Does the universe obey the holographic principle due to Stokes' theorem?

Question

\begin{equation} \int\limits_{\partial\Omega}\omega = \int\limits_{\Omega}\mathrm{d}\omega. \end{equation}

Can this theorem be enough proof of our Universe being a hologram – the choice of $\omega$ and $\Omega$ is completely arbitrary!

score 16 · Accepted Answer · answered Dec 16 '12 at 09:52

16

No, it cannot be enough. Stokes' theorem says that the volume ($\Omega$) integral of $d\omega$, a form that is the exterior derivative of another one (of $\omega$), may be written as a surface integral. But it doesn't allow us to rewrite the volume integral of a general integrand (which isn't the exterior derivative of anything) such as the Lagrangian density ${\mathcal L}$ as a surface integral. So the Stokes' theorem is useless for dealing e.g. with the action $S$ that defines the dynamics of a general theory in the volume.

One should mention that when the action is topologically invariant, ${\mathcal L}$ may indeed be locally written as a "total derivative", and in that case, the theory has indeed a provable relationship with lower-dimensional theories (a major example is Chern-Simons theory in 3 dimensions and the related WZNW theories in 2D). But the general theories we know – the Standard Model coupled to gravity – aren't of this special type, at least not manifestly so. What's happening in the volume is general – we surely do care about values of some fields such as the electric field in particular places of the volume – and there apparently isn't any "counterpart degree of freedom" on the surface that we could associate it with.

Some people including Leonard Susskind and Steve Shenker etc. do suspect that there exists some "conceptually simple" proof of the holography in which almost all the degrees of freedom in the volume would be unphysical or topological – some huge gauge symmetry that allows one to eliminate all the bulk degrees of freedom except for some leftovers on the surface. But such a proof of holography remains a wishful thinking. Meanwhile, we have several frameworks – especially the AdS/CFT – that seem to unmask the actual logic behind holography. The surface theory is inevitably "strongly coupled" (i.e. strongly dependent on quantum corrections) if the volume description appears at all so things can't be as simple as you suggest, it seems.

answered Dec 16 '12 at 09:52

Luboš Motl

179,018

"Some people including Leonard Susskind and Steve Shenker etc. do suspect that there exists some "conceptually simple" proof of the holography in which almost all the degrees of freedom in the volume would be unphysical or topological – some huge gauge symmetry that allows one to eliminate all the bulk degrees of freedom except for some leftovers on the surface."

