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  #151  
Old 05-08-2019, 04:24 PM
Max S. is offline
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Originally Posted by Half Man Half Wit View Post
But this isn't going to lead anywhere, is it. I've shown you quotes, arguments, and experiments, all of which explicitly disagree with you, and yet, to you, it still 'seems like' the second law should hold inviolable. You don't give any reason for that, beyond your own belief.
I think I see where we misunderstand each other. I do not believe the second law is inviolable because it is the second law of thermodynamics. I think the second law is inviolable because I know of no examples of a violation, theoretical or observed.

You say it has been violated, but in each case I can't see the violation without assuming some stochastic fundamental reality which also leaves the door open for violations of every other law of physics.

For example the random hopping in your first hypothetical means microscopic particles don't obey physics at all. I never saw a violation to begin with for the billiard table or the laser experiment.

~Max

Last edited by Max S.; 05-08-2019 at 04:28 PM. Reason: examples
  #152  
Old 05-08-2019, 04:39 PM
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I meant to say your analogy of flipping coins ceased to apply to the current discussion at step 5.

I will concede that one might formulate a law to the effect that 'greyness only increases', and further, that law can be violated. What is your point?

~Max
Don't worry about that for now. This is going to be a bit of a journey, and you keep running into confusions by getting ahead of yourself. But for now, the first step has been taken: we agree that it's possible to have a law, valid to all appearances in the macroscopic realm, concerning a certain quantity ('greyness') such that once that quantity has been understood on a microscopic level, we understand that the original law can only hold in an approximate sense. We can build from here.

Now, to the next step.
  1. Suppose you have two boxes, A and B.
  2. Box A is filled with white marbles, and box B is filled with black ones.
  3. Both boxes are placed on a vibrating plate, such that the marbles in them bounce around.
  4. The walls of the boxes are removable.
  5. Suppose you put both boxes next to one another, and remove the now adjacent walls, creating one big box.
  6. Marbles from the white box A will bounce into the black box, and marbles from the black box B will bounce into the white box.
  7. There are more ways of realizing a state that's pretty uniformly grey, than there are to realize a state that's (say) all white in box A, and all black in box B.
  8. Consequently, there are more ways to go from a state that's slightly inhomogeneous to one that's more homogeneous, than there are ways to go to a state that's even more inhomogeneous.
  9. To any observer who, as before, is only capable of seeing gross colors, the formerly black-and-white separated box will gradually tend to a shade of even gray.
  10. That observer might formulate a law, stating that black and white, if brought into contact, eventually even out to a uniform gray.
  11. Knowing the microscopic description, we know that this is just, again, a law of averages: there is nothing that prohibits, say, all or a sizable fraction of the white marbles from bouncing back into box A.
  12. Given a long enough timescale, the even grey will, eventually, separate into black and white patches.

I expect greater resistance with this example. But again, try not to think ahead to the rest of this discussion; just consider the above system, as I have presented it. Do you agree that the conclusion is reasonable, here? That there is once again a law that appears valid thanks to the limited observations made at the macroscale, which we can see must be violated once we know about the microscopic level?
  #153  
Old 05-08-2019, 04:59 PM
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If statistical and classical entropy really are the same thing, classical non-equilibrium thermodynamics says dQ implies dS. There's no reason to get into stochastic or chaotic microscopic models to make that point.
ok...? Not sure what you mean here (nor what is "classical non-equilibrium thermodynamics"); are you talking about Gibbs's canonical ensemble?

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Actually I was wondering how entropy can increase (or decrease!) in an isolated system as Half Man Half Wit seems to imply it does.
You start out in some improbable state, but then randomness and/or chaos mixes shit up.

Last edited by DPRK; 05-08-2019 at 04:59 PM.
  #154  
Old 05-08-2019, 05:20 PM
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ok...? Not sure what you mean here (nor what is "classical non-equilibrium thermodynamics"); are you talking about Gibbs's canonical ensemble?
No, I just meant applying Clausius's definition of entropy to a system with external net heat flow, that is, a system that is not isolated.


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You start out in some improbable state, but then randomness and/or chaos mixes shit up.
But that's just my point, assuming microscopic randomness or chaos means a particle can violate any number of laws. Not just the second law of thermodynamics.

~Max
  #155  
Old 05-08-2019, 05:37 PM
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Re: Marbles


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Originally Posted by Half Man Half Wit View Post
Don't worry about that for now. This is going to be a bit of a journey, and you keep running into confusions by getting ahead of yourself.
Got it. I'll go with you, one step at a time.

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Originally Posted by Half Man Half Wit View Post
But for now, the first step has been taken: we agree that it's possible to have a law, valid to all appearances in the macroscopic realm, concerning a certain quantity ('greyness') such that once that quantity has been understood on a microscopic level, we understand that the original law can only hold in an approximate sense. We can build from here.
First I want to point out that the law from the coin example is not valid, was not formulated on a valid basis, and need not even appear to be valid. It is possible to observe greyness decreasing on a macroscopic scale in the previous example. The law was formulated nonetheless.

