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?

  • $\begingroup$ 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? $\endgroup$
    – pppqqq
    Jan 10 '19 at 18:17
  • $\begingroup$ $\Lambda$ is dark energy, not dark matter. You can add dark energy to any model. Your lecturer probably left it out to simplify things. $\endgroup$
    – G. Smith
    Jan 10 '19 at 18:18
  • $\begingroup$ @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 ? $\endgroup$
    – bhapi
    Jan 10 '19 at 18:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – G. Smith
    Jan 10 '19 at 18:26
  • $\begingroup$ @G.Smith ah okay , so my lecturer must just be simplifying things, ironically I always find that more confusing . $\endgroup$
    – 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

  • Can a matter dominated universe have any spatial curvature?
  • Was our universe ever matter dominated?

But if you're going to be working under the assumption of matter domination, you set $\Lambda=0$ by definition.

  • $\begingroup$ Perhaps, in the last sentence, it would be useful to briefly explain why setting $\Lambda = 0$ is a reasonable definition? $\endgroup$ Jan 10 '19 at 19:56
  • $\begingroup$ @N.Steinle: if I'm not misunderstanding your question, I already explained that in my first paragraph: "matter" dominated means that all other forms of energy are negligible in comparison to regular matter. $\endgroup$
    – Javier
    Jan 10 '19 at 19:58
  • $\begingroup$ 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 $\Lambda$ $\endgroup$
    – bhapi
    Jan 10 '19 at 21:02
  • $\begingroup$ @Javier I like your answer, and my comment was meant more as a suggestion than a question. Indeed you said "We take Λ=0 because otherwise the universe wouldn't be matter dominated," but i.e. why does any nonzero value of lambda mean that the universe is dominated by the energy of expansion? Just in the interest of completeness, but the OP has already accepted.... $\endgroup$ Jan 10 '19 at 21:45
  • 1
    $\begingroup$ @exodius By definition, in a X-dominated universe you ignore everything except X. The density of radiation would not be low in a radiation dominated universe. $\endgroup$
    – Javier
    Jan 10 '19 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.