# How does universe expand when cosmological constant is zero?

From what I learned, Einstein believed in a static universe but from his general relativity equations universe must collapse under gravity. Hence Einstein adjusted this gravity with cosmological contact which is a kind of anti-gravitational effect. But later it was discovered that the universe is expanding. Einstein was ashamed by his cosmological constant and threw it out from his equations. My doubt is how the removal of cosmological constant agrees with an expanding universe ?. When there's nothing to counteract gravity, shouldn't the universe be contracting?.

• Jan 1, 2021 at 4:37
• This post by @benrg might also be helpful. Jan 1, 2021 at 5:15

Assuming the universe to be spatially homogeneous and isotropic, and combining this with the Einstein field equations produces the two Friedmann equations$$\frac{\dot{a}(t)}{a(t)} = \frac{8\pi G}{3}\rho - \frac{k}{a^2(t)}+\frac{\Lambda}{3}$$ and $$\frac{\ddot{a}(t)}{a(t)}=-\frac{4\pi G}{3}(\rho+3p)+\frac{\Lambda}{3}$$ where $$k=+1,0,-1$$ depending on curvature. $$\Lambda$$ is the cosmological constant.
if we want $$\ddot{a}(t)=\dot{a}(t)=0$$ (no expansion) and $$\Lambda=0$$, then the first equation implies $$\frac{8\pi G}{3}\rho a^2(t) = k$$. This will not work if $$k=0, -1$$ since the left side is nonzero and positive. The second equation leads to $$\rho+3p=0$$: for any positive density there has to be negative pressure even if we are just thinking of the contents of the universe as pressure-free dust. So it looks like $$\dot{a}(t) \neq 0$$... unless one adds a suitable nonzero value of $$\Lambda$$ to make things stand still.