Isn't this essentially guaranteed to be true? Not that it must be conceptually simple, but that in order for two theories to describe the same physics, they must have the same # of degrees of freedom => gauge symmetry must exist
– reductionista Sep 11 '22 at 05:08
I don't think it is true. Having the same entropy or number of states is very far from having two equivalent theories. I think that AdS is the only example where the theory on the boundary is "local" and it is only true because the warp factor is infinite at the AdS boundary. I think that the description for finite regions doesn't exist and even if it did, it is in no way a simple local theory on the boundary. – Luboš Motl Sep 15 '22 at 17:22
Ok, I suppose my argument fails for infinite systems, since different dimensions all have the same cardinality. And it sounds like you don't think there will ever be an actual holographic duality between finite systems. But even so, I think a gauge symmetry is still necessary even for the Bousso bound to be true... otherwise you'd be stuck with too much entropy & too many degrees of freedom. – reductionista Sep 18 '22 at 05:58
First, I am near certain that "cardinality" in the sense of set theory isn't a physically relevant notion. Cardinality is the number of elements (points) which are counted with arbitrary fineness but physics always involves an explicit or de facto cutoff and distances shorter than the cutoff aren't fully distinguished, become fuzzy etc., so you just can't count points this accurately in the sense of set theory. On the other hand, the amount of structure that physics puts in the spacetime is much greater than you imagine, and much greater than counting the points (the latter is meaningless). – Luboš Motl Sep 19 '22 at 06:32
Second, and it is related, your "proof" of a gauge symmetry is clearly wrong. You can't prove the "existence of gauge symmetry" by any counting of degrees of freedom. What is true is that holography or Bekenstein (Bousso's?) bounds eliminate the description of the physics in terms of local field theory whose entropy scales like the volume. But that doesn't mean that the right theory is a local theory with a gauge symmetry. Almost certainly, the correct theory cannot have this form. – Luboš Motl Sep 19 '22 at 06:33
On top of that, "what is the gauge symmetry of a physical system" is a physically meaningless question, too. A gauge symmetry is just a piece of a formalism but the same physics may admit many possible descriptions, with or without gauge symmetries, or with different symmetries. The space of possible physical theories is much wider than you are imagining (some local theories) and quantum gravity almost certainly requires a theory that you assume not to exist. – Luboš Motl Sep 19 '22 at 06:35
"You can't prove the "existence of gauge symmetry" by any counting of degrees of freedom." You must have a different definition of what a "gauge symmetry" is then. The definition as I understand it is a Hilbert space where some of the degrees of freedom are redundant / correlated / dependent. Is there an alternate definition I'm unaware of? – reductionista Sep 22 '22 at 07:06
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I think I understand your last comment though: we agree that quantum gravity is a non-local theory. But you are skeptical that there is even any way of writing it in a way that looks local (by introducing extra gauge freedom). That makes more sense than I thought. – reductionista Sep 22 '22 at 07:09
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Yes, gauge symmetry is just a redundancy. But I am just saying that the original, larger Hilbert space with the unphysical states isn't given by physics, that is why these extra states are unphysical. And constructing either Hilbert space has more structure than just counting states. – Luboš Motl Sep 23 '22 at 03:09
At the level of counting, all Hilbert spaces are the same. Infinite dimensional complex vector spaces. This is not enough for making predictions. You need some structure, observables. Know how the observables are associated with points in the bulk or the surface. – Luboš Motl Sep 23 '22 at 03:12
I guess if you believe in the String Theory Landscape, all backgrounds must be part of the same Hilbert space and therefore all infinite dimensional. But personally, I tend to agree with our mutual advisor Tom that an asymptotically deSitter space like ours ought to be finite dimensional, due to black hole complimentarily. I recall a conversation once where Michael Dine asked him why Lenny doesn't agree. His reply: "He doesn't have any argument, he just throws up his hands and says 'I don't believe it!"' – reductionista Sep 23 '22 at 19:01
Hi, I have had interactions with Tom about the finite dimensional Hilbert space for de Sitter since tasi 99 in Colorado. It is a dogma for him and all wouldbe arguments are spurious. In particular, Tom incorrectly assumes that a finite entropy demands a finite number of states. That is wrong. A volume of photon gas at some temperature also has a finite entropy even though the Fock space is infinite dimensional. Most of the states are just guaranteed by the distribution to have decreasing and low probabilities. – Luboš Motl Sep 25 '22 at 05:57
At any rate, I didn't even want to assume that Tom was wrong. I just defined the Hilbert space to be infinite dimensional. Sometimes I allow the finite dimensional ones to be called Hilbert spaces, too. It is just terminology. And one that has nothing to do with the previous questions in this thread. – Luboš Motl Sep 25 '22 at 05:58
Tasi 2007 for me, "Dawn of the LHC Era!" The phase space of a photon gas is only infinite-dimensional because there is no UV cutoff. Given both an IR cutoff (deSitter horizon) and a UV cutoff (Planck scale) I don't see how the number of orthogonal states could be anything but finite. The only way entropy can be less than the log of the number of microstates is if part of phase space is inaccessible. But the holographic bound is not entropy at a specific energy or temperature, it's maximum possible allowed entropy! Fundamentally inaccessible states would be unobservable. – reductionista Sep 26 '22 at 22:59
This is getting long though, maybe I should ask this as a separate question? – reductionista Sep 26 '22 at 23:04
Correction: Tasi 2008 – reductionista Sep 27 '22 at 20:22
