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Bianchi's entropy result-what to ask, what to learn from it

by marcus
Tags: bianchi, entropy, learn, resultwhat
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Physics Monkey
#19
Apr25-12, 08:35 PM
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Quote Quote by fzero View Post
In that paper, he uses [itex]A_f \sim \sqrt{j_f(j_f+1)}[/itex] and the entropy still has the coefficient of [itex]\gamma[/itex]. The troublesome thing is that, in the new paper

[tex] E \sim \sum_f j_f [/tex]

is not proportional to

[tex] A \sim \sum_f \sqrt{j_f(j_f+1)}.[/tex]

Using the Clausius relation gives a correction to the area law. At first order, the correction is proportional to [itex]N[/itex], the number of facets. The 2010 paper, if it applies here, suggests in eq (19) that [itex] N \sim A[/itex], so this corrects the coefficient of the leading term (away from 1/4).
I didn't even notice this at first, but it looks like bianchi is either doing the large j limit or made an important mistake?
Physics Monkey
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Apr25-12, 08:39 PM
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Quote Quote by atyy View Post
That's fascinating. In the original case of a Lorentz invariant field theory on the half space, are there also "edge states"?
There certainly can be. Not all Lorentz invariant theories have protected physical edge states on a half space, but we showed that if they do then the half space entanglement spectrum (with no physical edge) also has the universal aspects of these physical edge states.
Physics Monkey
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Apr25-12, 08:46 PM
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Quote Quote by marcus View Post
I've been appreciating your comments, since you know a lot about this. I'm glad you took an interest and read the paper. Part of the confusion could be due to problems with notation. I think of what we have now as a draft to which more explanation could be added.

I could be wrong but I don't think it says "E and A don't commute". The OPERATORS for energy and area are denoted H and A, are they not?
The letter E seems to denote a quantity. At one point he says E = <s|H|s>, so as a quantity it would commute with everything I suppose.

The energy operator H is defined by eqn (8) and seems to be composed of boost pieces. The area operator seems to be composed of rotation pieces. Correct me if I'm wrong these don't commute as operators, do they? Equation (6) just says they have the same matrix element form. Let me know if I'm saying something really stupid. So anyway I think on page 3, middle of first column, where he says "the energy does not commute with the area of the quantum horizon" what he means is "H and A don't commute."

Is this right? You are by far the expert in this context.
You're certainly right that what I mean is H and A. However, as we are discussing with fzero, it seems like Bianchi has used a strange expression for A (roughly just Lz) and Lz and Kz do commute (according to the lorentz algebra). Besides this issue I'm also confused about the state space, because if Eq. 6 holds for matrix elements between physical states, then it also follows that H and A commute. Now for some reason one seems to be using a vastly expanded set of sets where K and L are independent to analyze the physics, but I don't understand all these extra states and what they mean geometrically. I thought the physical Hilbert space was specified by the spin network states, that is the state of geometry at a fixed time. This also seems to be crucial for the detector analysis below where, under my naive reading, Bianchi uses both states of geometry and states of "energy" independently i.e. as a tensor product, which suggests K and L act on different spaces and hence commute.
fzero
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Apr25-12, 08:49 PM
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Quote Quote by Physics Monkey View Post
I didn't even notice this at first, but it looks like bianchi is either doing the large j limit or made an important mistake?
Well I made a mistake. The corrections go like

[tex] \sum_f \frac{1}{j_f^p},[/tex]

so presumably these converge and go to zero in the large [itex]j_f[/itex] limit. So maybe it's not so bad, but it would help to clarify the role of the limit.
Physics Monkey
#23
Apr25-12, 08:53 PM
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Quote Quote by fzero View Post
Well I made a mistake. The corrections go like

