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Dark energy, vacuum, and uncertainty

  1. Dec 1, 2008 #1
    Well, the title says the most, but the question is: If Virtual particles exist... Could it be dark energy?, if it really can, would it be the same idea of a Cosmological Constant??

    Last edited: Dec 1, 2008
  2. jcsd
  3. Dec 1, 2008 #2


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    From the standpoint of particle physics this is the natural thing to suppose.

    And this was the standard idea floating around since the 1990s, to explain the Cosmological Constant. So you are in good company.

    The basic tool of particle physics is Quantum Field Theory and using QFT you can calculate what the intrinsic energy of the vacuum ought to be (given that it is always seething with a vast swarm of virtual particles)

    So when, in 1998, the supernova evidence of accelerated expansion was reported, the particle physicists said Ahah! This could be the effect of the Vacuum Energy arising from virtual particles!

    The only hitch is that if you measure the acceleration, by astronomical observtion, you get a very small value for the dark energy----about 0.62 joules per cubic kilometer. Because the observed acceleration is very slight.

    And when the particle physicists calculate what the Vacuum Energy should be according to QFT, they get a figure which is many many powers of ten bigger than that.

    So we have to accept that the question of what is the cosmological constant and dark energy remains unanswered so far.

    What you have proposed is a common speculation which so far has not checked out.
  4. Dec 1, 2008 #3

    Does it mean we don't really know how to calculate it (obvius), but what I mean is that if as long as we keep observing we we'll reach an answer.
    I've also heard that cosmologists are trying to find the state equation and i think a professor told me it could be someghing like [tex]-p[/tex] if we accorded light speed was 1 or something like that but I don't really remember, sorry.
    Could somebody explain this please?

  5. Dec 1, 2008 #4


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    the equation of state (EOS) coefficient of dark energy is most often assumed to be w=-1

    that sounds like confusing jargon, maybe. It is actually very simple.

    the whole of cosmology runs on two very short simple equations called Friedmann-equations which govern the evolution of the scalefactor a(t), a measure of universe size. If a(t) increases the U expands. If a(t) increases faster, that means the expansion accelerates.
    If you know introductory calculus, acceleration here just means a'' (t) is positive, the second derivative is positive.

    Well the Friedmann-equation that governs a"(t) is a short simple formula involving both the energy density and the pressure--of the matter/energy---basically all the stuff--in the universe.
    The conventional symbol for the energy density is rho and the symbol for pressure is p and the EOS connecting them is
    p = w rho

    So if w = -1
    p = - rho

    The reason this is a big help is that if you can write the pressure as - rho
    then you simplify the equation even further. Because instead of having two terms to worry about you now just have one term, the rho term.

    You can learn about the two basic equations at Wiki if you want
    but if you aren't looking for details it is pretty simple.

    The first equation shows that the dark energy density (the rhoDE) has to be around 0.62 nanojoules per cubic meter (to make the model fit data)
    and if the EOS coefficient w really is equal to -1
    then that shows the dark energy pressure has to be
    around -0.62 nanopascal
    (you probably know the Pascal, it is a metric unit of pressure that goes with joule).

    The reason your teacher would be talking about putting c = 1 is because it is convenient to convert back and forth between mass density and energy density, by E = mc2 because some of the stuff is kilograms per cubic meter and some is joules per cubic meter and it's a pain in the butt to have all these different units. So they tend to use units that are adapted to the problem, to make it simpler.

    I may be getting you in trouble with your teacher by telling you the dark energy density is 0.62 nanojoules per cubic meter but that's actually what it's estimated to be, namely 73% of the total (also the critical) energy density which is 0.85 nanojoules per cubic meter.
    I just like to use ordinary metric units for this kind of stuff, because everybody knows them and you can waste a lot of time just handshaking with the other guy and finding out what his units are, which why bother if we can avoid it.

