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Using the Wikipedia version of the Bekenstein bound, and substituting the Wikipedia values for electron mass and radius, one obtains 0.0662 bits. Does this really mean that a system, any system, placed inside a sphere the size of an electron, and weighing no more than an electron does, is almost determinate? How about an electron itself? Wouldn't one need at least a few bits to characterise the behavior of an electron in magnetic space?

(I am a professional mathematician but I know very little about physics, I'm sure I'm missing something obvious here...)

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  • $\begingroup$ It only means, that a physicist has come up with another "It's not even false!" statement. Until someone drops 16 electrons into a black hole and can prove experimentally, that that's the lowest number to store a whole bit in the system, it's simply nothing but a meaningless statement. $\endgroup$ – CuriousOne Aug 22 '14 at 23:27
  • $\begingroup$ The "classical electron radius" isn't classical and isn't an electron radius. As far as we know, the electron is a pointlike particle. There are empirical upper bounds on its size (if it has internal strructure) which are far smaller than the classical electron radius. $\endgroup$ – Ben Crowell Aug 23 '14 at 19:10
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You have found an elaborate way of calculating $2\pi \alpha/ \ln 2 \approx 0.0661658$. Here, $\alpha \approx 1/137$ represents the fine-structure constant.

The points to note is that:

A) Bekenstein's bound defines the maximum number of nats of information that can be contained in a spherical region as the circumference of that region divided by the reduced Compton wavelength associated with the total energy contained within that region,

and

B) the classical electron radius is equal to the fine structure constant times the reduced Compton wavelength of the electron.

Would you redo your calculation using the electron mass and the reduced Compton wavelength of the electron, you would obtain a value of $9.0647$ bits. However, you would obtain exactly the same value for a proton or whatever other elementary or composite particle you might chose. I would not attach any physical significance to these results.


Added: We currently don't have a consistent quantum gravity theory, and we don't even have an idea what would be the fundamental degrees of freedom in such a theory. Therefore any statement in response to questions like "how many bits/nats of information can be associated with an electron mass" runs the risk of leading to nonsense. Having said this, the holographic (Bekenstein-Hawking / black hole) bound seems more capable of providing reasonable leads. Using $4\pi$ times the square of the reduced Compton wavelength of the electron as area in the BH bound leads to an information content of $S/k = \pi \hbar c /G m^2$ nats. Here $m$ denotes the electron mass. This result for "the information content of a volume large enough to contain an electron" is in essence the square of the ratio of the Planck mass over the electron mass. That's a lot of nats.

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  • $\begingroup$ I was using the third equation in the WP article en.wikipedia.org/wiki/Bekenstein_bound. I understand that the ln 2 comes from the nat/bit conversion, but that's already there in WP, and can't account for the two orders of magnitude between the 9.06 bits that you computed and the 0.066 bits that the WP formula yields. When you say "don't attach any physical significance" are you saying, perhaps in more polite language, the same thing that @Jerry Schirmer said, namely that the bound is not valid at this scale? $\endgroup$ – StudentT Aug 23 '14 at 14:31
  • $\begingroup$ @StudentT - the two orders of magnitude come from the fine-structure constant (the difference between using the classical electron radius and the Compton radius of the electron). Bottom line is: the computation leads to a circular reasoning void of physics. $\endgroup$ – Johannes Aug 23 '14 at 16:47
  • $\begingroup$ Dear @Johannes, let me ask the question in a non-circular way: given a physical system that fits into an electron, and has no more mass/energy than an electron, what is the maximum number of distinguishable states it can have? Perhaps physics cannot (yet) provide a bound. I was originally interested in a simpler question: given a system that takes exactly 1 bit to characterise, how small can it be? But then I thought it would be a good sanity check to look at the Bekenstein formula for some existing system, and found the rather surprising outcome that I posted above. $\endgroup$ – StudentT Aug 23 '14 at 17:50
  • $\begingroup$ @StudentT - it seems you are looking for an estimate based on the BH bound. Have appended some text to my answer above. Hope it helps. $\endgroup$ – Johannes Aug 23 '14 at 18:35
  • $\begingroup$ Dear @Johannes, thank you! It helps of course, but it also adds somewhat to my confusion, in that the answer comes out as $2.587\cdot 10^{45}$ bits, larger than what wikipedia has for a sphere of 6.7cm radius (see the section "The human brain" in en.wikipedia.org/wiki/Bekenstein_bound). This is not to say that WP is always 100% accurate, but in the math section that I'm more familiar with generally a lot of knowledgeable people look articles over and don't let outrageous stuff slip by. Anyway, your effort to clarify this is greatly appreciated! $\endgroup$ – StudentT Aug 23 '14 at 19:30
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One can't take results like that too seriously at the scale at which an electron would apply. In particular, the classical general relativistic model, applied naively to a point mass electron would tell you that the electron has too large a charge and angular momentum to have a black hole horizon, and would instead be the exotic type of object called a naked singularity.

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  • $\begingroup$ Before asking the question I first checked out Bekenstein's explanation at Scholarpedia. His method of deriving the bound is by dropping the object (in this case the electron) in a black hole. It is not clear to an outsider such as myself which part of this derivation not to take seriously. $\endgroup$ – StudentT Aug 22 '14 at 20:18
  • $\begingroup$ @StudentT: he's dropping it into a black hole's horizon. If you take general relativity to be true all the way down to an electron's scale, there is no horizon, so none of Bekenstein's equations make any sense, since they are all based on crossing the horizon. $\endgroup$ – Jerry Schirmer Aug 23 '14 at 3:40
  • $\begingroup$ Great, thank you! Does the same logic apply to Hawking radiation? It seems to be the same scale issue: you look at pair creation (presumably the members of the pair are not far from one another on a quantum scale) when one member is inside and the other outside the event horizon, a sphere whose radius is measured on a cosmic scale? Anyway, the original question is closed, and thanks again. $\endgroup$ – StudentT Aug 23 '14 at 6:47

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