Talk:Cosmological constant

I think what the author meant is that is unclear that current observations are due to a cosmological constant, or a cosmological "variable", i.e , not a constant term, is that so?

What are the units of Λ?

The article first says sec² and then later J⁴. AxelBoldt

The units are sec^-2. The article is correct now (I realize that the above question was asked a long time ago). Merenta 16:44, 16 Nov 2004 (UTC)

Actually the original units are m^-2.

Originally Λ, that was equal 4πGρ/c², where ρ was density of space in [kg/m³], was the curvature of space of stationary universe, or Λ=1/R², where R was so called "Einstein's radius of the universe" (in [m] and that's why Λ was in [m^-2]). However, since the nature of Λ is still a subject of debate, the units are irrelevant and everybody is free to express Λ in units supplied by his/her pet theory. People who don't believe Einstein's theory of gravitation even interpret it as "repulsive gravitational force" (and then they probably express it in [kgm/s²]). This interpretation is not possible in Einstein's gravitation since there are no gravitational forces acting at distance (repulsive or otherwise) and the gravitation is just an expression of geometry of spacetime. But if one likes ρ to be density of energy rather than density of mass then of course the units are s^-2. Of course it requires dividing Λ by c² and the rest of Einstein's equations is nicely recovered. Jim 20:41, 24 Nov 2004 (UTC)

Pushing around factors of the speed of light and Planck's constant the most popular units for the cosmological constant seem to be Planck units, GeV⁴ and g/cm³. --Joke137 01:20, 5 Feb 2005 (UTC)

Inaccuracy in article?

Because the cosmological constant has negative pressure, according to general relativity a positive cosmological constant--which means empty space has positive energy--causes the expansion (or contraction) of empty space to accelerate.

(Emphasis mine.) I don't know enough about the topic to remove the bolded part, nor can I find good enough material online (grumble). But from my understanding, isn't a positive CC expansion and a negative CC contraction? -- Wisq 16:57, 2005 Feb 23 (UTC)

It's alright, but perhaps not totally transparent:

• positive cosmologicla constant <=> positive energy <=> negative pressure

Pjacobi 01:00, 2005 Feb 24 (UTC)

Okay, but my question is, could a positive cosmological constant equal contraction? It seemed to me that positive constant = negative vacuum (expansion), while negative constant = positive vacuum (contraction)... if so, then saying "positive constant means expansion (or contraction)" would be inaccurate. -- Wisq 03:32, 2005 Feb 24 (UTC)

postive constant gives an extra negative pressure term, but if the normal and dark matter outweight that term, the universe would still collapse (it would never reach the regime of exponential expansion, characteristci for positive c.c. (this is thought to be ruled out by current observations)

Pjacobi 10:01, 2005 Feb 24 (UTC)

I think I put that in. It is both inaccurate and confusing. A universe with a c.c. can either be expanding or contracting (although it won't switch from one to the other). Both the expansion and contraction are exponential, but it doesn't really make sense to say that the contraction is "accelerating." --Joke137 16:20, 24 Feb 2005 (UTC)

What you are saying applies to universe without any matter. A matter with a positive cosmological constant small compared to the matter content, can expand and than contract. --Pjacobi 19:03, 2005 Feb 24 (UTC)

No, because the matter will red-shift away and the c.c. will eventually dominate. That is what is happening in our own universe. --Joke137 19:22, 24 Feb 2005 (UTC)

Just add enought matter, eventuell the solution will show recollapse before the c.c. has a chance to dominate, see [1] (http://relativity.livingreviews.org/Articles/lrr-2001-1/node7.html) for some calculations. Of course, as said above, this setting is pretty musch ruled out for our universe. --Pjacobi 20:02, 2005 Feb 24 (UTC)

Oh, if you allow curvature I agree. I was assuming <math>\Omega_k=0<math> --Joke137 20:35, 24 Feb 2005 (UTC)