Cosmological Constant and Accelerated Expansion

Accelerated expansion is determined by the active gravitational combination $\rho+3p$, not by energy density alone. A positive cosmological constant can be kept as geometry or represented exactly as a stress tensor with $p_\Lambda=-\rho_\Lambda$; the two descriptions are a dictionary, not two independent sources. For a mixture, acceleration begins when the total pressure is negative enough. In a matter–vacuum model this occurs before matter and vacuum have equal densities.

Required background. The Einstein Equation supplies the Einstein Equations and the geometric cosmological term; Flat-Background Hilbert Stress-Energy from Metric Variation supplies the Stress–Energy and Conservation interpretation of a source.

Helpful background. Relativistic Matter and Compact Objects supplies perfect-fluid equations of state and their physical limitations.

The active gravitational combination

For an FLRW spacetime, the acceleration equation is

$$\frac{\ddot a}{a} =-\frac{4\pi G}{3}(\rho+3p)+\frac{\Lambda}{3},$$

where $\rho$ and $p$ here exclude the geometrically displayed $\Lambda$. On an expanding branch, accelerated expansion means $\ddot a\gt0$, so

$$\Lambda\gt4\pi G(\rho+3p).$$

If $\Lambda=0$ and one separately conserved component dominates, the criterion becomes

$$w\equiv\frac{p}{\rho}\lt-\frac13$$

for positive $\rho$. Negative pressure is therefore not sufficient by itself: the weighted sum over every dynamically important component must satisfy the criterion. Curvature does not appear explicitly in $\ddot a/a$, although it still constrains $H$ and the allowed history through the Friedmann equation.

The deceleration parameter

$$q\equiv-\frac{\ddot a}{aH^2}$$

packages the same statement as $q\lt0$ only where $H\ne0$. It is undefined at a bounce, turnaround, or static event even though $\ddot a/a$ remains meaningful.

Accelerated expansion does not require the Hubble parameter to increase. For $a(t)=C t^p$ with $C\gt0$ and $p\gt1$, one has $\ddot a/a=p(p-1)/t^2\gt0$ but $H=p/t$ and $\dot H=-p/t^2\lt0$. The separation $D=a\chi$ of a fixed comoving pair accelerates because $\ddot D=\chi\ddot a\gt0$, even while its fractional growth rate $\dot D/D=H$ falls. Exponential de Sitter expansion is the limiting example with constant $H$, not increasing $H$.

Geometric and effective-fluid descriptions

Begin with

$$G_{\mu\nu}+\Lambda g_{\mu\nu} =8\pi G\,T_{\mu\nu}.$$

Moving the constant term to the right defines

$$T^{(\Lambda)}_{\mu\nu} \equiv-\frac{\Lambda}{8\pi G}g_{\mu\nu}, \qquad G_{\mu\nu} =8\pi G\left(T_{\mu\nu}+T^{(\Lambda)}_{\mu\nu}\right).$$

For any unit timelike observer $u^\mu$,

$$\rho_\Lambda =T^{(\Lambda)}_{\mu\nu}u^\mu u^\nu =\frac{\Lambda}{8\pi G}, \qquad p_\Lambda=-\rho_\Lambda.$$

Thus $w_\Lambda=-1$, and the continuity equation gives $\dot\rho_\Lambda=0$. The stress tensor is Lorentz invariant because it is proportional to the metric; it does not select a material rest frame.

One may now write the acceleration equation entirely in fluid language,

$$\frac{\ddot a}{a} =-\frac{4\pi G}{3} \sum_i(\rho_i+3p_i),$$

provided the sum includes $\rho_\Lambda$ and $p_\Lambda$. Alternatively, keep $\Lambda/3$ outside the sum. Combining the two conventions would double count the same term. The dictionary and vacuum equation of state are given in Particle Data Group 2026, §22.1.3, p. 2, text following eq. (22.6) and eqs. (22.8)–(22.9), and §22.1.5.4, pp. 4–5, eqs. (22.20)–(22.22) (PDF).

Matter, radiation, and the onset of acceleration

For pressureless matter plus vacuum energy,

$$\frac{\ddot a}{a} =-\frac{4\pi G}{3} \left(\rho_m-2\rho_\Lambda\right).$$

The transition $\ddot a=0$ occurs at

$$\rho_m=2\rho_\Lambda.$$

Matter–vacuum equality instead means $\rho_m=\rho_\Lambda$. Since $\rho_m\propto a^{-3}$ while $\rho_\Lambda$ is constant, acceleration begins first. With $a_0=1$ and present density parameters,

$$a_{\rm acc} =\left(\frac{\Omega_{m0}}{2\Omega_{\Lambda0}}\right)^{1/3}, \qquad a_{m\Lambda} =\left(\frac{\Omega_{m0}}{\Omega_{\Lambda0}}\right)^{1/3},$$

so $a_{\rm acc}=2^{-1/3}a_{m\Lambda}$. These are model relations, not current numerical measurements.

