Scale Factor Evolution
Solving the Friedmann equation for single-component and multi-component universes, deriving the deceleration parameter, and computing the age of the cosmos
1. Introduction: The Scale Factor as the Fundamental Dynamical Variable
In the previous chapter, we derived the Friedmann equations from Einstein's field equations applied to the FLRW metric. The central object that emerged is the scale factor \(a(t)\), a dimensionless function of cosmic time that encodes the entire expansion history of the homogeneous and isotropic universe. Every physical distance between comoving observers scales as:
By convention, we normalize \(a(t_0) = 1\) at the present epoch. The scale factor at earlier times is related to the cosmological redshift by \(a = 1/(1+z)\). The Hubble parameter is the logarithmic derivative:
The goal of this chapter is to solve the Friedmann equation for \(a(t)\) under various assumptions about the energy content. We will derive analytic solutions for single-component universes (radiation, matter, cosmological constant), then consider mixed-component models, define the deceleration parameter, and finally compute the age of the universe for the standard\(\Lambda\)CDM cosmology.
The First Friedmann Equation (Recap)
For a spatially flat universe (\(k = 0\)), the first Friedmann equation reads:
where \(\rho\) is the total energy density. Each component \(i\) obeys the continuity equation \(\dot{\rho}_i + 3H(1 + w_i)\rho_i = 0\), giving \(\rho_i \propto a^{-3(1+w_i)}\) for constant equation of state parameter \(w_i\).
2. Derivation 1: Scale Factor for Radiation, Matter, and Λ Domination
We begin with the simplest case: a spatially flat universe dominated by a single perfect fluid with equation of state \(p = w\rho c^2\). The energy density evolves as:
Substituting into the Friedmann equation and writing \(H = \dot{a}/a\):
General single-component ODE
$$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3c^2}\,\rho_0\,a^{-3(1+w)}$$
$$\dot{a}^2 = \frac{8\pi G\rho_0}{3c^2}\,a^{-3(1+w)+2} = \frac{8\pi G\rho_0}{3c^2}\,a^{-(1+3w)}$$
$$\dot{a} = \sqrt{\frac{8\pi G\rho_0}{3c^2}}\;a^{-(1+3w)/2}$$
This is a separable ODE. We separate variables and integrate, treating each value of \(w\) below.
2.1 Radiation-Dominated Universe (\(w = 1/3\))
For radiation (photons and relativistic neutrinos), \(w = 1/3\), so the energy density scales as \(\rho_r \propto a^{-4}\). The extra factor of \(a^{-1}\) beyond the matter dilution \(a^{-3}\) arises from the cosmological redshift of each photon's energy.
Step-by-step derivation
Step 1: With \(w = 1/3\), the exponent is \(-(1+3w)/2 = -(1+1)/2 = -1\):
$$\dot{a} = C\,a^{-1} \qquad \text{where } C = \sqrt{\frac{8\pi G\rho_0}{3c^2}}$$
Step 2: Separate variables: \(a\,da = C\,dt\)
Step 3: Integrate from \(a = 0\) at \(t = 0\):
$$\int_0^a a'\,da' = C\int_0^t dt' \quad\Longrightarrow\quad \frac{a^2}{2} = Ct$$
Step 4: Solve for \(a(t)\):
$$\boxed{a(t) = \sqrt{2Ct} = (2H_0 t)^{1/2} \propto t^{1/2}}$$
The Hubble parameter is \(H = \dot{a}/a = 1/(2t)\), so the Hubble time equals twice the age:\(t = 1/(2H)\). The temperature of the radiation bath scales as \(T \propto a^{-1} \propto t^{-1/2}\).
2.2 Matter-Dominated Universe (\(w = 0\))
For pressureless dust (non-relativistic matter: baryons + cold dark matter), \(w = 0\), so \(\rho_m \propto a^{-3}\). The energy density simply dilutes with the expanding volume.
Step-by-step derivation
Step 1: With \(w = 0\), the exponent is \(-(1+0)/2 = -1/2\):
$$\dot{a} = C\,a^{-1/2}$$
Step 2: Separate variables: \(a^{1/2}\,da = C\,dt\)
Step 3: Integrate:
$$\int_0^a a'^{1/2}\,da' = C\int_0^t dt' \quad\Longrightarrow\quad \frac{2}{3}\,a^{3/2} = Ct$$
Step 4: Solve for \(a(t)\):
$$\boxed{a(t) = \left(\frac{3Ct}{2}\right)^{2/3} = \left(\frac{3H_0 t}{2}\right)^{2/3} \propto t^{2/3}}$$
The Hubble parameter is \(H = 2/(3t)\). This is the Einstein–de Sitter model. The age of the universe would be \(t_0 = 2/(3H_0) \approx 9.3\) Gyr — significantly less than the observed 13.8 Gyr, indicating that matter alone cannot be the whole story.
