Set a ~ ~ = a a 0 , ρ ρ c = 3 H 0 2 8 π π G , Ω Ω = ρ ρ ρ ρ c , t = t ~ ~ H 0 , Ω Ω k = − − k c 2 H 0 2 a 0 2 , where a 0 and H 0 are separately the scale factor and the Hubble parameter today. Then we can have 1 2 ( d a ~ ~ d t ~ ~ ) 2 + U eff ( a ~ ~ ) = 1 2 Ω Ω k where U eff ( a ~ ~ ) = − − Ω Ω a ~ ~ 2 2 .