[tex] \sum_f \frac{1}{j_f^p},[/tex]

so presumably these converge and go to zero in the large [itex]j_f[/itex] limit. So maybe it's not so bad, but it would help to clarify the role of the limit.
I'm not sure what mistake you're referring to, but I think it could still be an issue. As I recall, in the very old days I heard that j=1/2 states played an important role for which obviously the different is substantial. Also in Fig. 1 Bianchi uses j=1 as an example, so I'm not convinced that Bianchi just meant for us to assume large j.
fzero
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Apr25-12, 09:05 PM
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Quote Quote by Physics Monkey View Post
I'm not sure what mistake you're referring to, but I think it could still be an issue. As I recall, in the very old days I heard that j=1/2 states played an important role for which obviously the different is substantial. Also in Fig. 1 Bianchi uses j=1 as an example, so I'm not convinced that Bianchi just meant for us to assume large j.
Yes, I don't have an authoritative reference handy, but as I recall from earlier discussions, the large j limit (perhaps with j/N held fixed) is supposed to be a semiclassical limit. Bianchi does remark that his result does not include quantum corrections (below (3) and in the last paragraph of the conclusion).
marcus
#25
Apr25-12, 09:08 PM
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Quote Quote by Physics Monkey View Post
... As I recall, in the very old days I heard that j=1/2 states played an important role for which obviously the different is substantial. Also in Fig. 1 Bianchi uses j=1 as an example, so I'm not convinced that Bianchi just meant for us to assume large j.
I agree (with Phy. Monk. post #23) As I recall j=1/2 predominates.
We still have to find out how |Letc| is defined. I don't see it. There could be a discrepancy that needs to be fixed. Possibly just a typo. Or it could be all right and I'm just missing something.

What we want is for |L_f| = j_f
(rather than the sqrt of j(j+1).) Could things have actually been defined so it comes out that way?
DrDu
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Apr26-12, 07:17 AM
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Quote Quote by marcus View Post
Likewise black hole horizon temperature is highly dependent on how far away the observer is hovering.
Why? Temperature is defined by two systems being in thermal equilibrium when they have the same temperature and by transitivity of the equilibrium. So if I put thermometer at large distance from the horizon and it is in thermal equilibrium with the black hole horizon, then any intermediate system which is in thermal equilibrium will be assigned the same temperature.
Physics Monkey
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Apr26-12, 07:44 AM
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I reached out to bianchi for clarification about his area formula. In the interest of keeping his privacy, I will just summarize the main points of his brief reply that are apparently common knowledge.

In short, both [itex] \sqrt{j(j+1)} [/itex] and [itex] j [/itex] are acceptable area operators (they differ by an operator ordering ambiguity that vanishes as [itex] \hbar \rightarrow 0 [/itex] (which I guess here means something like [itex] j \rightarrow \infty [/itex] as fzero and others suggested).

The two criteria for an area operator are apparently 1) that its eigenvalues go to j in the large j limit and 2) that its eigenvalue vanish for j=0.

More systematically, bianchi is using a Schwinger oscillator type representation where we have two operators [itex] a_i [/itex] and the spins are [itex] \vec{J} = \frac{1}{2} a^+ \vec{\sigma} a [/itex]. The total spin of the representation can be read off from the total number [itex] N = a_1^+ a_1 + a_2^+ a_2 = 2j [/itex]. On the other hand, you can work out [itex] J^2 [/itex] for yourself to find [itex] J^2 = \frac{1}{4}( N^2 + 2N) [/itex] which one easily verifies gives [itex] J^2 = j(j+1) [/itex]. Thus by [itex] |\vec{L}| [/itex] bianchi appears to mean [itex] N/2 [/itex].

It is again interesting to see this kind of representation appearing in a useful way since it is quite important in condensed matter.
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Apr26-12, 08:04 AM
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Quote Quote by DrDu View Post
Why? Temperature is defined by two systems being in thermal equilibrium when they have the same temperature and by transitivity of the equilibrium. So if I put thermometer at large distance from the horizon and it is in thermal equilibrium with the black hole horizon, then any intermediate system which is in thermal equilibrium will be assigned the same temperature.
At the most basic level, this is gravitational red shift at work. There is a notion of local temperature given by [itex] T_\ell(r) = T/\sqrt{-g_{tt}(r)} [/itex] which in something like the scharzchild metric [itex] g_{tt}(r) = -(1-r/r_S) [/itex] gives a diverging temperature as the horizon is approached.