2.3 Dark Energy Dominated Universe (\(w = -1\))
For a cosmological constant \(\Lambda\), the equation of state is \(w = -1\), so \(\rho_\Lambda = \text{const}\). The energy density does not dilute — the vacuum energy is a property of spacetime itself.
Step-by-step derivation
Step 1: With \(w = -1\), the exponent is \(-(1-3)/2 = 1\):
$$\dot{a} = C\,a^{1} = H_\Lambda\,a$$
where \(H_\Lambda = \sqrt{8\pi G\rho_\Lambda/(3c^2)} = \text{const}\) is the constant de Sitter Hubble rate.
Step 2: This is \(\dot{a}/a = H_\Lambda\), which is a simple exponential ODE:
$$\frac{da}{a} = H_\Lambda\,dt$$
Step 3: Integrate:
$$\ln a = H_\Lambda\,t + \text{const}$$
Step 4: Exponentiate:
$$\boxed{a(t) = a_i\,e^{H_\Lambda(t - t_i)} \propto e^{H_\Lambda t}}$$
This is the de Sitter solution: eternal exponential expansion. The Hubble parameter\(H = H_\Lambda = \text{const}\) remains fixed. This solution describes both the inflationary epoch at \(t \sim 10^{-36}\) s and the far-future fate of our universe as matter and radiation dilute away.
Summary: Single-Component Solutions
| Component | \(w\) | \(\rho(a)\) | \(a(t)\) | \(H(t)\) |
|---|---|---|---|---|
| Radiation | \(1/3\) | \(\propto a^{-4}\) | \(\propto t^{1/2}\) | \(1/(2t)\) |
| Matter | \(0\) | \(\propto a^{-3}\) | \(\propto t^{2/3}\) | \(2/(3t)\) |
| Cosmological const. | \(-1\) | \(\text{const}\) | \(\propto e^{Ht}\) | \(\text{const}\) |
The general power-law solution for \(w \neq -1\) is \(a(t) \propto t^{2/[3(1+w)]}\). For stiff matter (\(w=1\)), \(a \propto t^{1/3}\).
3. Derivation 2: General Solutions with Mixed Components
Real cosmology involves multiple energy components simultaneously. We write the first Friedmann equation in terms of the density parameters \(\Omega_i = \rho_i / \rho_{\text{crit}}\) where the critical density is \(\rho_{\text{crit}} = 3H^2 c^2/(8\pi G)\):
Friedmann equation in dimensionless form
Step 1: Express each component's density at scale factor \(a\) in terms of its present value:
$$\rho_r(a) = \rho_{r,0}\,a^{-4}, \quad \rho_m(a) = \rho_{m,0}\,a^{-3}, \quad \rho_\Lambda = \rho_{\Lambda,0}$$
Step 2: Divide the Friedmann equation by \(H_0^2\):
$$\frac{H^2}{H_0^2} = \frac{\Omega_r}{a^4} + \frac{\Omega_m}{a^3} + \Omega_\Lambda + \frac{\Omega_k}{a^2}$$
where \(\Omega_k = 1 - \Omega_r - \Omega_m - \Omega_\Lambda\) accounts for spatial curvature, and all \(\Omega\)'s are evaluated at \(a = 1\) (today).
Step 3: Define the dimensionless Hubble function:
$$\boxed{E(a) \equiv \frac{H(a)}{H_0} = \sqrt{\frac{\Omega_r}{a^4} + \frac{\Omega_m}{a^3} + \frac{\Omega_k}{a^2} + \Omega_\Lambda}}$$
The constraint \(\Omega_r + \Omega_m + \Omega_k + \Omega_\Lambda = 1\) is automatically satisfied at \(a = 1\), where \(E(1) = 1\) by construction.
3.1 Radiation–Matter Equality
A key epoch in cosmic history is when the radiation and matter energy densities are equal. At this point, the expansion transitions from the radiation-dominated power law \(a \propto t^{1/2}\) to the matter-dominated law \(a \propto t^{2/3}\).
