Now WAIT just a minute (claims of scale, checked)
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December 7, 2017 at 12:36 am #23537
My dear brother just wrote me to inform that this site exists. I, however, being somewhat math-inclined, wanted to see if the 10e4617 possibilities conjecture (with over 1,300,000 characters) was fair.
here’s my reply to him.
Kind of funny.
It turns out that a number like 10^80 or so (100 followed by 26 triads of “,000”) is the number of atoms in the universe. A countable number. Indeed, something like “The Sun” which weighs in at about 2,000,000,000,000,000,000,000,000,000,000 kg, being made mostly of hydrogen (say 90% by mass) has about
2e30 kg * 1,000 kg/g / 1 g/mol * 6.022e23 atom/g = 12e56 atoms of hydrogen. How’s that compare to 1e80? The beauty of multiplying and dividing exponents is that they just add when multiplying, or subtract when dividing.
1e80 / 12e56 (1.2e57) = 0.8e23
So the mass of the universe is roughly 100,000,000,000,000,000,000,000 solar masses. Our galaxy has about 2 trillion solar masses, which is only 2e12 masses.
So 8e22 / 2e12 = 4e10 Milky-Way galaxies in total universe mass.
40,000,000,000 or forty billion Milky Ways.
Observationally, with our very best telescopes, we can “see” about 1% of those. Actually quite a bit less, due to distance.
The point? 1e80 – the number of atoms in the Universe – is an incredibly huge number. Already.
They talk about 1e4716 or 10 to the 4716th power. That’s not just a ridiculously huge number, its actually hard to comprehend. If (say, for conceptualization) EACH hydrogen atom in the universe were to support a whole universe’s worth of itty-bitty whatchmahoozies … per atom, thats still just
1e80 * 1e80 = 1e160. We’re far, far, far away from 1e4617.
You’d have to encapsulate whole universes of whatchmahoozie on EACH sub-whatchamahoozie, to 57 layers.
And to tell you the truth, I’ve got no way of conceiving of that. None.
EVEN when we work with Dr Professor Max Planck’s hypothesized “smallest unit volume of space” (which is truly incredibly small),
… the Planck Length is 1.6e-35 meters. The Planck volume, that cubed or 4.2e-105 cubic meters, even then with the observable volume of the universe
… 13.7 billion light years, in ALL directions
… Hubble dilated to about 47 billion LY in all directions
It turns out that there are “only” about 1e188 cubic Planck volumes in the entire universe. You’d need to have a WHOLE UNIVERSE of planck-volume bitlets inside EACH “our universe” bunch of them individually … and in each of those sub-bitlets, yet another universe worth … to 24 levels of subdivision.
Sorry, but I still ain’t feeling the concept-love.
BUT… BUT… BUT… here IS something that’s kind of irritating. If the “Library of Babel” is supposedly capable of showing snippets of absolutely EVERY possibility for all 1,300,000 known human language symbols, for all texts possible, but a universe-full-of-universes-full-of-monkeys typing at random, then I was forced to check the premise.
Turns out that the LOG10( 1,300,000 ) = 6.11 or so. Since we cannot raise one point 3 million to the twenty-five-thousandth power (in Excel), which would be the total number of possible combinations of the characters for “average” 25,000 character books, we need to work in another space. Log space.
Where 1,300,000 raised to 25,000 can be replayed as 6.11 * 25,000. Which equals 153,000. Thinking in reverse, this is therefore 10 raised to the 153,000th power. Quite a bit LARGER than their estimate of 10 raised to 4617th power.
Likewise, I can then work backwards from their 4617th power of 10 and determine that their library is really only capable of keeping track of all possible combinations of 771 characters from the whole character set. Not quite War and Peace, now is it?
In fact its puny.
And that’s bothersome.
Their promise isn’t being met with mathematical reality.
Your brother “Bung” the Well Hung.December 13, 2017 at 12:28 am #23620
According to the About page:
“At present it contains all possible pages of 3200 characters, about 10e4677 books.”
As I understand it, the online library of babel does NOT contain every possible book. Only every possible page (3200 characters), which apparently are then collected together in ‘books’.
Could Bob Lynch re-run his figures to reflect the accuracy of 3200 character combinations resulting in 10e4677 possibilities?