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I "Dark energy" may be nothing more than baseline curvature

  1. Feb 22, 2016 #1

    marcus

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    I know of no scientific reason to suppose that "dark energy" is anything more than the cosmological curvature constant identified by Einstein in 1917 as occurring naturally in the GR equation for spacetime curvature.

    It might eventually turn out to be related to some type of energy. That's possible. But so far there seems to be no reason to imagine that it is is connected with anything we'd normally consider an energy---it is simply the universe's baseline curvature in the absence of matter.

    The phrase "dark energy" tends to get newcomers confused because they try to understand something much simpler (a small pervasive constant baseline curvature) in terms of energy. The phrase should probably be replaced by something less misleading.
     
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  3. Feb 22, 2016 #2

    mfb

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    It is purely a matter of interpretation. A constant dark energy density and an explicit cosmological constant lead to exactly the same physics.
    If the "constant" is observed to vary with time or space in the future, then a cosmological constant does not work any more.

    A different name might have been better, in particular to avoid confusion with dark matter. But we cannot change it here.
     
  4. Feb 22, 2016 #3

    fresh_42

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    Wouldn't that still leave the possibility that the cosmological constant is a cosmological function of time?
    I like to ask how "dark energy" contributes to the overall balance. I usually read of about 0.75. Is such a notation compatible with the interpretation of curvature which I also like the best since it seems somehow natural? Wouldn't this apply an energy to a geometric object?
     
  5. Feb 22, 2016 #4

    Haelfix

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    You could in principle play the same game with a time dependent dark energy term. Normally something like quintessence is interpreted as living on the right hand side of the field equations as the varying potential of a scalar field, but nothing stops you from bringing it over to the left hand side and interpreting it as having to do with geometry.

    The dynamics are the same, provided Einstein's equations hold exactly, its just a matter of terminology*.

    *Bad terminology incidentally, in cosmology the word 'curvature constants' means something very different.
     
  6. Feb 22, 2016 #5

    mfb

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    Well, we wouldn't call it constant then I guess. On the other hand, the Hubble "constant" is not constant in time either.
    The addition to approximately 1 needs an interpretation as energy density or something similar.
     
  7. Feb 23, 2016 #6

    timmdeeg

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    Two questions:

    Do you say that the cosmological constant / dark energy is not vacuum energy? And what would this mean regarding the Friedmann equation?

    What means "universe's baseline curvature"?
     
  8. Feb 23, 2016 #7

    marcus

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    Those questions are pretty well answered in the 2010 article "Why all these prejudices against a constant?" Google the title.

    You just write the Friedmann equation with a constant spacetime curvature on the lefthand side (like a constant of integration). It does not involve any matter term on the right hand.

    There is no successful calculation correctly predicting a corresponding "vacuum energy". If you want, call the cosmo constant "vacuum curvature" and forget about energy.

    The cosmo curvature constant, as a curvature is unavoidable. Einstein introduced it with the Lambda notation, on the lefthand side of GR where it appears naturally as any term would that is allowed by the symmetries of the theory. It had not been proven to be zero so he put it in. Bianchi and Rovelli explain this adequately, I think.
    http://arxiv.org/abs/1002.3966
    http://inspirehep.net/record/846447?ln=en
    https://www.google.com/?gws_rd=ssl#q=Why+all+these+prejudices+against+a+constant
     
  9. Feb 23, 2016 #8

    timmdeeg

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    Ok, thanks. So, either one has to postulate vacuum energy connected with some understanding of how it curves spacetime or one postulates an ad hoc existing "vacuum curvature" without assuming the cause. Would that be correct?
     
  10. Feb 23, 2016 #9

    marcus

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    You couldn't do better than read the Bianchi Rovelli paper and paraphrase that in summary, rather than paraphrasing what I said. But yeah. If spacetime geometry exists, i.e. if spacetime exists, then it has some inherent curvature. No reason that should be zero. It has to have some inherent intrinsic baseline curvature, if it exists. So you don't have to make up anything. It is forced on you.

    If you want to speculate that it "comes from" some kind of "energy", then you have to
    1. postulate that the baseline inherent curvature is ZERO and then
    2. you have to make up some "vacuum energy" that bends the spacetime just the right observed amount.

    Nothing like that has been calculated starting from an accepted theory of qg. One would not want to start from MINKOWSKI space the way e.g. QFT people do.
    Any "vacuum energy" you calculate based on Minkowski space (not quantum, i.e. not realistic geometry) is just silly. And you can see it gives a silly answer many OOM off the mark.

    Simplest thing is just to not make anything up. Spacetime behaves as if it has a basic baseline curvature prior to anything else affecting it. So accept that.
    It has been measured, it seems constant at a definite value. Like Planck's constant.
     
