Talk:Cosmological constant

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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?

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[edit] What are the units of Λ?

Basic SI unit analysis does not support s2 as valid units for this equation.
The SI units are as follows:
G = m3 kg-1 s-2
ρvac = N m m-3
c2 = m2 s-2
When this is all cancelled out you get m-2 AH 21:27 28th August 2006 (GMT)


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)

Safe enough, of course, since relativists typically set c=1 and don't bother distinguishing between metres and seconds any more!

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)

okay so i might be wrong since i normally work in natural units until it comes time for data analysis, but shouldnt it be c^2 instead of c^4. Now i know it says energy density, but in my workive only seen it written in cgs units of g/cm^3; regardless of its radiation or matter or whatever(shouldnt matter). So even though we are talking about energy density its usually "spelled out" in these units. However i can see if you state the density to the equivalent of eV/m^3 you would need c^4. its just odd since i never see it written like that but maybe im in a weird nitch of cosmology research ?? its not important since c is just a factor but i thought i might point it out --Blckavnger 23:06, 27 November 2006 (UTC)

If it is being called an energy density then it must have c^4.

Sounds good to me ... its not important since its not engineering or something like that. Just like mass of electrons is always reported in terms of MeV even though its mass. It just looked wrong at first glance since ive never seen it to the 4th power in any paper or textbook ive used; of course ive just might have forgotten with all papers ive stumbled through in the past.
the fact that people still seem to want to change the power of c means we need to clarify the units. just cuz we are saying energy density versus mass density doesnt mean much, especially to people who work with relativity. (you could argue mass is more fundamental since its scalar where as energy is component of momentum 4 vector). whether it be to the 4th or 2nd doesnt really matter, in terms of the physics. we just need to be clear so we avoid back-and-forth edits.--Blckavnger 16:47, 6 December 2006 (UTC)
looks better but still a little confusing becuase the subscript should still be the same since its the density of the vaccuum, what changes is the units ... i just see a possiblity for confusion, id change it but im still a newbie to wiki. —The preceding unsigned comment was added by Blckavnger (talk • contribs) 17:04, 7 December 2006 (UTC).
I took care of it. — DAGwyn 23:37, 8 December 2006 (UTC)

[edit] 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] 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 Ωk = 0 --Joke137 20:35, 24 Feb 2005 (UTC)

[edit] Fine-tuning

I find it dishonest to speak about fine-tuning without qualifying the statements, but I don't know how to add this properly. I am watching "What we still don't know" documentary right now and they go out of their way to express how unlikely coincidence would that be to have the cosmological constant fine-tuned to 10^120th. But what we must note is that our understanding of this area of physics is still not final. It's not like we are already sure about the way expansion works. Even more important is that there may be no fine-tuning involved at all - we are just looking at the problem in the wrong way.

For example, I can claim that there is a remarkable finetuning of a certain variable for each couple - if that variable wasn't correctly set to 1 part in 10 million, there would be no conception or the mother would die. What that variable is? It's the number of dead sperms. If the difference between that number and the number of total produced sperms is more than, say, 10, the mother would explode, as its womb tries to produce 10 babies. :) If the difference is 0, no child will be born.

That BS, because the limiting factor to the number of babies conceived is the number of ova ready to be
fertilized, not the number of sperms. In fact there are millions of live sperms in each ejaculation, but since
only one sperm can fertilize each ovum, and usually only one (rarely two, and more than that usually only due
to hormonal treatment) ova are available in each period, only one (rarely two or more) sperm succeeds. No danger
of explosions here at all! 142.3.164.195 23:12, 6 March 2006 (UTC)

This is essentially the same as claiming that one term of the cosmological constant should be fine-tuned to differ from another term by a particular small amount. This is a wrong way of looking at the picture. For example, the number of people who are not Queen of the United Kingdom needs to be fine-tuned to the precision of 1 part in 6.4 billion. If that number was smaller than needed, there would be several Queens of the UK, which would lead to a significan confusion. If that number was bigger than needed, there would either be no Queen, which would cause chaos (at least in the British yellow press) or a negative number of them, which would be a huge violation of decorum and a scandal (which might cause chaos too, because of the coverage in the yellow press). Just like with the cosmological constant, I can say that a huge "Queen constant" is predicted by the world demographics and it needs to "be cancelled almost, but not exactly, by an equally large term of the opposite sign". Paranoid 8 July 2005 22:38 (UTC)


[edit] Decreasing for a trillion years?

