FR06 · L1 · Chapter 2 of 9

Follow the clock carried by the expanding matter

An expanding slice loses energy density in two ways: its volume grows, and its longitudinal pressure does work on its neighbours. The useful clock follows the matter. With boost symmetry, this turns a spacetime problem into one ordinary differential equation.

Before you beginChapter 1 defines the local rest frame and directional pressures. You need the derivative as a rate of change and the work formula: pressure times volume change.

Sandbox Physics · 2026-10-27

Replace simultaneous laboratory snapshots

A patch farther along the beam moves faster. If every observer related by a longitudinal boost sees the same local evolution, the density should depend on the age measured by each patch, not on laboratory time alone. This idealisation is useful near central rapidity; it discards the edges of the collision and longitudinal fluctuations.

Set light speed to one. Proper time $\tau$ labels the age and spacetime rapidity $\eta_s$ labels which patch we follow. The latter is a position coordinate, distinct from a measured particle’s momentum rapidity.

\tau=\sqrt{t^2-z^2},\quad \eta_s=\frac12\log\frac{t+z}{t-z},\quad t=\tau\cosh\eta_s,\quad z=\tau\sinh\eta_s

The construction applies inside the future light cone. A fixed spacetime rapidity follows a straight outward-moving worldline.

Straight rays follow fixed spacetime rapidity. Each coloured hyperbola joins patches at the same proper age; their laboratory times differ. The geometry is schematic.

The Bjorken velocity is $v_z=z/t$ . At fixed $\eta_s$ , the proper-time derivative follows the matter. Boost invariance assigns each rest-frame scalar the same value at every spacetime rapidity on a constant-proper-time slice.

Derive the dilution law from work

Take a transverse area $A_\perp$ and a small fixed rapidity interval $\Delta\eta_s$ . On a constant-proper-time slice its comoving volume is $V=A_\perp\tau\Delta\eta_s$ . The transverse area remains fixed in this reduction. If the slice’s rest energy is $eV$ , longitudinal expansion gives $d(eV)=-P_L\,dV$ .

\frac{d(e\tau)}{d\tau}=-P_L\quad\Longrightarrow\quad \frac{de}{d\tau}=-\frac{e+P_L}{\tau}

The product rule supplies one term from changing energy density and one from changing volume. The extra pressure term is energy passed to neighbouring slices by work.

This equation follows from energy–momentum conservation under the stated symmetry. It remains meaningful for unequal pressures. To predict the future, however, we still need a rule for $P_L$ . Giving that rule is called closing the evolution equations.

Use dimensionless time without losing units

The Labs use $\tau_0=1$ and $e_0=1$ as reference scales. For a conformal equation of state $e=aT^4$ , energy matching gives $T/T_0=(e/e_0)^{1/4}$ . The constant $a$ fixes how the energy scale maps to temperature.

In natural units $\hbar=c=k_B=1$ , the product $w=\tau T$ is dimensionless. To multiply a time in femtometres by a temperature in GeV, divide by $\hbar c\simeq0.1973\,\mathrm{GeV\,fm}$ . Mapping that curve to a real collision also requires an absolute temperature scale and the equation-of-state constant.

Worked example

Two observers can select the same age

At laboratory coordinates $t=5$ and $z=3$ in consistent light-travel units, the patch has age $\tau=\sqrt{25-9}=4$ , rapidity $\eta_s=\log2\simeq0.693$ and velocity $v_z=0.6$ . The central patch reaches this same proper age at $t=4,z=0$ .

Suppose a patch has $P_L/e=0.1$ . The conservation equation gives $d\log e/d\log\tau=-1.1$ . If that ratio remained constant until twice the current proper time, the density would be multiplied by $2^{-1.1}\simeq0.467$ . In a relaxation model the ratio changes, so this is a local slope estimate, not its full trajectory.

Predict, then try the model

Which model should lose energy density fastest if it maintains the largest longitudinal pressure?

In Bjorken expansion, compare the three energy curves at the same proper time. Use the time control to compare their longitudinal pressures.

Bjorken expansion ↗

Compare your explanation

A larger longitudinal pressure transfers more energy through work, on top of the dilution shared by every expanding slice.

Exercises with explanations

1. If $P_L=0$ for the entire evolution, what quantity remains constant?

Read the reasoning

Conservation gives $d(e\tau)/d\tau=0$ , so $e\tau$ stays constant. Energy per fixed rapidity interval is conserved because there is no longitudinal pressure work.

2. Convert $\tau=0.5\,\mathrm{fm}/c$ and $T=0.35\,\mathrm{GeV}$ to the dimensionless product.

Read the reasoning

Use $w=(0.5)(0.35)/0.1973\simeq0.887$ . Treating the number as 0.175 would mix laboratory units with natural units.

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.