How Does Penrose Calculate Entropy in Cycles of Time?

[bob012345]

Sir Roger Penrose in his book Cycles of Time on page 19 states the result of a calculation of probability of mixing red and blue balls as an illustration of entropy as state counting and the Second Law. He assumes an equal number of each. There is a cube of 10^8 balls on an edge subdivided into smaller cubes of 10^5 on an edge. He states each smaller cube looks uniformly purple if the ratio of red/blue balls is between 0.999 and 1.001. Then he states there are around 10^23,570,000,000,000,000,000,000,000 different arrangements of all the balls that give the appearance of uniform purple and some 10^46,500,000,000,000 different arrangements of the "original configuration in which the blue is entirely on top and the red entirely on the bottom".

If anyone has read the book, I am seeking help understanding how he gets to those numbers. Not necessarily a complete solution but a hint on where to start. Thanks and Happy New Year!

 

[bob012345]

Sir Roger Penrose in his book Cycles of Time on page 19 states the result of a calculation of probability of mixing red and blue balls as an illustration of entropy as state counting and the Second Law. He assumes an equal number of each. There is a cube of 10^8 balls on an edge subdivided into smaller cubes of 10^5 on an edge. He states each smaller cube looks uniformly purple if the ratio of red/blue balls is between 0.999 and 1.001. Then he states there are around 10^23,570,000,000,000,000,000,000,000 different arrangements of all the balls that give the appearance of uniform purple and some 10^46,500,000,000,000 different arrangements of the "original configuration in which the blue is entirely on top and the red entirely on the bottom".

If anyone has read the book, I am seeking help understanding how he gets to those numbers. Not necessarily a complete solution but a hint on where to start. Thanks and Happy New Year!

My plan is to start by adapting the random walk problem to the sub blocks and then use Sterlings approximation for the large factorials.

 

[BvU]

where to start

Start with 32 white balls and 32 black balls in a 4x4x4 box ...

Sterlings approximation

The guy's name is Stirling, James Stirling

 

[bob012345]

Start with 32 white balls and 32 black balls in a 4x4x4 box ...
The guy's name is Stirling, James Stirling

Thanks. Don't know why I used an 'e'!

The number of states for 32/32 in a 64 ball cube is 64!/(32!32!) or 1.822X 10^18. I don't know how to compare that to all other possibilities. If I don't know any sub cube is mixed evenly, which is what I'm trying to show is most probably as N gets large, there could be also the other combinations such as 31/33, 30/34, 29,35 ...1/63. Then wouldn't I have to sum all the combinations. Or is there a straightforward formula for that?

 

[bob012345]

I reproduced Penrose's number which is just the Stirling approximation of 1x10^24! That seems a slight oversimplification from how he set up the problem but he did say the number of states was 'around' that number. The difference between 24 and 23.57 in the exponent comes from the 1/e in Stirlings formula. Penrose ignored the multiplicative factor of ~ 10^12 since that would just bury a 1 halfway among the zeros.

 

[matsu]

I don't understand the same part either, did you find out how the value 10^46,500,000,000,000 was calculated?

 

[BvU]

Not good to start at the tail end of a four year old thread. Better to start your own fresh one...

Am I right guessing you can find the 10^23.... number but not the 10^46... number.?

$\$

 

[matsu]

Not good to start at the tail end of a four year old thread. Better to start your own fresh one...

Am I right guessing you can find the 10^23.... number but not the 10^46... number.?

$\$

Thank you for your replying.
Yes.
10^23… must be approximate value of (10^24)!

But I have no idea for 10^46…

I will follow your advice and start a new thread.