I am trying to relate the surface-area-to-volume-ratio of a sphere to the Bekenstein bound. Since the surface-area-to-volume-ratio decreases with increasing volume, one would surmise that, per unit of volume, a small space is richer in information than a large one. How can this be and how can this bound work for black holes of various sizes?

Thank you very much Mr. Rennie. I appreciate and have investigated your answer. It turns out that I am familiar with the AdS/CFT correspondence and have sufficient understanding of the math (just barely) to be intrigued with the conjecture and, of course, the holographic theory. If the correspondence only works for a certain diameter black hole, the conjecture seems, to me, weak because of the changing surface-area-to-volume-ratio of a sphere. For myself, it would appear to be, likely, a mathematical curiosity or fluke. However, if, through some aspect that I do not understand, the correspondence holds for varying diameters, in fact, all diameters of black holes, then it seems quite astonishing, indeed. After searching for some time, I have once seen the amount described as trivial and possibly in another instance, that it may have something to do with informational redundancy. I’m afraid I cannot site these references as they were far too brief to be of any help.

  • $\begingroup$ You ask "How can this be?". If I had a really good answer to that I would be writing off to Stockholm for my Nobel prize. As far as I know (which is not much) the AdS/CFT conjecture (en.wikipedia.org/wiki/AdS/CFT_correspondence) is the best currently known approach to the issue. $\endgroup$ Commented Oct 4, 2013 at 14:20
  • $\begingroup$ blogs.discovermagazine.com/outthere/2013/08/20/… I think this answers my question. Apparently I understand the concept and can follow the math but I fail to believe it. As a black hole gets bigger and bigger, it is less and less dense, until eventually it isn't particularly dense at all. Faith in mathematics would seem to be my issue. Thank you for your help. Jim McKenzie. $\endgroup$ Commented Nov 30, 2013 at 18:19
  • $\begingroup$ I may be throwing in a red herring here, but how is the volume of a black hole actually even defined? Because of the curvature of space-time, the volume contained within for example the surface of the Sun and the Earth is greater than the volume enclosed by a similar surface in Euclidean space. Presumably for a black hole this would be pathologically extreme. $\endgroup$
    – Robotbugs
    Commented Oct 27, 2015 at 1:49


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