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Cosmological constant from first principles

  1. Oct 18, 2012 #1

    Chronos

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    Can the cosmological constant be derived from first principles? The answer appears to be - YES, according to this paper by Padmanabhan - 'The Physical Principle that determines the Value of the Cosmological Constant', http://arxiv.org/abs/1210.4174. This is, in part, an extension of Padmanabhan's earlier paper 'Emergent perspective of Gravity and Dark Energy', http://arxiv.org/abs/1207.0505.
     
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  3. Oct 18, 2012 #2

    BillSaltLake

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    There's a possible problem here: he's saying (I think) that λLF2 is ~1/nμ where n is the # of phase space cells within the Hubble radius (and μ turns out to be ~1.2). However, during the matter era, n [itex]\propto[/itex] ρ-3/4 [itex]\propto[/itex] t3/2, which would make λ variable. This is not allowed in GR. (When I say "# of phase space cells", I mean the # of photons that would result if all energy in the observable universe were converted to BB radiation.)
     
    Last edited: Oct 18, 2012
  4. Oct 19, 2012 #3
    I'm not sure if I understood correctly, so please explain if I didn't...

    But so it seems to me he says that there are three different phases of expansion: first de Sitter, then radiation dominated, then de Sitter again. If the Hubble parameter at first de Sitter phase is of order Planck mass, then the current Hubble parameter should be
    [tex] H_{now} = \frac{a_{then}^2}{a_{now}^2} H_{then} = \frac{a_{then}^2}{a_{now}^2} L_P^{-1}[/tex]
    and since in de Sitter, cosmological constant is related to H, one gets
    [tex] \Lambda = 3H_{now}^2 = 3 \frac{a_{then}^4}{a_{now}^4} L_P^{-2} [/tex]

    Then he goes about calculating [itex]Q = a_{now}/a_{then} [/itex]. I don't understand the calculation. There has to be some clear assumption for when the second de Sitter phase starts, and it has to be put in by hand. Where does the value fundamentally come from?
     
  5. Oct 19, 2012 #4

    Chronos

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    The Hubble parameter during the inflationary epoch [1st de Sitter phase] is the Planck length [Lp]. The inflationary epoch is assumed to end when the de Sitter temperature is reached, defined as Tp = 1/(2piLp). This occurs at point D on p3 graph, the beginning of the radiation epoch. The radiation epoch ends when the number of comoving wave vectors that reenter the Hubble radius is the same as the number that exited during the inflationary epoch. This occurs at point B on p3 graph, which also marks the beginning of the second de Sitter phase. Q is the expansion factor, which is expected to be the same during all three epochs. It appears to me you can use the point when accelerated expansion began as the start of the second de Sitter phase.
     
    Last edited: Oct 20, 2012
  6. Oct 20, 2012 #5

    Jorrie

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    I suppose this depends on the chosen units and will be correct with everything expressed in Planck units, but then H_then = 1, not so?

    Would Lp not be the Hubble radius, rather than the Hubble parameter, which would be extremely large, i.e. [itex]H_{then} = 1/T_{Planck} \approx 10^{43} \, \, sec^{-1}[/itex]?
     
    Last edited: Oct 20, 2012
  7. Oct 20, 2012 #6

    Chronos

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    Agreed, the initial Hubble radius appears to be Lp, which expands by H~a during the inflationary epoch, followed by H~a^2 during the radiation epoch - which appears consistent with the LCDM model.
     
  8. Oct 20, 2012 #7

    Jorrie

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    I thought that during inflation (which is the 1st de Sitter phase), the Hubble radius remained constant and only started to grow when inflation ended (point D in Padmanabhan Fig.1). It is [itex]\dot{a}[/itex] that initially increased exponentially, but [itex]H = \dot{a}/a[/itex] remained constant. Or am I mixing things up the wrong way here?
     
  9. Oct 20, 2012 #8

    Chronos

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    He calls the Hubble radius 'constant asymptotically' during inflation [p2], which lead me to assume H could increase linearly while 'a' went wild. It seemed logical, the modes within the initial Hubble radius would be whisked away, unable to reenter the Hubble radius until the radiation epoch commenced. The change in the Hubble radius during inflation may, however, be too trivial to be of any consequence.
     
  10. Oct 21, 2012 #9

    Jorrie

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    Thanks, makes sense. Sharp slope changes on log-log plots are really gradual changes on linear plots.

    From his page 7, second bullet:
    attachment.php?attachmentid=52145&stc=1&d=1350798844.jpg

    I don't quite catch the meaning of the last sentence. Does he mean that all the radiation and matter (energy) that we observe emerged around point D, or did it gradually emerge during the middle phase (D to B), i.e. migrated from the left side of the parallelogram to the right side? We are presumable situated very near point B, busy entering phase B to C.
     

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  11. Oct 21, 2012 #10

    Chronos

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    Me neither. The emergent phase is not well characterized. It appears he asserts a quantum gravity solution is required on that count.
     
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