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Originally Posted by Half Man Half Wit View Post
Now, to the next step.
  1. Suppose you have two boxes, A and B.
  2. Box A is filled with white marbles, and box B is filled with black ones.
  3. Both boxes are placed on a vibrating plate, such that the marbles in them bounce around.
  4. The walls of the boxes are removable.
  5. Suppose you put both boxes next to one another, and remove the now adjacent walls, creating one big box.
  6. Marbles from the white box A will bounce into the black box, and marbles from the black box B will bounce into the white box.
  7. There are more ways of realizing a state that's pretty uniformly grey, than there are to realize a state that's (say) all white in box A, and all black in box B.
  8. Consequently, there are more ways to go from a state that's slightly inhomogeneous to one that's more homogeneous, than there are ways to go to a state that's even more inhomogeneous.
  9. To any observer who, as before, is only capable of seeing gross colors, the formerly black-and-white separated box will gradually tend to a shade of even gray.
  10. That observer might formulate a law, stating that black and white, if brought into contact, eventually even out to a uniform gray.
  11. Knowing the microscopic description, we know that this is just, again, a law of averages: there is nothing that prohibits, say, all or a sizable fraction of the white marbles from bouncing back into box A.
  12. Given a long enough timescale, the even grey will, eventually, separate into black and white patches.

I expect greater resistance with this example. But again, try not to think ahead to the rest of this discussion; just consider the above system, as I have presented it. Do you agree that the conclusion is reasonable, here? That there is once again a law that appears valid thanks to the limited observations made at the macroscale, which we can see must be violated once we know about the microscopic level?
I take issue with step 9. I have no reason to believe the observer will observe a gradual change in color. You say the observer can only see gross color. If before he could distinguish two boxes of different colors, as soon as you remove the walls I take it he can only observe the combined box as a whole. Therefore the instant you remove the wall the observer sees flat grey.

~Max
  #156  
Old 05-08-2019, 05:43 PM
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Originally Posted by Half Man Half Wit View Post
  1. Suppose you have two boxes, A and B.
  2. Box A is filled with white marbles, and box B is filled with black ones.
  3. Both boxes are placed on a vibrating plate, such that the marbles in them bounce around.
  4. The walls of the boxes are removable.
  5. Suppose you put both boxes next to one another, and remove the now adjacent walls, creating one big box.
  6. Marbles from the white box A will bounce into the black box, and marbles from the black box B will bounce into the white box.
  7. There are more ways of realizing a state that's pretty uniformly grey, than there are to realize a state that's (say) all white in box A, and all black in box B.
  8. Consequently, there are more ways to go from a state that's slightly inhomogeneous to one that's more homogeneous, than there are ways to go to a state that's even more inhomogeneous.
  9. To any observer who, as before, is only capable of seeing gross colors, the formerly black-and-white separated box will gradually tend to a shade of even gray.
  10. That observer might formulate a law, stating that black and white, if brought into contact, eventually even out to a uniform gray.
  11. Knowing the microscopic description, we know that this is just, again, a law of averages: there is nothing that prohibits, say, all or a sizable fraction of the white marbles from bouncing back into box A.
  12. Given a long enough timescale, the even grey will, eventually, separate into black and white patches.

I expect greater resistance with this example. But again, try not to think ahead to the rest of this discussion; just consider the above system, as I have presented it. Do you agree that the conclusion is reasonable, here? That there is once again a law that appears valid thanks to the limited observations made at the macroscale, which we can see must be violated once we know about the microscopic level?
Actually I'm not sure if I agree with step 8 either. You can't necessarily "go" from any arbitrary configuration of marbles to another, certainly not if you include a succession of three states in order.

~Max
  #157  
Old 05-08-2019, 07:45 PM
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It's historical verbiage.
And we still call Pluto a "planet," but science books have a big fat asterisk next to that.
  #158  
Old 05-08-2019, 11:34 PM
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First I want to point out that the law from the coin example is not valid, was not formulated on a valid basis, and need not even appear to be valid. It is possible to observe greyness decreasing on a macroscopic scale in the previous example. The law was formulated nonetheless.
It's formulated on exactly the same basis as every other law: by generalization from observational regularities. You do the coin example experiment a couple of hundred times, and you'll always find the law to hold; absent knowledge of the microscopic dynamics, you have no justification to assume that in some cases, it'll be violated. So it's on exactly the same grounds as any other law of physics.

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I take issue with step 9. I have no reason to believe the observer will observe a gradual change in color. You say the observer can only see gross color. If before he could distinguish two boxes of different colors, as soon as you remove the walls I take it he can only observe the combined box as a whole. Therefore the instant you remove the wall the observer sees flat grey.

