What is the calculation for value B in Penrose's entropy model?

In summary, the book "Cycles of Time" by Roger Penrose has a part of the explanation of entropy that I don't understand. There are 10^24 balls, half of which are red and the other half blue. The model is to arrange the balls in a cube with 10^8 balls on each edge. It also divides the cube into smaller cubes of 10^5 on a edge, and defines a cube as uniformly purple if the blue/red ratio in the cube is between 0.999 and 1.001. Then comes the description of the values A and B below, which are approximate values that are unclear how they were calculated. (Value A)Number of states in which the entire 10^24 cube is
  • #1
matsu
5
0
In the book "Cycles of Time" by Roger Penrose, there is a part of the explanation of entropy that I don't understand.
There are 10^24 balls, half of which are red and the other half blue.
The model is to arrange the balls in a cube with 10^8 balls on each edge.
It also divides the cube into smaller cubes of 10^5 on a edge, and defines a cube as uniformly purple
if the blue/red ratio in the cube is between 0.999 and 1.001.
Then comes the description of the values A and B below, which are approximate values that
are unclear how they were calculated.

(Value A)Number of states in which the entire 10^24 cube is uniformly purple: 10^23,570,00
0,000,000,000,000,000,000,000
(Value B)Number of states where the top half of the 10^24 cube is blue and the bottom half
is red: 10^46,500,000,000,000,000,000

I guess that the value A would be the value of (10^24)! but I have no idea how value B is calculated.
Can anyone give me a hint or help?

Note: The same question was asked in a previous thread, but the value B was not resolved.
https://www.physicsforums.com/threads/entropy-as-state-counting.963356/
 
Science news on Phys.org
  • #2
Can you write out the Stirling expression evaluation for the A number?

Should be something with log( 1024 ! )

Then do the same for half that number and take the square....

[edit] two things:
1. You have three zeroes too many. In the book he has
23 570000 000000 000000 000000, not
23,570,000,000,000,000,000,000,000,000

2. My recipe doesn't seem to work :cry: -- have to dig a little deeper

## \ ##
 
Last edited:
  • Like
Likes matsu
  • #3
Thanks for your reply and advice!!!

BvU said:
1. You have three zeroes too many. In the book he has
23 570000 000000 000000 000000, not
23,570,000,000,000,000,000,000,000,000
Sorry, I wrote it wrong.
2.357 * 10^25 is the correct value A.

BvU said:
Can you write out the Stirling expression evaluation for the A number?

I did not use Stirling's approximation, but used WolframAlpha to get the approximation below.

wa20230317.png

BvU said:
2. My recipe doesn't seem to work :cry: -- have to dig a little deeper
I did the same calculation but it did not work.
The number is too large for the value B.
wa20230317_2.png

Could you please let me know if there is any progress?
 
  • #4
matsu said:
I did not use Stirling's approximation
Why not ? Missed a nice opportunity ...
##N\log N - N = 10^{ 2.2357e25}\ ## for ##N=1e24## (number A)

With ##N=5e23## I get ##N\log N - N = 10^{1.163e25}\ ## and the square of that is ##10^{2.326e25}## which is a factor ##3.01e23## lower. Enormously lower, but it gets us nowhere near the ##10^{4.65e19}##

I did the same calculation but it did not work.
So now we're both stuck :smile: and we need some help. I'll ask around.

##\ ##
 
  • Like
Likes matsu
  • #5
BvU said:
Why not ? Missed a nice opportunity ...
##N\log N - N = 10^{ 2.2357e25}\ ## for ##N=1e24## (number A)

With ##N=5e23## I get ##N\log N - N = 10^{1.163e25}\ ## and the square of that is ##10^{2.326e25}## which is a factor ##3.01e23## lower. Enormously lower, but it gets us nowhere near the ##10^{4.65e19}##
Thanks for the clear expression about the Stirling approximation.
I can now do the calculations myself.

BvU said:
So now we're both stuck :smile: and we need some help. I'll ask around.
I appreciate your help so much!
 
  • #6
BvU said:
...
and the square of that is ##10^{2.326e25}## which is a factor ##3.01e23## lower. Enormously lower, but it gets us nowhere near the ##10^{4.65e19}##
It's even lower: I posted the base 10 logarithm. The factor itself is ##10^{\,10^{\,3.01\times 10^{23}}}\ ## as @mfb pointed out.

