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Bianchi's entropy resultwhat to ask, what to learn from it 
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#55
Apr2912, 08:59 PM

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Thanks for the comment on Pranzetti's paper! Not sure exactly what you mean by "semiclassical" though.
You are right that the derivation of S = A/4 (without Immirzi dependence) happens in the first 3 pages. Indeed the grand canonical ensemble of quantum states of horizon's geometry is set up in the first few equations on page 2, where he gives the "canonical partition function" of quantum states, e.g. equation (3). There is no hint of "QFT on a curved spacetime" there. Punctures are simply where spin network states go through the horizon. Their edges carry area quantum numbers j. A class of spin network states is specified by {s_{j}} in his notation. There s_{j} is the number of spin network edges with spin label j, which pass thru the horizon. So the approach is fully background independent. There is no prior geometry. All the geometry is in the spin networks which are quantum states of geometry. Standard in Loop gravity. So he sets up to derive S = A/4 with those equations (2) (3) (4) .....(9) The fully quantum conclusion is equation (9) where you see the quantum corrections as well as the leading term which has the 1/4 coefficient. Then to recover the Bek Hawk. result he of course takes a limit so that the quantum corrections go away. That is equation (10) in the next paragraph. But already before that in equation (9), which is not semiclassical, you see there is no dependence on Immirzi. Anyway that is how I read it. Do you see equations (2  9) as in any way semiclassical? For me they come entirely within ordinary spin network Loop gravity. 


