Generic instabilities in the relativistic Chapman–Enskog heat conduction law

Abstract

We address the well-posedness of the Cauchy problem corresponding to the relativistic first-order fluid equations, coupled with the Chapman–Enskog heat-flux constitutive relation. We show that the system of equations that results by considering linear perturbations with respect to a generic time direction is non-hyperbolic, since there are modes that may arbitrarily grow as wave-number increases. Then, using a result provided by Strang (J Differ Equ 2:107–114, 1966), we conclude that the full non-linear first-order theory is also non-hyperbolic, thus admitting an ill-posed initial-value formulation. Unlike Eckart’s theory, these instabilities are not present when the time direction is aligned with the fluid’s direction. However, since in general the fluid velocity is not surface-forming, the instability can only be avoided in the particular case where no rotation is present.

A

In this appendix we give a brief review about divergence-type fluid theories, and then we compute the most general constitutive tensor for first-order DTT, considering the Jüttner distribution function. Finally, we get an expression of the heat-flux constitutive relation predicted from this formalism by taking the appropriate projection for the divergence of the constitutive tensor.

1.1 A.1 Brief detour on DTT

Roughly speaking, a set of dynamic equations is considered a divergence- type theory (DTT) if it can be written as a set of equations on the divergence of the corresponding dynamical variables. Any fluid theory which is governed by a set of conservation laws (particle number density, energy and momentum densities, etc) constitutes, indeed, a divergence-type theory. However, constitutive equations, which play a major role in the assessment of causality and hyperbolicity through Geroch’s criterion need to be also expressed in divergence form. In other words, an additional tensor containing the thermodynamic forces is needed, and the equation for its divergence should lead to the corresponding constitutive equations.

Formally speaking, divergence-type fluid theories are the type of theories which satisfy the following three conditions:

(i):

The dynamical variables are given by the energy-momentum tensor $T^{\mu \nu }$ of the fluid and the particle number current, $N^{\mu }$;

(ii):

The dynamical equations are given by

$$\begin{aligned} \nabla _{\mu }N^{\mu }= & {} 0 \end{aligned}$$

(A.1)

$$\begin{aligned} \nabla _{\mu }T^{\mu \nu }= & {} 0 \end{aligned}$$

(A.2)

$$\begin{aligned} \nabla _{\mu }A^{\mu \nu \sigma }= & {} I^{\nu \sigma } \end{aligned}$$

(A.3)

Here, both the constitutive tensor $A^{\mu \nu \sigma }$ and the source tensor $I^{\nu \sigma }$ are algebraic functions of the dynamical variables $N^{\mu }$ and $T^{\mu \nu }$, and $I^{\nu \sigma }$ is symmetric and traceless.

(iii):

There exists a four-current $S^{\alpha }$, which is also a local algebraic function of $T^{\mu \nu }$ and $N^{\mu }$ that satisfies, as a consequence of the dynamical equations,

$$\begin{aligned} \nabla _{\alpha }S^{\alpha }=\sigma , \end{aligned}$$

with $\sigma \ge 0$ an algebraic function of $T^{\mu \nu }$ and $N^{\mu }$ and $\sigma = 0$ if and only if $I^{\nu \sigma }\equiv 0$.

In order to close the system and to have the same quantity of variables and equations, condition $g_{\mu \nu }A^{\alpha \mu \nu } = 0$ is required. Conditions (ii) and (iii) imply the existence of a generating function $\chi (\xi ,\xi _{\mu },\xi _{\mu \nu })$ which contains all the information of the theory. In particular, particle flux and energy-momentum tensor can be obtained by taking derivatives of $\chi$ with respect to the corresponding variables, as it is shown in Eqs. (2.6) and (2.7). The study of hyperbolicity properties of the theory in terms of this new formulation is much more direct. In fact, by introducing a collective abstract variable $\xi ^A := (\xi , \xi _{\mu },\; \xi _{\mu \nu })$, Eqs. (A.1), (A.2) and (A.3) can be set into the form

$$\begin{aligned} \mathcal {K}^{\mu }{}_{AB}\nabla _{\mu } \xi ^B = J_A, \end{aligned}$$

(A.4)

where

$$\begin{aligned} \mathcal {K}^{\mu }{}_{AB} := \frac{\partial ^3 \chi }{\partial \xi _{\mu } \partial \xi ^A \partial \xi ^B} \end{aligned}$$

(A.5)

is the principal symbol of (A.4) (which by construction is symmetric in the capital indices) and $J_A := (0,0,\;I_{\mu \nu })$. Then, following Geroch’s formalism [20], we say that the system is symmetric-hyperbolic is there exists a time-like vector field $t^a$ such that the symmetric form $h_{AB} = t_a\mathcal {K}^a{}_{AB}$ is a norm; that is, it is positive-definite.