So now we're both stuck :smile: and we need some help. I'll ask around.
No-progress report: so far I have not found a way to come closer to the
##10^{\,4.65\times 10^{19}}\ ## Penrose states

1679143243425-png.png
which is as good as zero.

##\ ##
 
  • #7
matsu said:
(Value A)Number of states in which the entire 10^24 cube is uniformly purple: 10^23,570,00
0,000,000,000,000,000,000,000

I guess that the value A would be the value of (10^24)!
(10^24)! would be all the ways of placing 10^24 balls in the cube, wouldn't it?
But that would include configurations where a large part of the cube is just blue or just red, not only uniformly purple configurations.
 
  • #8
matsu said:
(Value B)Number of states where the top half of the 10^24 cube is blue and the bottom half
is red: 10^46,500,000,000,000,000,000
For B I suggest ((0.5 x 10^24)!)^2
 
  • #9
Philip Koeck said:
For B I suggest ((0.5 x 10^24)!)^2
As worked out in #4.
(with a slight correction in #6)

But Penrose B is hideously smaller....

## \ ##
 
  • Like
Likes Philip Koeck
  • #10
BvU said:
As worked out in #4.
(with a slight correction in #6)

But Penrose B is hideously smaller....

## \ ##
Ah, I missed that you'd already worked out B.

I can't see any other way to calculate B. Maybe Penrose just got it wrong?

It's also obvious that (1024)! can't be right for A, isn't it?
 
  • #11
Philip Koeck said:
It's also obvious that (1024)! can't be right for A, isn't it?
True, but the point is to show that subtracting not-purple leaves us with the same number -- at least down to the first 23 or 24 decimals.

##\ ##
 
  • Like
Likes Philip Koeck
  • #12
BvU said:
True, but the point is to show that subtracting not-purple leaves us with the same number -- at least down to the first 23 or 24 decimals.

##\ ##
What about looking at something simpler: If we require the smaller cubes to have exactly the ratio 1 between blue and red balls we should get a value A' that's smaller than Penrose's A:

A' = ((0.5 x 1024)!)2 x ((1015)! / ((0.5 x 1015)! (0.5 x 1015)!))1 000 000 000
 
Last edited:
  • #13
Don't understand how this A' comes about.

Penrose introduces intermediate-sized boxes (your 'smaller cubes', I assume)

1679352802274.png

and has ##N = 10^8,\ k = 10^3,\ n = 10^5##. 'purple' means red/blue in ALL intermediate boxes is between 0.999 and 1.001. He finds (without explaining) there are ##10^{10^{ 2.357\times 10^{25}}} \ ## different arrangements are 'purple' . Our number A .

As it happens, ##\left ( 10^{24} \right)\,! ## -- the way to arrange 1024 balls on 1024 positions -- is also this number. Well... the whole treatment is supposed to explain (make plausible) that A is extremely close to ##\left ( 10^{24} \right)\,! ## . We all know A ##<< \left ( 10^{24} \right)\,! ## of course, but the whole story is aimed at how much smaller.

Penrose then does NOT work out a number for not-uniformly-all-'purple' configurations, but instead states (see #6) a number for 'all blue at the top' that is much too small (we found ##10^{10^{2.326\times 10^{25}}}\ ## in #6 -- meaning the number for not-uniformly-all-'purple' configurations is frighteningly close to ##\left ( 10^{24} \right)\,! ## as well)

(and once more: sorry for the many mishaps due to boggling mind :nb) )​

So, for me the inevitable conclusion is that Penrose does not do a good job here.

a while ago I proposed
BvU said:
Start with 32 white balls and 32 black balls in a 4x4x4 box ...
but this too doesn't help us to end up at the ##10^{4.65\times 10^{19}}## in the book -- sorry @matsu !

##\ ##
 
  • #14
matsu said:
(Value A)Number of states in which the entire 10^24 cube is uniformly purple: 10^23,570,00
0,000,000,000,000,000,000,000
(Value B)Number of states where the top half of the 10^24 cube is blue and the bottom half
is red: 10^46,500,000,000,000,000,000

I guess that the value A would be the value of (10^24)! but I have no idea how value B is calculated.
Can anyone give me a hint or help?
It would not be 10^24)! Because they are not 10^24 different balls, but are two groups of 0.5x10^24 equivalent balls.