#56
Apr3012, 01:47 AM

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I was a bit too fast to comment on the Pranzetti paper. This is for two reasons: 1. I was swayed by his own reference to a semiclassical analysis to fix [itex]\beta \bar{\kappa} = 2\pi/\hbar[/itex], and 2. My own thermodynamics is quite rusty.
It appears to me that Pranzetti's calculation in the grand canonical ensemble (GCE) is equivalent (there is a question of the role of the large [itex]N[/itex] limit that I comment on below) to the calculation Bianchi made in the microcanonical ensemble (MCE) with the area/energy constraint. In both cases you find the correct leading term in the entropy, but for a fixed value of the Immirzi parameter. In the Bianchi calculation, the fixing of the Immirzi parameter is explicit. In the Pranzetti calculation it is hidden, but he refers to it below his equation (11). I will discuss this more in a bit. Now, using the MCE is rather transparent. We're aiming to count microstates and not worry about dynamical processes like emission or absorption, so we can fix the number of quanta and energy. To ensure that we're using the correct mixed state, we impose the area constraint and extremize the entropy. In the GCE, we allow the number of quanta and energy to fluctuate but we use a heat bath to fix the temperature of the system state. We end up getting the same answer for the entropy as before when we fix the average energy to be the appropriate multiple of the energy. The setup is a bit unphysical for a real BH, but for our purposes we can always imagine feeding the right amount of matter in to balance the radiation coming out. It also seems to me to be nicer to impose the area constraint dynamically, rather than by hand, but this is more opinion than a serious objection. However, what the GCE also seems to do for us is let us avoid the large [itex]N[/itex] limit. For Bianchi, the large [itex]N[/itex] limit was not just important to allow us to use the Stirling approximation, but it was also important in obtaining a manageable form for the number of states with the same occupation numbers ([itex]\Omega[/itex] in B10). Pranzetti's use of the GCE seems to remove the need for us to take this limit, at the expense of an extra free parameter, the chemical potential. This leads us naturally to the new conundrum surrounding the Immirzi parameter. We will work with an area operator that is a mix of the ones that Pranzetti and Bianchi use: [tex] \hat{H}  \{ s_j\}\rangle = \hbar \bar{\kappa} \gamma \sum_j s_j j  \{ s_j\}\rangle .[/tex] I am using Pranzetti's notation, but I have set [itex]\hbar G = \ell_p^2[/itex] for convenience, as well as chosen the Schwingertype basis of Bianchi to simplify some calculations later. Now, it could be that I don't understand the state space well enough and there is some inequivalence between the Bianchi and Pranzetti pictures. This would presumably address the large [itex]N[/itex] questions above. Somehow the difference would go away in the large [itex]N[/itex] limit, explaining why they agree. In any case, I will keep going under the assumption that I understand the mechanics of the states, if not their complete motivation. It will be convenient to define a parameter [itex]\nu = \hbar \beta \bar{\kappa} \gamma.[/itex] If we set [itex]\beta \bar{\kappa} = 2\pi/\hbar[/itex], then we can write the Immrizi parameter as [tex] \gamma = \frac{\nu}{2\pi}.[/tex] Now, we can write all thermodynamic quantities in terms of the function [tex] f(\nu) = \sum_j (2j+1) e^{\nu j} \longrightarrow \frac{4}{\nu^2} (\nu+1) e^{\nu/2} ~\mathrm{as}~N\rightarrow \infty.[/tex] This is the same expression that turned up in the result for the occupation numbers in B10, so I've included the value for the sum that we find in the large [itex]N[/itex] limit. The relevant equation from Pranzetti is (10), which we write as [tex] f(\nu) = \frac{\bar{N}}{\bar{N}+1} e^{\beta \mu} . [/tex] In general, this is a transcendental equation that determines [itex]\nu[/itex] (equivalently [itex]\gamma[/itex]) in terms of [itex]\mu[/itex] and [itex]\bar{N}[/itex]. In the large [itex]\bar{N}[/itex] limit, the explicit [itex]\bar{N}[/itex] dependence drops out. If we did not wish to take the limit, we could just eliminate [itex]\bar{N}[/itex] using the energy constraint, which is [itex] \frac{\bar{\kappa}}{8\pi G} A = \bar{E} =  \hbar \bar{\kappa} \gamma (\bar{N}+1) \frac{d}{d\nu} \log f.[/itex] Now the chemical potential represents the cost in energy to take a particle from the heat bath and place it into the system (the isolated horizon or BH). Physically, we might think that this is related to the surface gravity of the horizon. Perhaps we might think that it is zero, since this is precisely the notion of energy that is ambiguous in a gravitational system. It might be possible to address this by thinking more carefully about how we have to define the GCE. I will probably think a bit more about it, but in any case, the Immrizi parameter now depends implicitly on the chemical potential. If we set the chemical potential to zero and take the large [itex]\bar{N}[/itex] limit, then we will recover the same numerical solution as in the modified Bianchi calculation [itex]\nu \sim 2.086,~~~\gamma \sim 0.322.[/itex] For other values of [itex]\mu[/itex], we will obtain some other value of [itex]\gamma[/itex]. This is either strange or expected. On the one hand, it might seem strange that we need to change the quantum of area to accommodate a change in chemical potential. On the other hand, we might think that whatever change we made to the system represents some sort of change in the natural energy scale of the problem and the Immirzi parameter must run. Anyway, the upshot of all of this is that B10 and Pranzetti look like correct computations for the state spaces they are using. Their results seem to agree qualitatively and quantitatively in the common regime of validity. I still think that B12 is pulling a bit of a fast one by using a pure state, but I understand it to be correct as a semiclassical computation, rather than a purely quantum one. 


#57
Apr3012, 10:08 AM

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I would expect there to be some dependence on Immirzi in the second term. Perhaps both, I haven't thought much about it. But it is the leading term that is the area term, where you see the proportionality of the entropy with the area, and that is where you don't get dependence on Immirzi. I should have made that clearer. BTW just for clarification the μ that appears in the second term of (9) is called the chemical potential. Doubtless familiar to you, Fzero, but others might be reading. The Nbar that appears in the second term is the average number of punctures. I write it Ñ to avoid having to resort to LaTex. I really like this Pranzetti paper! For convenience here's the link: http://arxiv.org/abs/1204.0702 The second term in equation (9) is quite interesting. I think I first saw it in the Ghosh Perez paper last year, but I'm not sure. It is μβÑ. The β, as he says, can be interpreted as a "generic temperature β for the preferred local observer O hovering outside the horizon at proper distance l, as result..." It seems to be this second term which you are scrutinizing in your above post #56. More power to you . 


