Is it ever physically possible for a broken egg to reassemble into an unbroken one?

[polymath257]

And in my estimate, I was *underestimating* the factorial by using n!<n^n. In the case of a deck of cards, 52! <52^52 <100^52 =10^104. A rather big over-estimate of the factorial, I think.


NO, That is NOWHERE close to a googolplex. It barely scratches the surface, in fact. Now try (10^80)!, which is less than (10^80)^(10^80), which is less than 10^10^83.

Let's do nanometer scale for a second at nanosecond scale with a pint of beer (water). Now, a pint is approximately half a liter, which is 500cc. There is 1 gram of water per cc, and the molecular weight of water is 18, so there are 500/18 <30 moles of water in a pint. This gives 30*6*10^23<2*10^25 <10^26 molecules of water.

We'll call 500cc~500*10^21 =5*10^23 cubic namometers. And a second is 10^9 ns. That gives5* 10^32<10^33 nm.ns in a pint of beer for a second.

Notice that at each stage I over-estimated the numbers involved, sometimes quite drastically.

Next, the number of possible arrangements of 10^26 molecules is, as you say, (10^26)!<(10^26)^(10^26) <(10^100)^(10^26)=10^(100*10^26)=10^10^28.

Again, the estimate of 10^26 <10^100 is a HUGE over-estimate as is the estimate (10^26)! <(10^26)^(10^26). In general n! <n^n.

Next, let's do (10^10^28)^(10^33) for a random arrangement for each nanometer.nanosecond. This gives
(10^10^28)^(10^33) =10^(10^28 * 10^33)=10^10^61 possibilities.

This is still a LONG way away from a googolplex.


The problem is that expressions like (10^n)! <(10^n)^(10*n)=10^(n*10^n) don't tend to increase the second exponent by much. For n<100, you get AT MOST 10^10^(n+2).

You can add another 7 to that second exponent by considering that pint for a full year. And then another 22 or so if you consider the whole earth. That gets you up to 10^10^90, but I think that already violates your original claim that we see such odds all the time.

Now, if you do femtometers and femtoseconds for the whole Earth for a full year, you can, indeed get above a googolplex, managing odds of 1 in 10^10^102.

OK, check my thinking on this:

n! doesn't represent the probabilities correctly. That's just the number of permutations you'd get if you shuffled the system's components and put them into a linear array like a deck of cards. We need to account for spatial distribution because there's a big difference between two molecules a light year apart and the same two molecules a micrometer apart.

To account for space we need something more like a 'seat n people in m chairs' problem but seating particles in potential locations. That yields m!/(m-n)! combinations, which can be nasty to calculate. Happily, so long as the number of potential locations is much, much larger than the number of particles, the difference between m and m-n is trivial and we can approximate it as m^n, which is a much simpler calculation. It's simpler than that since, as you pointed out, the value of m doesn't influence the uppermost exponent much. The spatial resolution doesn't matter much so long as it's fine enough to satisfy m >> n. In that case the calculations get really simple because a system with 10^x components has roughly 10^10^x+2 possible arrangements. So a pint of water with ~10^25 molecules wouldn't have more than 10^10^27 possible arrangements, which is impressively large but a long way short of a googolplex.

But none of that accounts for time. We need to set this all in motion so what we need is less of a 'seat n people in m chairs' problem and more of a 'musical chairs' problem but with a lot more chairs than people. For this we simply view each iteration as a single m^n snapshot and multiply the results for i iterations. That will be valid so long as each m^n snapshot is unique, which it will be since the laws of thermodymanics preclude take-backsies. That produces m^n^i possible combinations, and that will get very large very quickly.

No, it produces (m^n)^i =m^(ni). For this, you *multiply* exponents. This is very different than m^n^i=m^(n^i). You are doing the parentheses the wrong way and it makes a difference. (x^y)^z=x^(yz) is very different than x^(y^z).

For example, (10^3)^2 =(1000)^2=1000000=10^6, but 10^(3^2)=10^9. Or, another, (10^3)^10=10^30, but 10^3^10 =10^59049, which is considerably bigger.

Using nanosecond resolution, a pint of water with 10^25 molecules passes the googolplex mark in 4 nanoseconds, beats 10^10^200 in 8 nanoseconds, and passes 10^10^25,000,000,000 in the first second. I know, those are absurdly large numbers, but that's what iterating in 4 dimensions gets you.

No, it passes through the *googol* mark in 4 nanoseconds (10^25)^4=10^100. That is a googol. It gets to 10^200 in 8 seconds. In a second, it gets to (10^25)^(10^9) =10^(25*10^9)<10^(100*10^9)=10^10^11

In a year, it would get to (10^25)^(3*10^16)=10^(75*10^16)<10^(100*10^16)=10^10^18

In 10 billion years, it would get to (10^25)^(3*10^26)<10^10^28

Hmmm....the main difficulty with complex exponentials is that they are multiple valued. The complex number in the base has infinitely many logarithms, so the definition of the expression gives infinitely many different values, depending on which value of the logarithm you use. It isn't going to simplify algebraically in any nice way (but neither does sqrt(2)^sqrt(2) ).

Nonetheless, we can say that all values are transcendental numbers (not the root of any polynomial with integer coefficients).

They spawn an infinity of little horrors.

Some people enjoy such horrors. I'm a professional mathematician. This is my wheelhouse.