• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,728 comments
1,470 users with positive rep
818 active unimported users
More ...

  Could the cosmological constant $\Lambda$ decay with time and still have $w=-1$?

+ 0 like - 0 dislike

One of the larger puzzles (coincidences) of cosmology is that the cosmological constant $\Lambda$ has roughly (within a factor of order one) the value determined by the (inverse square) radius of the present observable universe (about 3.3 times the Hubble radius).

At the same time, cosmological data states that dark energy obeys $w=-1$ with a high precision of only a few percent. This is usually interpreted as meaning that dark energy / the cosmological constant is really *constant* in time.

Could it be that $\Lambda$ decays with time (and therefore would *always* be given by the inverse square radius of the observable universe (or the Hubble radius, with the same factor of order one), thus avoiding the coincidence issue) and that $w=-1$ holds nevertheless?

asked May 23, 2020 in Astronomy by Gina [ revision history ]
edited May 25, 2020

Can you give a link to a source where it says that the cosmological constant Λ has roughly the value given by the radius of the present universe?
I have never heard that before ...

Lambda is roughly the inverse square of the radius of the universe. The fact is mentioned "practically everywhere"... You can also check the numbers yourself.

@Gina: I also would like to ask you to provide a source. What the radius of the present universe is in numbers I do not know. For a flat or open universe the Robertson-Walker coordinate \(r\) is unlimited (unless artificial boundaries are introduced), and therefore, as soon as the scale factor is positive, the universe is infinite. The radius of the observable universe is a different thing, of course.

Using a value for the Hubble parameter \(H=70{km\over sMpc}\)and \(0.7\) as the ratio of the contribution of \(\Lambda\) to the total energy density of the universe should give \({1\over\sqrt{\Lambda}}\approx3000Mpc\). There appear to be observations of objects at larger distances, e.g. https://en.wikipedia.org/wiki/List_of_the_most_distant_astronomical_objects

@Gina: You have edited your question, but still not provided a reference. Furthermore, I suggest you state the numbers you are working with. For example, based on the numbers in my previous comment, I find a Hubble radius of about 4300 Mpc, which is about 1.4 times \(1/\sqrt\Lambda\). Is this a coincidence, or a "larger puzzle"? Not necessarily.

In an empty space with cosmological constant you have for the scale factor \(a(t)\propto\exp(\sqrt{\Lambda/3}ct)\), giving \(H(t)=\dot a(t)/a(t)=c\sqrt{\Lambda/3}\), and a Hubble radius \(c/H= \sqrt{3/\Lambda}\)

You just showed that the Hubble radius is the inverse square of Lambda, within a factor of order one. That was the first point in the question. So let's go on to the next point, until we reach the question...

My above argument for empty space with cosmological constant is valid for all \(t\) and constant \(\Lambda\). The stated expression for \(a(t)\) is just a solution of the Friedmann equations for the case considered. So I fail to see a special coincidence or "puzzle".

Your question suggests a decaying \(\Lambda\) to avoid the "coincidence issue". Above I have shown that there is no coincidence issue for constant \(\Lambda\). Thus there is no need for a decaying \(\Lambda\), rendering your question void.

Note that you do not have to answer. You assumed that we live in a de Sitter space/universe. When we look at the sky at night we all have proof that we do not: we see stars. So your avoidance of the question "Can we have decaying Lambda and still w=-1?" is not correct. I repeat, you do not have to answer. Maybe somebody else can.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights