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234. Continuing dS/CFT. Why it is so hard to prove?

ASTRO, HEP-TH/PH — By Dmitry Podolsky on February 6, 2009 at 9:04 pm
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Dmitry Podolsky has got his PhD from Landau Institute for Theoretical Physics. He currently works as postdoc at Case Western Reserve University. He is also one of the editors of NEQNET.

I continue today the discussion of dS/CFT correspondence started a week ago.

As you probably remember, I finished last time pointing out the discrepancy between non-perturbative and perturbative values of the dimension of the dS Hilbert space. Namely, since the entropy of dS space is finite, the dimension of of the Hilbert space is finite as well since

234. Continuing dS/CFT. Why it is so hard to prove?.

On the other hand, perturbation theory in dS space is defined as expansion w.r.t. the small parameter 234. Continuing dS/CFT. Why it is so hard to prove?, where 234. Continuing dS/CFT. Why it is so hard to prove? is the value of the cosmological constant. The total number of states is clearly infinite at 234. Continuing dS/CFT. Why it is so hard to prove? (since this situation corresponds to just Minkowski space). Naively, it seems to be impossible to make the number of states finite making 234. Continuing dS/CFT. Why it is so hard to prove? infinitely small but non-zero.

How to reconcile non-perturbative value of 234. Continuing dS/CFT. Why it is so hard to prove? (finite) with infinite perturbative value? This question is clearly reduced to the question how the number of states 234. Continuing dS/CFT. Why it is so hard to prove? behaves as a function of the perturbation theory parameter 234. Continuing dS/CFT. Why it is so hard to prove?.

A satisfying answer would be that the number of states behaves as

234. Continuing dS/CFT. Why it is so hard to prove?,

where 234. Continuing dS/CFT. Why it is so hard to prove? is some constant – if 234. Continuing dS/CFT. Why it is so hard to prove?, then entropy diverges. As we will see, this is indeed the case (well, you know what is the entropy of de Sitter space 234. Continuing dS/CFT. Why it is so hard to prove? ), and we will explicitly calculate the value of the constant 234. Continuing dS/CFT. Why it is so hard to prove?.

But to get ready for the quantitative analysis, let us first quickly go through the paper of the most prominent criticizers of dS/CFT – Dyson, Lindesay and Susskind.


1. Why DLS are unhappy with dS/CFT?

The reason is of course the finite number of states that dS supports. Let us recall how the AdS/CFT correspondence works. If the curvature of the background is small (gauge coupling is large), good degrees of freedom are strings (attached to 234. Continuing dS/CFT. Why it is so hard to prove? branes) living in the bulk. If the curvature of the background is large, the only relevant degrees of freedom are gauge fields (corresponding to Chan-Paton factors of strings attached to 234. Continuing dS/CFT. Why it is so hard to prove?-branes; if there are N coincident 234. Continuing dS/CFT. Why it is so hard to prove?-branes, the symmetry group of the gauge fields is 234. Continuing dS/CFT. Why it is so hard to prove?). Since the physics of the latter is the physics of CFT, that is, quantum field theory (sic! stupid statement), then the number of states is infinite in the vicinity of horizon.

The same should hold for dS geometry, if duality holds and the physics of gravitational degrees of freedom on dS background can be effectively described by QFT. The number of short wave length modes in the vicinity of dS horizon is diverging – so is the associated entropy. However, as we have found out, non-perturbatively entropy is supposed to be finite! So, DSL conclude, bulk physics of dS cannot be dual to near-horizon QFT physics.

The argument goes deeper than that. According to proponents of dS/CFT, one point correlation functions in the static patch of dS should exponentially decay. DSL show that this is impossible in the situation when the number of states that the theory supports is finite.

2. What DLS analysis is based on?

It is based on the key proposition of dS complemetarity – there are no independent degrees of freedom, no non-trivial physics beyond dS horizon. In a sense, dS can be considered as closed thermal cavity. This is closely related to so called membrane paradigm – dS horizon is considered as impenetrable thermal membrane that can absorb, thermalize and then reemit the information via Hawking radiation.