1.2 A.2 Details of calculations of Section 2

We now compute the most general constitutive tensor field that can be constructed from first-order divergence-type fluid theories, using a Jüttner distribution function in local equilibrium, $f^{(0)}$. As was pointed out in Section 2, such a constitutive tensor is made up by means of the third moment of $f^{(0)}$.

From the macroscopic point of view, the constitutive tensor field must be an algebraic function of the dynamical variables $N^{\mu }$ and $T^{\mu \nu }$. On the other hand, since we are considering first-order theories and $A^{\mu \alpha \beta }$ includes, by definition, first derivatives with respect to the dissipative tensor, we conclude that it must be of zeroth-order (for that reason we are considering just the Jüttner distribution function). Up to this order, both $N^{\mu }$ and $T^{\mu \nu }$ are made up in terms of the fluid four-velocity $u^{\mu }$ and the background metric $g_{\mu \nu }$. It is rather straightforward to see that the most general tensor field that satisfies these requirements, and has the symmetries imposed by the theory, is the one given in Eq. (2.8). In fact, recalling that the possible $p^{\mu }$ are those restricted to the mass-shell $p^{\mu }p_{\mu }=-1$ (where the mass of each fluid component is normalized to $m = 1$), we get

$$\begin{aligned} A_o= & {} - \frac{2}{3} A^{\mu \alpha \beta }\left( 2u_{\mu }u_{\alpha } + h_{\mu \alpha }\right) u_{\beta } \nonumber \\= & {} 2 \int {f^{(0)}\left( -u_{\alpha }p^{\alpha }\right) \left[ \left( -u_{\alpha }p^{\alpha }\right) ^2 - \frac{1}{2}\right] \,d\Omega }. \end{aligned}$$

(A.6)

By the change of variables

$$\begin{aligned} \varepsilon = -u_{\alpha }p^{\alpha }, \end{aligned}$$

we get

$$\begin{aligned} f^{(0)} = \frac{n}{4\pi z K_2\left( 1/z\right) }\,e^{-\varepsilon /z}, \end{aligned}$$

and $d\Omega = \sqrt{\varepsilon ^2 - 1}\, d\varepsilon \, dS_2$, where $dS_2$ is the area element in the unit sphere of momentum directions. Thus, integral (A.6) reduces to

$$\begin{aligned} A_o= & {} \frac{2n}{zK_2\left( 1/z\right) }\int _1^{\infty }{e^{-\varepsilon /z}\varepsilon \left( \varepsilon ^2 - \frac{1}{2}\right) \sqrt{\varepsilon ^2 - 1}\;d\varepsilon }\nonumber \\= & {} \frac{2n}{zK_2\left( 1/z\right) }\left( \frac{1}{2}\mathcal {I}_1(z) + \mathcal {I}_2(z)\right) , \nonumber \end{aligned}$$

where, for $\ell \in \mathbb {N}$,

$$\begin{aligned} \mathcal {I}_{\ell }(z) := \int _{1}^{\infty }{e^{-\varepsilon /z}\varepsilon \left( \varepsilon ^2 - 1\right) ^{\ell -1/2}d\varepsilon }. \end{aligned}$$

(A.7)

The integral in Eq. (A.7) can be easily computed by means of some properties of modified Bessel functions, as shown in the following

Proposition 1

Let $K_{\ell }$ be the $\ell$-th modified Bessel function. Then, the following identities hold for any $\ell \in \mathbb {N}$:

• (i)
  $$\begin{aligned} \int _{1}^{\infty }{e^{-\varepsilon /z}\varepsilon \left( \varepsilon ^2 - 1\right) ^{\ell -1/2}d\varepsilon } = \frac{\Gamma \left( \ell +\frac{1}{2}\right) }{\Gamma \left( \frac{1}{2}\right) }\,\left( 2z\right) ^{\ell }K_{\ell +1}\left( \frac{1}{z}\right) . \end{aligned}$$

