Continuity Equation and Adiabatic Expansion

Covariant stress-energy conservation fixes how a homogeneous cosmic fluid changes as its comoving volume expands. For a perfect fluid the result is $\dot\rho+3H(\rho+p)=0$: density changes through volume dilution and pressure work. The equation follows locally from $\nabla_\mu T^{\mu\nu}=0$ and does not by itself assert conservation of a global cosmological energy. Calling the expansion adiabatic requires additional thermodynamic assumptions—an ideal nondissipative fluid, a conserved particle current when one is used, and a local-equilibrium Gibbs relation.

Required background. The Einstein Equation supplies the Einstein Equations and Bianchi consistency; Flat-Background Hilbert Stress-Energy from Metric Variation supplies the Stress–Energy and Conservation interpretation of local energy balance.

Helpful background. Relativistic Matter and Compact Objects supplies perfect-fluid thermodynamics and closure.

Covariant energy projection

Let $u^\mu$ be the comoving four-velocity, normalized by $u^\mu u_\mu=-1$, and define

$$\dot X\equiv u^\mu\nabla_\mu X, \qquad \theta\equiv\nabla_\mu u^\mu .$$

For a perfect fluid,

$$T^{\mu\nu} =(\rho+p)u^\mu u^\nu+p\,g^{\mu\nu}.$$

Project conservation along the fluid velocity:

$$\begin{aligned} 0 &=u_\nu\nabla_\mu T^{\mu\nu}\\ &=-\dot{(\rho+p)} -(\rho+p)\theta+\dot p\\ &=-\dot\rho-(\rho+p)\theta . \end{aligned}$$

The normalization identity $u_\nu\dot u^\nu=\tfrac12u^\mu\nabla_\mu(u_\nu u^\nu)=0$ removes the acceleration term. Thus the covariant energy equation is

$$\dot\rho+(\rho+p)\theta=0.$$

In FLRW, $\theta=3H$ and a comoving scalar has $\dot\rho=d\rho/dt$, so

$$\boxed{\dot\rho+3H(\rho+p)=0}.$$

The projection orthogonal to $u^\mu$ gives the relativistic Euler equation. Exact FLRW homogeneity sets the spatial pressure gradient to zero, and the fundamental congruence is geodesic; the energy projection is therefore the only nontrivial perfect-fluid conservation equation for the background. Ellis and van Elst derive the general projection and its FLRW reduction in Ellis and van Elst 1999, §2.2.2, pp. 10–11, eqs. (35), (37), and (38) (PDF).

Energy in a comoving volume

Take a fixed comoving region of reference volume $V_c$. Its physical volume is

$$V(t)=a^3(t)V_c, \qquad \dot V=3HV.$$

The energy measured by comoving observers is $E=\rho V$. Using the continuity equation,

$$\begin{aligned} \dot E &=V\dot\rho+\rho\dot V\\ &=-3HV(\rho+p)+3H\rho V\\ &=-p\dot V. \end{aligned}$$

Therefore

$$d(\rho V)=-p\,dV.$$

For pressureless matter, energy in the comoving region is constant. For positive pressure, $\rho$ falls faster than $V^{-1}$, while $E$ decreases by the work done during expansion. In a generic time-dependent spacetime there is no timelike Killing field that would turn this local law into a conserved total energy of “matter plus the gravitational field.” The pressure-work statement is the controlled result.

Tong derives the same comoving-volume relation from the first law in Tong 2019, §1.2.2, pp. 27–29, eqs. (1.38)–(1.41) (PDF). That thermodynamic derivation and the covariant projection agree once their fluid assumptions are matched.

When expansion is adiabatic

Stress-energy conservation alone does not prove zero entropy production. Suppose additionally that a particle current

$$N^\mu=n u^\mu$$

is conserved, so $\dot n+n\theta=0$, and let $s$ be entropy per particle. For matter in local thermodynamic equilibrium, the Gibbs relation is

$$T\,ds=d\left(\frac{\rho}{n}\right) +p\,d\left(\frac1n\right).$$

Taking a derivative along $u^\mu$ gives

$$nT\dot s =\dot\rho-(\rho+p)\frac{\dot n}{n}.$$

Particle conservation and the perfect-fluid energy equation make the right-hand side vanish. For $n\gt0$ and $T\gt0$, hence $\dot s=0$, and the entropy $S=nsV$ in a comoving region is constant. This is the precise adiabatic statement under the declared assumptions.

