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Empirical verification of conditions for Hawking singularity theorem when Λ≠0?

  1. Jul 17, 2011 #1

    bcrowell

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    As far as I know, the classic paper applying the Hawking singularity theorem to our universe is this one: Hawking and Ellis, "The Cosmic Black-Body Radiation and the Existence of Singularities in Our Universe," Astrophysical Journal, vol. 152, p. 25, 1968, http://articles.adsabs.harvard.edu/full/1968ApJ...152...25H The version of the theorem they use is one that says that if the SEC holds, there are no CTCs, and a trapped surface exists, then an incomplete geodesic exists.
    The paper assumes Λ=0. For Λ≠0, the SEC is violated. Does anyone know of a paper that connects the dots between observation and a singularity theorem for Λ≠0 in the same way that this paper did for Λ=0? E.g., this review article http://www.livingreviews.org/lrr-1998-11 doesn't seem to mention the issue.
     
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  3. Jul 17, 2011 #2

    atyy

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    Is the SEC violated even if Λ is a cosmological constant, and not dark energy?
     
  4. Jul 17, 2011 #3

    bcrowell

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    Yeah, I'm pretty sure it is. For example, see p. 2 of http://arxiv.org/abs/gr-qc/0205066 Actually, checking that this is the case seems like a good exercise. I'll try it and if I can get it to work out, I'll post my calculation.
     
  5. Jul 17, 2011 #4

    bcrowell

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    OK, it turns out to be a pretty trivial calculation. In the One True Signature (+---), the Einstein field equations are [itex]G_{ab}=8\pi T_{ab}+\Lambda g_{ab}[/itex]. That means that a cosmological constant is equivalent to [itex]\rho=(1/8\pi)\Lambda[/itex] and [itex]P=-(1/8\pi)\Lambda[/itex]. This gives [itex]\rho+3P=(1/8\pi)(-2\Lambda)[/itex], which violates the SEC for [itex]\Lambda>0[/itex], since part of the SEC is [itex]\rho+3P \ge 0[/itex].
     
  6. Jul 17, 2011 #5

    bcrowell

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    Ah, I see. In the present-day universe, the SEC is violated. But in the early universe, radiation made contributions to both ρ and P that diverged to infinity, so the SEC still held. A correct statement would be that the SEC is violated if Λ>0 in a vacuum.
     
  7. Jul 17, 2011 #6

    atyy

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    Doesn't that treat [itex]\Lambda g_{ab}[/itex] as part of [itex]T[/itex]?

    What if you treat [itex]\Lambda[/itex] as geometry, and write [itex]G_{ab}-\Lambda g_{ab}=8\pi T_{ab}[/itex]?
     
  8. Jul 17, 2011 #7

    bcrowell

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    Right. I'm pretty sure that's what you've got to do, though. I think the physical justification is that if you had a form of matter that acted like a cosmological constant, you would have to include it in T, but it would represent all the same physics as if you put it in the cosmological constant term. Looking around at various references online, they do all state energy conditions in terms of T alone, but I think they simply date back far enough that there was no real interest in including a Λ. I could be wrong about this, but I can't see what else you could do that would be reconcilable with the physical jusitification I gave above, or with the statements in papers like Visser's that Λ violates the SEC.

    Algebraic manipulations don't affect the validity or interpretation of an equation, and geometry isn't something that you can just redefine.
     
    Last edited: Jul 17, 2011
  9. Jul 17, 2011 #8

    atyy

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    How about like in http://www.scholarpedia.org/article/Dark_energy, where the SEC is said to be violated in the 3rd scenario (outside the standard model), but no mention of it seems to be made in the 2nd scenario (gravitational origin)?
     
  10. Jul 17, 2011 #9

    atyy

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    Anyway, I see you got your answer directly from the latter pages of the paper itself.

    So I think on technicalities, the SEC defined only using T is not necessarily violated by a cosmological constant. However, the theorem requires a joint condition on the SEC and the cosmological constant, which can be satisfied by 0 cc and SEC. In the case of a positive cc, they argue that in reality, T is such that a small positive cc would still lie within the joint condition on T and the cc.

    I wonder if the known cc lies within what they assumed as bounds on reality at that time? Looks like yes, they assume no greater than 5E-29 g/cm3 and http://www.scholarpedia.org/article/Dark_energy says it's 7E-30 g/cm3.
     
    Last edited: Jul 17, 2011
  11. Jul 20, 2011 #10

    George Jones

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    But the 1970 Hawking-Penrose theorem does not have the Strong Energy Condition as one of its hypotheses, its has [itex]R_{a b} U^a U^b \geq 0[/itex] for all timelike and lightlike [itex]U[/itex]. When used with the Raychaudhuri's equation, this geometrical condition produces focusing of geodesics.
    In this case,
    [tex]R_{ab} = 8 \pi \left(T_{ab} -\frac{1}{2} g_{ab} - \frac{\Lambda}{8 \pi} g_{ab} \right).[/tex]
    The Strong Energy Condition is
    [tex]\left(T_{ab} -\frac{1}{2} g_{ab} \right) U^a U^b \geq 0[/tex]
    for all timelike [itex]U[/itex].

    Consequently, if the cosmological constant (on the geometrical left side of Einstein's equation) "dominates" the Ricci tensor, then the hypotheses of the 1970 Hawking-Penrose singularity theorem are violated even if the strong energy condition is satisfied. As Ben mentioned, during the radiation dominated phase of the early universe, the (normal) stress-energy part of the Ricci tensor dominates, so this hypothesis of the theorem is met.

    Only if there is no cosmological constant, or if it is included in the stress-energy tensor as dark energy, is this geometrical hypothesis equivalent to the Strong Energy Condition.

    If there was inflation before the radiation-dominated phase, then the above geometrical hypothesis is violated, but there has been a singularity theorem put forward for this case,

    http://arxiv.org/abs/gr-qc/0110012.
     
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