~Max
Huh? How on Earth could that happen? The observer's powers of observation are equal to the previous experiment; they're able to perceive (what comes down to) the fraction of black vs. white marbles in a region, but not the actual marbles. This is really the same thing your computer screen does---by mixing three colors in different proportions, because you can't see the individual pixels, different gross colors emerge, like in this picture. Or take something like this ASCII art: the different grey regions are just different fractions of black, mixed with white. That's the way the observer sees the system: the individual marbles are too small to be individually resolved, but different mixtures of them realize different levels of grey.

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Originally Posted by Max S. View Post
Actually I'm not sure if I agree with step 8 either. You can't necessarily "go" from any arbitrary configuration of marbles to another, certainly not if you include a succession of three states in order.

~Max
But that's just counting. If there are more 'more grey' states than 'less grey' states, then a system in a 'less grey' state has more ways of evolving towards a 'more grey' state.

So assume some 'less grey' state G< can be realized by means of white and black marbles in ten different ways (corresponding to microstates G<1 through G<10). Assume there are more ways to realize the 'more grey' state G> than there are ways of realizing the 'even less grey' state G<<. Then, more of these ten states will evolve towards the more grey state than will evolve towards the less grey one.

Say there are eight states G>1 through G>8, and the states G<<1 and G<<2. These are the states available from either of the states realizing G< on the next time-step. Then, eight out of ten times, if the system is in the state G<, we will see it evolve into G>. (Perhaps it helps to recall, here, that a reversible evolution always takes different states to different states, as if it didn't, i. e. by taking two different states to the same following state, you can't tell which was the original state from looking at the later one, and thus, can't reverse the evolution.)

Thus, at each time-step, we're more likely to observe an increase in grey-ness, simply because there are more ways to get more grey.

Last edited by Half Man Half Wit; 05-08-2019 at 11:35 PM.
  #159  
Old 05-08-2019, 11:35 PM
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And we still call Pluto a "planet," but science books have a big fat asterisk next to that.
It's the same with the second law of thermodynamics---textbooks generally point out that it's only valid statistically.
  #160  
Old 05-09-2019, 12:37 AM
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But that's just my point, assuming microscopic randomness or chaos means a particle can violate any number of laws. Not just the second law of thermodynamics.
It's not about assuming randomness or chaotic behaviour, it's that models that include these features make good predictions. e.g. we can predict very precisely the frequency with which electrons will tunnel through a barrier.

(OK, I know with "randomness" there's a degree of interpretation involved and it's possible there could be non-local hidden variables. Half Man Half Wit will know better than me.
But I'm not aware of any effort to interpret away chaotic behaviour. Or anyone that claims it breaks physical laws.)

Last edited by Mijin; 05-09-2019 at 12:38 AM.
  #161  
Old 05-09-2019, 04:15 AM
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I continue to feel that my claim ...
Unlike the Laws of Newton, Maxwell and Einstein, the Second Law isn't a dynamical law; it's just a statistical fact, closely akin to the Law of Large Numbers.

... is correct, and should have quiesced this bickering!

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... What does free energy have to do with the second law? ...
Everything.

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Originally Posted by https://en.wikipedia.org/wiki/Gibbs_free_energy
According to the second law of thermodynamics, for systems reacting at STP (or any other fixed temperature and pressure), there is a general natural tendency to achieve a minimum of the Gibbs free energy.
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Another way of looking at the Helmholtz free energy is realizing that in equilibrium a system at constant T and constant V will be in a minimum of the Helmholtz free energy. Indeed this can be easily derived from the second law ...
(Google to detail the difference between Helmholtz's and Gibbs' free energy.)
  #162  
Old 05-09-2019, 08:38 AM
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I continue to feel that my claim ...
Unlike the Laws of Newton, Maxwell and Einstein, the Second Law isn't a dynamical law; it's just a statistical fact, closely akin to the Law of Large Numbers.

... is correct, and should have quiesced this bickering!



Everything.





(Google to detail the difference between Helmholtz's and Gibbs' free energy.)
But I disagree. Why is the second law but a statistical fact? I asserted and continue to assert that there have been no theoretical or observed violations of the second law of thermodynamics, as defined in the three primary sources in the original post. None of those formulations involve entropy, although Clausius in a corollary said entropy can never decrease. But he was using dS=dQ/T, not free energy or microstates.

So far the only violations of the second law have been violations of Clausius's corollary, using different definitions of entropy. And it seems to me that said definitions of entropy also imply the possibility that the laws of Newton, Einstein, and Maxwell can also be violated, although the probability is vanishingly small.