  (A.8)
• (ii)
  $$\begin{aligned} \frac{dK_{\ell }(x)}{dx} = \frac{\ell K_{\ell }(x)}{x} - K_{\ell +1}(x). \end{aligned}$$

  (A.9)

Proof. Identity (ii) is a direct consequence of the derivative formula

$$\begin{aligned} \frac{d}{dx}\left[ \frac{K_{\ell }(x)}{x^{\ell }}\right] = - \frac{K_{\ell + 1}(x)}{x^{\ell }}. \end{aligned}$$

Identity (i) is also consequence of the above formula and the following important one:

$$\begin{aligned} K_{\ell }(x) = \left( \frac{x}{2}\right) ^{\ell }\frac{\Gamma (1/2)}{\Gamma (\ell +1/2)}\int _{1}^{\infty }{e^{-xy}\left( y^2 - 1\right) ^{\ell -1/2}\,dy}. \end{aligned}$$

                                                                                                                                          $\Box$

By applying the proposition above we get, then, $\mathcal {I}_1 = zK_2(1/z)$, $\mathcal {I}_2 = 3z^2 K_3(1/z)$ and

$$\begin{aligned} A_o = n\left( 1 + 6z\mathcal {G}(z)\right) . \end{aligned}$$

Analogously, for $A_1$ we get

$$\begin{aligned} A_1= & {} - \frac{1}{3}A^{\mu \alpha \beta }u_{\mu }u_{\alpha }u_{\beta } \nonumber \\= & {} \frac{n}{3zK_2(1/z)}\int _1^{\infty }{e^{-\varepsilon /z}\varepsilon \left( \varepsilon ^2 - \frac{1}{4}\right) \sqrt{\varepsilon ^2 - 1}\;d\varepsilon } \nonumber \\= & {} \frac{n}{3zK_2\left( 1/z\right) }\left( \frac{3}{4}\mathcal {I}_1(z) + \mathcal {I}_2(z)\right) \nonumber \\= & {} n\left( z\mathcal {G}(z) + \frac{1}{4}\right) . \end{aligned}$$

(A.10)

Now, in order to obtain the Chapman–Enskog constitutive relation, we proceed to project the constitutive tensor in the space perpendicular to $u^{\mu }$. In order to do so, we find it useful to express the constitutive tensor as a sum of three contributions, namely

$$\begin{aligned} A^{\mu \alpha \beta } = A_1^{\mu \alpha \beta } + A_2^{\mu \alpha \beta } + A_3^{\mu \alpha \beta }, \end{aligned}$$

(A.11)

where

$$\begin{aligned} A_1^{\mu \alpha \beta }= & {} A_o u^{\mu }u^{\alpha }u^{\beta } \nonumber \\ A_2^{\mu \alpha \beta }= & {} A_1 u^{\mu }g^{\alpha \beta } \nonumber \\ A_3^{\mu \alpha \beta }= & {} (A_o - 4A_1)g^{\mu (\alpha }u^{\beta )} \nonumber \end{aligned}$$

Then,

$$\begin{aligned} h_{\alpha }{}^{\gamma }\nabla _{\mu }A_{1}^{\mu \alpha \beta }= & {} h_{\alpha }{}^{\gamma }\left[ u^{\alpha }u^{\beta }\dot{A}_o+A_o\nabla _{\mu }\left( u^{\mu }u^{\alpha }u^{\beta }\right) \right] \nonumber \\= & {} A_o u^{\beta }\dot{u}^{\gamma }, \end{aligned}$$

(A.12)

yielding

$$\begin{aligned} u_{\beta }h_{\alpha }{}^{\gamma }\nabla _{\mu }A_{1}^{\mu \alpha \beta } = -A_o \dot{u}^{\gamma }. \end{aligned}$$

