I Age of the Universe and the Cosmological Constant

[Ranku]

The presence of the cosmological constant means that the universe is older than it otherwise would have been if it were a purely matter dominated universe. The physical rational is that "a cosmological-constant universe is older because it took longer to reach its present rate of expansion". I am having a hard time understanding quite what that means. Since the age of the universe is the inverse of the Hubble constant, as in 1/H = D/V, so one would expect a higher age would require a lower value of V. However, the cosmological constant accelerates the expansion of the universe, which would lead to a higher V than would be the case if the universe were expanding purely gravitationally deceleratingly. So, what is the missing piece here that leads to a higher age with higher V?

 

[PeterDonis]

the age of the universe is the inverse of the Hubble constant,

No, it isn't.

 

[Ibix]

Since the age of the universe is the inverse of the Hubble constant

In a $\Lambda=0$ universe, $1/H$ is the age of the universe. Not in general.

I suppose one way to think of it is if you think of two galaxies on opposite sides of us at the same redshift, if you calculate backwards assuming they had constant recession velocity then they'd be coincident with us at some time in the past. But if their recession velocity was lower in the past they'd need more time to get from coincident with us to where they are now. So factoring in dark energy the singularity is longer ago than if you assume constant recession speed.

 

[PeterDonis]

In a $\Lambda=0$ universe, $1/H$ is the age of the universe.

I'm not sure even that is true. I think it's only true in the edge case of the Milne universe.

 

[javisot]

The age of the universe was calculated with Friedmann's equations involving the scale factor?, or do I remember wrong?

 

[PeterDonis]

The age of the universe was calculated with Friedmann's equations involving the scale factor?, or do I remember wrong?

The actual age of the universe is calculated using our best current model of the time dependence of the scale factor. This model involves Friedmann's equations, yes, but with multiple functions for the density and pressure (one each for radiation, matter, and dark energy--our best current model is spatially flat so there is no term for spatial curvature in it, although one does see that term in more general presentations), and those functions have different time dependences. So the behavior of the scale factor over time is not just one simple function.

Note that this model is decelerating (radiation dominated until a few hundred thousand years after the Big Bang, then matter dominated) until a few billion years ago, and then accelerating (dark energy dominated) after that. But it still gives a longer age of the universe than a purely matter dominated model with the same Hubble constant at the current epoch would give.

 

[Ranku]

In a $\Lambda=0$ universe, $1/H$ is the age of the universe. Not in general.

I suppose one way to think of it is if you think of two galaxies on opposite sides of us at the same redshift, if you calculate backwards assuming they had constant recession velocity then they'd be coincident with us at some time in the past. But if their recession velocity was lower in the past they'd need more time to get from coincident with us to where they are now. So factoring in dark energy the singularity is longer ago than if you assume constant recession speed.

So does this logic also apply to a universe where the recession velocity was less than constant, such as in a gravitationally decelerating expanding universe?

 

[PeterDonis]

does this logic also apply to a universe where the recession velocity was less than constant, such as in a gravitationally decelerating expanding universe?

The opposite would apply: the singularity would be a shorter time ago than if you assume constant recession velocity.

 

[Ranku]

The opposite would apply: the singularity would be a shorter time ago than if you assume constant recession velocity.

So when you add the cosmological constant, to a universe that was expanding deceleratingly, the singularity would be even a longer time ago, compared to a universe that was expanding with uniform velocity.

 

[PeterDonis]

when you add the cosmological constant, to a universe that was expanding deceleratingly

You don't do that. You're not taking a universe with matter in it and adding a cosmological constant. You're comparing different models, one of which has a cosmological constant and one of which doesn't. (Actually, if you include the "constant recession velocity" model, you have two which don't, one matter dominated--the decelerating one--and the other of which has nothing in it at all--the "constant recession velocity" one.)

 

[Ranku]

You don't do that. You're not taking a universe with matter in it and adding a cosmological constant. You're comparing different models, one of which has a cosmological constant and one of which doesn't. (Actually, if you include the "constant recession velocity" model, you have two which don't, one matter dominated--the decelerating one--and the other of which has nothing in it at all--the "constant recession velocity" one.)

In the Friedman equations with the cosmological constant, doesn’t the energy density rho include the energy density of the cosmological constant?

 

[PeterDonis]

In the Friedman equations with the cosmological constant, doesn’t the energy density rho include the energy density of the cosmological constant?

With the cosmological constant, yes. But that doesn't mean you can take a model without a cosmological constant, and ask what happens if you add one. That's a meaningless question. The only meaningful question you can ask is about a comparison between a model that has a cosmological constant and a model that doesn't, and how well each model explains the data we have.

 

[Jaime Rudas]

However, the cosmological constant accelerates the expansion of the universe, which would lead to a higher V than would be the case if the universe were expanding purely gravitationally deceleratingly. So, what is the missing piece here that leads to a higher age with higher V?

Perhaps what you're not taking into account is that the parameter we know (because we've measured it) is the current expansion rate. For a constantly accelerating universe, all previous rates are less than the current one, while for a decelerating one, the opposite is true.
(To avoid confusion, I clarify that the most widely accepted current model doesn't fit either of the two described above.)

 

[Ranku]

Anyone wants to put the standard formula for calculating the age of the universe using the Friedman equations - l am seeing different versions of it strewn around.

 

[Orodruin]

In a $\Lambda=0$ universe, $1/H$ is the age of the universe. Not in general.

It is not. Assuming one component dominates the entire expansion history it is a numerical factor multiplied by 1/H. Iirc, this factor is 2/3 for a matter dominated universe, but don’t quote me in that …

 

[Orodruin]

Anyone wants to put the standard formula for calculating the age of the universe using the Friedman equations - l am seeing different versions of it strewn around.

There is no standard formula except if you assume domination by one component. Generally you have to solve the Friedman equations to find out.

 

[Jaime Rudas]

Anyone wants to put the standard formula for calculating the age of the universe using the Friedman equations - l am seeing different versions of it strewn around.

The current age of the universe $T$ can be calculated using the following formula:
$$T = \frac c {H_0} \int_0^1 {\frac {a \hspace{2mm} da}{\sqrt{\Omega_{r_0} + \Omega_{m_0} a + \Omega_{k_0} a^2 + \Omega_{Λ_0} a^4}}}$$

 

[Orodruin]

The current age of the universe $T$ can be calculated using the following formula:
$$T = \frac c {H_0} \int_0^1 {\frac {a \hspace{2mm} da}{\sqrt{\Omega_{r_0} + \Omega_{m_0} a + \Omega_{k_0} a^2 + \Omega_{Λ_0} a^4}}}$$

To connect to what I said above, I meant that there is no algebraic closed form expression for the general case. The integral here is in essence the solution to and results from integrating the Friedman equations. The general case has no closed form solution, but of course the integral can be solved directly if only one $\Omega$ is non-zero.

For example, if $\Omega_{m_0} = 1$ we obtain
$$
H_0 T = \int_0^1 \sqrt a \, da = \frac 23
$$ as mentioned earier for a matter dominated universe.