Derivation of \(a_{\text{eq}}\)
Step 1: Set \(\rho_r(a_{\text{eq}}) = \rho_m(a_{\text{eq}})\):
$$\rho_{r,0}\,a_{\text{eq}}^{-4} = \rho_{m,0}\,a_{\text{eq}}^{-3}$$
Step 2: Divide both sides by \(\rho_{\text{crit},0}\) and simplify:
$$\Omega_r\,a_{\text{eq}}^{-4} = \Omega_m\,a_{\text{eq}}^{-3}$$
Step 3: Solve for \(a_{\text{eq}}\):
$$\boxed{a_{\text{eq}} = \frac{\Omega_r}{\Omega_m}}$$
Step 4: Convert to redshift:
$$z_{\text{eq}} = \frac{1}{a_{\text{eq}}} - 1 = \frac{\Omega_m}{\Omega_r} - 1$$
Using Planck 2018 values (\(\Omega_m = 0.315\), \(\Omega_r = 9.1 \times 10^{-5}\)):\(a_{\text{eq}} \approx 2.9 \times 10^{-4}\), corresponding to \(z_{\text{eq}} \approx 3400\). This occurred roughly 47,000 years after the Big Bang, well before recombination (\(z \approx 1100\)).
3.2 The Full \(\Lambda\)CDM Evolution
For a flat \(\Lambda\)CDM universe (\(\Omega_k = 0\)), the scale factor obeys:
This equation has no closed-form solution when all three components are present. However, in the matter–\(\Lambda\) limit (\(\Omega_r \to 0\)), an exact solution exists:
Exact matter + \(\Lambda\) solution
Step 1: For \(\Omega_r = 0\), \(\Omega_k = 0\), \(\Omega_\Lambda = 1 - \Omega_m\):
$$\dot{a}^2 = H_0^2\left(\frac{\Omega_m}{a} + \Omega_\Lambda\,a^2\right)$$
Step 2: Separate and integrate using the substitution \(u = a^{3/2}\):
$$t = \frac{1}{H_0}\int_0^a \frac{da'}{a'\sqrt{\Omega_m/a'^3 + \Omega_\Lambda}}$$
Step 3: The result is:
$$\boxed{a(t) = \left(\frac{\Omega_m}{\Omega_\Lambda}\right)^{1/3}\left[\sinh\!\left(\frac{3}{2}\sqrt{\Omega_\Lambda}\,H_0 t\right)\right]^{2/3}}$$
At early times (\(H_0 t \ll 1\)), \(\sinh(x) \approx x\), recovering the matter-dominated solution \(a \propto t^{2/3}\). At late times, \(\sinh(x) \approx e^x/2\), giving exponential de Sitter expansion. The transition between deceleration and acceleration occurs at the inflection point of this function.
4. Derivation 3: The Deceleration Parameter
The deceleration parameter \(q\) quantifies whether the expansion is speeding up or slowing down. It is defined as:
The sign convention is historical: \(q > 0\) means deceleration (the expansion is slowing down), while \(q < 0\) means acceleration. Let us derive \(q\) in terms of the density parameters.
Derivation of \(q\) from the Friedmann equations
Step 1: Start with the second Friedmann equation (acceleration equation):
$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3c^2}\left(\rho + \frac{3p}{c^2}\right) = -\frac{4\pi G}{3c^2}\sum_i \rho_i(1 + 3w_i)$$
Step 2: Divide by \(H^2 = 8\pi G\rho_{\text{tot}}/(3c^2)\):
$$\frac{\ddot{a}}{aH^2} = -\frac{1}{2}\sum_i \frac{\rho_i}{\rho_{\text{tot}}}(1 + 3w_i) = -\frac{1}{2}\sum_i \frac{\Omega_i(a)}{\sum_j \Omega_j(a)}(1 + 3w_i)$$
Step 3: Therefore \(q = -\ddot{a}/(aH^2)\) becomes:
$$q = \frac{1}{2}\sum_i \Omega_i(a)(1 + 3w_i)$$
Step 4: Substitute the three components (\(w_r = 1/3\), \(w_m = 0\), \(w_\Lambda = -1\)):
$$\boxed{q = \Omega_r(a) + \frac{\Omega_m(a)}{2} - \Omega_\Lambda(a)}$$
Here \(\Omega_i(a)\) denotes the density parameter at scale factor \(a\), not at the present epoch. We use \(\Omega_i(a) = \Omega_{i,0}\,a^{-3(1+w_i)} / E^2(a)\).