There's some interesting news out today [2] that Paul Steinhardt at Princeton University, and Neil Turok at Cambridge University are hypothesizing the age of the universe to be about a trillion years (after it's been through a series of big bangs and crunches), allowing for the cosmological constant to decrease over time. Of course, I'm not sure how to work the theory into the article ;-) -- ke4roh 20:40, 5 May 2006 (UTC)

[edit] Whither the vacuum energy ?

When people talk about the vacuum energy (aka the Zero Point energy) they're referring to an effect and the formulae that come from understanding it. Or, rather, figuring out that it exists at all. I wonder if anyone has considered and discussed a few facts relevent to the topic ?

All "particles" have an associated field. These fields extend to the edges of the universe. All sorts of energy go into this field (a. Space is large. b. space is really large. c. all "fields" does not mean some "fields"). Interactions of "particles" that produce new "particles" do not create new fields, simply rearrange existing fields (else why would all "particles" decay into other "particles"?). Note: Part of this problem is people are still thinking about "particles", which don't really exist. Replace "particle" with "probability locus" and you're closer.

This makes up the energy of the "Virtual Field" Prof. Hawking mentions as where particles near black holes are produced as well as the source of what we perceive as gravity and inertia.

The "energy" we see in this field can't be drawn upon without some localized energy input (like a black hole in the process of swallowing matter, say) and then only what the gradient allows.

It seems to me that if energy can't be drawn from the field because it has no pressure behind it (think of this energy like a really large room with a few weak fountains of water in it and a floor covered with a shallow layer of water - this shallow layer of water is the energy referred to here), then there is nothing left over to power a bigger CC. It could explain the perception of so called "dark matter" and be responsible for the apparent "accelleration" of spatial expansion.

[edit] Static universe solution

The "General relativity" section of the article states:

Also, a static-universe solution to the original field equations was discovered, so it was not necessary after all to add the extra term in order to achieve such a solution.

Is this correct? What is the name of the solution, as I haven't heard it referred to elsewhere? --Christopher Thomas 23:06, 28 August 2006 (UTC)

Neither have I. In fact, I am sure no such thing exists, so I removed the sentence. I think I have run into this particular misconception before on Wikipedia, but I can't remember where... –Joke 03:07, 29 August 2006 (UTC)
It's not a misconception, but what Einstein himself stated. I restored the reference in a different way that ties in better to the rest of the text, along with the links to relevant existing articles. Don't be misled by the occasional modern formulation of the solution showing a cosmological constant, as the constant can be exactly zero and you still get an exact static solution (also expanding or contracting, depending on another parameter).DAGwyn 20:39, 29 August 2006 (UTC)
Hm, perhaps you mean a static universe with positive spatial curvature (which is also unstable)? Otherwise, there would either have to be matter with negative energy, or the universe would have to be empty. –Joke 20:45, 29 August 2006 (UTC)
Perhaps DAGwyn refers to Einstein's misconception that a static solution existed, but was later shown to be in error (i.e. it was shown to be unstable (by Friedman?)). --Michael C. Price talk 21:14, 29 August 2006 (UTC)

The text attributing Einstein's "biggest blunder" comment to Friedman's contribution is in error. The comment was in response to Hubble's observations of the cosmic expansion:

"When Albert Einstein applied his theory of general relativity to the universe the paradigm was that the universe was static. Since matter and energy gravitate, they drive the universe to collapse on itself. This was physically unacceptable, so Einstein introduced a cosmological constant term in his equations to balance the attractive force of gravity. It was latter discovered by Edwin Hubble that other galaxies appear to be moving away from us, that the universe was actually expanding. It was these observations that caused Einstein to claim that the inclusion of the cosmological constant was his biggest blunder, and was subsequently dropped from cosmological theories."[3]

--Michael C. Price talk 21:24, 29 August 2006 (UTC)

