People like to say this is the "information age". But just how much information are we talking about? I recently found a nice source of... umm... information about this question:
You have to be careful interpreting these figures.
First, while some are precise, others are rough order-of-magnitude guesses, like "all the words ever spoken by human beings".
Second, the information in a year's worth of the New York Times is far less than the information capacity of all the newsprint put out by the New York Times during this year — because there are thousands of identical copies of each newspaper. I have tried to distinguish between these two by saying "capacity" for the latter sort of figure. For example, when I say 130 petabytes is the capacity of all audio tapes produced in 2002, this includes thousands of identical copies of some album by the Backstreet Boys, and also lots of blank tape. It's quite possible that I screwed up on some of the items above, because my source makes it a bit hard to tell what's what.
A third problem is that not everyone agrees on the definition of 'kilobyte', 'megabyte', 'gigabyte', and so on! Originally people used 'kilobyte' to mean not 1000 but 1024 bytes, since 1024 = 2^{10}, and powers of 2 are nicer when you're using binary. The difference is just 2.4% percent. But by the time you get to 'zettabytes' it makes a bigger difference: 2^{80} is about 20.8% more than 10^{24}.
There have, in fact, been lawsuits over hard drives that contained only 10^{9} bytes per gigabyte, instead of 2^{30}. And at some point, defenders of the metric system tried to crack down on the evil practice of using prefixes like 'kilo-', 'mega-', giga-' to mean something other than powers of ten. 2^{10} bytes is now officially a kibibyte, 2^{20} bytes is a mebibyte, and so on. But I haven't heard people actually say these words.
Finally, the figures above don't count the fact that information is typically not compressed as much as possible. For example, there are almost 32 = 2^{5} letters in English, which would suggest an information of roughly 5 bits per letter. But some letters are used more than others, and taking advantage of this lets us cut the information down to about 4.1 bits per letter. We can compress text further using correlation between letters. Taking correlations within each 8-letter block into account, Shannon estimated in 1948 that there are just 2.3 bits per letter. In 1950 he went further and estimated that taking correlations in 100-letter blocks into account, each letter holds only 1 bit of information.
So, there's not as much info around as this chart suggests. For some figures that try to take compression into account, see How Much Information? 2003.
Each of us has chromosomes with a total of about 5 billion base pairs. Each base pair can hold 2 bits of information: A, T, C, or G. That's 10 billion bits, or 1.25 gigabytes.
(Here and in everything that follows, I'll use the decimal system, so a gigabyte will be exactly 10^{9} bits, and so on.)
Multiplying this by the roughly 7.5 billion people in the world now, we get about 9 × 10^{18} bytes, or 9 exabytes. They only built 2 exabytes of hard disks in 2002. But, if we wanted to store the complete genetic identity of everyone on hard drives, we could easily do it, using data compression, because a lot of genes are the same from person to person.
Furthermore, we can compress any one particular genome! For starters, the genetic code codes for 22 different amino acids using triplets of base pairs. That reduces the information to at most ln(22)/ln(2) ≈ 4.5 bits per triplet, or 1.5 bits per base pair.
There are also long patches of highly repetitive DNA, such as long terminal repeats. On top of that, at least 20% of human DNA consists of numerous copies of parasitic genes that do nothing but make copies of themselves, like long interspersed nuclear elements! So you could compress the human genome a lot more if you wanted.
I read somewhere else that the human brain has 10^{14} synapses. If each stored just one bit — which is far from true — that would be about 10 terabytes.
... limited, of course, by the Heisenberg uncertainty principle! That's what makes the amount of information finite.
How can we calculate this? It sounds hard, but it's not if you look up a few numbers.
First of all, the entropy of water! At room temperature (25 degrees Celsius) and normal atmospheric pressure, the entropy of a mole of water is 69.95 joules per kelvin.
To understand this, first you need to know that chemists like moles. By a mole, I don't mean that fuzzy creature that ruins your lawn. I mean a certain ridiculously large number of molecules or atoms, invented to deal with the fact that even a tiny little thing is made of lots of atoms. A mole is very close to the number of atoms in one gram of hydrogen.
A guy named Avogadro figured out that this number is about 6.022 × 10^{23}. People now call this number the Avogadro constant. So, a mole of water is 6.022 × 10^{23} molecules of water. And since a water molecule is 18 times heavier than a hydrogen atom, this is 18 grams of water.
So, if we prefer grams to moles, the entropy of a gram of water is 69.95/18 = 3.89 joules per kelvin. By the way, I don't want to explain why entropy is measured in joules per kelvin — that's another fun story.