By similar calculations, we get

$$\begin{aligned} h_{\alpha }{}^{\gamma }\nabla _{\mu }A_{2}^{\mu \alpha \beta }=\left( \dot{A}_1 + A_1 \nabla _{\mu }u^{\mu }\right) h^{\gamma \beta }, \end{aligned}$$

which implies $u_{\beta }h_{\alpha }{}^{\gamma }\nabla _{\mu }A_{2}^{\mu \alpha \beta } = 0$. Finally, defining

$$\begin{aligned} A_3 := \frac{A_o}{2} - 2A_1, \end{aligned}$$

we have

$$\begin{aligned} u_{\beta }h_{\alpha }{}^{\gamma }\nabla _{\mu }A_{3}^{\mu \alpha \beta }= & {} 2u_{\beta }h_{\alpha }{}^{\gamma }\left[ g^{\mu (\alpha }u^{\beta )}\nabla _{\mu } A_3 + A_3\nabla ^{(\alpha }u^{\beta )}\right] \nonumber \\= & {} - h^{\gamma \mu }\left( \nabla _{\mu }A_3\right) + A_3\dot{u}^{\gamma } \nonumber \end{aligned}$$

Plugging all together, we get

$$\begin{aligned} q^{\gamma }\propto & {} u_{\beta }h^{\gamma }{}_{\alpha }\nabla _{\mu }A^{\mu \alpha \beta } \nonumber \\\propto & {} h^{\gamma \mu }\left( \nabla _{\mu }A_3 + A_3\frac{\nabla _{\mu }n}{n}\right) + \left( \frac{A_o}{2} + 2A_1\right) \dot{u}^{\gamma } \nonumber \\= & {} zh^{\gamma \mu }\left( \left( z\mathcal {G}(z)\right) ' \frac{\nabla _{\mu }T}{T} + \mathcal {G}(z)\frac{\nabla _{\mu }n}{n}\right) \nonumber \\+ & {} \left( 1+5z\mathcal {G}(z)\right) \dot{u}^{\gamma } \end{aligned}$$

(A.13)

Then, we use the local expression (2.13) for the acceleration at the leading order, and the following formula for the derivative of $\mathcal {G}(z)$:

$$\begin{aligned} \mathcal {G}'(z) = -\frac{1}{z^2}\left[ \mathcal {G}^2(z)-5z\mathcal {G}(z)-1\right] , \end{aligned}$$

(A.14)

which implies that

$$\begin{aligned} (z\mathcal {G}(z))'= \mathcal {G}(z)\left[ 1-\frac{\bar{\mathcal {G}}(z)}{z}\right] , \end{aligned}$$

(A.15)

where

$$\begin{aligned} \bar{\mathcal {G}}(z):=\mathcal {G}(z)-\frac{1+5z\mathcal {G}(z)}{\mathcal {G}(z)}. \end{aligned}$$

Finally, recalling the contribution to the heat flux $\delta q^{\gamma }$ given in Eq. (2.18) by taking into account first-order corrections in the four-acceleration, we get

$$\begin{aligned} q^{\gamma }\propto & {} h^{\gamma \mu } \mathcal {G}(z) \left[ \left( 1 - \frac{\bar{\mathcal {G}}(z)}{z}\right) \frac{\nabla _{\mu }T}{T} + \frac{\nabla _{\mu }n}{n}\right] - \frac{1+5z\mathcal {G}(z)}{\mathcal {G}(z)} h^{\gamma \mu }\left( \frac{\nabla _{\mu }T}{T} + \frac{\nabla _{\mu }n}{n}\right) + \delta q^{\gamma } \nonumber \\= & {} h^{\gamma \mu } \mathcal {G}(z) \left[ \left( 1 - \frac{\bar{\mathcal {G}}(z)}{z}\right) \frac{\nabla _{\mu }T}{T} + \frac{\nabla _{\mu }n}{n}\right] + h^{\gamma \mu }\left[ \bar{\mathcal {G}}(z) - \mathcal {G}(z)\right] \left( \frac{\nabla _{\mu }T}{T} + \frac{\nabla _{\mu }n}{n}\right) + \delta q^{\gamma } \nonumber \\= & {} h^{\gamma \mu }\bar{\mathcal {G}}(z)\left[ \left( 1-\frac{\mathcal {G}(z)}{z}\right) \frac{\nabla _{\mu }T}{T} + \frac{\nabla _{\mu }n}{n}\right] + \delta q^{\gamma } \nonumber \\= & {} -h^{\gamma \mu }\bar{\mathcal {G}}(z)\left( \frac{\kappa }{\lambda }\frac{\nabla _{\mu }T}{T} - \frac{\nabla _{\mu }n}{n}\right) + \delta q^{\gamma }. \end{aligned}$$

(A.16)