# The Bekenstein Bound

1. Feb 14, 2013

### HowardTheDuck

Hi Guys, I've been struggling over a problem with the Bekenstein Bound, and I wonder if someone can help, please.

The Bekenstein Bound is derived from the entropy of black holes, and says that the maximum information content of a region of space is proportional the area of that region, not the volume of that region (which might be expected).

So if I have a cube whose sides are 2x2x2, the area (and hence the maximum information content of the cube) is going to be 4x6= 24 units.

But I could divide that cube into 8 smaller cubes with sides 1x1x1 and hence the area of all those cubes is going to be 6x8 = 48 units.

So it seems like the maximum information content of the region is not 24 at all, it could be 48 or higher. Can anyone see where I am going wrong? Thanks.

2. Feb 14, 2013

### bapowell

Dividing the cube into smaller cubes doesn't change the total surface area: you now have a total area of (1x1)x4x6 = 24. You seem to be summing the surface areas of each of the smaller cubes, but 3 of the faces of each little cube are 'inside' the larger cube that they form (so you need to subtract 3x8 from your calculation, and you recover 48-24 = 24.) If you were to continue your procedure to ever smaller cubes, summing the surface areas of each, you'd succeed in obtaining the volume of the larger cube!

3. Feb 14, 2013

### HowardTheDuck

Well, yes, I agree with what you say. But I don't see that it matters if a surface of a smaller cube is facing outwards or facing inwards. We're only interested in the information content. Those smaller cubes could be lying about on my desk, not touching each other. In which case, I sum the total information content of those 8 cubes as 6x8=48. But somehow if I arrange those 8 cubes into the large cube then the maximum information content of that equivalent region now drops to 24. Surely that can't be right. Or is it right? Is that what the Bekenstein bound says?

If so, then it really is highly-counterintuitve.

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4. Feb 14, 2013

### Staff: Mentor

The "area" that appears in the Bekenstein bound is the horizon area of a black hole with the same mass as the object under consideration; it's not the surface area of the object itself.

So, for example, if I have a 2x2x2 cube of some substance with density 1, it has a mass of 8; the horizon area associated with this mass is 16 pi (8x8) = 1024 pi. Eight 1x1x1 cubes of the same substance have a mass of 1 each, for a total horizon area of 8 x 16 pi (1x1) = 128 pi. So the maximum information content of the 2x2x2 cube is larger than the maximum information content of the 8 1x1x1 cubes.

5. Feb 14, 2013

### HowardTheDuck

So you're treating the cubes like they have to be black holes? Working out the Schwarzschild radius? OK. But the Bekenstein Bound is not specifically about black holes - it is about the information content of any region of space. It genuinely says that the maximum information content of any region of space is proportional to its area.

At least, I think that's what it means.

6. Feb 14, 2013

### haael

It's right. The total information capacity of 8 cubes is smaller than 8 times the information capacity of each cube alone.

The key word here is: correlation.

Suppose you have two uncorrelated data arrays. When you join them into one array, it's information capacity will be sum of information capacities of the original arrays.

Now suppose you have two correlated data arrays. When you join them into one array it's information capacity will be sum of information capacities of the original arrays minus correlation.

That's what Bekenstein bound says. The world exhibits correlations on some basic level. When you set some bits in your neighbourhood, you automatically set some bits in distant regions of universe. When considering your neighbourhood alone, one does not see anything suspicious, as well when considering the distant region. But when one wants to look at your place and the distant place as a whole, he will see that the regions were not independent.

In my opinion, the Bekenstein bound is the same phenomenon as quantum entanglement, Bell's inequalities violation, apparent noncocality of space. This is the thing we call "quantum". The space is nonlocal in some sense, but it doesn't transfer information but only correlation of information. You can not send "0" and "1" with superluminal speed. You can only toss a coin and know that some alien in different galaxy got the same result on his coin. No information transfer, only correlation.

7. Feb 14, 2013

### HowardTheDuck

Thanks very much, Haael.

I have never seen that explained before. That's bizarre.

8. Feb 14, 2013

### haael

Read "The Black Hole War" by Leonard Susskind.

9. Feb 14, 2013

### HowardTheDuck

I'm reading it at the moment, actually. I'm not that far, though! A good book, really simple.

10. Feb 14, 2013

### Staff: Mentor

No, I'm saying that the "area" that appears in the Bekenstein bound is a black hole horizon area.