~Max
  #163  
Old 05-09-2019, 08:49 AM
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Re:Re:Re: Flipping Coins


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Originally Posted by Half Man Half Wit View Post
It's formulated on exactly the same basis as every other law: by generalization from observational regularities. You do the coin example experiment a couple of hundred times, and you'll always find the law to hold; absent knowledge of the microscopic dynamics, you have no justification to assume that in some cases, it'll be violated. So it's on exactly the same grounds as any other law of physics.
No, if you do the coin example experiment a couple hundred of times, you might find the law to hold. It is not even given that the law will probably hold.

I agreed that the law could be formulated but I did not agree that the steps between 5 and 18 of that example were valid, logical steps. I never agreed with your logic past step 5, because you missed a very important premise. See my post on dice logic, #150.

~Max
  #164  
Old 05-09-2019, 09:00 AM
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Huh? How on Earth could that happen? The observer's powers of observation are equal to the previous experiment; they're able to perceive (what comes down to) the fraction of black vs. white marbles in a region, but not the actual marbles. This is really the same thing your computer screen does---by mixing three colors in different proportions, because you can't see the individual pixels, different gross colors emerge, like in this picture. Or take something like this ASCII art: the different grey regions are just different fractions of black, mixed with white. That's the way the observer sees the system: the individual marbles are too small to be individually resolved, but different mixtures of them realize different levels of grey.
Very well, so the observer can still distinguish between the color of the left box and the color of the right box after you have removed the wall separating them. I still assume he can only see the solid color of each box - he sees the entire left box as one color, and the entire right box as one color too. In that case, I predict he will likely see a never ending fluctuation of monochromatic colors. It is possible but highly unlikely that he will see a solid color in either box, unchanging over time. More probably, the color of each box will usually be a middle shade of grey, constantly flickering from lighter to darker tones. Eventually both boxes will return to their original black and white, if only for a moment.

~Max
  #165  
Old 05-09-2019, 09:46 AM
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Greyness


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Originally Posted by Half Man Half Wit View Post
But that's just counting. If there are more 'more grey' states than 'less grey' states, then a system in a 'less grey' state has more ways of evolving towards a 'more grey' state.

So assume some 'less grey' state G< can be realized by means of white and black marbles in ten different ways (corresponding to microstates G<1 through G<10). Assume there are more ways to realize the 'more grey' state G> than there are ways of realizing the 'even less grey' state G<<. Then, more of these ten states will evolve towards the more grey state than will evolve towards the less grey one.

Say there are eight states G>1 through G>8, and the states G<<1 and G<<2. These are the states available from either of the states realizing G< on the next time-step. Then, eight out of ten times, if the system is in the state G<, we will see it evolve into G>. (Perhaps it helps to recall, here, that a reversible evolution always takes different states to different states, as if it didn't, i. e. by taking two different states to the same following state, you can't tell which was the original state from looking at the later one, and thus, can't reverse the evolution.)

Thus, at each time-step, we're more likely to observe an increase in grey-ness, simply because there are more ways to get more grey.
Let's say we have a number line from negative ten to positive ten, where negative ten is black, zero is grey, and ten is white. On this number line is a point whose initial position is random, and whose position at every time step is random. The absolute value of the point constitutes the entropy of the system you describe.

We can see now that a point on the leftmost position, -10, has a 19/21 probability of becoming "more grey" or "less entropic" at the next step, and a 2/21 probability of remaining at the same distance from grey or of having no change in entropy. So does a point on the rightmost position. Here are the other values (you may need to switch forum themes at the bottom left of the page):

Code:
   Position to Entropy    
Pos | Less  | Same |  More
-10 | 19/21 | 2/21 |  0/21
- 9 | 17/21 | 2/21 |  2/21
- 8 | 15/21 | 2/21 |  4/21
- 7 | 13/21 | 2/21 |  6/21
- 6 | 11/21 | 2/21 |  8/21
- 5 |  9/21 | 2/21 | 10/21
- 4 |  7/21 | 2/21 | 12/21
- 3 |  5/21 | 2/21 | 14/21
- 2 |  3/21 | 2/21 | 16/21
- 1 |  1/21 | 2/21 | 18/21
  0 |  0/21 | 1/21 | 20/21
  1 |  1/21 | 2/21 | 18/21
  2 |  3/21 | 2/21 | 16/21
  3 |  5/21 | 2/21 | 14/21
  4 |  7/21 | 2/21 | 12/21
  5 |  9/21 | 2/21 | 10/21
  6 | 11/21 | 2/21 |  8/21
  7 | 13/21 | 2/21 |  6/21
  8 | 15/21 | 2/21 |  4/21
  9 | 17/21 | 2/21 |  2/21
 10 | 19/21 | 2/21 |  0/21
Total all of those probabilities together and we can see that after one instant, there is a 200/441 probability that the system will move to a less entropic state, a 41/441 probability that the entropy will not change, and a 200/441 probability that entropy will increase. In percentages, that's about 45% for entropy to decrease, 9% to stay the same, and 45% to increase.