I don't know why you prefer guesswork from a secondary source over Einstein's own account. Anyway, there is a serious misconception in the article in general, whether or not it reflects consensus among current workers in this field, in confusing instability of the space-time cosmological model with an assumed inability of the model to describe a stationary state. The model embeds the time dimension and is perforce not able to change, so the instability doesn't cause the cosmology to do anything. If a solution exists, then it is what it is and describes what it describes. The changes I made were both historically and technically correct and should be restored since they actually explain something relevant, but if you prefer to spread misconceptions, have it your way. DAGwyn 20:48, 30 August 2006 (UTC)
To say GR is static because it embeds time as a dimension is a serious misuse of language and will mislead virtually all readers. I'd be interested to see a direct quote by Einstein where he clearly attributes his "biggest blunder" confession Friedman's input. --Michael C. Price talk 21:03, 30 August 2006 (UTC)
That's not at all what I said! I said that the particular cosmological model (be it Friedman's, deSitter's, or whatever) describes the whole time-evolution of the (model) universe in itself, and whether or not there are other "nearby" solutions to the field equations has no bearing on whether the particular "unstable" (more accurately, critical) solution is a possible universe. The statement often seen (also here and in Bernstein's biography of Einstein) that the least little disturbance would cause the universe to move away from the static solution is literally nonsensical, since that whole way of talking presumes yet another time axis unrelated to the one in the model universe.
As to the blunder/Friedman relationship, probably the most accessible source is the Appendix for the Second Edition of "The Meaning of Relativity", which was reproduced in all subsequent editions. In that particular exposition Einstein emphasizes the connection with nonzero average density of matter rather than with a static universe, because by then the Hubble effect had been discovered. (Much more could be said about that Appendix, but the point is that it definitely ties the removal of the cosmological term to Friedman's model.) — DAGwyn 04:43, 31 August 2006 (UTC)
This is nonsense. You say I have misrepresented your views on "static time" and then you go an repeat exactly what I complained about! Any serious cosmologist can simultaneously grasp the concept of time as a dimension AKA "static" time and talk of unstable cosmological models. --Michael C. Price talk 06:39, 31 August 2006 (UTC)
Excuse me, but I was working in this field since the 1960s, and did my graduate work on it. After I got my degrees, I left Physics for another profession largely because I saw that otherwise I would be continually battling this kind of lack of understanding. If you would make the effort to understand what I did say (with careful choice of words), rather than filtering it through a contrary preconception, then perhaps we could have an intelligent discussion of the matter. — DAGwyn 05:10, 1 September 2006 (UTC)
I note that whilst I am accused of not understanding your point you make no further effort to explain how I have misrepresented you. --Michael C. Price talk 05:40, 1 September 2006 (UTC)
Since I already stated it twice in different ways, a third time doesn't seem likely to help. The fellow in the office next to mine, who isn't a physicist, understood immediately what I was saying, so I don't think the problem lies in my explanation. Anyway, one more try using a simple-minded analogy: Suppress a spatial dimension or two and construct the model universe described by some solution of the GR field equations, and set in on the desk before you. Now try to apply the claim "given a slight perturbation, the universe would expand/contract/oscillate/whatever": what does that mean? All those verbs pertain to the time axis in the room containing the desk, not to the time axis marked on the model universe. Yet they are taken as if they did apply to the universe itself. That's just wrong! If the model sitting on the desk was properly constructed then it does represent a "possible" universe, whether or not other models could be constructed. Note that, being a universe, it already contains all phenomena that exist; there can be no external forces such as some total-energy minimization principle "acting on" the universe to "change" it.
There is a lot of really sloppy description and analysis going on in theoretical physics these days; I had to stop reading Phys Rev D because it was raising my blood pressure. — DAGwyn 07:40, 2 September 2006 (UTC)
Yeah, right. Following this logic we shouldn't say the universe is expanding, or indeed that anything happens at all. --Michael C. Price talk 09:44, 2 September 2006 (UTC)
I see you actually haven't followed the logic. There can be expanding/contracting/oscillating/static models for the (entire) evolution of a universe, and some of them are compatible with the GR field equations, indeed some even with the source-free equations. There are also some interesting non-GR based models, such as E. A. Milne's "kinematic relativity", which supports a static universe with a Hubble-like red shift (not due to expansion!). But correct analysis of such models requires considerably more care than many are willing to invest. — DAGwyn 05:42, 4 September 2006 (UTC)
I see once again you claim I'm wrong without actually addressing the issues raised. No matter, your views are original research by your own admission. That's all we really need to know to exclude them from Wikepedia. --Michael C. Price talk 09:14, 4 September 2006 (UTC)
Far from it; the cited work by Einstein himself completely supports the mention of Friedman that is currently in the article. (The comment by Pervect below supports that.) And I did address the stability issue, but you haven't bothered to understand the argument. Note anyway that that is an educational matter and this isn't the proper forum for it. Maybe you could cut and paste the whole discussion to an appropriate newsgroup and get somebody else to explain it to you. (Don't try to merely paraphrase it, since you can't do that properly when you don't understand it.) — DAGwyn 01:34, 5 September 2006 (UTC)
Also note the original statement to which I originally objected (I think this may have gotten lost in the argument):
"It is now thought that adding the cosmological constant to Einstein's equations does not lead to a universe at equilibrium because the equilibrium is unstable: if the universe expands slightly, then the expansion releases vacuum energy, which causes yet more expansion. Likewise, a universe which contracts slightly will continue contracting. These sorts of small contractions are inevitable, due to the uneven distribution of matter throughout the universe."
Einstein mentions both Friedmann and Hubble in that appendix, which is available online. I would suggest mentioning them both. While I suspect that Hubble was very much more influential than Friedmann, your source possibly justifies including Friedmann's name as well as Hubble's. I do not believe that your reference can be interpreted in such a manner as to leave Hubble out entirely, which seems to be your (DAGwyn)'s current position. Pervect 05:28, 31 August 2006 (UTC)
No, I never implied that Hubble was irrelevant, just that that wasn't the only reason why Einstein was happy to drop the cosmological constant. — DAGwyn 05:10, 1 September 2006 (UTC)
Do you have the online link to hand? --Michael C. Price talk 06:45, 31 August 2006 (UTC)
http://www.gutenberg.org/etext/5001. Bartleby.com also has a version, but it's missing the necessary appendix where this is discussed (appendix iv). Pervect 08:04, 31 August 2006 (UTC)

