FR06 · L2 · Chapter 4 of 9

Let shear stress relax over a finite time

A viscous stress needs time to respond when the flow changes. Treating it as a dynamical variable gives the expanding patch a memory that ideal flow lacks. A finite response time also belongs to the causal formulation of relativistic dissipative fluid dynamics.

Before you beginChapters 1–3 define the shear sign and derive energy conservation. Here a coupled differential equation means that each quantity’s rate depends on the current values of both quantities.

Sandbox Physics · 2026-10-27

Promote shear to a second dynamical variable

Keep $p=e/3$ , $P_L=e/3-\phi$ and $P_T=e/3+\phi/2$ . The first-order viscous relation would set the shear instantly to $\phi_{\rm NS}=4\eta/(3\tau)$ , with shear viscosity $\eta$ . In a general relativistic spacetime that instantaneous constitutive law does not by itself give a causal initial-value theory.

The selected Müller–Israel–Stewart closure lets the shear approach its target in a finite time. It also contains the dilution term required by this conformal expanding reduction. We freeze the quadratic transport coefficient to zero, matching the simplified system in Heller and Spaliński’s 2015 benchmark.

\tau\dot e=-\frac43e+\phi,\qquad \tau_\pi\dot\phi=\frac{4\eta}{3\tau}-\frac{4\tau_\pi\phi}{3\tau}-\phi

Dots mean proper-time derivatives. The last term relaxes the shear; the middle term accounts for the expansion of the stress.

Define $\eta=C_\eta s_{\rm eq}$ and $\tau_\pi=C_\tau/T$ . Thus $C_\eta=\eta/s_{\rm eq}$ sets viscous strength and $C_\tau$ sets the relaxation time in temperature units. These are constants in this teaching closure. The entropy entering this parameterisation is the equilibrium reference at the energy-matched temperature.

A ratio makes the numerical variables transparent

Let $r=\phi/e$ . Differentiate the quotient, substitute both evolution equations, and use $\eta=4C_\eta e/(3T)$ . The dilution terms proportional to $4r/3$ cancel. This gives the two equations actually integrated in the Lab.

\frac{d\log e}{d\tau}=\frac{-4/3+r}{\tau},\qquad \frac{dr}{d\tau}=\frac{16C_\eta}{9C_\tau\tau}-\frac{rT}{C_\tau}-\frac{r^2}{\tau}

Logarithmic energy keeps positive energy density numerically. The temperature is obtained from the same evolving energy, not advanced as a separate independent field.

For this conformal shear sector the longitudinal characteristic speed squared near equilibrium is $c_{\rm char}^2=1/3+4C_\eta/(3C_\tau)$ . Requiring it not to exceed light speed gives $C_\tau\ge2C_\eta$ . The default $C_\tau=5C_\eta$ satisfies that check. This is a local linear condition. Far-from-equilibrium trajectories also need a check of their physical stress domain.

Pressure positivity supplies a visible warning

Nonnegative directional pressures require $-2/3\le r\le1/3$ . A numerical solution can cross this interval even though the differential equations remain finite. The Lab marks that portion as a closure extrapolation. It must not be read as a positive massless-particle distribution.

This is why increasing viscosity or starting very early can produce a warning before it produces a numerical crash. The equations have a mathematical solution outside the domain where their stress tensor has the intended microscopic interpretation.

Worked example

Inspect the initial slope before running the solver

For this worked state use $\tau_0=e_0=1$ , $T_0=0.6$ , $C_\eta=0.16$ , $C_\tau=0.8$ and $r_0=0.25$ . Then $\tau_{\pi,0}=0.8/0.6=1.333$ and $\dot{\log e}_0=-1.08333$ .

\dot r_0=\frac{16(0.16)}{9(0.8)}-\frac{(0.25)(0.6)}{0.8}-(0.25)^2\simeq0.10556

The initial shear ratio grows. Expansion can drive anisotropy faster than the stress relaxes; a finite relaxation time alone does not guarantee monotonic isotropisation.

Predict, then try the model

Predict how longer relaxation changes the memory of the initial stress.

Pause Bjorken expansion. Open Relaxation time and model and vary only Relaxation coefficient ratio. Keep viscosity, initial expansion scale and initial shear fixed; compare early pressure ratios. Then restore the ratio and change only the initial shear.

Bjorken expansion ↗

Compare your explanation

The stress retains an initial-state dependence for a finite interval. Its later trajectory also depends on the continuing expansion, which keeps driving anisotropy.

Exercises with explanations

1. At $r=1/3$ , what are the longitudinal and transverse pressures?

Read the reasoning

$P_L=0$ and $P_T=e/2$ . Increasing the shear ratio further makes the longitudinal pressure negative, outside the positive-pressure domain used for a massless gas.

2. Why is solving only the energy equation insufficient to specify a viscous evolution?

Read the reasoning

The energy slope contains longitudinal pressure, which contains shear. A choice of how shear responds is additional physics. Different relaxation laws can conserve energy and still predict different stress trajectories.

Sources and reading scope

Baier et al., Relativistic viscous hydrodynamics, conformal invariance, and holography, arXiv:0712.2451v3
Version 3 (15 July 2008), §§3.5 and 6: Bjorken reduction, relaxation and causality. The implemented closure sets the nonlinear lambda-one coefficient to zero.
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.