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A Cosmological constant in the semiclassical limit of quantum gravity

  1. Jun 17, 2017 #1
    Why is it the case that, in a semiclassical description of the Einstein-Hilbert action, the cosmological constant is small in Planck units?

    Why does this mean that

    $$\ell \gg G$$

    for ##\Lambda = - 1/\ell^{2}##?
     
  2. jcsd
  3. Jun 17, 2017 #2

    PeterDonis

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    Nobody knows. This is an open area of research.
     
  4. Jun 17, 2017 #3
    Ah! Right!

    For the second part, we have, in Planck units,

    ##\Lambda \ll M_{P}^{4}##

    ##-1/\ell^{2} \ll M_{P}^{4}##

    ##-1/\ell^{2} \ll 1/G^{2}_{N}##

    ##1/\ell^{2} \gg 1/G^{2}_{N}##

    ##G^{2}_{N} \gg \ell^{2}##

    ##\ell^{2} \ll G^{2}_{N}##

    ##\ell \ll G_{N}##

    What have I done wrong here?
     
  5. Jun 17, 2017 #4

    PeterDonis

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    Why did you have ##\Lambda = - 1 / \ell^2## with a minus sign?
     
  6. Jun 17, 2017 #5
    Let's say that we have ##AdS## spacetime.
     
  7. Jun 17, 2017 #6

    PeterDonis

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    Then your derivation is only valid for ##AdS## spacetime. Do you have a reference that claims that ##\ell \gg G## for ##AdS## spacetime?
     
  8. Jun 17, 2017 #7
    It is in page 2 of this paper - https://arxiv.org/pdf/hep-th/9712251.pdf.

    See equation (2.2) and the text above the equation.
     
  9. Jun 17, 2017 #8

    PeterDonis

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    Hm. I think their statement that ##\ell \gg G## is only meant as a heuristic, because the two don't even have the same units; the units of ##G## are length squared (inverse mass squared), while the units of ##\ell## are, of course, length. So really it should be ##\ell^2 \gg G##.

    Regarding your derivation, you should not be including the minus sign regardless of the sign of ##\Lambda##, since the inequality only refers to relative magnitudes, not signs. In other words, the strictly correct way of writing the above inequality (with units corrected) is ##\vert \ell \vert^2 \gg \vert G \vert##. Or, if you write it in terms of ##\Lambda## and ##M_P##, it is ##\vert \Lambda \vert \ll \vert M_P \vert^2##. (Note the exponent, btw; it's the Planck mass squared, since the units of ##\Lambda## are mass squared, or inverse length squared.) The fact that the sign of ##\Lambda## is negative in ##AdS## doesn't change any of the above--its magnitude is still small compared to the magnitude of ##M_P##, which is the necessary requirement for the semiclassical approximation.
     
  10. Jun 17, 2017 #9
    I find that the reviews of important results in section 2 and section 3 of the paper could have better written.

    Thanks for pointing this out!
     
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