|WikiProject Physics||(Rated Start-class, Mid-importance)|
- 1 Energy-mass limit in given finite space
- 2 Comparison with applied information density?
- 3 Conditions of Application
- 4 Citation overkill
- 5 Silly numbers
- 6 Contradicts Clausius's Principle that the Entropy of the Universe Tends to a Maximum
- 7 Introduction -- Does Bekenstein bound really make an infinite Turing machine tape impossible?
- 8 Comparison with actual hardware
- 9 Examples Section
- 10 Cellular automata
- 11 Bekenstein non-computational limit
- 12 Should the word "bits" in the article be replaced by "qubits"?
- 13 Assessment comment
Energy-mass limit in given finite space
Need a little more clarification for laymen: there's a bound for the amount of information in a finite region of space (Bekenstein bound, which depends on the amount of energy-mass in that region. Does it mean if there's an infinite amount of energy in a region of space then there's an infinite amount of information? Should there be an upper bound for energy-mass if Pauli exclusion principle really holds true? Also, if that's the case then singularity is ruled out?Mastertek (talk) 14:50, 23 October 2011 (UTC)
Comparison with applied information density?
Would it be appropriate to state the maximum information density in bits per cubic meter (or cubic millimeter)? This looks like a straightforward calculation from the S = A/4 formula, but I don't know if there are gotchas in making this kind of comparison. I assume the applied kinds of densities in places like Computer storage density are many orders of magnitude from what is discussed here, but are there even deeper reasons why they are talking about two different things, or is it potentially comparable? (either a "yes" or a "no" might be good to add to the article). Kingdon 19:53, 25 May 2007 (UTC)
- You sure would think this would be possible, but it's not. Information capacity is constrained not by volume, as you'd think, but by the area surrounding the volume. The consequence are deep: the holographic principle. PhysPhD 22:25, 25 May 2007 (UTC)
- Hmm. Thanks for the answer. Kingdon 00:41, 27 May 2007 (UTC)
- Integrate gravitational potential properly defined over the surface and it is proportional to max entropy or information. Nothing holographic required.
- Kingdon, it is clear from the formulae that the maximum information density is proportional to both the radius (R), and the mass-energy (E). If you think about it, you get more quantum states with more mass, (the dense and dominant form of mass-energy), and it takes more information to describe their locations if the space in which they are located is larger. I think the kA/4 limit is a special case for black holes, and perhaps other other cases of "unity."
- Funny that you should mention maximum information storage capacity, though. When I was in school in the 1960s, there was a window display in the lobby of the UIUC Computer Sciences building. Sample modules from computers the department had either built or used were arranged, museum-like, from left to right in chronological order with placards identifying them, so you could see the progress of technology. Some prankster had placed a large rock rightmost in the display. Next to it was a placard displaying what is now apparently being referred to as the "Bekenstein bound" for the rock, and an explanation that it was the most powerful computer, but lacked I/O. Entertaining & educational, eh?
- But this example predates all literature on this subject referred to in this article, as well as on Bekenstein's wikipedia biography page. Perhaps this may be part of the reason for the "citation overkill" alluded to, below. Good luck! SJGooch (talk) 11:46, 7 January 2011 (UTC)
Conditions of Application
I am no expert, but when it says in the first paragraph, "... or the information necessary to perfectly describe that system, must be finite if the region of space or the energy is finite," doesn't the author actually mean 'AND', as in, "if the region of space AND the energy is finite?" The Berkenstein bound is proportional to both the radius, R, and the mass-energy, E, so if either quantity is unbounded, the resulting Berkenstein calculation will be without bound. Right? If I haven't misunderstood, would someone knowledgeable please fix this? Thanks. SJGooch (talk) 10:40, 7 January 2011 (UTC)
- I think they do, but I'll check the context.-- cheers, Michael C. Price talk 11:11, 7 January 2011 (UTC)
- That's correct. If either the energy or radius are unbounded then the information can be unbounded, even if one of those two variables are kept constant.--Jamie Michelle (talk) 03:30, 8 January 2011 (UTC)
- Likely better to have one good ref that clearly states the facts, and then have a nother section on Berkenstein's work using the concept, and references each statement about the papers. Graeme Bartlett (talk) 19:37, 7 November 2010 (UTC)
- It doesn't qualify for the definition given by Wikipedia of "citation clutter": i.e., "citations that are either mirror pages or just parrot the other sources". Each cited work addresses different aspects of the Bekenstein bound rather than parroting them (indeed, this is one of the requirements for peer-reviewed papers), and the papers cited are each important for what they contribute to the understanding of the Bekenstein bound, so one cannot simply do away with some of the citations without detracting from the article.
