FR06 · L3 · Chapter 5 of 9

Ask when fluid evolution works before pressures agree

Fluid evolution can predict a stress tensor before that tensor becomes isotropic. The test compares an evolving stress with a specified constitutive prediction at the same state. Pressure equality asks another question, and a full equilibrium particle distribution asks more than either stress test can answer.

Before you beginChapters 2–4 introduce energy-matched temperature, the shear ratio and its relaxation equation. A gradient expansion orders corrections by the size of slow variations.

Sandbox Physics · 2026-10-27

Choose what the prediction must match

The dimensionless expansion variable $w=\tau T$ compares the macroscopic expansion age with the temperature scale. In the selected constant-coefficient closure, expanding the solution at large $w$ gives a first-order shear term and a second-order correction.

r_{\rm grad}^{(2)}(w)=\frac{16C_\eta}{9w}+\frac{32C_\eta C_\tau}{27w^2}

This expression follows from the large-w series in Eq. (10) of the frozen 2015 source, using the change of variables derived in the next chapter. It is a truncation, not an exact curve.

A small residual means this chosen prediction tracks the numerical MIS stress. It is a useful within-model diagnostic. A microscopic test requires a separately computed microscopic stress evolution for comparison with the hydrodynamic prediction.

Measure isotropy separately

The pressure ratio comes directly from the same shear ratio: $P_L/P_T=(1/3-r)/(1/3+r/2)$ . It reaches unity only when the shear vanishes. An expanding viscous solution usually carries a nonzero shear even after its initial transient has faded.

\mathcal I=\left|\frac{P_L}{P_T}-1\right|,\qquad \mathcal R=|r-r_{\rm grad}^{(2)}|

The first diagnostic measures pressure inequality. The second measures agreement with a declared gradient truncation; the residual is absolute in the dimensionless shear ratio.

A tolerance for the residual must be stated before naming a crossing time. Changing the tolerance, the order of the constitutive expression or the transport coefficients can move that time. The isotropy diagnostic uses a different criterion and will generally cross its own threshold elsewhere.

Temperature and entropy retain their definitions

The energy-matched temperature assigns one number to the current energy density. Many nonequilibrium particle distributions have that same energy and different pressure anisotropies. Even identical stress tensors can hide differences in higher moments of the distribution.

Thermal equilibrium requires a particular local distribution, with the appropriate conserved quantities and statistical form. Neither a small stress residual nor equal directional pressures proves that full statement. For this reason the Lab displays its temperature as an energy scale and its entropy as an equilibrium reference.

Worked example

Good agreement can coexist with a large pressure imbalance

At $w=3$ , take $C_\eta=0.16$ and $C_\tau=0.8$ . The second-order formula predicts $r_{\rm grad}^{(2)}\simeq0.11167$ . Suppose a numerical trajectory has precisely that value at this point.

\mathcal R=0,\qquad \frac{P_L}{P_T}=\frac{1/3-0.11167}{1/3+0.11167/2}\simeq0.570

The declared constitutive comparison is exact at this point, yet longitudinal pressure is only about 57 percent of transverse pressure. This worked state illustrates the distinction; it is not a reported measurement.

At smaller expansion variables the gradient approximation can itself predict a negative pressure. That failure is a reason to inspect the truncated approximation, not a reason to reinterpret a negative pressure as successful isotropisation.

Predict, then try the model

Will the initial-state trajectories become close to their common late-time behaviour before their pressure ratios approach one?

Open the attractor Lab. Compare the spread of trajectories, the gradient residual and the pressure ratio at the same scaled time. Repeat with a slower-relaxation preset.

Hydrodynamic attractor ↗

Compare your explanation

The decay of initial-state memory and pressure equality have different rates. The plot identifies the model and comparison used for each diagnostic.

Exercises with explanations

1. Can a collisionless distribution have equal longitudinal and transverse pressures at one instant?

Read the reasoning

Yes. An initially isotropic collisionless distribution does. The later free-streaming evolution drives unequal pressures. A single isotropic stress snapshot cannot determine the collision history or prove thermal equilibrium.

2. Why must a hydrodynamisation time include the chosen prediction and tolerance?

Read the reasoning

It is defined by acceptable predictive error. Without the constitutive order and error criterion, two quoted crossing times can refer to different tests even for the same trajectory.

Sources and reading scope

Heller & Spaliński, Hydrodynamics beyond the Gradient Expansion: Resurgence and Resummation, arXiv:1503.07514v2
Version 2 (28 August 2015), Eqs. (4)–(10), Fig. 1. Equation (9) and the zero-lambda-one choice define the limited numerical benchmark; no Borel resummation is reproduced.
Florkowski, Heller & Spaliński, New theories of relativistic hydrodynamics in the LHC era, arXiv:1707.02282v3
Version 3 (19 October 2017), §§5.1, 5.3, 6 and 7.4; notation and limits of kinetic and hydrodynamic descriptions. This is a review, not an independent experimental dataset.