I'm having trouble working thru it, but consider the simplified 2x2 matrix. There are 6 possible states:
r|r r|b r|b b|r b|r b|b
b|b b|r r|b b|r r|b r|r

4! would be 24 possible states.

I was pondering a 2x2x2 cube with 8 positions. and the number of states is quite large. But again, smaller than the factoral.

It was not clear to me whether the blue-red split between top and bottom was based on the 0.999 to 1.001 is purple, else blue or red.
 
  • #15
votingmachine said:
It would not be 10^24)! Because they are not 10^24 different balls, but are two groups of 0.5x10^24 equivalent balls.

I'm having trouble working thru it, but consider the simplified 2x2 matrix. There are 6 possible states:
r|r r|b r|b b|r b|r b|b
b|b b|r r|b b|r r|b r|r

4! would be 24 possible states.

I was pondering a 2x2x2 cube with 8 positions. and the number of states is quite large. But again, smaller than the factoral.

It was not clear to me whether the blue-red split between top and bottom was based on the 0.999 to 1.001 is purple, else blue or red.
The way I read it Penrose is considering distinguishable balls. So you get a new state by swapping two balls of the same color.
Otherwise B would be 1.
 
  • Like
Likes BvU
  • #16
BvU said:
Don't understand how this A' comes about.

Penrose introduces intermediate-sized boxes (your 'smaller cubes', I assume)
Not sure it's right, but here's my thinking:

There are 1 000 000 000 smaller cubes (intermediate boxes) in the big box.

First I fill the lower halves (exactly) of all the smaller cubes with red and the upper halves with blue balls.
For this there are ((0.5 x 1024)!)2 possibilities, the same as our value for B.

Now I have to take into account that it doesn't matter how the red and blue balls are placed in the smaller cubes as long as they are equal in number in each box.
I have to multiply with the number of permutations within each smaller box, but avoid re-counting the permutations I've already considered (the ones only involving balls of the same color).
For each smaller box that gives a factor (1015)!/((0.5 x 1015)!)2 and there are 1 000 000 000 smaller boxes.
 
Last edited:
  • #17
Philip Koeck said:
The way I read it Penrose is considering distinguishable balls. So you get a new state by swapping two balls of the same color.
Otherwise B would be 1.
Hmm. I was reading it as a model of say air, where there are indistinguishable nitrogen molecules and indistinguishable oxygen molecules, and then what is the probability of even a slight deviation from a perfect mix.

And the easiest approximation I thought of was using the idea of replacing each ball selection with a coin flip, and then using the binomial distribution for determining the states outside the very nearly 1:1 ratio. But that is a different entropy problem.
 
  • Like
Likes Philip Koeck
  • #18
votingmachine said:
Hmm. I was reading it as a model of say air, where there are indistinguishable nitrogen molecules and indistinguishable oxygen molecules, and then what is the probability of even a slight deviation from a perfect mix.
I don't really know what Penrose is modelling, but I don't think it's a good model of a gas mixture.
In the model there is only one available state per particle and each state can only take one particle.
That's why there is only one way of putting all the red balls in the lower half and all the blue in the upper half if balls of the same color are indistinguishable from each other. B = 1 for indistinguishable balls.

In a gas there are always many more states than atoms (or molecules).
That means that, for a mixture of two gases, there are very many ways of putting all atoms of one kind in one half of a container even if atoms of the same kind are indistinguishable from each other.
 
  • #19
votingmachine said:
Hmm. I was reading it as a model of say air, where there are indistinguishable nitrogen molecules and indistinguishable oxygen molecules, and then what is the probability of even a slight deviation from a perfect mix.

And the easiest approximation I thought of was using the idea of replacing each ball selection with a coin flip, and then using the binomial distribution for determining the states outside the very nearly 1:1 ratio. But that is a different entropy problem.
The story is about paint, the treatment is as if it's about a solid. Distinguishable balls, half of them blue, the other half red.
A coin flip is supposed to have the same probability every time -- you could end up with all blue.

So ##M## coin flips yield ##P = \displaystyle \left (1\over 2\right )^M## probability of all heads. If ##M = {1\over 2}N\ ## then ##\ \log P = -{1\over 2}N\log 2## .

Placing ##N## balls in ##N## locations can be done in ##A' = N\,!## ways.