Radiation contributes $\rho_r+3p_r=2\rho_r$. Retaining it changes the transition condition to

$$\rho_m+2\rho_r=2\rho_\Lambda.$$

More generally, a component with $w\lt-1/3$ may be present without producing acceleration if matter or radiation still dominates the active gravitational sum.

Limiting solutions and interpretation

In spatially flat vacuum with $\Lambda\gt0$, the Friedmann equation gives

$$H^2=\frac{\Lambda}{3}.$$

The expanding branch is de Sitter,

$$a(t)=a_\star \exp\left[\sqrt{\frac{\Lambda}{3}}\,(t-t_\star)\right],$$

and indeed $\ddot a/a=\Lambda/3$. By contrast, a flat pressureless model with $\Lambda=0$ has $a\propto t^{2/3}$ and $\ddot a\lt0$. These limits check both the sign and the factor of three. Tong works through the vacuum and matter–vacuum solutions in Tong 2019, §1.3.3, pp. 40–44, eqs. (1.59)–(1.64) (PDF).

Acceleration is a statement about the second derivative of the scale factor with respect to comoving proper time. It is not itself evidence for a particular microscopic origin. A cosmological constant, a dynamical field, modified gravity, or an inference systematic can require different tests even if they reproduce a similar background $H(a)$. This page establishes only the general-relativistic FLRW criterion and the exact $\Lambda$ dictionary.

Common pitfalls

Equality is not the acceleration transition. In a matter–vacuum model, the active vacuum contribution is $-2\rho_\Lambda$, so the transition occurs at $\rho_m=2\rho_\Lambda$, not at equal densities.

A negative $q$ needs an expanding context. The formula remains algebraic, but the phrase “accelerated expansion” presumes $H\gt0$. At $H=0$, use $\ddot a/a$ rather than $q$.

Vacuum energy is not counted twice. Choose either geometric $\Lambda$ or the effective fluid $T^{(\Lambda)}_{\mu\nu}$ before assembling the source budget.

Exercises

For a flat, separately conserved mixture of radiation, matter, and a positive cosmological constant, express the acceleration transition in terms of present density parameters and $a$.

Solution

Use

$$\rho_r=\rho_{r0}a^{-4}, \qquad \rho_m=\rho_{m0}a^{-3}, \qquad \rho_\Lambda=\rho_{\Lambda0}.$$

The condition $\rho_m+2\rho_r=2\rho_\Lambda$ becomes

$$\Omega_{m0}a^{-3} +2\Omega_{r0}a^{-4} -2\Omega_{\Lambda0}=0.$$

Multiplying by $a^4$ gives the regular algebraic equation

$$2\Omega_{\Lambda0}a^4 -\Omega_{m0}a -2\Omega_{r0}=0.$$

Its positive root is the transition scale factor. Neglecting radiation recovers $a_{\rm acc}=(\Omega_{m0}/2\Omega_{\Lambda0})^{1/3}$. For fixed matter and vacuum densities, the positive radiation term delays the transition to a larger scale factor because its pressure adds to deceleration. Only a positive root lying on the modeled expansion branch has physical relevance.

Show directly that moving $\Lambda$ to the source side gives $p_\Lambda=-\rho_\Lambda$ for the site signature.

Solution

Define $T^{(\Lambda)}_{\mu\nu}=-(\Lambda/8\pi G)g_{\mu\nu}$. In a comoving orthonormal frame, $g_{\hat0\hat0}=-1$ and $g_{\hat i\hat j}=\delta_{ij}$. Therefore

$$T^{(\Lambda)}_{\hat0\hat0} =\frac{\Lambda}{8\pi G} =\rho_\Lambda, \qquad T^{(\Lambda)}_{\hat i\hat j} =-\frac{\Lambda}{8\pi G}\delta_{ij} =p_\Lambda\delta_{ij}.$$

Hence $p_\Lambda=-\rho_\Lambda$. Substituting this pair into $-(4\pi G/3)(\rho+3p)$ produces $+\Lambda/3$, exactly matching the geometric form. The three equal spatial components also show that the effective source is isotropic for every inertial observer; no preferred material velocity had to be introduced.

References

• 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.