#58
Apr3012, 12:23 PM

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[tex] \mu \beta \bar{N} =\bar{N} \log z = \bar{N} \log f + \bar{N} \log\frac{\bar{N}+1}{\bar{N}},[/tex] [tex]\log \mathscr{Z}=\log(\bar{N}+1).[/tex] So we can write [tex] S = \frac{A}{4G\hbar} + \bar{N} \log f + \log \frac{(\bar{N}+1)^{\bar{N}+1}}{\bar{N}^\bar{N}}.[/tex] Now, remember that [tex](\log f)' = c \frac{A}{\bar{N}+1}[/tex] by the energy constraint. We can't explicitly integrate this because [itex]\bar{N}[/itex] is (defined transcendentally as) a function of [itex]\nu[/itex]. It is interesting to try to compare this to the large N result in B10: [tex] S = \frac{A}{4G\hbar}  \frac{3}{2} \log \frac{A}{G\hbar}. [/tex] Naively, it doesn't appear possible to reproduce the [itex]\log A[/itex] correction, since [itex]\log f\rightarrow 0[/itex]in the [itex]\mu=0[/itex], large [itex]\bar{N}[/itex] limit. Perhaps there is some subtlety in taking these limits. 


#59
May112, 11:19 PM

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I don't think people are entirely sure what the correction terms are. They all seem to agree that the MAIN term for the entropy is A/4.
And that obviously does not depend on the Immirzi. And from what i see they expect the correction terms, which arise when you do a full quantum treatment of the BH entropy, to involve the Immirzi. Bianchi says as much in his recent paper. And if I remember right this is explictly the case both with Ghosh Perez and with Pranzetti. So this is the picture that is emerging more or less across the board with these young researchers' work on Loop BH. Supposing they are right and the work is born out, then I think this allows for the Immirzi to run. Run with what? With scale? with energy density? Then the size of the BH would affect the slight corrections in the formula by which the entropy was calculated. Maybe the "bare" UV value of Immirzi is, say, 0.274. And she runs to zero when you go to larger and larger scale. Just speculating ======================== I should copy Pranzetti's equation (9) since that seems to be the main equation we are discussing. S = (βκ/8πG)A μβÑ + log curlyZ ==endquote== curlyZ is defined in equation (2) as a function of the local observer's temperature β and is the grand canonical partition function for the gas of punctures. These are the links of spin networks sticking out thru the BH horizon. The term (βκ/8πG) turns out to be 1/4, in the appropriate units, with G = hbar = c = 1. This is when β is seen to be the Unruh temperature associated with the acceleration which the observer must maintain in order to continue hovering at a fixed distance above the horizon, which has surface gravity κ. I have no particular reason to copy in equation (9) at this point. We have been discussing it for the past Idon'tknowhowmany posts. But I just wanted to finally write it in for completeness, for the record so to speak. 


#60
May212, 02:44 PM

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1. SU(2) ChernSimons with level [itex]k\rightarrow \infty[/itex], microcanonical ensemble: entropy depends on [itex]\gamma[/itex] as [tex]S = \frac{c_1}{\gamma} \frac{A}{4G}.[/tex] 2. Microcanonical ensemble with area/energy constraint: entropy depends on [itex]\gamma[/itex] as [tex]S = \frac{c_2}{\gamma} \frac{A}{4G},[/tex] where [itex]c_2[/itex] is set at a critical value via the constraint. Value agrees with approach 1 above. 3. SU(2) CS with finite level, MCE: entropy does not depend explicitly on [itex]\gamma[/itex], but after extremizing [itex]\gamma[/itex] depends on the level (see, for example, the discussion around fig 6 in http://arxiv.org/abs/1103.2723v1). Value quickly converges to [itex]k=\infty[/itex] result for [itex]k\geq 4[/itex], so consistent with 1 and 2. 4. Grand canonical ensemble with area/energy constraint, chemical potential [itex]\mu[/itex] introduced: entropy does not depend on [itex]\gamma[/itex]. Thermodynamics, including energy constraint, fix [itex]\gamma[/itex] in terms of [itex]\mu[/itex]. In [itex]\mu =0[/itex], large [itex]A[/itex] limit, recover value of [itex]\gamma[/itex] consistent with approaches 1 and 2. "As we have just seen, k does modify the leading but does not modify the subleading corrections of the entropy. In that sense, the logarithmic corrections seems to be universal and independent of the Immirzi parameter" But I think they're making the mistake of forgetting that their critical exponent [itex]\alpha[/itex] is defined in terms of [itex]\gamma[/itex]. Essentially the ratio of [itex]\alpha[/itex] and [itex]\gamma[/itex] must take a critical value when computing the entropy. 