    Last edited: Feb 23, 2016
  11. Feb 23, 2016 #10
    The notion of curvature of nothing is based upon Einstein's formulation of space as static. Using Friedmann's equation rho = [q][3H^2/4piG)] and taking q per de Sitter [c^2/R], the dynamic solution for a zero energy universe [(rho)(c^2)/3 = -P] leads to the same result as lambda = G as originally proposed by Einstein. No dark energy needed ....because for the zero energy universe the solution exponential?
     
  12. Feb 23, 2016 #11

    marcus

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    Hi Yogi, I'll propose a Friedmann equation context you can use or not as you like. In units with c=1, spacetime curvature has units either inverse length squared (inverse area) or inverse TIME squared.
    The latter is convenient and it means that Einstein's cosmological curvature constant Λ is the square of a reciprocal time. Same units as H^2 in the Friedmann.

    So we can write

    H2(t) - Λ/3 = [some constant] ρ(t)

    where ρ rho is the energy density including all usual types of matter and energy, so that is how the Hubble rate H(t) evolves.

    Eventually ρ → 0 in an expanding universe, as things thin out. So Λ/3 is the limit of the square of H(t) as t→∞

    So you can define H as the limit that H(t) approaches. H2 = Λ/3

    H2(t) - H2 = [some constant] ρ(t)

    This gives the cosmo constant some real tangible meaning. Expressed as H, it turns out that H is about
    1/173 percent per million years. that is the asymptotic expansion rate---the present rate H being 1/144 percent per million years.
     
    Last edited: Feb 23, 2016
  13. Feb 23, 2016 #12

    marcus

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    The constant in the Friedmann equation (that I called "some constant" earlier) is either 8πG/3, if you are using c=1 units,

    or if you like to include c it is that same thing divided by c2, namely 8πG/3c2

    This is the spatially flat case of the Friedmann, which is what mostly gets used since the observed spatial curvature is so close to zero.
    I hope you like this context for working with the Friedmann equation. It's fairly clean and easy to use. The critical energy density does not involve any "dark energy" component of course. But it includes dark matter---there is pretty good scientific evidence that exists.

    The Bianchi Rovelli 2010 paper is really worth studying.
    http://arxiv.org/abs/1002.3966
    http://inspirehep.net/record/846447?ln=en
    https://www.google.com/?gws_rd=ssl#q=Why+all+these+prejudices+against+a+constant
    I see Google Scholar says it has been cited 65 times, listed here. Included are some citations in books but most are in research publications
    https://scholar.google.com/scholar?...2&um=1&ie=UTF-8&lr&cites=10452416241992441383
     
  14. Feb 24, 2016 #13

    timmdeeg

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    I've read the Bianchi Rovelli paper and if I understand them correctly they say if the cosmological constant was a "great mystery" then all natural constants are a great mystery. Hard to oppose. And yes, the prediction of QFT is in deed far off. Hopefully I got your (interesting) point.
     
  15. Feb 24, 2016 #14

    Haelfix

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    Classically, the cosmological constant is just that.. A constant of integration, and indeed it can be anything you want as there is no measure over the real numbers (1.59 is just as likely or unlikely as 1e-120 in natural units). It matters not one wit if you put it on the left hand side or the right hand side, that's just a matter of how you want to group things and it's perfectly well understood and trivial at this level.

    The cosmological constant problem is a problem about the quantum field side of this thing, really the effective field theory of gravity at large distances. Note that Rovelli's paper is completely specious on this point, b/c it doesn't even try to address the fundamental problem. And the fundamental problem is that the cosmological constant in the quantum regime apparently takes contributions from ALL physics, everything we know about from the Hubble scale all the way down to the Planck scale. Forget about the Planck scale, the problem is already extreme at the level of things we understand very well for instance the physics from the Hubble scale to the physics of quantum electrodynamics. Already there you have a buildup of quantum field theoretical modes from objects like electrons that we understand well, and clearly something is very wrong with the calculations (more on this in a bit).

    Viewed in this light, it cannot and should not be seen as 'just a constant', anymore than the hierarchy problem in particle physics can be seen as just 1 finetuned number. Technically what's going on, is that the cosmological constant is radiatively unstable. So for instance, you might take the point of view of Rovelli and argue that there is identically zero contribution from quantum mechanics (and instead have it just remain perfectly classical). But first problem, you have to show this explicitly, you can't just wave your hand (which is completely circular logic). Second problem, this is backwards (one derives classical mechanics from quantum mechanics, not the other way around) and third problem, if you adjust the physics at one scale to cancel the quantum contributions, you now have to figure out a way or a reason to adjust the same physics to the next scale. This would be like if you had an unexplained problem with the trajectory of baseballs being hit on Alpha Centauri that you wanted to explain, but found out that the answer depended very sensitively on some tiny detail about the unknown physics of the quarks within the ball (except some 15 orders of magnitude worse than that). That sort of scale sensitivity almost never happens in physics absent some symmetry that would force it upon us, so when we see a problem that requires it, it seems unnatural to us.

