FR06 · L2 · Chapter 3 of 9

Compare ideal flow with particles that never collide

Collisions redistribute momentum among directions. With instantaneous isotropisation, longitudinal pressure keeps doing substantial work. With no collisions, longitudinal momenta redshift and the pressure weakens. Starting the two models from the same isotropic state lets us see that difference without changing the initial energy.

Before you beginUse the conservation law from chapter 2 and the conformal pressure relation from chapter 1. An integral here is a weighted sum over the initial particle directions.

Sandbox Physics · 2026-10-27

The ideal conformal curve is an analytic reference

An ideal conformal fluid keeps $P_L=P_T=e/3$ . Substitute that pressure into the conservation law, divide by energy density, then integrate the logarithmic derivative. The temperature follows by taking a fourth root.

e(\tau)=e_0\left(\frac{\tau_0}{\tau}\right)^{4/3},\qquad T(\tau)=T_0\left(\frac{\tau_0}{\tau}\right)^{1/3}

These powers are exact within boost-invariant, transversely homogeneous ideal conformal flow. They provide an independent check of numerical time evolution.

At zero chemical potential the equilibrium entropy density is $s_{\rm eq}=(e+p)/T=4e/(3T)$ . Its ideal evolution satisfies $s_{\rm eq}\tau=\mathrm{constant}$ . Far from equilibrium this expression remains an equilibrium reference; the physical kinetic entropy must be computed separately from the full distribution.

Free streaming retains its initial angular distribution

Now take massless particles with an initially isotropic momentum distribution. Write $q=\tau_0/\tau$ and let $\mu$ be the cosine of the initial momentum angle to the beam. The longitudinal momentum at the later time is $p_z=q p_{z,0}$ , whereas transverse momenta are unchanged. The momentum-space volume element supplies a further factor of $q$ .

\frac{e(\tau)}{e_0}=\frac q2\int_{-1}^{1}\sqrt{1-(1-q^2)\mu^2}\,d\mu

This angular integral is obtained by adding each particle’s redshifted energy. The radial momentum integral cancels in the density ratio for an initially isotropic massless distribution.

\frac{P_L(\tau)}{e_0}=\frac{q^3}{2}\int_{-1}^{1}\frac{\mu^2\,d\mu}{\sqrt{1-(1-q^2)\mu^2}},\qquad P_T=\frac{e-P_L}{2}

Pressure weights each momentum by its directional momentum flux. At the starting time the integrals recover equal pressures.

The limiting inverse-time density law appears only after the longitudinal pressure has become small. Applying it from a finite initial isotropic state would silently replace that state with zero longitudinal pressure. The Lab’s free-streaming curve keeps the declared initial pressure distribution.

What the comparison leaves for dynamics

The ideal solution assumes directional momentum has equilibrated instantly. The collisionless solution never redistributes it. Neither contains a finite relaxation time. A causal relaxation model introduces that extra timescale and lets directional stress carry memory of how the system began.

Matching the same initial energy is always possible. Matching all initial pressures as well is possible for the isotropic preset; an ideal fluid cannot share an arbitrary anisotropic initial stress. When using another preset, compare each model’s stated initial pressure before attributing every difference to its later dynamics.

Worked example

Double the age from an isotropic initial state

Set $\tau/\tau_0=2$ . The ideal solution gives $e/e_0=2^{-4/3}\simeq0.39685$ . In the free-streaming integral $q=1/2$ ; its analytic value is below.

\frac{e}{e_0}=\frac q2\left[q+\frac{\arcsin\sqrt{1-q^2}}{\sqrt{1-q^2}}\right]\simeq0.42730

The collisionless density is larger because it has done less longitudinal work after the pressures begin to separate. The asymptotic expression would give 0.5 and is already inaccurate at this finite time.

Predict, then try the model

Predict whether a collisionless cloud can approach equal pressures simply by expanding for longer.

Choose the isotropic-start comparison. Advance proper time and watch both the energy curves and the pressure ratio.

Bjorken expansion ↗

Compare your explanation

The longitudinal momenta keep redshifting. Collisionless expansion makes the pressure distribution more anisotropic even as the energy density falls.

Exercises with explanations

1. How much of the ideal initial temperature remains after proper time increases eightfold?

Read the reasoning

The temperature power is one third, so $T/T_0=8^{-1/3}=1/2$ . Energy density falls by $8^{-4/3}=1/16$ .

2. Why can a falling energy density by itself not show that particles have thermalised?

Read the reasoning

Both free streaming and ideal hydrodynamics dilute the system. Their difference lies in momentum redistribution and pressure evolution. An energy density is only one moment of the full distribution.

Sources and reading scope

J. D. Bjorken, Highly relativistic nucleus–nucleus collisions: The central rapidity region (1983)
Original boost-invariant spacetime construction; proper time and spacetime rapidity. The course derives the pressure-work equation explicitly.
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.