# Why is $\Lambda=0$ for a flat matter dominated universe?

I was looking over some of my cosmology notes and was considering the Einstein de-sitter model of a matter dominated universe. I know that in this case that k=0 and also furthermore that $$\rho(t)$$ is proportional to $$a(t)^{-3}$$. Then when I looked at some solutions I saw that my lecturer assumes also that $$\Lambda=0$$. My question is , is this always true for a flat universe and if so why?

Note: one thought I have is that if the universe is matter dominated instead of dark matter dominated then we can take $$\Lambda$$ to be negligible is that the reason , or is it to do do with the universe being flat?

• Hint: how do the various contributions to $H ^2$ scale with the expansion factor $a$? In particular, how does matter compare to vacuum energy in this respect? – pppqqq Jan 10 '19 at 18:17
• $\Lambda$ is dark energy, not dark matter. You can add dark energy to any model. Your lecturer probably left it out to simplify things. – G. Smith Jan 10 '19 at 18:18
• @pppqqq ah okay think I get you. we can ignore it because matter density is much greater than vacuum density which is what we use to define the cosmological constant with . By a similar token then in a radiation dominated universe we would usually not neglect the cosmological constant because radiation density is more similar in magnitude to vacuum density ? – bhapi Jan 10 '19 at 18:23
• Dark energy always dominates at late times. The energy density of matter and radiation both decrease as the universe expands. The energy density of dark energy stays the same. You can ignore dark energy at early times but not at late times. – G. Smith Jan 10 '19 at 18:26
• @G.Smith ah okay , so my lecturer must just be simplifying things, ironically I always find that more confusing . – bhapi Jan 10 '19 at 18:48

In a matter dominated universe, matter is by definition the dominant contribution and all others are approximated to zero. We take $$\Lambda = 0$$ because otherwise the universe wouldn't be matter dominated. Of course, one may ask further questions, such as
But if you're going to be working under the assumption of matter domination, you set $$\Lambda=0$$ by definition.
• Perhaps, in the last sentence, it would be useful to briefly explain why setting $\Lambda = 0$ is a reasonable definition? – N. Steinle Jan 10 '19 at 19:56
• okay so I had been right to assume that in my note, I got mixed messages in the comments , but your answer had been what I assumed to be reasonable . Thank you . By the way to reiterate the question I posed to ppqqq , does that mean that in a radiation dominated universe we would usually not ignore the cosmological constant term because the density of radiation is closer to the density of a vacuum, at early times I'm not sure because maybe then the radiation density way actually quite large in comparison to the vacuum density and hence $\Lambda$ – bhapi Jan 10 '19 at 21:02