#61
May212, 03:28 PM

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Let's suppose Pranzetti's equation is right. Here's his equation (9)
===quote http://arxiv.org/abs/1204.0702 page 3 equation (9)=== S = (βκ/8πG)A μβÑ + log curlyZ ==endquote== Here is the first term, (βκ/8πG)A Are you saying that this term depends on the Immirzi? 


#62
May212, 04:17 PM

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I don't see how you could possibly be saying that and i don't see anything in your posts that implies it. So a simple "no" answer would suffice.
Just to be clear, I'd like to be sure of that. So that I know we both agree that the first term in Pranzetti's eqn (9) does not depend on the Immirzi. And in that case we can look at the other two terms, try to estimate their size etc, if you are so inclined. But first let's be sure we understand each other about the leading term. 


#63
May212, 06:03 PM

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[tex]S = \frac{A}{4G},[/tex] where [itex]A[/itex] is a macroscopic parameter. In terms of the microscopic parameters, [tex] A = F(\gamma,\mu).[/tex] So [itex]A[/itex] is the macroscopic area at a specific value of [itex]\gamma[/itex], much the same way that the microcanonical result [itex]c A/\gamma[/itex] is the area at a specific [itex]\gamma[/itex]. I've been looking at whether or not there's some way to derive an expression for the entropy that makes sense without appealing to the area constraint. I haven't found anything useful so far. I worked out what was confusing me about the [itex]\log A[/itex] term. What had happened was B10 partially reproduces the "quantum" corrections from the CS theory (they at least agree at large N). These corrections have been ignored in Pranzetti, so there's no point in looking for them. 


#64
May212, 08:13 PM

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So I would say your first statement is right. The leading term coefficient has no Immirzi dependence. S = A/4 But your second statement A = F(gamma, mu) does not connect with how I've seen things done in Loop gravity. I think it's pretty clear that the leading term in (9) need not change as gamma runs, as, for example, γ → 0. It would be interesting, though, to learn something about the dependence of the other two terms, and their sizes relative to the leading term. Various papers by Bianchi, Magliaro, Perini exemplify this socalled "double scaling limit" it makes sense to keep the overall region of space the same size as you vary parameters. I suspect that the proven usefulness of this type of limit is one of the motivations here: i.e. reasons for interest in the new work giving Immirzi parameter greater freedom. 


