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Dark Energy's effect on the curvature of space time

  1. Mar 16, 2006 #1
    Given that Dark energy supplies negative pressure:

    Does it have mass? Does it curve space? If so, does it help to close the universe? Can something that has negative pressure also make space more closed? Does it affect time dilation? How do you achieve a flat universe when there seems to be no end to the acceleration? If everything were dark energy, and if dark energy had a concentration of the critical density, would it stop the acceleration?
     
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  3. Mar 18, 2006 #2

    pervect

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    Dark energy has a positive energy density, and a negative pressure.

    If we take the Komar mass intergal for a region of dark energy, we get an effective negative mass for a region of dark energy- the intergal is essentially the intergal of rho+3P, and P = -rho. Thus I'm inclined to say that the "mass" of dark energy is negative, but on further thought I'm not positive about the validity of this statement. In order for the concept of Komar mass to be meaningful, the system must be static. I don't see how this is possible if the pressure is not zero at the boundary of the system.

    Perhaps we can get someone else to comment on this point.

    A closed universe requires that rho = 3H^2/8*pi*G because of the Friemdann equations.

    http://en.wikipedia.org/wiki/Friedmann_equations

    Because the energy density for dark energy is positive, it helps to meet this requirement. (The Lambda-CDM model has 70% of the energy in the universe being dark energy).

    http://en.wikipedia.org/wiki/Lambda-CDM_model

    The closure of the universe is related only to the geometry of space - it has nothing to do with whether or not the universe is expanding, or how rapidly the expansion is accelerating or deaccelerating.
     
  4. Mar 18, 2006 #3

    Garth

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    The density, and hence, integrated over a representative volume, the mass of DE is taken to be positive.

    Its pressure is negative and allows the universe to accelerate against gravitational forces. Negative pressure (tension) acts as an 'anti-gravity' force.

    The positive density adds to the total density of the unverse and in the standard theory is the largest contributor to that total density, helping to keep it just greater than the critical density for closure.

    Garth
     
    Last edited: Mar 19, 2006
  5. Mar 19, 2006 #4

    pervect

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    Integrating the energy density over a volume to get the mass only works when the pressure terms are small.

    In this case we have pressure terms that are equal to the energy density terms.
     
  6. Mar 19, 2006 #5

    Garth

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    The concept of Komar mass is not applicable to the cosmological solution, it is a definition of the total mass of a static, asymptotically flat space-time which is vacuum in the exterior region.

    If the concept of the total mass of a volume of DE is difficult to define then just work with density and pressure. If DE is the cosmological constant, [itex]\omega = -1[/itex], then as the universe expands the negative pressure will produce more 'total mass' as the density remains constant and the volume increases.

    One problem with this model is the value of that density [itex]\Omega_{DE} \succapprox \Omega_{DM} \succapprox \Omega{b}[/itex] and they will diverge as time progresses, so why are they approximately equal? This is called the cosmological coincidence problem.

    Garth
     
  7. Mar 20, 2006 #6

    pervect

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    I was hoping that we could put a "lump' of dark energy in a true vacuum (no cosmological constant) so that we could find its mass - either with the Komar mass, or perhaps the ADM mass.

    But it seems that the idea is controversial/complicated enough that it's not good to convey it as an explanation to a newcommer.

    OTOH, while it's true that the universe doesn't have a mass, it's a rather unsatisfying explanation to give when someone asks about the mass of dark energy.
     
  8. Mar 31, 2006 #7
    What Dark Energy May Really Be

    Hi,

    Is it possible that what we call Dark Energy is really a manifestation of geometry, rather than energy?

    Picture a balloon that is inflating. There are two ink dots on the surface of the balloon. The ink dots are moving away from each other as the balloon expands, PLUS the RATE at which they move away from each other increases as the size of the balloon increases.

    So is it possible that when we observe the galaxies moving apart at an accelerating rate, we are really just seeing the result of the geometry of the higher dimensions, not an actual force in the conventional sense?

    Rich
     
  9. Mar 31, 2006 #8

    ZapperZ

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    That would be a very strange and obvious oversight, since these are the very same people who are aware of the comological constant AND the GR scaling factor. When you go to a seminar on such things, these types of expansion are the first thing they talk about. It would be silly for them to suddenly "forget" to include such a thing when they talk about dark energy.

    Zz.
     
  10. Mar 31, 2006 #9

    Garth

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    Hi Rich and welcome to these Forums!

    Keep asking questions, that is how we learn.

    One possible solution to the DE question is that if [itex]\omega = -1[/itex], (which is a matter of determination from observation) then DE could be simply a non-zero Cosmological Constant. If this is the case then the CC can be interpreted as a sort of integration constant, I saw 'sort of' because this is tensor calculus and not ordinary calculus we are dealing with, and it would be a property of empty space. As gravitational fields are interpreted as the geometry of space-time in GR, then you could say that DE, if it is non-zero CC, is a manifestation of geometry. So you are not too far off the mainstream theory!
    Well that depends on how you blow up the balloon, and the balloon is only an analogy in thge first place so be careful. If the balloon expands at a steady rate then the dots will recede from each other at a constant velocity. If the balloon slows down in its expansion, because you are 'running out of puff' then the will mutually decelerate. A decelerating expansion was the standard model of the universe until about a decade ago. If the balloon accelerates in its expansion then so will the mutual velocity of the dots. You will need extra 'puff', extra energy to make it do this and that is where DE comes in.
    We are observing the effects of various forms of matter and energy of varying equations of state ([itex]p = \omega\rho[/itex]) acting on the geometry of space-time as described by Einstein's GR field equation:

    [tex]R_{\mu \nu} - \frac12 g_{\mu \nu}R - g_{\mu \nu} \Lambda = 8\pi GT_{\mu \nu}[/tex]

    Garth
     
    Last edited: Mar 31, 2006
  11. Mar 31, 2006 #10

    George Jones

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    Yes.

    But then empty space would not be flat Minkowski space, it would still have some curvature. I don't see a problem with this - why should empty space be flat?

    However, if the vacuum does have energy, then it can come into the equation in exactly the same mathematical form as this background curvature. Also, most physicists think that, due to quantum theory, the vacuum should have energy.

    I wouldn't say this. I would say that gravity can be attractive at small distances and while being repulsive at large cosmo0logical distances. It's consistent with the data that gravity is a theory that depends on 3 fundamental parameters - the Newtonian gravitational constant G, the speed of light c, and the cosmological contant Lambda.

    However, because of what I said above, most physicists think that the cosmological constant is a measure of dark (vacuum) energy, not a fundamental parameter of gravity.

    Regards,
    George
     
  12. Mar 31, 2006 #11
    Wow, thanks guys!

    Hello,

    Thanks, for your responses, Gentlemen! I appreciate you taking the time to consider my question seriously. Most of the time when I try to talk to people about my questions, they just give me blank stares. :frown:

    I look forward to thoughtful dialogs in the future. Thanks!
     
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