One physical meaning one can attach to this expression is the following. If an observer of four velocity [itex] u [/itex] (with [itex] u^2 = -1[/itex]) measures the energy of a particle with four momentum [itex] p [/itex] then the energy measured is [itex] E = u\cdot p [/itex]. An observer hovering about the black hole horizon has [itex] u = (1/\sqrt{-g_{tt}}) \partial_t [/itex]. On the other hand, the existence of the Killing vector [itex] \xi = \partial_t [/itex] means that the quantity [itex] \xi\cdot p [/itex] is conserved independent of r provided the particle of momentum p is following a geodesic. This quantity [itex] \xi \cdot p [/itex] is the energy of the particle measured at infinity [itex] E_\infty = \xi \cdot p [/itex]. Thus one sees that the energy measured by a hovering observer near the horizon is [itex] E = E_{\infty} /\sqrt{-g_{tt}} [/itex] which diverges as the hovering observer approaches the horizon.

Another perspective is that the hovering observer must fire her engines harder and harder to keep from falling as the horizon is approached. From the perspective of an inertial infalling observer the hovering observer is uniformly accelerated and hence experiences unruh radiation. The temperature of this radiation gets hotter and hotter as the hovering observer approaches the horizon (while the infalling observer sees nothing).

Still another point of contact is Luttinger's old idea in condensed matter to model position dependent temperature by adding a spatially varying gravitational field e.g. for computing heat currents using Kubo formulae. Of course, this is most obvious in the Lorentz invariant context when one computes variations with respect to a background metric to evaluate correlators of the stress tensor i.e. [itex] \delta S = \int d^D x \sqrt{-g} \delta g_{\mu \nu} T^{\mu \nu} [/itex].
DrDu
#29
Apr26-12, 08:08 AM
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Yes, I know this local temperature concept. However I think it has no relevance in defining entropy.
Physics Monkey
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Apr26-12, 08:21 AM
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Well, I think it is a subtle notion, but it can be useful for studying entropy.

For example, consider Rindler space [itex] ds^2 = - \rho^2 d \eta^2 + d\rho^2 + dx_\perp^2 [/itex] where [itex] \eta [/itex] is like a time and [itex] \rho= 0[/itex] is the horizon. Imagine a CFT living in this space. Then one gets an entropy density of roughly [itex] s \sim T_\ell^d [/itex] (with d the space dimension). Integrating this over all space gives [tex] \int d\rho dx^{d-1}_\perp \rho^{-d} \sim \frac{A_\perp}{\rho^{d-1}_c} .[/tex] [itex] A_\perp [/itex] is the cross sectional area and [itex] \rho_c [/itex] is some small distance cutoff. Putting something like the planck length in as a fundamental cutoff gives something that looks an awful lot like black hole entropy.

Viewing the CFT in rindler space as an approximation to the near horizon of a black hole with matter, it looks like this calculation is giving a quantum correction due to the matter fields, the CFT, to the entropy of the black hole.
Demystifier
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Apr27-12, 06:19 AM
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Quote Quote by atyy View Post
If there's no state counting, isn't this just a semiclassical calculation, like Hawking's?
I would put this way: If there's no state counting, then the entropy is merely a thermodynamic entropy (like Hawking's), not a statistical entropy.
francesca
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Apr27-12, 04:37 PM
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Quote Quote by Demystifier View Post
I would put this way: If there's no state counting, then the entropy is merely a thermodynamic entropy (like Hawking's), not a statistical entropy.
That's right, but it would not say "merely": the microstate counting was already done in LQG, and Eugenio contributed to this with his nice paper "BH Entropy, Loop Gravity, and Polymer Physics", identying these microstates with the quantum geometry at the horizon. SO the "thermodynamical side" was the aspect somehow lacking... lacking in which sense? in the sense that Hawking calculation was done in the context of QFT in curved space, so spacetime was classical, while Eugenio's one is the thermodynamics of a quantum theory of gravity.
marcus
#33
Apr27-12, 06:23 PM
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Francesca, these are very instructive replies to Demy and to Atyy's question.
Quote Quote by atyy View Post
If there's no state counting, isn't this just a semiclassical calculation, like Hawking's?
Quote Quote by francesca View Post
No, it's not semiclassical. In fact in the paper all the ingredients are derived from the full quantum theory. The relation energy-area was found by Frodden Gosh Perez using Einstein equations; here it is found using the boost generator given by Spinfoam Theory. The calculation in order to find Unruh temperature is done here again using the boost generator, it's completely new. And finally there is the remarkable demonstration that the Spinfoam amplitude implies the right distribution, that yields Hawking entropy.
Quote Quote by Demystifier View Post
I would put this way: If there's no state counting, then the entropy is merely a thermodynamic entropy (like Hawking's), not a statistical entropy.
Quote Quote by francesca View Post
That's right, but it would not say "merely": the microstate counting was already done in LQG, and Eugenio contributed to this with his nice paper "BH Entropy, Loop Gravity, and Polymer Physics", identying these microstates with the quantum geometry at the horizon. SO the "thermodynamical side" was the aspect somehow lacking... lacking in which sense? in the sense that Hawking calculation was done in the context of QFT in curved space, so spacetime was classical, while Eugenio's one is the thermodynamics of a quantum theory of gravity.
Indeed we are talking about Black Hole thermodynamics and the Bekenstein Hawking entropy SBekHaw was thermodynamic. So likewise should Bianchi black hole entropy SBianchi be thermodynamic according to Clausius. This was one of the aims and achievements of the paper. And he evaluates it in a purely quantum way. Not in the context of QFT on a fixed curved space. So definitely not semiclassical . And as you said earlier, it is completely new!
The researchers might be able to learn something by comparing Bianchi's thermodynamic entropy with the statistical state-counting done earlier. I imagine some are studying this comparison and it may prove fruitful.