#65
May212, 10:08 PM

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We start with [tex]\log \mathscr{Z} =  \log ( 1  z \sum_j (2j+1) e^{\beta E_j} ), ~~z = e^{\beta\mu}.[/tex] We use the Schwinger basis and [itex]G\hbar =\ell_p^2[/itex], then [tex]E_j = \bar{\kappa} \hbar \gamma j .[/tex] Using [itex]\beta\bar{\kappa} = 2\pi/\hbar[/itex], we can write [tex]\beta E_j = 2\pi \gamma j.[/tex] The ensemble energy is [tex] \bar{E} =  \frac{\partial}{\partial \beta} \log \mathscr{Z} = \frac{z\sum_j (2j+1)E_j e^{\beta E_j}}{1  z \sum_j (2j+1) e^{\beta E_j}}.[/tex] Let's get a neater expression from this by noting that [tex] \sum_j (2j+1) E_j e^{\beta E_j} = \frac{ \gamma f'(\gamma) }{\beta} ,[/tex] where [tex] f(\gamma) = \sum_j (2j+1) e^{2\pi\gamma j}.[/tex] Now the energy constraint is [tex] \frac{\bar{\kappa} A}{8\pi G} = \bar{E} =\frac{1}{\beta} \frac{\gamma f'(\gamma)}{1z f(\gamma)}, [/tex] so we can write [tex] A = 4G\hbar \frac{\gamma f'(\gamma)}{1z f(\gamma)}. [/tex] The righthand side of this expression is what we mean by [itex]F(\gamma,\mu)[/itex]. The area [itex]A[/itex] is a fixed input, so it is a transcendental equation that relates [itex]\gamma[/itex] and [itex]\mu[/itex]. We can also note immediately that the leading contribution to the entropy is [tex] S = \frac{\gamma f'(\gamma)}{1z f(\gamma)} +\cdots. [/tex] In terms of microscopic quantities, this looks [itex]\gamma[/itex] dependent, but the area constraint sets it to a macroscopic constant. As for the other terms, there are a variety of ways to express them using the expressions [tex] \bar{N} = \frac{zf}{1zf}, ~~~ zf = \frac{\bar{N}}{\bar{N}+1}.[/tex] In particular [tex] S = \beta \bar{E}  \beta \mu \bar{N} + \log\mathscr{Z} ,[/tex] [tex] = \frac{\gamma f'(\gamma)}{1z f(\gamma)} \beta\mu \frac{zf}{1zf}  \log (1zf),[/tex] [tex]=(\bar{N}+1) \gamma f'  \beta\mu \bar{N} + \log(\bar{N}+1).[/tex] To try to examine these terms, it's useful to work at large [itex]\bar{N}[/itex], for which [tex] e^{\beta\mu} \approx f(\gamma) \approx \frac{2\pi\gamma + 1}{\pi^2\gamma^2} e^{\pi\gamma}.[/tex] One thing to note about this expression is that there doesn't seem to be any limiting value of [itex]\mu[/itex] for which [itex]\gamma\rightarrow 0[/itex]. In any case, we can use this to write [tex]S\approx \bar{N}\gamma f' +\bar{N} \log f + \log\bar{N}.[/tex] The first two terms are roughly of the same order for [itex]\gamma = O(1)[/itex]. The relation between [itex]\gamma[/itex] and [itex]\mu[/itex] is too unwieldy to do much analytically, but maybe some rough numerics could prove insightful. Edit: Actually, when [itex]\mu =0[/itex], [itex]f\approx 1[/itex], so [itex]\log f\approx 0[/itex]. From the 1st term, it turns out that [tex]\bar{N} \approx 0.4227 \frac{A}{4G},[/tex] so the 3rd term goes like [itex]\log A[/itex]. However, as I mentioned earlier, there are other corrections to the partition function that have not been taken into account that contribute to the log. 