But what does all this have to do with information? Well, Boltzmann, Shannon and others figured out how entropy and information are related. The formula is pretty simple: one nat of information equals 1.3808 × 10^{-23} joules per kelvin of entropy. This number is called Boltzmann's constant.
What's a 'nat' of information? Well, bits of information are a good unit when you're using binary notation — 0's and 1's — but trits would be a good unit if you were using base 3, and so on. For physics the most natural unit is a nat, where we use base e. So, 'nat' stands for 'natural'. A bit equals ln(2) nats, or approximately .6931 nats.
Don't get in a snit over the fact that we can't actually write numbers using base e — if you do, I'll just say you're nitpicking, or natpicking! The point is, information in the physical world is not binary — so base e turns out to be the best.
Okay: so, by taking the reciprocal of Boltzmann's constant we see that one joule per kelvin of entropy equals 7.24 × 10^{22} nats of information. Or, dividing that by ln(2), it equals about 1.04 × 10^{23} bits.
That's all we need to look up. We can now just multiply and see that a gram of water (at room temperature and pressure) holds $$ 3.89 \times 7.24 \times 10^{23} \; = \; 2.81 \times 10^{24} $$
nats of information. In other words, this is how much information it takes to completely specify the state of one gram of water!
Or if you prefer bits, use the fact that a bit is .6931 nats. Dividing by this, we see a gram of water holds 4.05 × 10^{24} bits. Dividing by 8, that's about 5 × 10^{23} bytes, or 500 zettabytes, or 500,000,000 petabytes.
One thing this goes to show is that the sheer amount of information has little to do with how useful or interesting it is. It also shows that all our magnificent methods of information storage come nowhere close to handling the information packed in even the simplest things of nature.
Of course, the problem is that there are a lot of water molecules in a gram of water. As I mentioned, a mole of water has 6.022 × 10^{23} molecules, and its entropy is 69.91 joules per kelvin; to convert that to bits of we divide by Boltzmann's constant and ln(2). So, the number bits of information it takes to describe one molecule of liquid water — its position, momentum, angular momentum, its precise shape and so on on — is $$ \frac{69.95}{6.022 \times 10^{23} \times 1.3808 \times 10^{-23} \times 0.6931} \; = \; 12.14 $$
bits. That's a more manageable number! By the way, for ice the figure is 7.11 bits per molecule, while for water vapor it's 32.76 bits per molecule.
Here's what I find most cool, though: 500 zettabytes capture everything about what's happening in the cubic centimeter of space and the gram of liquid water in it — unless there are things going on that have almost no interaction with anything we can observe. The great thing about entropy is that it's comprehensive: the entropy of water did not change when we discovered protons were made of quarks, and it will not change if we discover quarks are made of sub-quarks.
Only with quantum mechanics is this 'backwards-compatibility' of entropy possible. In classical mechanics it doesn't work: the more little parts something has, the more entropy it has. Classicaly, even the entropy of something so simple as an ideal gas works out to be infinite, since it takes an infinite amount of information to precisely specify the position and velocity of a point particle in a box!
You see, classical mechanics is sick when it comes to thermodynamics. Planck got famous for curing the 'ultraviolet divergence' that makes the energy of a classical box of light infinite in thermal equilibrium. The infinity you get when you calculate the entropy of a classical ideal gas is almost as disturbing, though for some reason less well-known.
I can't look up the entropy of red blood cells, but while hemoglobin and lipids are a lot more interesting than an equivalent mass of water, their entropy is probably not vastly different: this is just how chemistry works. So, if we just want an order-of-magnitude estimate, we can do an easier problem: compute the information in a red-blood-cell sized droplet of water.
A red blood cell is about 10^{-2} centimeters, so its volume is roughly 10^{-6} cubic centimeters. So, let's consider an equivalent volume of water. This has a mass of 10^{-6} grams — a microgram — since a cubic centimeter of water weighs (almost exactly) a gram.
We computed the information in gram of water just a minute ago. So, we just need to divide by a million to get the information content of a microgram of water. Easy The answer is: 500 petabytes!
About 440 petabytes of email were sent in 2002. So, all the emails sent that year would be almost, but not quite, enough to completely describe a red blood cell down to the subatomic level.