Thanks for the Gutenberg link. I believe the text justifies the sole inclusion of Hubble for the removal of the cosmological constant, since Einstein states he was originally strongly motivated to find a static solution with positive average density, which required the cosmological constant. Friedman, according to Einstein, demonstrated the existence of a non-static solution with positive average density, but which didn't require a cosmological constant -- but Einstein gives no indication that he accepted this as a model of the actual universe (which Einstein, along with consensus at the time, still believed to be static) until Hubble's data emerged. So it was Hubble who caused Einstein to reject the cosmological comstant, not Friedman. --Michael C. Price talk 09:29, 31 August 2006 (UTC)

Einstein's own presentation in the Appendix for the Second Edition heavily emphasizes Friedman's contribution. Certainly, the shift to an apparently expanding universe was instrumental in Einstein's change of mind about the need for a static universe, but if you at all understand his presentation you will see that it was the existence of a model (Friedman's), based on the original field equations, that supported a nonzero average mass density that was instrumental in convincing him that there was no need for the cosmological constant.
Having reviewed the source material, I'm prepared to grant that although Friedman's model does in fact have a flat (static) solution, as well as solutions for positive and negative curvature, it may not have been the static solution that Einstein really cared about by that time. I'll make another attempt at editing the article, that I hope you will be able to tolerate. As a general principle, additional relevant information should always be welcome. — DAGwyn 05:10, 1 September 2006 (UTC)
Another, far more plausible, interpretation of Einstein's comments is that it was Hubble's data that persuaded Einstein to drop the cosmological constant, after which Einstein was more inclined to look sympathetically upon Friedman's work, which he had previously regarded as an non-physical mathematical solution to the field equations. --Michael C. Price talk 05:55, 1 September 2006 (UTC)
That may be, although there is no way to know for sure at this point. Anyway, it is consistent with my most recent edit to the article. — DAGwyn 07:40, 2 September 2006 (UTC)

[edit] Steinhardt

Is Steinhardt's idea notable enough to be included here? With some effort, it should be possible find any number of ideas why the CC is that small, but none of them gathered a sizable followship so far. --Pjacobi 14:16, 24 November 2006 (UTC)

In "cosmological units" the constant is 1, which puts a different slant on the question. Eddington thought that it related the (inverse of the) radius of the universe to our laboratory units of length. — DAGwyn 23:41, 26 November 2006 (UTC)
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