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14.1 The “standard” relativistic models

It is natural to begin by recalling the equations that describe the evolution of a perfect fluid (see Section 6). For a single particle species, we have the number current μ n which satisfies
∇μn μ = 0, where nμ = nu μ. (270 )
The stress energy tensor
Tμν = p ⊥ μν +ρu μuν, (271 )
where p and ρ represent the pressure and the energy density, respectively, satisfies
∇ μT μν = 0. (272 )
In order to account for dissipation we need to introduce additional fields. First introduce a vector μ ν representing particle diffusion. That is, let
n μ −→ nuμ + ν μ, (273 )
and assume that the diffusion satisfies the constraint uμνμ = 0. This simply means that it is purely spatial according to an observer moving with the particles in the inviscid limit, exactly what one would expect from a diffusive process. Next we introduce the heat flux qμ and the viscous stress tensor, decomposed into a trace-part τ (not to be confused with the proper time) and a trace-free bit μν τ, such that
T μν −→ (p + τ) ⊥ μν + ρuμuν + q(μuν) + τ μν, (274 )
subject to the constraints
μ μ u qμ = τ μ = 0, (275 ) uμ τμν = 0, (276 ) τμν − τνμ = 0. (277 )
That is, both the heat flux and the trace-less part of the viscous stress tensor are purely spatial in the matter frame, and τμν is also symmetric. So far, the description is quite general. The constraints have simply been imposed to ensure that the problem has the anticipated number of degrees of freedom.

The next step is to deduce the permissible form for the additional fields from the second law of thermodynamics. The requirement that the total entropy must not decrease leads to the entropy flux sμ having to be such that

μ ∇ μs ≥ 0. (278 )
Assuming that the entropy flux is a combination of all the available vectors, we have
sμ = suμ + βq μ − λνμ, (279 )
where β and λ are yet to be specified. It is easy to work out the divergence of this. Then using the component of Equation (272View Equation) along uμ, i.e. 
μν uμ∇ νT = 0, (280 )
and the thermodynamic relation
1- p-+-ρ- n ∇ μxs = T ∇ μρ − nT ∇ μn, (281 )
which follows from assuming the equation of state s = s(ρ,n), and we recall that xs = s∕n, one can show that
( ) ( ) ∇ μsμ = qμ ∇ μβ − 1u ν∇ νuμ + β − 1- ∇μq μ ( T ) T p-+-ρ- μ μ τ- μ τμν- − xs + λ − nT ∇ μν − ν ∇ μλ − T ∇μu − T ∇ μu ν. (282 )
We want to ensure that the right-hand side of this equation is positive definite (or indefinite). An easy way to achieve this is to make the following identifications:
β = 1∕T, (283 )
and
p + ρ λ = − xs + -----. (284 ) nT
Here we note that λ = g∕nT, where g is the Gibbs free energy density. We also identify
μ 2 μν ν = − σT ⊥ ∇ νλ, (285 )
where the “diffusion coefficient” σ ≥ 0, and the projection is needed in order for the constraint μ u μν = 0 to be satisfied. Furthermore, we can use
μ τ = − ζ∇ μu , (286 )
where ζ ≥ 0 is the coefficient of bulk viscosity, and
( ) μ μν -1 α q = − κT ⊥ T ∇ νT + u ∇ αuν , (287 )
with κ ≥ 0 being the heat conductivity coefficient. To complete the description, we need to rewrite the final term in Equation (282View Equation). To do this it is useful to note that the gradient of the four-velocity can generally be written as
1- ∇ μuν = σμν + 3 ⊥ μν θ + ϖ μν − aνuμ, (288 )
where the acceleration is defined as
aμ = u β∇ βuμ, (289 )
the expansion is μ θ = ∇ μu, and the shear is given by
1( α α ) 1- σμν = 2 ⊥ ν ∇ αuμ+ ⊥ μ ∇ αu ν − 3 ⊥ μν θ. (290 )
Finally, the “twist” follows from5
( ) ϖ = 1- ⊥ α∇ u − ⊥ α ∇ u . (291 ) μν 2 ν α μ μ α ν
Since we want τμν to be symmetric, trace-free, and purely spatial according to an observer moving along μ u, it is useful to introduce the notation
( ) ⟨Aμν⟩ = 1-⊥ ρ⊥ σ A ρσ + Aσρ − 2-⊥ ρσ⊥ γδ A γδ (292 ) 2 μ ν 3
for any A μν. In the case of the gradient of the four-velocity, it is easy to show that this leads to
⟨∇ u ⟩ = σ (293 ) μ ν μν
and therefore it is natural to use
τ μν = − ησ μν, (294 )
where η ≥ 0 is the shear viscosity coefficient. Given these relations, we can write
μ qμqμ τ νμνμ τ μντμν T ∇ μs = κT---+ ζ-+ σT-2-+ --2η---≥ 0. (295 )
By construction, the second law of thermodynamics is satisfied.

The model we have written down is quite general. In particular, it is worth noticing that we did not yet specify the four-velocity u μ. By doing this we can obtain from the above equations both the formulation due to Eckart [39Jump To The Next Citation Point] and that of Landau and Lifshitz [66Jump To The Next Citation Point]. To arrive at the Eckart description, we associate μ u with the flow of particles. Thus we take μ ν = 0 (or equivalently σ = 0). This prescription has the advantage of being easy to implement. The Landau and Lifshitz model follows if we choose the four-velocity to be a timelike eigenvector of the stress-energy tensor. From Equation (274View Equation) it is easy to see that, by setting qμ = 0, we get

μν ν uμT = − ρu . (296 )
This is equivalent to setting κ = 0. Unfortunately, these models, which have been used in most applications of relativistic dissipation to date, are not at all satisfactory. While they pass the key test set by the second law of thermodynamics, they fail several other requirements of a relativistic description. A detailed analysis of perturbations away from an equilibrium state [54] demonstrates serious pathologies. The dynamics of small perturbations tends to be dominated by rapidly growing instabilities. This suggests that these formulations are likely to be practically useless. From the mathematical point of view they are also not acceptable since, being non-hyperbolic, they do not admit a well-posed initial-value problem.
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