}{2} (T_j - T_i) + \frac{1}{2} m_j (\Delta v)^2\right) . \end{aligned}$$

(133)

Here

$$\begin{aligned} \alpha _{ij} = \frac{64\pi \mu _{ij}^2 n_i n_j (m_i T_j + m_j T_i)^{1/2}}{3 (2\pi m_i m_j)^{3/2}} I_5^{ij} \end{aligned}$$

(134)

is the coefficient for energy transfer and the key quantity used in deriving BGK collision rates.

Using (133), we write the relaxation of temperature differences as

$$\begin{aligned} \frac{\partial (T_j - T_i)}{\partial t}&= -\alpha _{ij} \left[ \frac{n_i + n_j}{n_i n_j} (T_j - T_i) + \frac{1}{3} \frac{\rho _j - \rho _i}{n_i n_j} (\Delta v)^2\right] , \end{aligned}$$

(135)

For momentum transfer, a similar calculation for the integral in (125) gives

$$\begin{aligned} \frac{\partial (\rho _i \mathbf {v}_i)}{\partial t}&= \frac{32\pi n_i n_j \mu _{ij}(m_i T_j + m_j T_i)^{1/2}}{3 (2\pi )^{3/2} (m_i m_j)^{1/2} } I_5^{ij}\ \Delta \mathbf {v}, \end{aligned}$$

(136)

and the relaxation of the velocity difference is

$$\begin{aligned} \frac{\partial (\mathbf {v}_j - \mathbf {v}_i)}{\partial t}&= - \alpha _{ij} \left( \frac{\rho _i + \rho _j}{\rho _i \rho _j}\right) \frac{ (m_i + m_j) }{2} (\mathbf {v}_j - \mathbf {v}_i). \end{aligned}$$

(137)

Two Useful Lemmas

Below are two useful lemmas that are used in the computation of temperature and velocity relations in Sect. 5.1. The proofs are provided for the reader’s convenience.

Lemma 1

Let $A \in \mathbf {R}^{n \times n}$ be a symmetric matrix with strictly positive components. Then the linear system

$$\begin{aligned} -\Big ( \sum _j A_{ij} \Big ) w_i + \sum _j A_{ij} w_j = 0, \quad i=1,\dots ,n. \end{aligned}$$

(138)

has a one parameter family of solutions

$$\begin{aligned} w_i = \bar{w} = \frac{1}{n} \sum _j w_j, \quad i=1,\dots ,n. \end{aligned}$$

(139)

Furthermore, the matrix corresponding to the linear system in (138) is negative semi-definite.

Proof

Because A is symmetric, multiplying (138) by $w_i$ and summing over the index i gives

$$\begin{aligned} 0 = \sum _{i,j} A_{ij} \bigg (w_iw_j - w_i^2\bigg ) = \sum _{i,j} A_{ij} \bigg (w_i w_j - w_j^2\bigg ) = -\frac{1}{2} \sum _{i,j} A_{ij} (w_i - w_j)^2 \end{aligned}$$

(140)

Hence because the components of A are all strictly positive, $w_i$ must be independent of i, and the formula for $\bar{w}$ follows immediately. $\square$

The next Lemma is used to derive the symmetric form of the diffusion coefficient matrix D in (98).

Lemma 2

Let $A \in \mathbf {R}^{n \times n}$ be a symmetric matrix given by

$$\begin{aligned} A_{ij} = \gamma _{ij} > 0, \quad i \ne j, \quad \qquad A_{ii} = - \sum _{j,j\ne i} A_{ij}, \end{aligned}$$

(141)

and let $B \in \mathbf {R}^{n \times n}$ be the symmetric matrix

$$\begin{aligned} B = A - \alpha \mathbf {a}\mathbf {a}^T, \end{aligned}$$

(142)

where $\mathbf {a}$ is a vector such that $a_i > 0\ \forall i$ and $\alpha > 0$. Furthermore, let $\mathbf {b}$ satisfy $\sum _j b_j = 0$. Then B is a symmetric negative definite matrix and the linear system

$$\begin{aligned} B\mathbf {w}= \mathbf {b}\end{aligned}$$

(143)

has the same solution as the constrained system

$$\begin{aligned} A\mathbf {w}&= \mathbf {b}\nonumber \\ \mathbf {a}^T \mathbf {w}&= 0. \end{aligned}$$

(144)

Furthermore, the solution is independent of the magnitude of $\alpha$.