4.1 Physical Interpretation
\(q > 0\): Decelerating Expansion
For a matter-dominated universe: \(q = \Omega_m/2 = 1/2\) (Einstein–de Sitter). For radiation domination: \(q = \Omega_r = 1\). Gravity pulls back on the expansion.
Attractive gravity from ordinary matter and radiation always decelerates the expansion.
\(q < 0\): Accelerating Expansion
For a \(\Lambda\)-dominated universe: \(q = -\Omega_\Lambda = -1\). The negative pressure of dark energy drives acceleration. Today, \(q_0 \approx -0.53\).
This was the stunning 1998 discovery by the Supernova Cosmology Project and the High-z Supernova Search Team.
4.2 The Transition Redshift \(z_{\text{acc}}\)
The universe transitions from deceleration to acceleration when \(q = 0\). Let us find the redshift at which this occurs, neglecting radiation (valid for \(z \lesssim 100\)).
Derivation of \(z_{\text{acc}}\)
Step 1: Set \(q = 0\) with \(\Omega_r \approx 0\):
$$\frac{\Omega_m(a)}{2} = \Omega_\Lambda(a)$$
Step 2: Express in terms of present-day parameters. Since \(\Omega_i(a) = \Omega_{i,0}\,a^{-3(1+w_i)}/E^2(a)\):
$$\frac{\Omega_m\,a^{-3}}{2} = \Omega_\Lambda$$
(The \(E^2(a)\) factors cancel.)
Step 3: Solve for \(a\):
$$a_{\text{acc}}^3 = \frac{\Omega_m}{2\Omega_\Lambda}$$
Step 4: Convert to redshift using \(1+z = 1/a\):
$$\boxed{z_{\text{acc}} = \left(\frac{2\Omega_\Lambda}{\Omega_m}\right)^{1/3} - 1}$$
For \(\Omega_m = 0.315\), \(\Omega_\Lambda = 0.685\): \(z_{\text{acc}} = (2 \times 0.685/0.315)^{1/3} - 1 \approx 0.64\). This corresponds to roughly 6 billion years ago. Before this epoch, the expansion was decelerating; after it, dark energy won and the expansion has been accelerating.
5. Derivation 4: The Age of the Universe
One of the most fundamental predictions of any cosmological model is the age of the universe. We derive it by integrating \(dt\) from the Big Bang (\(a = 0\)) to today (\(a = 1\)).
Derivation of \(t_0\)
Step 1: From \(H = \dot{a}/a\), we have \(dt = da/(aH)\):
$$t_0 = \int_0^{t_0} dt = \int_0^1 \frac{da}{a\,H(a)}$$
Step 2: Substitute \(H(a) = H_0\,E(a)\):
$$t_0 = \frac{1}{H_0}\int_0^1 \frac{da}{a\,E(a)} = \frac{1}{H_0}\int_0^1 \frac{da}{a\,\sqrt{\Omega_r\,a^{-4} + \Omega_m\,a^{-3} + \Omega_\Lambda}}$$
Step 3: Simplify the integrand by multiplying numerator and denominator by \(a\):
$$\boxed{t_0 = \frac{1}{H_0}\int_0^1 \frac{da}{\sqrt{\Omega_r\,a^{-2} + \Omega_m\,a^{-1} + \Omega_\Lambda\,a^2}}}$$
Step 4: Equivalently, change variables to \(z\) using \(a = 1/(1+z)\), \(da = -dz/(1+z)^2\):
$$\boxed{t_0 = \frac{1}{H_0}\int_0^\infty \frac{dz}{(1+z)\,E(z)}}$$
where \(E(z) = \sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_\Lambda}\).
Numerical Evaluation
For the Planck 2018 best-fit parameters (\(\Omega_m = 0.315\), \(\Omega_\Lambda = 0.685\), \(\Omega_r = 9.1 \times 10^{-5}\), \(H_0 = 67.4\) km/s/Mpc):
$$\frac{1}{H_0} = \frac{1}{67.4\;\text{km/s/Mpc}} = 14.52\;\text{Gyr}$$
$$\int_0^1 \frac{da}{a\,E(a)} \approx 0.9506$$
$$\boxed{t_0 = 0.9506 \times 14.52\;\text{Gyr} \approx 13.8\;\text{Gyr}}$$
This is in excellent agreement with independent age estimates from globular clusters (\(\gtrsim 12\) Gyr), white dwarf cooling sequences, and radioactive dating of old stars. The concordance of these methods is a triumph of the \(\Lambda\)CDM model.