And this is after assuming the equiprobability of microstates at each measure of entropy, which I do not wish to assume.

~Max
  #166  
Old 05-09-2019, 09:57 AM
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It's not about assuming randomness or chaotic behaviour, it's that models that include these features make good predictions. e.g. we can predict very precisely the frequency with which electrons will tunnel through a barrier.

(OK, I know with "randomness" there's a degree of interpretation involved and it's possible there could be non-local hidden variables. Half Man Half Wit will know better than me.
But I'm not aware of any effort to interpret away chaotic behaviour. Or anyone that claims it breaks physical laws.)
But the models of statistical mechanics assume stochastic or chaotic behavior, thus the laws of statistical mechanics are actually heuristics, not laws. The second law of thermodynamics, as formulated using the statistical definition of entropy, has been violated by Half Man Half Wit's own citation. This is not to detract from the utility of statistical mechanics, which is much easier to calculate.

There are three (or four) forms of the second law of thermodynamics which are not heuristic, and these are the ones I was taught in school and cited in the first post. So in order to disprove those laws, you would need to show a contradiction. That's what this thread is about.

~Max

Last edited by Max S.; 05-09-2019 at 10:01 AM.
  #167  
Old 05-09-2019, 11:05 AM
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No, if you do the coin example experiment a couple hundred of times, you might find the law to hold. It is not even given that the law will probably hold.
Not a given in the sense that it's not a 100% certainty, that I agree with. But it's effectively certain in the sense that the probability of observing a violation can be made arbitrarily small, with a suitable choice of numbers.

Or, in other words, consider the following. You let a ball drop a hundred (a thousand, a million...) times. It always falls down. From this, you formulate a law: stuff falls down.

Now suppose that whatever deity has created the universe has made it so that the actual law is: stuff falls down, except once every sextillion times, when it just hovers in place.

You've got no data to support that the law is actually the latter, and never will observe any. Thus, you still formulate the law as 'stuff falls down'. You're completely justified in doing so; however, you happen to be wrong.

That's the situation we're in here: whenever we try, we will find greyness increasing with overwhelming probability. That is, in any concrete, reasonably large series of trials, we won't observe a violation. Thus, we formulate a law to the effect that greyness always increases. This law stands on equal footing with every other physical law: it's a generalization from finitely many observations.

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I agreed that the law could be formulated but I did not agree that the steps between 5 and 18 of that example were valid, logical steps. I never agreed with your logic past step 5, because you missed a very important premise. See my post on dice logic, #150.

~Max
There's no missing premise. I have explained how we can say that the probability is 1/6, without having to assume it: because one out of every six possible evolutions of the die from arbitrary initial conditions ends with it showing any given number.

Take a coin. Suppose you can only throw it in two different ways---way A and way B. Way A always lands heads up; way B always lands tails up. You don't have control over whether you've thrown it according to way A or way B (random initial conditions, you remember). Then, the probability that it comes up heads is 50% on each throw. No further assumptions necessary.

Suppose now you can throw the coin in 20 different ways, 10 of which come up heads, 10 of which come up tails. This yields the same conclusion. As does supposing that there are 100, or 1000, and so on different ways. What matters is that from the set of possible initial conditions, half of them yield heads, and half of them yield tails.

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Originally Posted by Max S. View Post
Very well, so the observer can still distinguish between the color of the left box and the color of the right box after you have removed the wall separating them. I still assume he can only see the solid color of each box - he sees the entire left box as one color, and the entire right box as one color too. In that case, I predict he will likely see a never ending fluctuation of monochromatic colors. It is possible but highly unlikely that he will see a solid color in either box, unchanging over time. More probably, the color of each box will usually be a middle shade of grey, constantly flickering from lighter to darker tones. Eventually both boxes will return to their original black and white, if only for a moment.

~Max
Yes, this is basically right. And for a large enough number of marbles, what'll happen is that the observer will observe a transition to uniform grey in every case they do the experiment, since the likelihood of the colors separating will decrease with the number of marbles. Thus, at some point, if they do the experiment 10, or a hundred, or a 1000, or a billion times, they won't observe a violation of the law 'the system tends to greyness' with any appreciable probability. Agreed?


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Let's say we have a number line from negative ten to positive ten, where negative ten is black, zero is grey, and ten is white. On this number line is a point whose initial position is random, and whose position at every time step is random. The absolute value of the point constitutes the entropy of the system you describe.
This is in contradiction to the system as I set it up. The position at every time step is not random; rather, it is determined (exactly and absolutely) by the microstate corresponding to the position of the previous time step.