Proof

We will first show that B is negative definite. We have

$$\begin{aligned} \sum _i \sum _j B_{ij} w_i w_j = -\frac{1}{2} \sum _i \sum _j A_{ij}(w_i - w_j)^2 - \alpha \left( \sum _i w_i a_i\right) ^2 < 0, \end{aligned}$$

(145)

where we have used the result of Lemma 1, and the assumption that $a_i > 0$.

We next show that (143) and (144) are equivalent. It is obvious that a solution of (144) is also a solution of (143). Thus we need only to show that (143) implies (144). We suppose that $\mathbf {w}$ solves (143) and write A in terms of its eigenvector decomposition

$$\begin{aligned} A = \sum _{j=1}^{n} \lambda _j \mathbf {q}_j \mathbf {q}_j^T. \end{aligned}$$

(146)

Because A is symmetric, these eigenvectors are orthogonal. In Lemma 1, we showed that the null space A is spanned by the vector $\mathbf e = \frac{1}{\sqrt{n}} (1\ 1\ \dots \ 1)^T$, which we identify with the eigenvector $\mathbf {q}_n$ having eigenvalue $\lambda _n = 0$. Using (146), we can write the system in (143) as

$$\begin{aligned} \left( \sum _{j=1}^{n-1} \lambda _j \mathbf {q}_j \mathbf {q}_j^T - \alpha \mathbf {a}\mathbf {a}^T\right) \mathbf {w}= \mathbf {b}. \end{aligned}$$

(147)

Left multiplying (147) by $\mathbf {q}_n^T$, we have

$$\begin{aligned} - \alpha \mathbf {q}_n^T \mathbf {a}\mathbf {a}^T \mathbf {w}= \mathbf {q}_n^T \mathbf {b}= 0, \end{aligned}$$

(148)

due to the orthogonality of the eigenvectors and the assumption on $\mathbf {b}$ that its components sum to zero. Because the elements of $\mathbf {a}$ are all strictly positive, $\alpha \mathbf {q}_n^T \mathbf {a}>0$. It follows then from (148) that $\mathbf {a}^T \mathbf {w}= 0$. Hence (147) reduces to

$$\begin{aligned} \sum _{j=1}^{n-1} \lambda _j \mathbf {q}_j \mathbf {q}_j^T \mathbf {w}= A\mathbf {w}= \mathbf {b}. \end{aligned}$$

(149)

Therefore $\mathbf {w}$ solves (144), and the two systems are equivalent.

Next, we show that the solution to (143) is independent of $\alpha$. To this end, suppose that $B_1 \mathbf {w}_1 = B_2 \mathbf {w}_2 = \mathbf {b}$, where $B_i = A - \alpha _i \mathbf {a}\mathbf {a}^T$, $i=1,2$, and $\alpha _1,\alpha _2 > 0$. Following the arguments above, we conclude that $\mathbf {a}^T \mathbf {w}_1 = \mathbf {a}^T\mathbf {w}_2 = 0$. Hence,

$$\begin{aligned} 0 = B_1 \mathbf {w}_1 - B_2 \mathbf {w}_2 = A(\mathbf {w}_1 - \mathbf {w}_2) = B_1 (\mathbf {w}_1 - \mathbf {w}_2). \end{aligned}$$

(150)

Since $B_1$ is invertible, it follows that $\mathbf {w}_1 = \mathbf {w}_2$. $\square$