    Now you could just say that this is a calculation error of humans. But this too is wrong, or at least its completely not obvious exactly whats going wrong. You see, the quantum vacuum does produce very real tangible effects. Effects that we have calculated already to quite exquisite precision. For instance in QED the famous Lamb shift is known to 9 or 10 decimal places, and is indeed one of the most accurate and successful predictions ever computed in the history of physics, we know this with much more precision than anything to do with gravity. Now, this same effect, 'gravitates' in atoms b/c it contributes to the total mass and precision tests of the equivalence principle shows that indeed it does (with some ridiculously small error bars). Therefore it naively seems like the effect of quantum mechanics is not identically zero, that they do have some contribution to the value of the CC, at least within the vicinity of atoms.

    This of course makes the problem tangibly worse, bc now instead of figuring out a way to make the contributions identically zero for all quantum mechanical processes (like Rovelli would like) we have to figure out a way to make the sum of all such processes (through all scales) ridiculously tiny. This sum of unknown quantities will look schematically something like this 2 - 5 + .4 + .003 - .00002 + .000000000985 + ... = 1 e -120. In other word each number in the term (corresponding to contributions arising from extremely different physical phenomena, many of which are unknown) somehow mysteriously cancels each other number to some incredible accuracy. Why this is the definition of the (new) cosmological constant problem.
     
  16. Feb 25, 2016 #15

    timmdeeg

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    Thanks, very interesting post!
    According to R.L. Jaffe "Casimir forces can be computed without taking reference to zero point energies." So, from this one couldn't argue that quantum fluctuations do contribute to the CC. On the contrary the Lamb shift does. And not only that, I wasn't aware of the conclusion that it "gravitates". Very interesting point, which seems to propose that there should be somehow a contribution to the CC. Is there any idea from researchers why that is so tiny?
     
  17. Feb 25, 2016 #16
    Except Occam's razor.

    QFT says that vacuum must have some intrinsic energy.

    You are right saying that currently we can't calculate it (calculations are divergent, and even with an energy cutoff they give insane huge values), but it exists, and hopefully one day we will be able to calculate it correctly. It's rather unlikely to turn out to be zero.

    IOW: we don't make up some "vacuum energy". Our theories say it's there.
     
  18. Feb 25, 2016 #17
    And many other things do not depend on zero point energy - as long as you are using Special Relativity framework. But when you use General Relativity...

    ...things change. GR's stress-energy tensor places energy into its T00 element, not its difference from "zero point". Thus, if zero point energy is nonzero, then stress-energy tensor of empty space is nonzero too. It gravitates.
     
  19. Feb 25, 2016 #18

    timmdeeg

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    I think if the CC acts like a vacuum energy density, it would exert negative pressure, represented by the components ##T##ii.
     
  20. Feb 25, 2016 #19

    Haelfix

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    Right, so there are definitely subtleties regarding what is or is not vacuum energy, or really what you call it. This is why people frequently pick a weakly coupled, Abelian theory to make the point where we have, to a great approximation, the ability to talk about things like vacuum polarization, mass renormalization and the nontrivial structure of the zero point energy of electrons and where we don't get into too many messy technicalities of quantum field theory. For instance, the case is in principle much stronger for QCD. The mass of the quarks is only a small fraction of the mass of the nucleus, therefore almost all of the mass of things we see around us arises from quantum effects eg the gluon kinetic and potential energies that are constantly fluctuating in the QCD vacuum. Clearly we don't need to worry about violating the equivalence principle in that case, but there of course the analysis is less clean b/c of the subtleties that Jaffe mentions and for other reasons (confinement etc). In any event, equivalence principle tests in the case of platinum and Aluminum atoms shows that the Lamb effect (in those atoms) satisfies the equivalence principle to one part in 10 ^6, which shows that at the very least something is going on.

    Anyway whatever 'it' is, clearly gravitates, and so part of a solution (if you want to explain it away) would be to precisely explain why those effective field theory arguments are wrong (which amounts to essentially rewriting large parts of quantum field theory and/or General relativity). Or if they are right, how do these effects cancel with each other to such fantastic accuracy? So you see, it is a very big and difficult problem no matter how you slice it.

    As for potential ideas for why it is tiny.. Again, many many ideas. All of which have many, many problems. There are many review articles out there, for instance (where you can see the arguments I gave flushed out in a little more detail):
    http://arxiv.org/abs/hep-th/0603249
    http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.61.1
    or a recent one
    http://arxiv.org/abs/1502.05296
     
  21. Feb 25, 2016 #20

    Haelfix

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    The CC term enters Einsteins equation with the metric tensor Guv. So you have something like Rhovac Guv (where I am forgetting factors of 8pi and G), so it will enter in along the diagonal matrix elements of the stress energy tensor.
     
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