The inference fails if particles are created or destroyed without enlarging the thermodynamic description, if viscosity or heat conduction produces entropy, or if the system is far from local equilibrium. Vacuum energy also needs no fictitious conserved particle current: its constant density follows directly from $p=-\rho$.

Density as a function of scale factor

On a monotonic branch with $H\ne0$, the chain rule converts the continuity equation into

$$\frac{d\rho}{d\ln a}=-3(\rho+p).$$

For a separately conserved component with constant $w=p/\rho$, integration gives

$$\rho(a) =\rho_\star \left(\frac{a}{a_\star}\right)^{-3(1+w)}.$$

Although using $\ln a$ as the independent variable required $H\ne0$, the constant-$w$ result itself does not. Directly,

$$\frac{d}{dt}\left[\rho a^{3(1+w)}\right]=0,$$

so one normalization continues through a smooth turnaround or bounce as long as $a\gt0$ and the same conserved fluid description remains valid.

Three immediate checks are

$$\begin{array}{c|c|c} \text{component} & w & \rho(a)\\ \hline \text{pressureless matter} & 0 & \rho\propto a^{-3}\\ \text{radiation} & \tfrac13 & \rho\propto a^{-4}\\ \text{vacuum energy} & -1 & \rho=\text{constant}. \end{array}$$

Radiation supplies the clearest pressure-work interpretation: $E=\rho V\propto a^{-1}$, the same extra factor produced by cosmological redshift. At an isolated turnaround the $t$-form remains regular even though $a$ is not a valid monotonic clock.

The Particle Data Group states the background conservation equation and constant-$w$ consequences in Particle Data Group 2026, §22.1.3, p. 3, eq. (22.10), and §22.1.5.1, p. 4, eq. (22.16) (PDF).

Interacting and nonideal components

If several components exchange energy, individual stress tensors need not be conserved. A useful background convention is

$$\dot\rho_i+3H(\rho_i+p_i)=Q_i, \qquad \sum_i Q_i=0,$$

where $Q_i\gt0$ means energy enters component $i$. Summing the equations recovers conservation of the total stress tensor. Applying the $a^{-3(1+w_i)}$ law separately when $Q_i\ne0$ would be inconsistent.

Nonideal stresses require the same care. Bulk viscosity changes the effective isotropic pressure; heat flux and anisotropic stress add terms to the energy projection outside exact FLRW. The perfect-fluid equation is not a license to hide such effects inside an arbitrary $w$ without specifying the underlying stress tensor and frame.

Exercises

Derive the entropy result from the Gibbs relation. Which assumptions enter after $\nabla_\mu T^{\mu\nu}=0$?

Solution

From

$$T\,ds=d(\rho/n)+p\,d(1/n)$$

one obtains

$$nT\dot s =\dot\rho-(\rho+p)\frac{\dot n}{n}.$$

The conserved particle current gives $\dot n/n=-\theta$, while the perfect-fluid energy projection gives $\dot\rho=-(\rho+p)\theta$. Therefore $nT\dot s=0$. Concluding $\dot s=0$ uses a meaningful conserved number density with $n\gt0$, a local-equilibrium Gibbs relation with $T\gt0$, and the absence of dissipative entropy sources. The zero-temperature limit requires separate care. Stress-energy conservation supplied only one part of the argument.

Two components exchange energy with $Q_1=-Q_2=Q(t)$. Show that the total density obeys a source-free continuity equation even though neither component generally has its uncoupled power law.

Solution

Adding

$$\dot\rho_1+3H(\rho_1+p_1)=Q, \qquad \dot\rho_2+3H(\rho_2+p_2)=-Q$$

cancels the transfer term. With $\rho=\rho_1+\rho_2$ and $p=p_1+p_2$,

$$\dot\rho+3H(\rho+p)=0.$$

The cancellation expresses internal exchange. Each component’s solution depends on the history of $Q(t)$, so its source-free constant-$w$ scaling need not hold.

References

• Ellis, George F. R., and Henk van Elst. “Cosmological Models (Cargèse Lectures 1998).” In Marc Lachièze-Rey, ed., Theoretical and Observational Cosmology, pp. 1–116. Dordrecht: Springer, 1999. DOI. Open PDF, arXiv:gr-qc/9812046v5.
• Particle Data Group. “Big-Bang Cosmology.” In Review of Particle Physics 2026, revised August 2025. Official PDF.
• Tong, David. Cosmology. University of Cambridge Part II Mathematical Tripos lecture notes, Michaelmas 2019. Official PDF.