    If you are at all interested you should really look at
    and there you will see that the second equation is the one that determines the acceleration a''(t)
    actually it tells you the ratio a"(t)/a(t)
    and that is the equation that involves the PRESSURE, you can see the little p there as well as the rho. And the only way that the acceleration a" can ever be positive is if the pressure is negative. You can see that easily from the equation, just pay attention to the minus sign. So right there that is the key to why dark energy has to have a negative pressure, in order to work right.

    So have a look at that Wikipedia page and get some more questions and come back, if you want.
    Last edited: Dec 1, 2008
  6. Dec 2, 2008 #5
    Hi again!
    I'm sorry but my maths aren't still good enough (this is my last year at High School) and I just know a little bit about derivatives, I understand some of the terms but it'd be great if you could explain me some things:

    1. Always we talk about [tex]a(t)[/tex] we're talking about the scale factor, or sometimes it means acceleration?

    2. When we derivate this term, what is what we have? I mean, if you derivate space you get velocity, and if you do it twice acceleration, so which is the meaning of derivating [tex]a(t)[/tex]?

    3. Another problem I have, is that I'm not english speaker, so there are some terms I don't understand nor to translate (and there is no article in wikipedia in Spanish), so when they talk of Comoving Distances, what do they refer to?

    Well, there are more things I don't really understand but I think I should know more physics and maths to understand them, so thanks for helping!
  7. Dec 2, 2008 #6


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    Here, when we write a(t) we mean scalefactor. I never use it to mean acceleration.

    Sometimes, to give it intuitive meaning, astronomy teachers will say that the scalefactor is proportional to the "average distance between galaxies". So if a(t) is increasing that means the distances between galaxies are increasing.

    But normally a(t) is made DIMENSIONLESS by normalizing it so that a(present) = 1.

    So it does not have the units of meters or lightyears. It has no units.

    I use a prime for the first derivative and a double prime for the second derivative.

    So a'(t) = da/dt and a''(t) = da'/dt

    If a(t) was not normalized, if it had the units of distance, like lightyears, then
    a'(t) would have the units of velocity, and
    a''(t) would have the units of acceleration, like meters per second2

    But since in the usual formalism a(t) has been normalized, the derivatives are
    only analogous to velocity and acceleration.

    a'(t) has the units of "per second" or per unit of time whatever the chosen unit.

    a''(t) has the units of "per second2" or whatever the chosen unit squared.

    In cosmology we have the idea of being at rest relative to the Hubble flow, or at rest with respect to the CMB, or at rest relative to the overall expansion process.

    The analogy is a white spot painted on a balloon which does not change its longitude and latitude as the balloon expands. In some sense it does not move.

    Only in real centimeters distance, the other spots are getting farther away from it.

    Comoving is another name for AT REST. Comoving objects are at rest with respect to the flow, or the expanding balloon.

    So one can have a system Comoving Coordinates with the origin, say, at our own galaxy. And if our galaxy is (very nearly) at rest and some other galaxy is (very nearly) at rest, then the comoving coordinates of that other galaxy do not change.

    And the comoving coordinates distance to that other galaxy does not change.

    Because of an artificial convention. The comoving distance is defined to be whatever the distance is TODAY and we take that to be the distance from the earliest times into the indefinite future.

    Comoving coordinates are a way to PARAMETRIZE all the objects at rest in the visible universe with numbers which do not change. And also parametrize all the objects which are very nearly at rest, giving them numbers which do not change very much.

    this is very convenient, because experience shows that most objects have very small proper motion (very small individual motion) and so they are almost at rest. So their comoving coordinate numbers stay almost the same.

    This means that having the comoving coordinates idea can be a big help in communication at the level of technical detail.

    But at the same time it is a very stupid-sounding terminology, if one is not accustomed. Because the comoving distance from here to a distant galaxy always stays the approximately the same----and yet everybody knows that the real distance is increasing, often at extremely rapid rate! So one immediately thinks what a stupid idea of distance.
    However it is useful as a parametrization of everything in our observable universe.

    similar to how longitude and latitude are still useful even though the balloon is expanding.
    Last edited: Dec 2, 2008
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