You assume that entropy gain and loss are equally likely. You get out that entropy gain and loss are equally likely. This isn't surprising.
  #168  
Old 05-09-2019, 11:12 AM
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Originally Posted by Half Man Half Wit View Post
Take a coin. Suppose you can only throw it in two different ways---way A and way B. Way A always lands heads up; way B always lands tails up. You don't have control over whether you've thrown it according to way A or way B (random initial conditions, you remember). Then, the probability that it comes up heads is 50% on each throw. No further assumptions necessary.
Bolding mine. That sentence is an assumption in and of itself, not implied by the rest of the paragraph. You have assumed that the probability of throwing way A or throwing way B is 50%. You are allowed to make that assumption, but you must acknowledge that you made it to complete your argument.

~Max
  #169  
Old 05-09-2019, 11:19 AM
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That's the situation we're in here: whenever we try, we will find greyness increasing with overwhelming probability. That is, in any concrete, reasonably large series of trials, we won't observe a violation. Thus, we formulate a law to the effect that greyness always increases. This law stands on equal footing with every other physical law: it's a generalization from finitely many observations.
I think we understood each other, right up until you said "in any concrete, reasonable large series of trials, we won't observe a violation." That's not true. You would have to say we probably won't observe a violation, and you can drop the rest of that sentence. But this is a minor issue of semantics. The person formulating the law does not think so, and that is why his law is flawed.

~Max
  #170  
Old 05-09-2019, 12:00 PM
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Yes, this is basically right. And for a large enough number of marbles, what'll happen is that the observer will observe a transition to uniform grey in every case they do the experiment, since the likelihood of the colors separating will decrease with the number of marbles. Thus, at some point, if they do the experiment 10, or a hundred, or a 1000, or a billion times, they won't observe a violation of the law 'the system tends to greyness' with any appreciable probability. Agreed?
If by 'the system tends to greyness' you mean "black and white, if brought into contact, eventually even out to a uniform gray", I disagree, the probability of a violation is exactly 100%. The colors of the two boxes never "eventually even out to a uniform gray", and our observer will notice a violation if he is allowed to observe the full period of the system. In the likely case that at least one marble from either box moves into the other box, one or both boxes will always be changing color. There can never be equilibrium at any particular shade of grey. If you assume equal aggregate velocities of marbles in each box, the system should cumulatively spend half of its period widening the color gap and half of its period shrinking the color gap between its two halves.

If by 'the system tends to greyness' you mean the disparity in color between the left and right boxes will always decrease over time, again this is disproven with 100% probability so long as the observer has enough time to watch the system.

If by 'the system tends to greyness' you mean 'after removing the wall, each box will usually be some shade of grey as opposed to absolute black or white', I have no qualms. But that is not a law.

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This is in contradiction to the system as I set it up. The position at every time step is not random; rather, it is determined (exactly and absolutely) by the microstate corresponding to the position of the previous time step.

You assume that entropy gain and loss are equally likely. You get out that entropy gain and loss are equally likely. This isn't surprising.
You did not prove to my satisfaction that, in any random configuration of marbles moving around in a box, entropy is more likely to increase than decrease over time. Without considering the effects of our assumptions about entropy gain and loss, mine is on equal footing with yours.

~Max
  #171  
Old 05-09-2019, 01:51 PM
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Bolding mine. That sentence is an assumption in and of itself, not implied by the rest of the paragraph. You have assumed that the probability of throwing way A or throwing way B is 50%. You are allowed to make that assumption, but you must acknowledge that you made it to complete your argument.

~Max
But we had already settled that issue, I thought:
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I am perfectly fine allowing for equiprobability of the initial microstate of a toy thermodynamic system.
Are you no longer fine with this? And, presuming that you are, do you agree that then, the probability of the coin comes out to 50% for each possibility? And likewise for the die?

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I think we understood each other, right up until you said "in any concrete, reasonable large series of trials, we won't observe a violation." That's not true. You would have to say we probably won't observe a violation, and you can drop the rest of that sentence. But this is a minor issue of semantics. The person formulating the law does not think so, and that is why his law is flawed.

~Max
Well, I could drag the extra verbiage through every statement I make, but I think it's good enough to say 'we won't observe any violation' instead of 'for any reasonable time scale, the probability of observing a violation is as small as we care to make it, by increasing the system size'. Because the outcome is the same: if you repeat the experiment some reasonable amount of times, you are astronomically unlikely to ever make an observation contradicting the law of increasing greyness, and thus, you will believe it holds.

You're right to point out that this is, ultimately, wrong; but the observer has no way to know that, without having the microscopic theory of greyness.

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If by 'the system tends to greyness' you mean "black and white, if brought into contact, eventually even out to a uniform gray", I disagree, the probability of a violation is exactly 100%. The colors of the two boxes never "eventually even out to a uniform gray", and our observer will notice a violation if he is allowed to observe the full period of the system.
Sure. But a full period, for any reasonably macroscopic system, is going to be fantastically huge. So you're not going to observe anything of the sort.