- The Bekenstein bound is a deep and important subject within physics. One can ignore all of the citations if one wishes, as most casual readers will do. For those readers who desire more information, the citations are there.--Jamie Michelle (talk) 02:56, 28 December 2010 (UTC)
In the end of the section on the human brain, how can 2^(n*10^42) be LESS THAN 10^42 ??? (something x 10^41) I.e. the author claims the number of STATES a string of bits can assume is LESS THAN the number of BITS. This is nonsense! — Preceding unsigned comment added by 188.8.131.52 (talk) 23:01, 12 October 2012 (UTC)
8 bits can have 2^8, ie 256 states. N bits can have 2^N states. — Preceding unsigned comment added by 184.108.40.206 (talk) 03:30, 8 September 2015 (UTC)
Contradicts Clausius's Principle that the Entropy of the Universe Tends to a Maximum
- The inequality for the ratio S/E has meaning only when the entropy is strictly proportional to the number of particles, i.e. an absolute entropy with no additive constants. [Otherwise, I could always add on an entropy constant so that the inequality is violated.] R must be interpreted as the (relativistic) thermal wavelength and the inequality must be inverted. Then it will say that the slope S/E from the origin to a point is always greater than that point on the entropy curve precisely because it is concave, i.e. it bends toward the energy axis. If I is the Shannon information measure the inequality makes no sense at all. It is also apparent that the inequality can not be valid for both concave and convex functions. Only the former are candidates for the entropy. See "Boltzmann's Heat Death versus Bekenstein's Bound at www.bernardhlavenda.com and references thereinBernhlav (talk) 16:44, 4 September 2012 (UTC)Bernhlav
Introduction -- Does Bekenstein bound really make an infinite Turing machine tape impossible?
Bekenstein's bound implies that roughly spherical objects like stars collapse into black holes if too much mass is packed within too small a radius. But a Turing machine could have an infinite rod-like tape, with a finite-sized head crawling back and forth along it, without any part of it exceeding the Bekenstein bound. Does anyone know of any argument why an extended rod-like object must suffer gravitational collapse if it gets too long? An easy calculation shows that the compressive force in an infinite rod due to self-gravitation remains finite, so the rod would not even need to be particularly strong. Maybe it would need to be infinitely stiff to avoid buckling, but that is a harder calculation. However that may be, the assertion that the Bekenstein bound prevents infinite Turing machines from existing requires further demonstration or documentation.CharlesHBennett (talk) 01:00, 11 April 2013 (UTC)
Agreed. Even if such as restriction exists, it is not implied by the bound itself. Added the "finite physical dimensions" qualification (that includes finite length, width, height, AND energy; I hope this meaning is clear, if not, the text should be reworded). Raven lv (talk) 16:20, 25 May 2013 (UTC)
Agreed too. The point is that if any law of physics makes Turing machines impossible this does not seem to be the Bekenstein bound. What one needs i just a large enough universe to create a tape that is as long as needed. If there is a problem with this it does not seem to stem from the Bekenstein bound. Moreover, the sentence is obscurely phrased: what does 'with finite dimension and unbounded memory' mean? Does it mean that the head-read head is finite but the tape can be as long as wanted (but still finite)? If yes, it should be said more clearly and avoid suggesting that there are Turing machines with infinite dimension (of what? of the head? why would one want the head to be infinite?) User:Mbtnt\Mbtnt (User talk:Mbtnt\Mbtnt) (11:46am Nov 7 2017) — Preceding unsigned comment added by Mbtnt (talk • contribs) 16:47, 7 November 2017 (UTC)
Comparison with actual hardware
Since this is included in Limits to computation, please include comparisons with the storage density of actual modern hardware, and with the theoretical maximum storage density of magnetic media — Preceding unsigned comment added by 220.127.116.11 (talk) 20:38, 5 May 2013 (UTC)
At the end of the Examples section, we have the lines "The existence of Bekenstein bound implies that the storage capacity of human brain is finite, although very large. This implication has important consequences on mind uploading, making it theoretically possible, given that physicalism is true."
