Imagining Graham's number in your head collapses your head to a black hole

Question

My question is how does the above phenomenon, as mentioned in Numberphile here, occur in a semi-quantitative way through physical laws. (i.e saying statistical mechanics implies that is not descriptive).

In general, I have seen information being mentioned in statistical mechanics/thermal mechanics courses vaguely, from time to time, but what it is is really elusive.

Note: The video does mention a formula - maximum amount of information your head can store $S_{\text{max}} = \frac{A}{4L^2}$ where $L = 1.616\times 10^{-35}m$, and $A = 4\pi r^2$, I think surface area of spherical head, but where does this equations come from is not clear.

Answer 1

Simply put, what's stated in the video is misleading to an extent that it's fair to say it's wrong.

Ignoring problems of how you define the word 'imagining' (this is physics stack exchange, after all), a generous interpretation of what they're saying is that if you tried to, say, encode all the digits of Graham's number into atoms and compress all these atoms to fit inside your head so that they could be accessed by your neurons somehow, then it would collapse into a black hole. A cartoon description of black hole formation is that if you try to compress a large amount of mass-energy into a small space, then it will collapse into a black hole.

But Graham's number is an abstract concept, so it doesn't make sense to talk about it containing any mass-energy. The assumption that to access the digits of Graham's number you need to actually encode all the digits explicitly is not justified. There's no reason this is necessary; you could instead just run an algorithm to compute the digits. Basic algorithms are known to do this, they just take a long time to run the more digits you need.

Remember that representing a number in terms of digits is really just as good as representing it any other way, it just depends what you want to do with the number. It does not necessarily give any better understanding to represent a number as, say 1024 vs representing it as $2^{10}$. After all, 1024 just means: $1 \times 10^3 + 0 \times 10^2 + 2 \times 10^1 + 4 \times 10^0$.

EDIT: A commenter asked why don't we compute the smallest volume inside which you could represent Graham's number anyway. The problem is that this depends on a lot of assumptions; too many to give a concrete answer. Graham's number is not a random sequence of digits, it's a number with a known formula. So depending on how we define it, the amount of information required could range from 0 to 'very large'.

In general though, if you want to store a certain amount of information, the lower bound of volume required to store this information in is given by the Bekenstein bound.

Answer 2

Bekenstein Bound

The Numberphile mathematicians were clearly describing an explicit representation of the bits in Graham's number, rather than any symbolic one (obviously, just writing "G" on a piece of paper could count if you go that route). So their claim, being more precise, would be something like: "Let $G$ be Graham's number. If you tried to fit $log(G)$ bits of entropy into a volume no larger than the size of your head, it would collapse into a black hole."

John Baez helpfully explains that a gram of water at STP contains about $10^{24}$ bits of entropy. So, if you could perfectly manipulate every atom of water in a 1 gram clump, you could theoretically explicitly represent a number with $10^{24}$ bits. Now, the amount of information entropy in a gram of water is a function of its degrees of freedom. Nobody claims that water is the ultimate information-dense substance. If you picked something else, you might be able to pack in more bits into the same volume.

But that just means if we are very clever, there is no end to the amount of information we can pack into a sphere the size of our heads, right? Not quite. The information content of ordinary matter is bounded per unit of mass, since each fundamental particle only has so many degrees of freedom, and massive particles can only get so small. But as you increase the mass density of your "storage medium", you reach a point where it collapses into a black hole. Then the information content becomes proportional to the surface area of the BH. More precisely, it is believed to be 1/4 nat per Planck area (also per Baez). Call that maybe 1/6 bit per "pixel". This limit is called the Bekenstein Bound.

So returning to the mathematics of the original question, we can pose it quite precisely. The human body has a total surface area around 2 $m^2$. The head comprises less than 10% of that. So it is fairly safe to say that the human head has no more than 1 $m^2$ of surface area. Thus, the Numberphile claim could be posed as: "log(G) > entropy of a BH with 1 $m^2$ event horizon". Hopefully, by studying the definition of Graham's number, you can convince yourself that this claim is trivially true.

EDIT

To get a sense for how big Graham's number is, I will provide some additional calculations, to address Michael's comment. Now a black hole is the densest way to store information in our universe (so far as our current understanding of physics is concerned), but not all BHs are created equal. Because the surface area only grows with the square of the radius, while the volume grows with the cube, very large BHs are much less information-dense than very small ones. So, for instance, a single black hole of mass equal to the purported mass of the observable universe would be much larger in volume than if we divided that mass up into many smaller BHs. Clearly, the limit then, must be the smallest possible BH, which would have a radius on the scale of the Planck length.

For convenience, let us assume these smallest black holes are cubes of length $l_P$ (the Planck length), which would mean they can encode about 1 bit of entropy each. We can fit about $10^{35}$ $l_P$ per meter, and about $10^{27}$ meters in the observable universe (approximating it to be a large cube). So the most information we could squeeze into the observable universe is about $10^{62*3}$ = $10^{186}$ bits. That's a lot of bits! Oh, and if we did manage to squeeze that much information into it, the entire universe would be at roughly big bang temperatures, which would make it a perhaps less-than-pleasant place to be. So I think it's safe to say that this is a very hard upper bound on how many bits we can squeeze into our corner of the universe.

Now, the problem with Graham's number is that it uses Knuth's up-arrow notation, which is a little obscure even by mathematician standards, since it is not needed in many contexts. But it is simply a generalization of our familiar sequence of addition -> multiplication -> exponentiation, etc. If you consider that each operation in the sequence is an iteration of the previous operation, you get a basic sense of what an "up arrow" is (and should notice how it is intended to be an extension of exponentiation, in a sense). So a number like $g_1$ = 3^^^^3 (where you should imagine the carets as up-arrows, since Physics MathJax does not render them natively) is already so large, it far exceeds the number of bits in our puny universe. How large is it? Well, it is roughly 3 raised to the power of itself a trillion times. And that's just the first term in a sequence which ends with Graham's number! The sequence iterates these "power towers" up to $g_{64}$, which is so incomprehensibly large that it is pretty meaningless to give any comparisons to it.

Answer 3

Without watching the video and going with your headline, my first guess would have been to use some black hole thermodynamics. If we trust a well known internet black hole calculator, then a 1.6e-35m black hole has a mass of 1e-8kg and a lifetime of 6e-41s (approx. 1000 Planck times). It would have a mass-energy of about 1GJ, which already sounds suspiciously high.

Using Landauer's principle at room temperature the mass of a bit of information can be given an equivalent mass of 3e-38kg. If we equate these masses, then the video would estimate the information content of the human brain as roughly 3e29 bits if I am not mistaken. Typical estimates for the information stored in the human brain in form of synaptic weights seem to be around 1Petabit, which only gets us to 1e15 bits. I seem to be missing some 15 orders of magnitude here, so I am not sure how they got to these numbers. Maybe somebody can plug them into the actual formulas for you, just in case the internet calculators and the Landauer bit mass estimates are wrong.

Answer 4

A simple answer which slightly expands on the other (excellent) answers here, coming at it from an information-theory perspective:

The Kolmogorov complexity of Graham's number is significantly smaller than the Bekenstein bound of our brains. This is why we we have no problem conceptualizing infinitely recursive structures despite them being literally infinitely larger than Graham's number. We don't need to expand Graham's number into a string of digits that we have to store in our brain simultaneously just to "imagine" it.

The video in question plays on the fact that massive numbers are hard to reason with because they are so massive that we cannot fall back on anything in our experience to get an intuitive feeling about its size. That does not mean we cannot comprehend or reason about it, just that we must think about it in abstract terms (such as Knuth' up-arrow notation).