We're talking about generalizations made from actually feasible observations. Given this, while there is an astronomically small probability of actually observing a violation of the law of increasing greyness, the far more probable course is going to be that, during the tens or hundreds or thousands of times that the experiment is repeated, no violation is observed, and thus, the law has the status of any other physical law ever formulated.

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In the likely case that at least one marble from either box moves into the other box, one or both boxes will always be changing color.
But not on any observable level. The observer has eyes that aren't significantly better than a human being's, so, while, again, you're right in principle, these deviations are not going to be observed. Remember, laws are formulated based on actual observations. And, with overwhelming probability, any actual observation is going to be the system evolving to an equal grey and staying that way, at least, for sufficiently large numbers of marbles.

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You did not prove to my satisfaction that, in any random configuration of marbles moving around in a box, entropy is more likely to increase than decrease over time.
That is proven simply by the fact that there are more states of high greyness than there are states of uneven color distribution. For if that's the case, and the microdynamics is reversible, then, if you take a state of intermediate greyness, there will be more ways ('more' meaning here 'astronomically many more') to evolve to a state of higher greyness than to evolve to a state of lower greyness. Thus, whenever you find the system in a state of intermediate greyness (i. e. uneven color distribution), then, with astronomic likelihood, the state at the next timestep will be one of higher greyness.

Take my introductory example. Here's again the distinguishable macrostates, together with the number of their microscopic realizations:
  • (A1B1C1): 6
  • (A2B1C0): 3
  • (A2B0C1): 3
  • (A1B2C0): 3
  • (A0B2C1): 3
  • (A1B0C2): 3
  • (A0B1C2): 3
  • (A3B0C0): 1
  • (A0B3C0): 1
  • (A0B0C3): 1

No matter what the microscopic dynamics are, each of the three microstates realizing, say, (A2B1C0) has 6 states corresponding to macrostates of higher entropy it can evolve to, 17 states of equal entropy, but only 3 states of lower entropy. Does that help clear things up?
  #172  
Old 05-09-2019, 02:37 PM
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If you are studying these marbles, once you have an ergodic Markov chain, then it will satisfy the asymptotic equipartition theorem; this is a mathematical result in information theory. You can compare the entropy rate of the forward and time-reversed process and it will satisfy the fluctuation theorem. At this point, you are talking about mathematical theorems, like the Law of Large Numbers, so there really isn't anything open to interpretation unless one wants to philosophize about it like Rosencrantz and Guildenstern.
  #173  
Old 05-09-2019, 02:44 PM
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Take a coin. Suppose you can only throw it in two different ways---way A and way B. Way A always lands heads up; way B always lands tails up. You don't have control over whether you've thrown it according to way A or way B (random initial conditions, you remember). Then, the probability that it comes up heads is 50% on each throw. No further assumptions necessary.
Bolding mine. That sentence is an assumption in and of itself, not implied by the rest of the paragraph. You have assumed that the probability of throwing way A or throwing way B is 50%. You are allowed to make that assumption, but you must acknowledge that you made it to complete your argument.

~Max
But we had already settled that issue, I thought:
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I am perfectly fine allowing for equiprobability of the initial microstate of a toy thermodynamic system.
Are you no longer fine with this? And, presuming that you are, do you agree that then, the probability of the coin comes out to 50% for each possibility? And likewise for the die?
I stand by both of my statements. In the bolded statement you claimed the probability of a coin throw is 50% each time. That is not in any way implied by the initial state of the coins. Besides, neither throw A nor throw B take into account the previous state of the coin because both throws always give heads or tails respectively. You haven't told me how A is chosen vs B, but you assume the probability is 50%.

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Well, I could drag the extra verbiage through every statement I make, but I think it's good enough to say 'we won't observe any violation' instead of 'for any reasonable time scale, the probability of observing a violation is as small as we care to make it, by increasing the system size'. Because the outcome is the same: if you repeat the experiment some reasonable amount of times, you are astronomically unlikely to ever make an observation contradicting the law of increasing greyness, and thus, you will believe it holds.

You're right to point out that this is, ultimately, wrong; but the observer has no way to know that, without having the microscopic theory of greyness.
I think we're agreed on the subject of phenomenological laws. You haven't convinced me that the observer is likely to see only increasing greyness, but if that's all he saw he could very well formulate a law of increasing greyness. I take no issue with the observer's logic.

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Sure. But a full period, for any reasonably macroscopic system, is going to be fantastically huge. So you're not going to observe anything of the sort.

We're talking about generalizations made from actually feasible observations. Given this, while there is an astronomically small probability of actually observing a violation of the law of increasing greyness, the far more probable course is going to be that, during the tens or hundreds or thousands of times that the experiment is repeated, no violation is observed, and thus, the law has the status of any other physical law ever formulated.
If we limit violations to observed violations rather than theoretical violations, I will concede the insurmountable difficulty in falsifying a law if violations are actually so improbable as to make observation unrealistic. But I do not yet concede the improbability of observing decreasing greyness.