I was interested by what you said in reference to the last section of the paper "Partition Function and Spinfoams".
That section goes beyond the main task of the paper and seems to point towards further work. Was that what you were referring to when you said "And finally there is the remarkable demonstration that the Spinfoam amplitude implies the right distribution, that yields Hawking entropy." For me, that section needs more elaboration--it does not explain enough what is being done--but I suspect that would be more appropriately done in a second paper, not to overextend this one.
fzero
#34
Apr27-12, 07:56 PM
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The polymer microstate calculation had an explicit dependence on the Immirzi parameter. The only reason the present calculation does not have this dependence is because Bianchi uses a single pure state to do the calculation. Since the BH is not a pure state, the correct way to do the computation is to compute the energy from (9) in an ensemble. This will reintroduce the factors of [itex]\mu^*[/itex] and [itex]\gamma[/itex] that were found in the polymer paper.
marcus
#35
Apr27-12, 08:43 PM
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Quote Quote by fzero View Post
... Since the BH is not a pure state, the correct way to do the computation is to compute the energy from (9) in an ensemble. This will reintroduce the factors of [itex]\mu^*[/itex] and [itex]\gamma[/itex] that were found in the polymer paper.
Why so? I see no reason that combining states of the form (9) to make a mixed state would need to introduce a [itex]\gamma[/itex]. Please explain.
marcus
#36
Apr27-12, 08:48 PM
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In case anyone is new to the discussion the "polymer" paper just referred to is from a years and a half ago and is:
http://arxiv.org/abs/1011.5628
Black Hole Entropy, Loop Gravity, and Polymer Physics
Eugenio Bianchi
(Submitted on 25 Nov 2010)
Loop Gravity provides a microscopic derivation of Black Hole entropy. In this paper, I show that the microstates counted admit a semiclassical description in terms of shapes of a tessellated horizon. The counting of microstates and the computation of the entropy can be done via a mapping to an equivalent statistical mechanical problem: the counting of conformations of a closed polymer chain. This correspondence suggests a number of intriguing relations between the thermodynamics of Black Holes and the physics of polymers.
13 pages, 2 figures

The main paper we are discussing is the one Bianchi just posted this week. For convenience, since we just turned a page, I will give the link and abstract again:
http://arxiv.org/abs/1204.5122
Entropy of Non-Extremal Black Holes from Loop Gravity
Eugenio Bianchi
(Submitted on 23 Apr 2012)
We compute the entropy of non-extremal black holes using the quantum dynamics of Loop Gravity. The horizon entropy is finite, scales linearly with the area A, and reproduces the Bekenstein-Hawking expression S = A/4 with the one-fourth coefficient for all values of the Immirzi parameter. The near-horizon geometry of a non-extremal 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 two-level 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


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