#66
May312, 12:24 PM

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Since we're on a new page I should probably recap what the main topic is. Haven't done that for a while.
http://arxiv.org/abs/1204.5122 Entropy of NonExtremal Black Holes from Loop Gravity Eugenio Bianchi (Submitted on 23 Apr 2012) We compute the entropy of nonextremal black holes using the quantum dynamics of Loop Gravity. The horizon entropy is finite, scales linearly with the area A, and reproduces the BekensteinHawking expression S = A/4 with the onefourth coefficient for all values of the Immirzi parameter. The nearhorizon geometry of a nonextremal black hole  as seen by a stationary observer  is described by a Rindler horizon. We introduce the notion of a quantum Rindler horizon in the framework of Loop Gravity. The system is described by a quantum surface and the dynamics is generated by the boost Hamiltonion of Lorentzian Spinfoams. We show that the expectation value of the boost Hamiltonian reproduces the local horizon energy of Frodden, Ghosh and Perez. We study the coupling of the geometry of the quantum horizon to a twolevel system and show that it thermalizes to the local Unruh temperature. The derived values of the energy and the temperature allow one to compute the thermodynamic entropy of the quantum horizon. The relation with the Spinfoam partition function is discussed. 6 pages, 1 figure ==quote first paragraph== There is strong theoretical evidence that Black Holes have a finite thermodynamic entropy equal to one quarter the area A of the horizon [1]. Providing a microscopic derivation of the BekensteinHawking entropy S_{BH} = A/(4G hbar) is a major task for a candidate theory of quantum gravity. Loop Gravity [2] has been shown to provide a geometric explanation of the finiteness of the entropy and of the proportionality to the area of the horizon [3]. The microstates are quantum geometries of the horizon [4]. What has been missing until recently is the identification of the nearhorizon quantum dynamics and a derivation of the universal form of the BekensteinHawking entropy with its 1/4 prefactor. This is achieved in this letter. ==endquote== Over the past year or so there have been several Loop gravity papers by various authors (Ghosh, Perez, Pranzetti, Frodden, Engle, Noui...) supporting this general conclusion. If it is sustained (and I think Bianchi's treatment of it will be, possibly among others) this will constitute a landmark. AFAIK no other approach to Quantum Gravity has achieved such a result at the equivalent level of generality. In stringy context the 1/4 prefactor was derived only for highly special extreme cases not expected to be observed in nature. So it would be natural if Bianchi's paper were to occasion an incredulous outcry from some quarters. We'll have to see if that happens. Anyway the story isn't over, Bianchi and Wieland have a followup paper in the works. Others I mentioned (or forgot to mention) may have as well. A nice choice of units is made in this paper. c = k_{B} = 1, so that at all times one sees the dependence on G and hbar and can immediately see what the effect of varying them would be. IOW time is measured in meters and temperature is measured in joules. In such units the planck area is Ghbar so A/Ghbar, as a ratio of areas, is dimensionless (a unitless number) and also, since k_{B}=1, entropy, which might otherwise be expressed as energy/temperature, turns out to be dimensionless. So the above equation is simply an equality of pure numbers. 


#67
May312, 01:06 PM

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Additional perspective on the significance of the LQG entropy result can be gleaned from this excerpt at the close of Bianchi's conclusion section.
==quote conclusions, page 5== The result obtained directly addresses some of the difficulties found in the original Loop Gravity derivation of BlackHole entropy where the areaensemble is used [3] and the Immirzi parameter shows up as an ambiguity in the expression of the entropy [20]. Introducing the notion of horizon energy in the quantum theory, we find that the entropy of large black holes is independent from the Immirzi parameter. Quantum gravity corrections to the entropy and the temperature of small black holes are expected to depend on the Immirzi parameter. ==endquote== 


#68
May412, 02:37 PM

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Now that we have the "Discrete Symmetries" paper (May 2012 Rov+WilsonEwing) a natural question to ask about Bianchi's entropy paper is what if any changes would follow from changing over to the proposed S' action?
In "Discrete Symmetries" RWE consider the effect of time and parity reversal on the conventional Holst action S[e,ω] that has so far been the basis of covariant LQG, i.e. of spinfoam dynamic geometry. They propose two alternative actions, since these are closely related I will just consider one (S') for simplicity. You can look up the other (S") in their paper if you wish. The classical basis for spinfoam QG is the Holst action. A 4D manifold M equipped with internal Minkowski space M at each point together with a tetrad e (oneforms valued in M) and a connection ω. The conventional Holst action is: S[e,ω]=∫e^{I}Λe^{J}Λ(∗ + 1/γ) F_{I J} Here the ∗ denotes the Hodge dual. Rovelli and WilsonEwing propose a new action S' that uses the signum of det e: s = sign(det e) defined to be zero if det e = 0 and otherwise ±1. S'[e,ω]=∫e^{I}Λe^{J}Λ(s ∗ + 1/γ) F_{I J} =========== http://arxiv.org/abs/1205.0733 Discrete Symmetries in Covariant LQG Carlo Rovelli, Edward WilsonEwing (Submitted on 3 May 2012) We study timereversal and parity on the physical manifold and in internal space in covariant loop gravity. We consider a minor modification of the Holst action which makes it transform coherently under such transformations. The classical theory is not affected but the quantum theory is slightly different. In particular, the simplicity constraints are slightly modified and this restricts orientation flips in a spinfoam to occur only across degenerate regions, thus reducing the sources of potential divergences. 8 pages So what if any effect would the modified simplicity constraints have on the BH entropy results of Bianchi and others? Here is the modified simplicity constraint for S': K+sγL=0 This seems to conflict with the idea in the Bianchi paper of a γsimple representation for which KγL=0 Section IV "Quantum Theory" starting on page 3 of the RWE paper is specifically about this kind of question: "Let us now study the effect of using the modified simplicity condition on the quantum theory. We refer the readers to [1, 9, 11, 12] for the general construction. In the quantum theory, π_{f}^{IJ} is promoted to a quantum operator which is identified as the generator of SL(2, C) over a suitable space formed by SL(2,C) unitary representations. K_{f} and L_{f} are then the generators of boosts and rotations respectively... ...Therefore the key effect of the introduction of the sign s is that the quantum theory now includes both positive and negative k representations..." This seems very interestingI'm just now trying to understand it. 