This implies that the Bekenstein bound has relevance for the human brain which is manifestly false. Human brains, indeed all matter, doesn't come even remotely close to approaching the bound. No matter what the bound is or indeed whether a bound exists at all has, I think, no relevance to human brains or to mind uploading. Is it ok with everyone if I delete the lines in question? Tmfs10 (talk) 21:47, 3 June 2013 (UTC)
Disagree. The mind uploading article explicitly mentions Bekenstein bound. At the same time, the brain example was removed from the Bekenstein bound page earlier, citing absence of justification as the reason. I readded the example and provided justification. I support that a clarification may be added, citing current research results about the estimated information storage capacity in brains and its comparison with the bound. However, perhaps the bound is more relevant to simulated reality hypothesis, and mentioning it would provide better justification. Raven lv (talk) 20:55, 18 June 2013 (UTC)
Listen what Connection Machine designer says recalls
The game of Life is an example of a class of computations that interested Feynman called "cellular automata". Like many physicists who had spent their lives going to successively lower and lower levels of atomic detail, Feynman often wondered what was at the bottom. One possible answer was a cellular automaton. The notion is that the "continuum" might, at its lowest levels, be discrete in both space and time, and that the laws of physics might simply be a macro-consequence of the average behavior of tiny cells. Each cell could be a simple automaton that obeys a small set of rules and communicates only with its nearest neighbors, like the lattice calculation for Lattice QCD. If the universe in fact worked this way, then it presumably would have testable consequences, such as an upper limit on the density of information per cubic meter of space.
Bekenstein non-computational limit
Does the Bekenstein limit only apply to computing power, or does it describe physical information, as in the disposition and energy states of the atoms within, say, a cubic millimeter? In terms of actual, physical entities (such as atoms) expressed in terms of various equations, and not an idea described by a series of 1s and 0s dictated by an electron's spin state, what would be the Bekenstein limit for a cubic millimeter? — Preceding unsigned comment added by 18.104.22.168 (talk) 19:30, 16 January 2015 (UTC)
Should the word "bits" in the article be replaced by "qubits"?
Since quantum data is measured in qubits rather than bits, shouldn't that be the unit used in this article? Storing n bits takes n qubits, but the converse is exponentially false. MvH (talk) 16:24, 23 April 2015 (UTC)MvH
The comment(s) below were originally left at several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section., and are posted here for posterity. Following
|Comments on article assessment. I think the article deserves a "low" status because 1. in the scientific community, the bound is still somewhat of a conjecture (perhaps proven for some matter contents, etc) and is a bit "hypothetical" (compare an article, on say, "atom") 2. it's generally of mostly specialist interest. I go for "start" class since there is a dead link or two (eg, "Planck area") and I think the article could do with a little more background, explanation for the non-expert reader, and perhaps links to other areas to help explain arguments -- especially in the last paragraph. The article could also be organized a little more: for example, a header followed by section(s). Wesino 00:45, 29 November 2006 (UTC)|
Last edited at 00:45, 29 November 2006 (UTC). Substituted at 09:21, 29 April 2016 (UTC)