You may say that the exponential of an entropy is the effective number of relevant states but it doesn't change anything about the fact that the theory needs infinitely many states in total. This is not just a fancy question about quantum gravity. Understanding thermodynamics of a harmonic oscillator is surely enough. I don't want to waste more time with this trivial point, who doesn't get it immediately is just dumb. – Luboš Motl Sep 28 '22 at 03:27
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The comment that the de Sitter entropy is the maximum possible entropy is correct but the conditions needed for the statement are obfuscated. It is the maximum entropy if you assume that the physical configuration looks like a de Sitter space of a given cosmological constant macroscopically. But that doesn't change about the unavoidable existence of further states in the same theory that can no longer be classified as objects in the preagreed de Sitter space of the given size. The real point is that the transition between "allowed states" and "not allowed states" is always continuous. – Luboš Motl Sep 28 '22 at 05:34
At the end, your or Tom's selection of a particular finite-dimensional space is nothing else than the idea that you may always construct an interesting theory with a microcanonical ensemble. But microcanonical ensembles are always physically unnatural. It is always the (grand) canonical ones that have natural mathematical formulae and the included microstates are always just those "near enough" the right energy etc. but there is no sharp separation between the included and forbidden microstates and the total number of states is always infinite. – Luboš Motl Sep 28 '22 at 05:36
The insistence on a finite-dimensional Hilbert space is ultimately the same dogma of the discrete physics pseudoscience favored by loop quantum gravity, Wolfram's cellular automatons, and all this ultrastupid nonsense. These things have no justification and have led to no confirmations, promising results, or interesting theories. In fact, it is almost certain that there exist no interesting predictive theories on finite-dimensional Hilbert spaces. Observables on finite-dimensional spaces are just the classes of all Hermitian matrices, none of them is more consistent than others. – Luboš Motl Sep 28 '22 at 05:37
It is only infinite-dimensional Hilbert spaces where predictive theories and special structures may emerge. Whether a spectrum of a localized object may be derived as a "spectrum of a string in string theory" is only meaningful in the context of infinite-dimensional Hilbert spaces, for example. All the arguments in favor of finite-dimensional spaces are a combination of sloppy reasoning and irrational dogmas but even more importantly, I think it can be almost rigorously proven that this philosophical "axiomatic framework" can never lead to any interesting mathematical structures. – Luboš Motl Sep 28 '22 at 05:38
Your point that most of the states in the Hilbert space may be finite deviations from pure deSitter space rather than small excitations within a fixed background with fixed $\Lambda$ is enlightening. Tom of course looks at it from exactly the opposite point of view, arguing vehemently that different asymptotic backgrounds are necessarily separate Hilbert spaces. But I'd have to agree, his arguments seem weak on that point. And certainly I would like to believe there is a more rich structure to the universe than just finite discrete math. But I wouldn't rule anything out based on aesthetics. – reductionista Sep 29 '22 at 08:04
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I have learned a lot from this conversation, thanks for not giving up on me! I will mull it all over. – reductionista Sep 29 '22 at 08:10
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Thanks for the chat and stimulating questions and feedback. – Luboš Motl Sep 29 '22 at 13:41
I could have said it more clearly. The finite dimensional Hil. space for a given value of the cosmological constant Lambda means that there is some quantization rule for Lambda that happens to accidentally produce a very large degeneracy of states, exp(S) states for a precise value of Lambda. I find it implausible. Because dS doesn't have an asymptotic region of infinite size, even the value of Lambda cannot be measured very precisely. So it is some observable that won't commute with most other observables and will have some uncertainty, that's why an interval of Lambda needed for many states – Luboš Motl Sep 29 '22 at 14:10
We need to allow a finite interval for a "microcanonical ensemble" of dS-looking states, or allow infinitely many of them with all Lambda and combine them in a grand canonical fashion. This is just a modest statement in the direction that Vafa believes. He believes that dS is entirely impossible in string theory and all dS-resembling states are excited quasistable states in sectors that must look like AdS of flat space if extended. I think that there is no full proof of this or its negation but it's possible. – Luboš Motl Sep 29 '22 at 14:12
But it is surely true that demanding Lambda to be fixed with exponential precision seems very unnatural and probably impossible, and even if this exponential precision for lambda were physically meaningful, there is no reason to expect an exponentially high degeneracy for a chosen value of Lambda that would cover all the states. So since 2000 when these de Sitter discussions in string theory started, I was inclined to think that all physics in dS space had some unavoidable minimal errors, like from the de Sitter temperature which is noznero and reduces predictivity. – Luboš Motl Sep 29 '22 at 14:13
Both Banks and Susskind, among many others, used to believe exactly the same but both men and many others have switched their opinions to clearly contradictory theses later and they have never quite explained why they did the switch. Postulating a very high integer as the dimension of some special de Sitter sector may look like a bold hypothesis but the boldness isn't enough to make it true and when looked as the structure of theories, it is actually a very cowardly assumption, not a bold one. ;-) – Luboš Motl Sep 29 '22 at 14:15

Does the universe obey the holographic principle due to Stokes' theorem?

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