#69
May612, 02:29 PM

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PF Gold
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If we try to discuss state counting, we should first note that the degenerate faces were always ignored in past calculations. There is a footnote on page 4 of the RovelliWilsonEwing (RWE) paper that claims that they can be erased from the spin network in canonical LQG. We can still allow degenerate edges, which we need in order to glue [itex]k_f>0[/itex] and [itex]k_f<0[/itex] faces together. The states that correspond to the BH entropy calculation can be determined from the usual prescription. We choose a triangulation [itex]\Delta[/itex] and then count the number of faces that pierce the surface of the horizon. The choice of orientation means that nondegenerate faces now come from two bins, so we have to sum over two species of spins. Suppose that we denote [itex]k_f>0[/itex] faces by [itex]N^+_j[/itex] and [itex]k_f<0[/itex] faces by [itex]N^_j[/itex]. If we also include [itex]N_0[/itex] degenerate faces, the number of states is now [tex] W = \frac{ N!}{(N_0)!} \prod_j \frac{(2j+1)^{N^+_j+N^_j}}{N^+_j!N^_j!}.[/tex] There are two constraints, namely [tex] N = N_0 + \sum_j(N^+_j+N^_j),[/tex] [tex] 8\pi G\hbar \gamma \sum_j j(N^+_j+N^_j) = A.[/tex] There is no other constraint on [itex]N_0[/itex]. Having [itex]N^\neq 0[/itex] means that we need to include degenerate edges, but only complete faces contribute to the state counting, not edges. Let's first consider the case that [itex]N_0=0[/itex]. Taking the large N limit and then extremizing the entropy subject to the constraints leads to the occupation numbers [tex] \frac{N^\pm_j}{N} = (2j+1) e^{\mu j},[/tex] where [tex] N = \frac{A}{8\pi G\hbar \gamma\alpha},~~~\mu\sim 2.753,~~~\alpha\sim 0.4801.[/tex] The entropy is [tex] S = \frac{\mu A}{8\pi G\hbar \gamma},[/tex] which results in [tex]\gamma = \frac{\mu}{2\pi} \sim 0.4382.[/tex] So we find the right entropy at a new value of the Immirzi parameter. Now, if we were to allow degenerate faces ([itex]N_0\neq 0[/itex]), we don't have enough information to fix the occupation numbers. In this case, [tex] \frac{N_0}{N} = \frac{1}{1+\sum_j(2j+1) e^{\mu j}} , [/tex] [tex] \frac{N_j}{N} = \frac{N_0}{N} (2j+1) e^{\mu j}.[/tex] The only constraint left is the area constraint and only the nondegenerate faces contribute to that. However, we have two unknowns, [itex]\mu[/itex] and [itex]N_0[/itex]. So we cannot compute the number of degenerate faces at this level of sophistication. There is a physical explanation for this. Namely, it costs very little entropy to replace a pair of spin states (faces) with a degenerate face and a single higher spin face in such a way to keep the area fixed. The amount of entropy is much smaller than the leading term in the large N limit. We can actually use the number of states to determine the change in entropy if we replace a spin [itex]j_1[/itex] and [itex]j_2[/itex] state with a spin [itex]j_1+j_2[/itex] state and a degenerate face. It is [tex] S (N_0+1)  S (N_0) = \ln \left[ \frac{2(j_1+j_2)+1}{(2j_1+1)(2j_2+1)} \frac{N_{j_1} N_{j_2}} {(N_0+1)( N_{j_1+j_2}+1)} \right].[/tex] In the large N limit, we can use the occupation numbers solved for above to find [tex] S (N_0+1)  S (N_0) \sim \mu ( j_1+j_2  (j_1+j_2)) \sim 0.[/tex] It might be useful to find a reference that explains why degenerate faces can be removed from the spin network. 


