On the Relativistic BGK-Boltzmann Model: Asymptotics and Hydrodynamics

Abstract

The generalization of the BGK relaxation model to the special relativity setting is revisited here. We deal with several issues related to this relativistic kinetic model which seem to have been overlooked in the previous physical literature, including the unique determination of associated physical parameters, classical, ultra-relativistic and hydrodynamical limits, maximum entropy principles and the analysis of the linearized operator.

Appendix

The aim of this Appendix is to compute some of the various quantities involved in this paper in order to make it easier to follow. For the sake of simplicity, it is enough to perform these computations for the dimensionless quantities as in Sects. 4 and 5, assuming the scaling (22) or, equivalently, to assume m=c=ω=η=1.

1.1 A.1 Lorentz Invariance

Let Λ be a Lorentz boost (i.e. a linear isometry with respect to the Minkowsky metric) in $\mathbb{R}_{q}^{4}$. As we are dealing with particles of unit rest mass (the so-called mass shell condition), this transformation Λ can be meaningfully seen as acting on $\mathbb{R}_{{\mathbf{q}}}^{3}$. Given any distribution function f, we can define a new distribution function f _Λ by means of

$$f_\varLambda(t,x,{\mathbf{q}}) = f(t,x,\varLambda{\mathbf{q}}). $$

As

$$ v_\mu z^\mu=( \varLambda v)_\mu(\varLambda z)^\mu\quad\mbox{for any}\ v, z\in\mathbb{R}^4, $$

(53)

we can check directly that n,e,p and σ are Lorentz invariant, i.e. $n_{f} = n_{f_{\varLambda}}$ and so on, for any Lorentz boost Λ. Moreover, making use of the fact that the ratio $\frac {d{\mathbf{q}}}{ q^{0} }$ is invariant under the action of Λ, we see that

This means that

$$u_{f_\varLambda} = \varLambda^{-1}u_f $$

(e.g. macroscopic boosts on the local velocity of the system are uniquely determined by the action of the same boosts—in a contravariant way—on the microscopic local velocities of the gas). The Lorentz invariance of the volume element $\frac{d{\mathbf{q}}}{ q^{0}}$ shows also that β is Lorentz invariant, as the ratio

$$\frac{K_1(\beta)}{K_2(\beta)} = \frac{1}{n_f} \int_{\mathbb{R}^3} f \frac{d{\mathbf{q}}}{q^0} $$

is Lorentz invariant too. We summarize these facts as:

Lemma 8

Given any distribution function f, the scalar quantities $n_{f_{\varLambda}}$, $e_{f_{\varLambda}}$, $p_{f_{\varLambda}}$, $\sigma_{f_{\varLambda}}$ and $\beta_{f_{\varLambda}}$ are Lorentz invariant. The vector u _f transforms according to $u_{f_{\varLambda}} = \varLambda^{-1}u_{f}$.

It is instructive to consider the special case in which the distribution function induces a local velocity that is found to be zero; that is, the physical objects that we are representing are at rest with respect to the reference frame that we use to describe them. This situation corresponds to distributions f having u _f =(1,0,0,0)—Lorentz rest frame. It is useful to display formulae for the macroscopic quantities of the gas in this case, as the computations are simpler than in the general case and the results can be related to a generic distribution by means of Lorentz boosts. These read now:

(54)

(55)

1.2 A.2 Computation of the Moments of the Jüttner Equilibrium

We will need a more precise information about the moments of the relativistic Maxwellian. First we list for convenience some of them that can be easily computed in the Lorentz rest frame. Notice that in this case the Jüttner equilibrium reduces to

$$J(n,\beta,0;{\mathbf{q}})= \frac{n}{M(\beta)} \exp\bigl\{- \beta\sqrt {1 + |{ \mathbf{q}} |^2} \bigr\}. $$

For future reference we point that, using modified Bessel functions for the non-negative integer number j

$$K_j(\beta) = \int_0^\infty\cosh(jr) \exp\bigl\{-\beta \cosh(r)\bigr\}dr, $$

we can simplify some of the related formulae. For instance, we can write the function M(β) given in (5) as

$$ M(\beta)= \frac{4\pi}{ \beta} K_2(\beta). $$

(56)

To simplify the notation we will introduce the function Ψ defined as follows

$$ \varPsi(\beta)= \frac{3}{\beta} + \frac{K_1(\beta)}{K_2(\beta)}. $$

(57)

Lemma 9

Let J=J(n,β,0;q). Then, the following identities are verified:

