# Is the cosmological constant constant?

1. Mar 31, 2010

### JollyOlly

In a flat, matter-dominated universe the cosmological constant is zero and the scale parameter a increases according to a two-thirds power law. In such a universe, the expansion of the universe gradually slows down (but never stops).

In recent years it has been discovered (?) that the rate of expansion of the universe is not slowing down but it is either constant or even increasing slightly. In order to account for this cosmologists have reintroduced the cosmological constant in the guise of dark energy which almost exactly counteracts the retarding effect of gravity.

In other words, the scale parameter is proportional to time. ie $$a \propto t$$. But $$a = a0 t$$ is not a solution to the Friedmann equations.

It follows that if the universe has expanded linearly since the Big Bang, the cosmological constant must have changed during that period. Alternatively, $$\Lambda$$ is constant but we just happen to live at the time when the total mass-energy density $$\Omega = 1$$

I have never liked the introduction of $$\Lambda$$. And I find the extraordinary coincidence that we happen to live at the exact time in the history of the universe when $$\Omega$$ = 1 too much to stomach.

What is going on?

2. Mar 31, 2010

### nicksauce

How is this scale parameter proportional to time?? This does not follow from what you said earlier, nor is it true at all. It is perfectly possible to solve the Friedmann equations with a constant $\Lambda$, though the solution won't be all that pretty.

3. Mar 31, 2010

### Ich

Didn't you look at the diagram I attached https://www.physicsforums.com/showthread.php?p=2648388#post2648388"?
a=a0 t is a solution to the Friedmann equations, e.g. for an empty universe. But it's not the LCDM solution.

Yeah, but it didn't.

Last edited by a moderator: Apr 24, 2017
4. Mar 31, 2010

### bapowell

The evolution of the scale factor for an empty universe is different from its evolution in a universe filled with $$\Lambda$$. As Ich says, an empty universe gives $$a \propto t$$. For the cosmological constant, one find $$a\propto e^{Ht}$$, where H is the Hubble parameter. nicksauce, you don't find that solution pretty?

5. Mar 31, 2010

### JollyOlly

Sorry to start a new thread, Ich. I thought the question was unrelated but I now see that it is exactly the same! Also I wanted to tempt some new people into the discussion.

If you look at the three lines on the image which Ich provided, you will see that The LCDM model cuts the axis at almost exactly the same point as the linear model. In other words, the gradient of the LCDM line happens to be exactly the same as the linear one at the moment.

A better set of curves is to be found http://www.astro.ucla.edu/~wright/cosmo_03.htm" [Broken] All the lines are drawn with the same gradient where they cross. The LCDM line then, by an extraordinary coincidence cuts the axis at the same point as the linear one.

Does anyone else find this to be an unacceptable coincidence?

Can anyone point me in the direction of some really obvious evidence that the cosmological constant is not zero? It is obvious from the three curves that in order to distinguish between the two models, you have got to look at objects which are at least 5 Gly away from us but although we can see such objects, measuring anything other than their red-shift accurately is an almost impossible task.

Last edited by a moderator: May 4, 2017
6. Mar 31, 2010

### JollyOlly

This is only true in an empty universe. In one with both matter and dark energy the analytical solution is not pretty (though the curve is!)

7. Mar 31, 2010

### bapowell

Yup. It's called the coincidence problem: http://dsc.discovery.com/news/2008/10/22/dark-energy-space.html [Broken].

It's not clear whether a cosmological constant is needed to satisfy cosmological observations. However, several complementary observations have pointed towards an accelerating universe: redshifts of type Ia supernovae as you mention, but also others like the Integrated Sachs-Wolfe effect in the cosmic microwave background data. The $$\Lambda$$CDM cosmology (here $$\Lambda$$ might not be a constant, but could be time varying -- we don't know yet...). It's also possible that the accelerated expansion is not due to some form of dark energy, but might be a result of modified gravity or even so-called back reaction of primordial perturbations (although the latter has been debated for some time).

So, the cosmological constant might be zero, however, the universe we observe today looks very much like it's accelerating, and several independent measurements suggest this.

Last edited by a moderator: May 4, 2017
8. Mar 31, 2010

### bapowell

As I stated, that is true in universe filled with a cosmological constant only. I would be careful with the word 'empty' -- as we know from inflation, a universe with vacuum energy is far from empty!

9. Mar 31, 2010

### Naty1

10. Mar 31, 2010

### nicksauce

There are three main ways, that I know of.

1) Type 1A Supernovae surveys allow us to measure redshifts and luminosity distances for distant objects. The best fit finds dark energy is needed. http://arxiv.org/abs/astro-ph/9805201

2) Fits to WMAP CMB data give a cosmological constant, consistent with that of method 1). http://arxiv.org/abs/1001.4744

3) The late time integrated Sachs-Wolfe effect. The idea of the Sachs-Wolfe effect is that a photon's energy changes as it moves through a gravitational potential. In a matter dominated universe, this effect is independent redshift, while if there is a cosmological constant it is redshift dependent. Recently, this has been measured, and dark energy is found to be present. http://iopscience.iop.org/1538-4357/683/2/L99/

In the future, it is hoped that mapping 21cm neutral hydrogen emission will give a good handle on dark energy. This is probably 5-10 years away.

11. Mar 31, 2010

### JollyOlly

Thanks for the reference to Rovelli's paper Naty1 - I found it most interesting but the coincidence problem he discusses refers to an 'order of magnitude' coincidence between $$\Omega$$m and $$\Omega$$$$\Lambda$$. (can anyone tell me how to do subscripts in tex?)To put it in my terms, why are we at the 'turning point on the LCDM curve where a slowing down becomes an acceleration - at least within an order of magnitude?'

My coincidence is a lot stronger. Of all the possible LCDM curves between $$\Lambda = 0$$ and $$\Lambda = \infty$$, why are we living in the one which predicts an age of the universe - 13.7 Gyr which is almost exactly equal to the one predicted by a linear expansion model 13.8 Gyr? This is not just within an order of magnitude.

12. Mar 31, 2010

### Ich

I don't think so. If we're somwhere in the early times of a universe, T0*H would always be of order unity. A "near miss" is not so unexpected, then.
So it comes back to "why have we just now (i.e. some 6 billion years ago) left the early stages?"