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This info is from Wikipedia

In physics, the Bekenstein bound is an upper limit on the entropy S, or information I, that can be contained within a given finite region of space which has a finite amount of energy—or conversely, the maximum amount of information required to perfectly describe a given physical system down to the quantum level.

Upon exceeding the Bekenstein bound a storage medium would collapse into a black hole.This finds parallels with the concept of a kugelblitz, a concentration of light or radiation so intense that its energy forms an event horizon and becomes self-trapped: according to general relativity and the equivalence of mass and energy.

My question is is there a known quantity of info or anything that is the limit of Bekenstein Bound or needed to overcome it?

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  • $\begingroup$ Is the question what the limit is (in terms of bits per meter per kilogram), or whether there are limits to the limit? $\endgroup$ – Anders Sandberg Jun 21 '18 at 15:16
  • $\begingroup$ The former, what is the limit for the Bekenstein Bound $\endgroup$ – C. Jordan Jun 21 '18 at 16:01
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The Bekenstein bound states that the maximum number of bits that can be stored inside a sphere of radius $R$ with total energy $E$ is $$I\leq \frac{2\pi}{\hbar c \ln(2)}RE = 2.8672\cdot10^{26} \, \mathrm{bits/J~m}$$ or, when expressed for mass, $$I\leq \frac{2\pi c }{\hbar \ln(2)}RM = 2.5769\cdot10^{43} \, \mathrm{bits/kg~ m}.$$

This bound is valid if self-gravity isn't too extreme and the spacetime is not curved so much that $R$ or $E$ becomes hard to define.

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  • $\begingroup$ Interesting, Thanks, so I would know what to do to find answers, Just to ask though, How would I type this equation into calculators like Google Calc? like, how to turn some of those symbols into numbers? $\endgroup$ – C. Jordan Jun 21 '18 at 23:43
  • $\begingroup$ You just multiply the constants above with the energy or mass (depending on which equation you use) and the radius. $\endgroup$ – Anders Sandberg Jun 22 '18 at 7:34
  • $\begingroup$ ok, thanks, one last question, what if I want to find out the energy/mass? Do I just do the same equation again but divide it by the number of bits/J/kg/m? $\endgroup$ – C. Jordan Jun 22 '18 at 17:21
  • $\begingroup$ Also I meant like what is the number for that (h) reduced Planck constant and what unit would be used for th speed of light? (Meters per second?) $\endgroup$ – C. Jordan Jun 22 '18 at 17:22
  • $\begingroup$ Also what would the “I <“ be in a calculator? $\endgroup$ – C. Jordan Jun 22 '18 at 21:39
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I am trying to put the formula for the Bekenstein Bound for energy into the calculator, and this is how I did it. I am trying to solve for energy.

((2*pi)/1.054571800(13)e−34*299792458*log(2))*1737400/2.8672e+26

  • 1.054571800(13)e−34 = h-bar
  • 299792458 = m/s speed of light
  • 1737400 = meters radius of the moon
  • log(2) = ln(2)

That is what I did, can someone verify If that is the correct way of doing it?

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