• 1. $\int_{\mathbb{R}^{3}} J d{\mathbf{q}}= n$,

• 2. $\int_{\mathbb{R}^{3}} q^{i} J d{\mathbf{q}}= \int_{\mathbb{R}^{3}} q^{i} J \frac{d{\mathbf{q}}}{q^{0} } = 0$,

• 3. $\int_{\mathbb{R}^{3}} |{\mathbf{q}}|^{2} J \frac{d{\mathbf{q}}}{q^{0} } = \frac{3 n}{\beta}$,

• 4. $\int_{\mathbb{R}^{3}} J \frac{d{\mathbf{q}}}{q^{0} } = \frac{n}{M(\beta)} \frac{4 \pi}{\beta} K_{1}(\beta)= n \frac {K_{1}(\beta)}{K_{2}(\beta)}$,

• 5. $\int_{\mathbb{R}^{3}} \sqrt{1 + |{\mathbf{q}}|^{2}} J d{\mathbf{q}}= n \varPsi(\beta)$.

Proof

The first relation follows from the very definition of M(β), and the second one just by a symmetry argument. To obtain the third one, we use integration by parts:

The fourth relation is a consequence of the following identity

$$\int_{\mathbb{R}^3} \frac{e^{-\beta\sqrt{1 + |{\mathbf {q}}|^2}}}{\sqrt{1 + |{\mathbf{q}} |^2}} d{\mathbf{q}}= 4 \pi\int _0^\infty r^2 \frac{e^{-\beta\sqrt{1 + r^2}}}{\sqrt{1 + r^2}} dr = \frac{4 \pi}{\beta}K_1(\beta). $$

The sum of the third and fourth relations yields the fifth one by using (59). □

The moments of the Jüttner distribution in general form can be obtained thanks to the following classical decomposition (see [42] for instance):

Lemma 10

The energy-momentum tensor T ^μν can be expressed as:

$$T^{\mu\nu}= \int_{\mathbb{R}^3} q^\mu q^\nu f \frac{d{\mathbf {q}}}{ q^0 }=-p_{f} g^{\mu\nu}+ (e_{f}+p_{f})u^\mu u^\nu. $$

Then all the moments of a given f that appear as components of T ^μν can be computed once we have the values of p _f and e _f . This can be combined with Lemma 8, which ensures that it suffices to compute the local energy and pressure in the Lorentz rest frame. These two are given by formulae (56) and (57).

For the special case of f=J we can go further as the computations in formulae (56) and (57) were already carried in Lemma 9. Then we get the following result.

Lemma 11

The quantities e _J and p _J are given by

$$e_{J}=n\varPsi(\beta), \qquad p_{J}= \frac{n}{\beta}. $$

Using the standard physical units,

$$e_{J}=c^2 n\varPsi(\beta), \qquad p_{J}= c^2 \frac{n}{\beta}. $$

A direct application of the program sketched above yields then

Lemma 12

Given any Jüttner distribution J, the following relations hold true:

• 1. $\int_{\mathbb{R}^{3}} q^{\mu}J \frac{d{\mathbf{q}}}{q^{0}} = n u^{\mu}$,

• 2. $\int_{\mathbb{R}^{3}} |{\mathbf{q}}|^{2} J \frac{d{\mathbf{q}}}{q^{0}} = e_{J} |{\mathbf{u}}|^{2} + p_{J} (3 + |{\mathbf{u}}|^{2}) =n \varPsi(\beta)|{\mathbf{u}}|^{2} + (3 + |{\mathbf {u}}|^{2})\frac{n}{\beta}$,

• 3. $\int_{\mathbb{R}^{3}} \sqrt{1 + |{\mathbf{q}}|^{2}} J d{\mathbf{q}}= p_{J} |{\mathbf{u}}|^{2} + e_{J} (1 + |{\mathbf{u}}|^{2}) = \frac{n}{\beta}|{\mathbf{u}}|^{2} + n \varPsi(\beta ) (1 + |{\mathbf{u}}|^{2})$,

• 4. $\int_{\mathbb{R}^{3}} q^{i} J d{\mathbf{q}}= (e_{J} +p_{J})\sqrt{1 + |{\mathbf{u}}|^{2}} u^{i} = (n \varPsi(\beta) + \frac{n}{\beta} ) \sqrt{1 + |{\mathbf {u}}|^{2}} u^{i}$,

• 5. $\int_{\mathbb{R}^{3}} J \frac{d{\mathbf{q}}}{q^{0}} = e_{J} - 3 p_{J}= n ( \varPsi (\beta) - \frac{3}{\beta} ) = n \frac{K_{1}(\beta)}{K_{2}(\beta)}$.

