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

In summary: I see. Anyway, I see you got your answer directly from the latter pages of the paper itself.In summary, the paper assumes Λ=0 and uses a theorem that states that if the SEC holds, there are no CTCs, and a trapped surface exists, then an incomplete geodesic exists. If Λ≠0, the SEC is violated.
  • #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.
To summarize, the observed isotropy of the black-body radiation indicates that the Universe is nearly homogeneous and isotropic back to at least the last time the radiation was scattered. If this occurred at a redshift greater than 100, we can conclude that the energy density of the black-body radiation alone would be sufficient to make all the past-directed timelike geodesics from p start converging again.
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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  • #2
Is the SEC violated even if Λ is a cosmological constant, and not dark energy?
 
  • #3
atyy said:
Is the SEC violated even if Λ is a cosmological constant, and not dark energy?

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.
 
  • #4
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].
 
  • #5
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.
 
  • #6
bcrowell said:
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].

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]?
 
  • #7
atyy said:
Doesn't that treat [itex]\Lambda g_{ab}[/itex] as part of [itex]T[/itex]?
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.

atyy said:
What if you treat [itex]\Lambda[/itex] as geometry, and write [itex]G_{ab}-\Lambda g_{ab}=8\pi T_{ab}[Λ/itex]?

Algebraic manipulations don't affect the validity or interpretation of an equation, and geometry isn't something that you can just redefine.
 
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  • #8
bcrowell said:
Algebraic manipulations don't affect the validity or interpretation of an equation, and geometry isn't something that you can just redefine.

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)?
 
  • #9
bcrowell said:
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.

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.
 
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  • #10
atyy said:
So I think on technicalities, the SEC defined only using T is not necessarily violated by a cosmological constant.

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.
atyy said:
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]?

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.
 

1. What is the Hawking singularity theorem?

The Hawking singularity theorem is a mathematical theorem proposed by physicist Stephen Hawking that states that under certain conditions, a singularity (a point of infinite density) must exist in the universe. This theorem is significant because it provides a mathematical basis for the existence of singularities in black holes and the Big Bang.

2. How is Λ (the cosmological constant) related to the Hawking singularity theorem?

The Hawking singularity theorem states that under certain conditions, a singularity must exist regardless of the value of Λ. However, the introduction of Λ into the equations of general relativity can affect the behavior of the singularity. When Λ is non-zero, it can potentially prevent the formation of singularities or even cause existing singularities to collapse.

3. What are the conditions for the Hawking singularity theorem to hold when Λ≠0?

The conditions for the Hawking singularity theorem to hold when Λ≠0 are still an active area of research. However, some proposed conditions include the presence of matter with negative energy density, the absence of closed timelike curves, and the existence of a non-zero expansion rate of the universe.

4. How has empirical verification been conducted for the Hawking singularity theorem?

Empirical verification of the Hawking singularity theorem has been primarily conducted through observations and measurements of cosmic phenomena. For example, the existence of a singularity in the center of a black hole has been indirectly confirmed through the detection of gravitational waves. Additionally, the cosmic microwave background radiation, which is a remnant of the Big Bang, provides evidence for the existence of a singularity in the early universe.

5. What are the implications of the Hawking singularity theorem when Λ≠0 for our understanding of the universe?

The Hawking singularity theorem, especially in the case of Λ≠0, has significant implications for our understanding of the universe. It provides a mathematical basis for the existence of singularities in black holes and the Big Bang, which are crucial concepts in modern cosmology. Additionally, it allows for the possibility of new physics beyond the current understanding of general relativity and may provide insights into the behavior of the universe at its earliest stages.

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