Trying to revive the dS/CFT correspondence, we will have to understand where these complementarity considerations can go wrong.

Both ideas (complemetarity and membrane paradigm) come from black hole physics. Although there are many similarities between physics of BHs and de Sitter space, there is also a crucial difference. BH horizon is “physical” (several billion years ago the gas in the center of our galaxy collapsed and formed a black hole – many observers living in different corners of the Milky way can certainly detect it by, say, studying behavior of nearby stars in the vicinity of the galactic center). On the other hand, dS horizon is observer-dependent.

To be continued.

7 Comments

  1. Lubos Motl says:
    February 7, 2009 at 10:29 am

    Well, it shouldn’t be that hard to calculate the exponent alpha. ;-)

    de Sitter horizons are observed-dependent but that is a simple consequence of the absence of the asymptotic region in the dS space.

    The “behind the cosmic horizon” volume is analogous to the black hole interior, while the observers outside the black hole become the observers inside the dS patch.

    Two observers outside the BH agree about the BH horizon because they define it relatively to “scri+” where the radiation can eventually end. Can a point P send signals to “scri+”? If it cannot, it is inside the BH, below the horizon.

    On the other hand, in dS, the horizon is defined by being able to send signals to a particular point at the top of the diagram which is horizontal, spacelike. There are many points like that, and the horizon only exists if you pick a specific one.

    But it’s still true that things behind this horizon are causally disconnected and will never get connected. It seems to be the key thing that makes the two cases analogous.

    So I feel that the dS space is bound to be “more fuzzy”, “more finite”, and “more undetermined” than the black hole horizon, but all the issues that existed in the latter also exist in the former.

    These are speculations because we don’t have any full description of dS physics. Whether such a description has a manifestly finite number of degrees of freedom, or whether the number is infinite and only “effectively finite” remains to be seen. At any rate, certain unpredictivity of dS space is probably guaranteed to exist because you never know what is exactly radiated towards you from the cosmic horizon.

    This problem is, of course, not a practical one because the dS horizon thermal radiation in the real cosmos is outrageously negligible.

    Reply
    • Dmitry says:
      February 7, 2009 at 11:19 pm

      Dear Lubos,

      I have a question: DLS say that BH complementarity picture is supported by AdS/CFT. How exactly does it work in the AdS/CFT setup?

      Cheers,
      Dmitry.

      Reply
  2. Lubos Motl says:
    February 8, 2009 at 10:07 am

    Dear Dmitry, I believe that they’re saying that the black hole complentarity is supported by AdS/CFT (and nothing about de Sitter horizon complementarity in AdS/CFT).

    I don’t know whether AdS/CFT really “proves” BH complementarity but in the pictures I can imagine, it directly incorporates it. It’s because a region on the boundary is only “dual” to some physics inside the spatially separated region in the bulk – from where it can receive returning signals, so to say.

    And only the black hole exterior is connected with the boundary in this way. The boundary contains all the information about all the bulk, however, so the BH interior is encoded in the boundary observables, too. But we just said that the BH exterior bulk is equivalent to the boundary, which means that the BH interior data are encoded in the BH exterior data.

    I am not aware of any more rigorous proof than the handwaving above but I surely do believe that their statement is true.

    Reply
    • Dmitry says:
      February 8, 2009 at 10:27 pm

      It is so much clearer to me now, thanks!

      Dmitry.

      Reply
  3. Max says:
    February 16, 2009 at 9:12 am

    Hi.

    What if to assume that the entropy product is c.c.?

    Reply
  4. Dmitry says:
    February 17, 2009 at 6:22 pm

    Hi Max,

    if I understand your question correctly, the present stage of the evolution of our Universe is quasi-de Sitter (albeit the cosmological constant is extremely small). If so, there is en entropy associated with superhorizon degrees of freedom in this (quasi) de Sitter Universe.

    Cheers,
    Dmitry.

    Reply

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