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Why "Dark Energy density" instead of CC?

  1. Mar 3, 2015 #1

    wabbit

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    Not worrying about the term DE exactly - my question really is, why is the CC described as an energy density, instead of just a nonlocal quantity ? Why bother with a constant scalar field when a constant would do ? After all we do not hear talk of a local gravitational strength density which by the way happens to be constant - so what is the fundamental difference between the two cases? Or is is just convention / usage, with no deep meaning ?

    Then, if this is indeed necessary, how should we interpret this in a relational view of spacetime ?

    Thanks

    Edit : I realize this allows an identification with QFT vacuum energy, but with 120 orders of magnitude to cross, this isn't really compelling - or is it ?
     
    Last edited: Mar 3, 2015
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  3. Mar 3, 2015 #2

    mfb

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    It is a mathematical equivalence, there is no physical difference between the two concepts in current theories.
    If we find some sort of interaction with such a field, the difference might become relevant.
     
  4. Mar 3, 2015 #3

    wabbit

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    Thanks - conceptually /intuitively it does change, as the density would be a local property of space (which happens not to vary, one more thing to explain), while a constant is well, just a constant.
    It's reassuring that this is just a choice of term so I can continue thinking only of CC : ).
     
  5. Mar 3, 2015 #4

    wabbit

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    Follow-up related question : in the statement "the universe is made of 70 % DE, 25 % DM and 5 % Matter" There is no particular meaning to the 100 %? It would be just as meaningful to leave the CC aside and say "80 % DM and 20 % matter", correct ? Or is there some kind of balancing betwwen CC and the rest that makes the first statement more meaningful ?
     
  6. Mar 3, 2015 #5

    phyzguy

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    I think the reason that people started talking of "dark energy" was because we don't really know that it is described by a cosmological constant, i.e. a value that is constant in space and time. It's true that today all of our measurements are consistent with a cosmological constant, but we should at least consider the possibility that this value varies in space and time. So we refer to it as "dark energy", and we are in the process of measuring just how constant it really is. If, after all of the data is in, we decide that it is truly a constant, then probably the term dark energy was a mistake. But if we always refer to it as the "cosmological constant", then we tend not to even entertain the possibility that it might vary.
     
  7. Mar 3, 2015 #6

    mfb

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    Well, we don't know, and we can search for deviations.

    The relative influence on the expansion of the universe is measurable. "More DE than other contributions" => the expansion is accelerating.
     
  8. Mar 3, 2015 #7

    wabbit

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    Presumably this would be more natural than a non-constant gravitational constant i.e, a less drastic change in the EFE, wouldn't break things as much ?

    Ah yes, fair enough. OTOH the M/DM split is constant in time so.it's a more structural aspect of our universe...but yes I see your point.
     
  9. Mar 3, 2015 #8

    wabbit

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    To add, about possible variation of CC this can either fall in the category "possible variation of fundamental constants" or in the category "measuring a field at different places to see if it is constant" - not necessarily a highly operational difference, but still at least intuitively, a different kind of question
     
  10. Mar 3, 2015 #9

    phyzguy

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    There is indeed a meaning to the 100%. They are typically quoted as fractions of the critical density, which is the density required to make the universe flat. Since our measurements today tell us that the universe is quite close to being flat, the densities add up to 100%. But this would not need to be the case.
     
  11. Mar 3, 2015 #10

    ChrisVer

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  12. Mar 3, 2015 #11

    wabbit

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    Cosmological Constant, sorry.
     
  13. Mar 3, 2015 #12

    wabbit

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    Thanks for the reminder, I thought it was expressed as % of total energy-matter content.
     
  14. Mar 3, 2015 #13

    ChrisVer

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    I think saying 80% DM and 20% matter will only destroy the universe we are observing....
    You can for example try setting the CC contribution to CMB background to 0, and the densities you proposed to check whether your result fits the actual data.
    Of course it will not:

    Your generated data ([itex]\Omega_{cdm}=0.8[/itex], [itex]\Omega_b =0.2[/itex] and [itex]\Omega_{\Lambda}=0[/itex]):

    camb1.png

    A close to measured data CMB angle spectrum ([itex]\Omega_{cdm}=0.224[/itex], [itex]\Omega_b =0.046[/itex] and [itex]\Omega_{\Lambda}=0.7[/itex]):

    camb2.png

    The age of the Universe can also be affected by making that assumption (drops from 13.72Gyr to 9.311Gyr).

    If on the other hand you want to put that much DM to the hot component, then you would destroy the structure formation.

    Figures generated by:
    http://lambda.gsfc.nasa.gov/toolbox/tb_camb_form.cfm
     
  15. Mar 3, 2015 #14

    wabbit

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    No no I said 80/20 leaving aside DE, by which I meant the same universe but refering to a different total, 80=25/30 etc.. I am not trying to set CC=0 sorry for the confusion:)
    But thanks, nice pics
     
  16. Mar 3, 2015 #15

    ChrisVer

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    The 100% corresponds to the cosmic sum rule.
    The Friedmann Eqs are:
    [itex] H^2 + \frac{k}{a^2} - \frac{\Lambda}{3} = \frac{8 \pi G}{3} ( \rho_m + \rho_r ) [/itex]
    At today [itex]t=t_0[/itex] this equation is:
    [itex] H^2_0 + \frac{k}{a^2_0} - \frac{\Lambda}{3} = \frac{8 \pi G}{3} ( \rho_{m0} + \rho_{r0} ) [/itex]
    And the critical density [itex]\rho_c= \frac{3H^2}{8 \pi G}[/itex] is given at time [itex]t_0[/itex] as: [itex]\rho_{c0} = \frac{3 H_0^2}{8 \pi G}[/itex] . Dividing with it you get:

    [itex] \Omega_{m} + \Omega_{r} + \Omega_\Lambda + \Omega_k = 1 [/itex]
    With [itex]\Omega_{m,r} = \frac{\rho_{m,r 0}}{\rho_{c0}}[/itex]
    [itex] \Omega_k = -\frac{3k}{8 \pi G a_0^2}= -\frac{k}{H_0^2 a_0^2}[/itex]
    [itex] \Omega_\Lambda = \frac{\Lambda}{8 \pi G \rho_{c0}}= \frac{\Lambda}{3H_0^2}[/itex]
    the 100% corresponds to the 1 on the RHS.
    If you say that DM+matter=1 then [itex]\Omega_\Lambda=0[/itex] (since the universe is spatially flat and [itex]\Omega_k \approx 0[/itex])
     
  17. Mar 3, 2015 #16

    wabbit

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    Thanks - that's not what I meant, hope it's clarified above. I am not touching the LCDM parameters, just refering to different ones, bad idea sorry
    To clarify again all my questions in this thread are about interpretation and context, not about the results of standard cosmology.
     
  18. Mar 3, 2015 #17

    ChrisVer

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    OK sorry...
     
  19. Mar 3, 2015 #18

    wabbit

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    Not at all it was my fault for beeing unclear, I appreciate your responding and trying to help.
     
  20. Mar 3, 2015 #19

    wabbit

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    Thanks everyone for the responses this was very helpful - or How I Learned to Stop Worrying and Love the CC :)
     
  21. Mar 3, 2015 #20

    Chalnoth

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    The cosmological constant is a specific idea that can be considered a parameter of the gravitational field equations, or a constant energy density.

    Dark energy is a concept that encompasses the cosmological constant as well as a variety of other models with similar properties.

    Saying "dark energy" instead of "cosmological constant" is the equivalent of saying, "Well, we don't really know what this is....maybe it's a cosmological constant, maybe it's something else."
     
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