The Planck area is a very small unit of area. It's the square of the Planck length, which is about 1.6162 × 10^{-35} meters. So, there's one nat of information for each $$ 4 \times (1.6162 \times 10^{-35})^2 \; = \; 1.0448 \times 10^{-69} $$ square meters of black hole event horizon. In other words, there is one bit per $$ \ln(2) \times 1.0448 \times 10^{-69} = 7.242 \times 10^{-70} $$ square meters of horizon. Or if you prefer, each square meter of a black hole's horizon takes $$ \frac{1}{7.242 \times 10^{-70}} \; = \; 1.381 \times 10^{69} $$ bits to describe.
This amazing fact suggests that information about matter that falls into a black hole is 'stored in the event horizon'. However, that's not something we really know. And Hawking didn't need to know this to do his calculation.
The history of this is very interesting. First, people discovered that black holes obey three laws that are very much like the three laws of thermodynamics. The second of these laws says that the surface area of the event horizons of black holes always increases. That's similar to the second law of thermodynamics, which says that entropy increases!
This made people suspect that black holes have entropy proportional to their surface area. And then, in a wonderful feat of ingenuity, Hawking calculated the constant of proportionality! This is why he's famous — along with the fact that he has a wheelchair and a cool-sounding electronic voice.
Anyway, let's use Hawking's result to calculate the information in a few black holes.
For this, it helps to know that for a nonrotating, uncharged black hole, the area of the event horizon is \(4\pi R^2\), where \(R\) is its radius. Moreover, its radius is $$ R = 2GM/c^2 $$ where \(M\) is the black hole's mass, \(G\) is the gravitational constant and \(c\) is the speed of light. Working this out in metric units, the radius \(R\) in meters is 1.4851 × 10^{-27} times its mass in kilograms.
So, the area of the event horizon in square meters is $$ 4 \pi (1.4851 \times 10^{-27} M)^2 \; = \; 2.7717 \times 10^{-53} M^2 $$ where \(M\) is its mass in kilograms. This means that the information required to fully describe this black hole is $$ 1.381 \times 10^{69} \; \times \; 2.7717 \times 10^{-53} M^2 \; = \; 3.827 \times 10^{16} M^2 $$
bits.
So, a 1-gram black hole has an information of 3.827 × 10^{10} bits, or about 4.78 gigabytes!
This is much less than the information in a gram of water. This is why a very small black hole, like 1-gram black hole, is highly unstable. You could increase the entropy a lot by letting it radiate away and then turning the energy from this radiation into water... and anti-water.
On the other hand, the mass of the Earth is 5.972 × 10^{24} kilograms, so if the Earth collapsed to form a black hole, the information needed to completely describe that black hole would be $$ 3.827 \times 10^{16} \; \times \; (5.972 \times 10^{24})^2 \; = \; 1.364 \times 10^{66} $$
bits. The information goes up as the square of the mass, so now the amount of information is huge. But the black hole would still be small: its radius would be just 0.88 centimeters!
What if the Sun collapsed to form a black hole? The Sun's mass is 1.989 × 10^{30} kilograms, so there are $$ 3.827 \times 10^{16} \; \times \; (1.989 \times 10^{30})^2 \; = \; 1.514 \times 10^{77} $$
bits of information in a solar-mass black hole.
What about the biggest black hole we know? The biggest ones have masses that aren't very accurately determined, but the one at the center of the supergiant galaxy Holmberg 15A may have a mass 170 billion times that of our Sun. That's about $$ 1.7 \times 10^{11} \; \times \; 2 \times 10^{30} \; = \; 3.4 \times 10^{41} $$ kilograms. So, there are roughly $$ 3.827 \times 10^{16} \; \times \; (3.4 \times 10^{41})^2 \; = \; 4.4 \times 10^{99} $$ bits of information in this black hole. That's almost a googol bits!
We can't answer this, not yet anyway, because we don't even know if the universe is finite or infinite in size.
So, let's talk about the observable universe. The universe may be infinite in size. But when we look back all the way to when the hot gas of the early universe first cooled down and became transparent, everything we see fits in a finite-sized ball centered at us. That ball has by now expanded to be much larger. This larger present-day ball is called the observable universe.
It should really be called the 'once-observed universe', since we can't see what distant galaxies are doing now. But regardless of what it's called, the radius of this ball is about 4.4 × 10^{26} meters. How much information does it take to describe everything in here?
There are a couple papers on this:
It's also fun to see Egan and Lineweaver's other estimates of the big contributors to entropy in the observable universe. They measure entropy in units of Boltzmann's constant k, so we we have to cross out the k and divide by ln(2) to get bits:
© 2022 John Baez
baez@math.removethis.ucr.andthis.edu