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

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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
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.

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