For special cases, the age integral can be evaluated analytically. In a matter-only Einstein–de Sitter universe, \(t_0 = 2/(3H_0) \approx 9.3\) Gyr, which is too short. In a pure \(\Lambda\) universe, \(t_0 \to \infty\) (eternal exponential expansion has no beginning). The \(\Lambda\)CDM age of 13.8 Gyr lies between these extremes.
6. Applications
6.1 Lookback Time
The lookback time to redshift \(z\) is the cosmic time elapsed between the emission of light at \(z\) and its observation today:
For example, light from a galaxy at \(z = 1\) was emitted about 7.9 Gyr ago — more than half the age of the universe. The CMB photons (\(z = 1100\)) were emitted 13.8 Gyr ago minus about 370,000 years, essentially at the beginning.
6.2 Cosmic Chronometers
The cosmic chronometer method provides a direct, model-independent measurement of \(H(z)\). It exploits the relation:
By measuring the age difference \(\Delta t\) of passively evolving galaxies at two close redshifts \(z\) and \(z + \Delta z\), one can estimate \(dz/dt \approx \Delta z/\Delta t\) and thereby reconstruct \(H(z)\)without assuming a particular cosmological model. Current measurements confirm the \(\Lambda\)CDM prediction to within 5–10%.
6.3 Expansion Rate History
The Hubble parameter as a function of redshift encodes the full expansion history:
Key features of \(H(z)\):
- At high \(z\): \(H(z) \approx H_0\sqrt{\Omega_r}\,(1+z)^2\) (radiation dominated)
- At intermediate \(z\): \(H(z) \approx H_0\sqrt{\Omega_m}\,(1+z)^{3/2}\) (matter dominated)
- At \(z \to -1\) (far future): \(H \to H_0\sqrt{\Omega_\Lambda}\) (de Sitter asymptote)
- Today (\(z = 0\)): \(H_0 = 67.4\) km/s/Mpc by definition
The minimum of \(H(z)\) (at low \(z\)) marks the transition from matter deceleration to\(\Lambda\)-driven acceleration. Baryon acoustic oscillation (BAO) surveys such as DESI measure \(H(z)\) through the radial BAO scale, providing percent-level constraints on dark energy.
7. Historical Context
Key Milestones in Understanding the Expanding Universe
8. Python Simulation: Scale Factor, Hubble Parameter, and Deceleration Parameter
The following simulation numerically integrates the Friedmann equation for four universes — radiation-only, matter-only, \(\Lambda\)-only, and the full \(\Lambda\)CDM model — then plots \(a(t)\), \(H(z)\), and \(q(z)\). Only numpy is used (no scipy).
Click Run to execute the Python code
Code will be executed with Python 3 on the server
Practice Problems
Problem 1:Solve the Friedmann equation for a radiation-dominated universe ($\Omega_r = 1$, flat) to find $a(t)$.
Solution:
Step 1: The Friedmann equation for radiation only: $H^2 = \left(\frac{\dot{a}}{a}\right)^2 = H_0^2\,\Omega_r\,a^{-4}$.
Step 2: Rearrange: $a\,da = H_0\sqrt{\Omega_r}\,dt$.
Step 3: Integrate: $\frac{a^2}{2} = H_0\sqrt{\Omega_r}\,t$, so $a(t) = (2H_0\sqrt{\Omega_r})^{1/2}\,t^{1/2}$.
Step 4: Normalizing so $a = 1$ at present gives $t_0 = 1/(2H_0)$.
Answer: $a(t) \propto t^{1/2}$. The Hubble parameter evolves as $H = 1/(2t)$, and the age of a purely radiation-dominated universe is $t_0 = 1/(2H_0)$.
Problem 2:Solve the Friedmann equation for a matter-dominated flat universe to find $a(t)$ and the age $t_0$.
Solution:
Step 1: Friedmann equation: $\left(\frac{\dot{a}}{a}\right)^2 = H_0^2\,\Omega_m\,a^{-3}$.
Step 2: Rearrange: $a^{1/2}\,da = H_0\sqrt{\Omega_m}\,dt$.