Proof

The first point follows from the definition of u ^μ in terms of N ^μ. For the remaining ones, we just take recourse on Lemma 10. From there we get that

This is to be combined with Lemma 11. To conclude, we notice that

$$\int_{\mathbb{R}^3} J \frac{d{\mathbf{q}}}{q^0 } = \int_{\mathbb{R}^3} \bigl(\bigl(q^0\bigr)^2 - |{\mathbf{q}}|^2 \bigr) J \frac{d{\mathbf{q}}}{q^0} $$

and the last relation follows. □

1.3 A.3 Entropy Fluxes

We can compute also the entropy densities and fluxes of the Jüttner equilibrium, which are used to obtain information in the hydrodynamical limit, thanks to the H-theorem.

Lemma 13

The following identities are verified:

(58)

(59)

Proof

Using Lemma 10 we get

$$\int_{\mathbb{R}^3} \frac{q^i {\mathbf{q}}\cdot{\mathbf{u}}}{q^0 }J d{\mathbf{q}}= u_j \bigl(p_{J} \delta^{ij} +(e_{J}+p_{J})u^i u^j\bigr)= u^i\bigl(p_{J}+(e_{J}+p_{J})|{ \mathbf{u}}|^2\bigr). $$

To prove (61), note that

and then using Lemma 12, items 1, 3 and 4 we obtain (61).

In the same way

and using Lemmas 11 and 12, items 1 and 4, combined with (60) we arrive to the last identity. This proves the lemma. □

1.4 A.4 Monotonicity of K ₁/K ₂

To begin with, let us recall the following recurrence relation:

$$ K_2(\beta)= \frac{2}{\beta} K_1(\beta) + K_0(\beta). $$

(60)

This can be used to show that

$$\frac{K_2(\beta)}{K_1(\beta)} \le\frac{2}{\beta}+1, $$

as K ₀(β)<K ₁(β). We also note that

(61)

which is strictly positive for β<2.

Next we analyze the case β≥2. For that we deal with the integral representations of the incomplete Bessel functions. Using the substitution x=sinh(s/2) we get

$$K_0(\beta)+K_1(\beta)= \int_0^\infty \bigl(1+\cosh(s)\bigr) e^{-\beta\cosh (s)} ds = 2 \int_0^\infty \frac{2+2x^2}{\sqrt{1+x^2}} e^{-\beta (1+2x^2)} dx. $$

By means of the inequality¹

$$\frac{1}{\sqrt{1+x^2}}\ge1-\frac{x^2}{2}\quad\mbox{for}\ x>0 $$

we obtain the estimate

$$K_0(\beta)+K_1(\beta) \ge e^{-\beta} \frac{\sqrt{2 \pi}}{\sqrt {\beta}} \biggl(1+\frac{1}{8 \beta}-\frac{3}{32 \beta^2} \biggr). $$

Arguing in a similar way but using this time the inequality

$$\frac{1}{\sqrt{1+x^2}} \le1-\frac{x^2}{2} + \frac{3}{8}x^4 \quad \mbox{for}\ x>0 $$

we arrive to

$$K_0(\beta) \le e^{-\beta} \frac{\sqrt{2 \pi}}{2\sqrt{\beta}} \biggl(1- \frac{1}{8 \beta}+\frac{9}{128 \beta^2} \biggr). $$

Therefore

$$\frac{K_0(\beta) + K_1(\beta)}{K_0(\beta)} \ge\frac{256 \beta^2+32 \beta-24}{128 \beta^2 -16 \beta+ 9} $$

and

$$\frac{K_1(\beta)}{K_0(\beta)} \ge\frac{128 \beta^2+48 \beta -33}{128 \beta^2 -16 \beta+ 9}. $$

Notice that both the numerator and the denominator are positive for β≥2. This estimate can be used in combination with (62) to get

$$\frac{K_2(\beta)}{K_1(\beta)} \le\frac{128 \beta^3+240 \beta^2+105 \beta-66}{128 \beta^3+48 \beta^2-33 \beta}. $$

We plug this into (63) so that

Using that β≥2 we conclude with

$$\biggl(\frac{K_1(\beta)}{K_2(\beta)} \biggr)' \ge\frac{3 (6656 \beta^4 + 2419 \beta^3 +726)}{(128 \beta^3+240 \beta^2+105 \beta -66)^2} >0. $$