## Accelerating expansion of the universe

• The accelerating expansion of the universe is the observation that the universe appears to be expanding at an increasing rate. In formal terms, this means that the cosmic scale factor a(t) has a positive second derivative, so that the velocity at which a distant galaxy is receding from the observer is continuously increasing with time.

The expansion of the universe has been accelerating since the universe entered its dark-energy-dominated era, at redshift z ≈ 0.4 (roughly 5 billion years ago). Within the framework of general relativity, an accelerating expansion can be accounted for by a positive value of the cosmological constant Λ, equivalent to the presence of a positive vacuum energy, dubbed "dark energy". While there are alternative possible explanations, the description assuming dark energy (positive Λ) is used in the current standard model of cosmology, known as ΛCDM (lambda cold dark matter).

The accelerated expansion was discovered in 1998, when two independent projects, the Supernova Cosmology Project and the High-Z Supernova Search Team simultaneously obtained results suggesting an acceleration in the expansion of the universe by using distant type Ia supernovae as standard candles. The discovery was unexpected, cosmologists at the time expecting a deceleration in the expansion of the universe, and amounts to the realization that the universe is currently in a "dark-energy-dominated era". Three members of these two groups have subsequently been awarded Nobel Prizes for their discovery. Confirmatory evidence has been found in baryon acoustic oscillations and other new results about the clustering of galaxies.

${\displaystyle H^{2}={\left({\frac {\dot {a}}{a}}\right)}^{2}={\frac {8{\pi }G}{3}}\rho -{\frac {{\mathrm {K} }c^{2}}{R^{2}a^{2}}}}$
${\displaystyle \rho _{c}={\frac {3H^{2}}{8{\pi }G}}}$
${\displaystyle \Omega ={\frac {\rho }{\rho _{c}}}}$
${\displaystyle H(a)=H_{0}{\sqrt {{\Omega _{k}a^{-2}+\Omega }_{m}a^{-3}+\Omega _{r}a^{-4}+\Omega _{\mathrm {DE} }a^{-3(1+w)}}}}$
${\displaystyle {\frac {\ddot {a}}{a}}=-{\frac {4{\pi }G}{3}}\left(\rho +{\frac {3P}{c^{2}}}\right)}$
${\displaystyle a(t)={\frac {1}{1+z}}}$
${\displaystyle t_{0}=\int _{0}^{1}{\frac {da}{\dot {a}}}}$
${\displaystyle P=wc^{2}\rho }$
Wikipedia