I think there are different versions of it, or at least the term "Bekenstein bound" is being used to refer to different things. But the one that has a rigorous proof is this one:

http://en.wikipedia.org/wiki/Bekenstein_bound

Note that what appears in the formula is E, the energy (or mass), times R, the radius; the area does not appear directly. To get an area in there, you have to convert the energy to a radius, and the "conversion factor" is that a given amount of energy/mass is equivalent to the horizon radius of a black hole with the same mass.

11. Feb 14, 2013

### HowardTheDuck

Hi Peter, the area is the area of the sphere with radius R. It's like a sphere which encloses the entire system. You have to provide the R value.

12. Feb 14, 2013

### Staff: Mentor

Yes, I know, but look at the formula on the Wikipedia page I linked to. It only has R in it, not R squared, which is what you would expect if the area of the sphere with radius R were what determined the bound. Instead you have E times R, where E is the energy/mass inside the sphere with radius R. That means the bound is *not* a function of the area of the sphere; it's a function of the radius of the sphere times the energy enclosed in the sphere. You can convert that energy into a length, so that the bound has units of area; but it won't be the area of the sphere.

[Edit: *Unless* the object is a black hole; then R is the horizon radius, and E, when converted into a length, is...the horizon radius. So in the case of a black hole, but *only* in that case, the bound is proportional to the surface area of the sphere with radius R.]

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13. Feb 14, 2013

### HowardTheDuck

Thanks for that - and thanks for the edit. That's made me mildly less confused.

We're always told that the information content of a region of space is proportional to the area of that region - I've read that in many places. But, as you say, it appears we have been misled. The information content appears proportional to the radius, not the area.

It seems that it is the Holographic Bound which is proportional to its area:

http://en.wikipedia.org/wiki/Bousso's_holographic_bound

It's all a bit confusing.

Last edited: Feb 14, 2013
14. Feb 14, 2013

### Staff: Mentor

If you read the fine print, this is actually saying the same thing I was saying: the bound is proportional to the area *for a black hole*. It just tries to generalize "black hole" to cover any circumstance where there is a horizon present; for example, it applies to a cosmological horizon. But it still doesn't say maximum information is proportional to area for *any* object at all; that still only holds for special kinds of "objects".

15. Feb 14, 2013

### HowardTheDuck

Thanks, Peter. Yes, it does say "black hole", but I actually think it does try to extend it to any region of space. It does say "the maximum entropy which can be enclosed by a spatial boundary".

I've seen an article written by Bekenstein in the Scientific American a while back about this, and I've pulled it out. His thinking is that if you have a region of space, radius R, it will have a specific amount of entropy. If you keep adding mass (information) then eventually that will collapse to form a black hole. The resultant black hole is the most efficient store of entropy (information), so before collapse, that region of space must have contained less information than the equivalent black hole. So, as the black hole information is proportional to the area of its event horizon, any region of space's information is proportional to its area.
This appears to be what they mean by holographic bound.

Actually, I have just found the article online:

http://sufizmveinsan.com/fizik/holographic.html

"A related idea, the holographic bound, was devised in 1995 by Leonard Susskind of Stanford University. It limits how much entropy can be contained in matter and energy occupying a specified volume of space.

In his work on the holographic bound, Susskind considered any approximately spherical isolated mass that is not itself a black hole and that fits inside a closed surface of area A. If the mass can collapse to a black hole, that hole will end up with a horizon area smaller than A. The black hole entropy is therefore smaller than A/4. According to the GSL, the entropy of the system cannot decrease, so the mass's original entropy cannot have been bigger than A/4. It follows that the entropy of an isolated physical system with boundary area A is necessarily less than A/4."

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16. Feb 14, 2013

### Staff: Mentor

Yes, this is true, but this...

...does not actually follow from it. Go back to the Bekenstein bound formula again. Note that it is an inequality; that means that, as long as E is less than R (remember we are measuring E in length units here), we can replace E with R and still have a valid inequality. In other words, if the total information stored is less than a constant times E times R, it will also be less than the same constant times R squared if E < R. However, this does *not* mean the total information is *proportional* to R squared, because an inequality is not the same as a proportionality; saying a < b is not the same as saying that a is proportional to b.

The same comment applies to the Susskind quote you give; yes, he derives a correct inequality, but that's not the same as deriving a proportionality. (Also, he fails to note that the inequality he derives is much *looser* than the actual Bekenstein bound, because E times R is much less than R squared for an ordinary object.)

17. Feb 14, 2013

### HowardTheDuck

Thanks very much, Peter. And thanks to everyone who posted.