#70
May712, 11:46 PM

Sci Advisor
P: 8,371

Motl points to an interesting paper by Sen: "we apply Euclidean gravity to compute logarithmic corrections to the entropy of various nonextremal black holes in different dimensions ... For Schwarzschild black holes in four spacetime dimensions the macroscopic result seems to disagree with the existing result in loop quantum gravity."



#71
May812, 07:06 PM

Astronomy
Sci Advisor
PF Gold
P: 23,102

A. it's completely speculative what the best QG formula for BH entropy is. I wouldn't guess or bet unless forced to. We don't know that any particular approach even has the right degrees of freedom to describe a BH quantum geometrically. That includes Sen with the "Euclidean" approach. And of course Nature has the last word. B. It doesn't matter much, but just "for the record" Sen does not accurately reflect what I think are the prevailing ideas of the log term among Loop researchers. He seems off by a factor of 2. It looks on first sight like a factor of 4, but half of that is a difference in notation. C. If I were forced to bet, I'd guess Bianchi (and others who find the areaterm coefficient to be 1/4 independent of Immirzi) are moving in the right direction. I expect followup papers to appear and it would be naive to assume that they will use the same methodology. Insights and methods don't stand still so one cannot predict the future course of research. My post #2 from the other thread says pretty much where I stand. On the other hand Sen says that in the Loop context the log term is log A. IOW off by a factor of two. I suppose he is depending mostly on older or marginal sources. What he actually says is let a be the linear scale of the BH, in other words essentially sqrt(A) then the Loop term is 2log(a). This amounts to the same thing as log(A). It's of little if any consequence. For clarity/completeness, I'll include the rest of my comment: ==quote post #2== These authors have a different log term (see table on page 30) from what Ashoke Sen refers to as characterizing the Loop BH entropy. They say (1/2)log a and he says (on page 28) 2log a. Superficially different at leastperhaps reconcilable but I don't see how. I'm not sure any of that will hold over the long termstill too much technical disagreement. As I guess you are well aware, the question of black hole entropy is not settled in LQG. Even in the pre2012 work, where the authors think that they must specify a value of the Immirzi parameter in order to recover Bek.Hawk semiclassical, they use different enough methods so that some get γ=0.237 and others get γ=0.274. Again see the table on page 30 of the Agullo et al paper. http://arXiv.org/abs/1101.3660 Crisp summary of differences. And then Bianchi posted a paper last month (April 2012) which finds the entropy to be quite different from either group. Basically proportional to area with coefficient 1/4 without fixing the value of Immirzi at all! If I had to bet, I'd guess that Bianchi is closer to being rightthat the BH entropy relation does not require fixing a particular value of Immirzi (a radical innovation in context of earlier work). And Bianchi has not yet worked out the quantum corrections, or any way not posted. His paper does not specifically mention a log term at all. So we'll just have to wait and see if there is a log term and if so what it is. ==endquote== 


#72
May812, 08:48 PM

Sci Advisor
HW Helper
PF Gold
P: 2,606

He also shows that the logarithmic term actually vanishes in the U(1) CS theory after converting to his measure. However this is consistent with completely averaging the SU(2) result over spins. 


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