Dark Energy

Dark Energy

In 1998, observations of distant Type Ia supernovae revealed that the expansion of the universe is accelerating — a discovery that earned the 2011 Nobel Prize. The cause is called dark energy, making up about 68% of the total energy budget of the universe.

The Cosmological Constant

The simplest model of dark energy is Einstein's cosmological constant Λ, with equation of state:

$w = \frac{p}{\rho c^2} = -1$

This means dark energy has constant energy density regardless of how much the universe expands:

$\rho_{\rm DE}(a) = \rho_{\rm DE,0} \cdot a^{-3(1+w)} = \rho_{\rm DE,0} \quad (w = -1)$

General Dark Energy

A more general dark energy model allows $w \neq -1$. The density evolves as:

$\frac{\rho_{\rm DE}(a)}{\rho_{\rm DE,0}} = a^{-3(1+w)}$

The Critical Density

The present-day dark energy density is:

$\rho_{\rm DE,0} = \Omega_\Lambda \cdot \rho_c = \Omega_\Lambda \cdot \frac{3 H_0^2}{8\pi G}$

where $H_0$ is the Hubble constant in SI units: $H_0 = H_{0,\rm km/s/Mpc} \times 1000 / (3.0857 \times 10^{22}$ m).

For $\Omega_\Lambda = 0.7$ and $H_0 = 70$ km/s/Mpc: $\rho_{\rm DE,0} \approx 6.4 \times 10^{-27}$ kg/m³.

Your Task

Implement three functions. All constants must be defined inside each function.

• dark_energy_density_ratio(a, w) — returns $\rho_{\rm DE}(a) / \rho_{\rm DE,0} = a^{-3(1+w)}$
• dark_energy_density_today_kg_m3(Omega_Lambda, H0_km_s_Mpc) — returns $\rho_{\rm DE,0}$ in kg/m³; use $G = 6.674 \times 10^{-11}$, $H_0^{\rm SI} = H_0 \times 1000 / 3.0857 \times 10^{22}$
• equation_of_state_pressure(w, rho_DE_kg_m3) — returns $p = w \rho c^2$ in Pa; use $c = 2.998 \times 10^8$ m/s