Step 3: Integrate: $\frac{2}{3}a^{3/2} = H_0\sqrt{\Omega_m}\,t$.
Step 4: Solve: $a(t) = \left(\frac{3}{2}H_0\sqrt{\Omega_m}\,t\right)^{2/3}$. At $a = 1$: $t_0 = \frac{2}{3H_0}$ (for $\Omega_m = 1$).
Answer: $a(t) \propto t^{2/3}$ (Einstein–de Sitter). The age is $t_0 = 2/(3H_0) \approx 9.3$ Gyr for $H_0 = 70$ km/s/Mpc, which is too young.
Problem 3:Solve the Friedmann equation for a $\Lambda$-dominated universe and show the expansion is exponential.
Solution:
Step 1: Friedmann equation: $H^2 = H_0^2\,\Omega_\Lambda$ = constant.
Step 2: Since $H = \dot{a}/a = H_0\sqrt{\Omega_\Lambda} \equiv H_\Lambda$ is constant, we have $\dot{a} = H_\Lambda\,a$.
Step 3: This is a first-order ODE with solution $a(t) = a_0\,e^{H_\Lambda(t - t_0)}$.
Step 4: The doubling time is $t_d = \ln 2/H_\Lambda$. For our universe ($\Omega_\Lambda = 0.685$): $H_\Lambda = 0.828\,H_0$, giving $t_d \approx 11.4$ Gyr.
Answer: $a(t) \propto e^{H_\Lambda t}$ (de Sitter expansion). The Hubble parameter remains constant and the universe expands exponentially, diluting all matter and radiation.
Problem 4:Evaluate the age of the universe integral $t_0 = \frac{1}{H_0}\int_0^1 \frac{da}{a\,E(a)}$ for a flat matter+$\Lambda$ model with $\Omega_m = 0.3$, $\Omega_\Lambda = 0.7$.
Solution:
Step 1: $E(a) = \sqrt{\Omega_m\,a^{-3} + \Omega_\Lambda}$, so the integrand is $\frac{1}{a\sqrt{0.3\,a^{-3} + 0.7}}$.
Step 2: Simplify: $\frac{1}{\sqrt{0.3\,a^{-1} + 0.7\,a^2}}$.
Step 3: This integral has an analytic solution: $t_0 = \frac{2}{3H_0\sqrt{\Omega_\Lambda}}\ln\!\left(\frac{1 + \sqrt{\Omega_\Lambda}}{\sqrt{\Omega_m}}\right)$.
Step 4: Substituting: $t_0 = \frac{2}{3H_0\sqrt{0.7}}\ln\!\left(\frac{1 + \sqrt{0.7}}{\sqrt{0.3}}\right) = \frac{2}{3 \times 0.8367\,H_0}\ln(3.264) = \frac{0.964}{H_0}$.
Answer: $t_0 \approx 0.964/H_0 \approx 13.5$ Gyr for $H_0 = 70$ km/s/Mpc. The cosmological constant makes the universe older than the matter-only prediction of $2/(3H_0)$.
Problem 5:Derive the deceleration parameter $q_0$ for a flat universe with $\Omega_m = 0.315$ and $\Omega_\Lambda = 0.685$, and find the transition redshift $z_{\text{acc}}$.
Solution:
Step 1: The deceleration parameter is $q = -\frac{\ddot{a}a}{\dot{a}^2}$. From the acceleration equation: $q = \frac{\Omega_m}{2} - \Omega_\Lambda$ (neglecting radiation).
Step 2: Today: $q_0 = \frac{0.315}{2} - 0.685 = 0.1575 - 0.685 = -0.5275$.
Step 3: Transition occurs when $q = 0$: $\frac{\Omega_m(1+z)^3}{2E^2(z)} = \frac{\Omega_\Lambda}{E^2(z)}$, simplifying to $\Omega_m(1+z)^3 = 2\Omega_\Lambda$.
Step 4: Solve: $1 + z_{\text{acc}} = \left(\frac{2\Omega_\Lambda}{\Omega_m}\right)^{1/3} = \left(\frac{2 \times 0.685}{0.315}\right)^{1/3} = (4.349)^{1/3} = 1.632$.
Answer: $q_0 \approx -0.53$ (accelerating) and $z_{\text{acc}} \approx 0.63$, corresponding to about 6 billion years ago when dark energy began to dominate.