Placing ##{1\over 2}N## blue balls ##{1\over 2}N## locations (say the top half of the cube) can be done in ##C = \left ( {1\over 2}N\right ) \,!## ways. All the red ones in the other half of the cube idem.

So ratio ##Q## (top half blue and bottom half red)/(totally random) is ##C^2/A'##.

Stirling: ##\ \log A'= N\log N - N \ ## and ##\ \log C= {1\over 2}N\log {1\over 2}N - {1\over 2}N \ ## so that
##\log Q = N\left(\log {1\over 2}N - \log N\right )= -N\log 2 \quad ##( with thanks to @mfb )

Meaning ##Q=P^2\ ##: with coin flips it's a lot easier than with balls :smile:

All this is slightly (well...) repetitive and doesn't get us anywhere in understanding how Penrose messes up, I'm afraid :frown:

##\ ##
 
  • Like
Likes Philip Koeck
  • #20
N! makes sense to me now. To calculate the number of states with the bottom uniformly red, you could calculate the probability and multiply by the number of states. I can't see how the calculation would go but the sequence is:
1st ball probability: 0.5N / N
2nd ball probability: (0.5N-1) / N-1
3rd ball probability: (0.5N-2) / N-2

The probabilities would be multiplied for the total

Someone proposed a simplification with a 4x4x4 cube and 32 balls.

The probability would then be:
Product (i from 0 to 15) of (0.5N-i) / N
Product (i from 0 to 15) of (16-i) / (32-i)
16/32 x 15/31 x ... x 1/17 = 16! / (32! / 16!) = (16!^2) / 32!Mutiply that times 32!
[(16!^2) / 32!] x 32! = (16!^2)

I think it would generalize to:

(0.5N! ^2) = number of states with half one color

I caught one math error as I re-read this ... it is entirely possible there is another in plane sight. I think the principle of calculating the number of states from the probability times the total number of states would work though.

EDIT: a 4x4x4 cube is 64 balls. I missed that math error.
EDIT-EDIT: I see now that this was worked out in post #7.
 
Last edited:
  • #21
Philip Koeck said:
What about looking at something simpler: If we require the smaller cubes to have exactly the ratio 1 between blue and red balls we should get a value A' that's smaller than Penrose's A:

A' = ((0.5 x 1024)!)2 x ((1015)! / ((0.5 x 1015)! (0.5 x 1015)!))1 000 000 000
An idea:

Assuming my value A' for exactly purple is correct we can allow for deviations from exactly purple like this.
There are S different shades of purple that fit the definition of "uniformly purple".
S should be 0.5 x 1012 in the example, right?
If each of them approximately has the same statistical weight (A') then we can get A from

A ≈ A' S1 000 000 000

Not sure, is that right?
 
Last edited:

Related to What is the calculation for value B in Penrose's entropy model?

What is Penrose's entropy model?

Penrose's entropy model is a theoretical framework proposed by physicist Roger Penrose to describe the entropy of the universe, particularly in the context of cosmology and the second law of thermodynamics. It emphasizes the role of gravitational entropy and the special initial conditions of the universe.

What is value B in Penrose's entropy model?

Value B in Penrose's entropy model often refers to a specific constant or parameter used in the calculation of the entropy associated with the gravitational field. This value is crucial for determining the overall entropy in the context of the universe's initial and boundary conditions.

How is value B calculated in Penrose's entropy model?

The calculation of value B typically involves complex mathematical formulations and assumptions about the initial state of the universe, including the distribution of matter and the curvature of space-time. It often requires advanced knowledge of general relativity and thermodynamics.

Why is value B important in Penrose's entropy model?

Value B is important because it helps quantify the gravitational entropy, which is a key component in understanding the overall entropy of the universe. This, in turn, has implications for the arrow of time and the evolution of the cosmos.

Can value B be measured or is it purely theoretical?

Value B is primarily a theoretical construct derived from Penrose's model. While it is based on physical principles and mathematical formulations, direct measurement is challenging due to the abstract nature of gravitational entropy and the vast scales involved in cosmological observations.

Similar threads

Replies
7
Views
7K
  • Beyond the Standard Models
Replies
21
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
6K
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
2K
  • Sci-Fi Writing and World Building
Replies
15
Views
3K
Replies
13
Views
3K
Replies
5
Views
5K
Back
Top