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But not on any observable level. The observer has eyes that aren't significantly better than a human being's, so, while, again, you're right in principle, these deviations are not going to be observed. Remember, laws are formulated based on actual observations. And, with overwhelming probability, any actual observation is going to be the system evolving to an equal grey and staying that way, at least, for sufficiently large numbers of marbles.
If you are to limit the observer's sight in such a way that he can only distinguish between almost black, almost white, and everything-else-is-grey, and if I was to assume significant enough differences in color for the observer to notice are rare (which I do not concede yet), I would concede that the observer can in fact observe grey over time despite microscopic fluctuations. He will assume the left and right boxes are equally grey not because they are, but because his senses are so dull as to fail to recognize the difference and constant fluctuation of lighter and darker shades in one box then the other. But I have not yet conceded the underlying premise.

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That is proven simply by the fact that there are more states of high greyness than there are states of uneven color distribution. For if that's the case, and the microdynamics is reversible, then, if you take a state of intermediate greyness, there will be more ways ('more' meaning here 'astronomically many more') to evolve to a state of higher greyness than to evolve to a state of lower greyness. Thus, whenever you find the system in a state of intermediate greyness (i. e. uneven color distribution), then, with astronomic likelihood, the state at the next timestep will be one of higher greyness.

Take my introductory example. Here's again the distinguishable macrostates, together with the number of their microscopic realizations:
  • (A1B1C1): 6
  • (A2B1C0): 3
  • (A2B0C1): 3
  • (A1B2C0): 3
  • (A0B2C1): 3
  • (A1B0C2): 3
  • (A0B1C2): 3
  • (A3B0C0): 1
  • (A0B3C0): 1
  • (A0B0C3): 1

No matter what the microscopic dynamics are, each of the three microstates realizing, say, (A2B1C0) has 6 states corresponding to macrostates of higher entropy it can evolve to, 17 states of equal entropy, but only 3 states of lower entropy. Does that help clear things up?
No, having more potential states with one property does not imply higher probability for 'evolution' towards a state with said property. Simply having more states says nothing about probability. You have not given me a basis for a probability distribution and, quite to the contrary, you denied that the distribution is random. In fact you say the underlying dynamics are deterministic. How do you come to the conclusion that a system in a state of intermediate grayness will probably evolve into a state of higher greyness? It seems like you are pulling a postulate from thin air.

~Max
  #174  
Old 05-09-2019, 03:01 PM
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If you are studying these marbles, once you have an ergodic Markov chain, then it will satisfy the asymptotic equipartition theorem; this is a mathematical result in information theory. You can compare the entropy rate of the forward and time-reversed process and it will satisfy the fluctuation theorem. At this point, you are talking about mathematical theorems, like the Law of Large Numbers, so there really isn't anything open to interpretation unless one wants to philosophize about it like Rosencrantz and Guildenstern.
What is an "ergodic Markov chain"? The internet definitions I found assume some sort of stochastic behavior, whereas neither Half Man Half Wit or I have backed down from the premise that microscopic dynamics are deterministic.

~Max
  #175  
Old 05-09-2019, 04:14 PM
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I stand by both of my statements. In the bolded statement you claimed the probability of a coin throw is 50% each time. That is not in any way implied by the initial state of the coins.
Huh? Sure it is! If the initial state can be either A or B with 50% probability (equiprobability of initial states, remember!), A always yields heads, and B always yields tails, then each throw yields heads with 50% probability.

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You haven't told me how A is chosen vs B, but you assume the probability is 50%.
Again, that's just going from what you said: initial states can be chosen equiprobably.

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If we limit violations to observed violations rather than theoretical violations, I will concede the insurmountable difficulty in falsifying a law if violations are actually so improbable as to make observation unrealistic.
Theories are build on observation, so of course, the latter must be what we start with.

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If you are to limit the observer's sight in such a way that he can only distinguish between almost black, almost white, and everything-else-is-grey, and if I was to assume significant enough differences in color for the observer to notice are rare (which I do not concede yet), I would concede that the observer can in fact observe grey over time despite microscopic fluctuations. He will assume the left and right boxes are equally grey not because they are, but because his senses are so dull as to fail to recognize the difference and constant fluctuation of lighter and darker shades in one box then the other.
I'm really having trouble believing you're sincere here. I am not restricting the observer's vision, or anything. Rather, that he only sees shades of grey, so to speak, is for the same reason as that you only see shades of grey in pictures like the ones here. If the density of black and white pixels were slightly different (say, a black pixel added here and there), you would not observe any difference. If there are some regions with higher density of black pixels, the gray there is darker, if there are regions with whiter pixels, they are brighter.

That's what our experimenter sees: exactly what you would see.


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No, having more potential states with one property does not imply higher probability for 'evolution' towards a state with said property.
Of course. The number of possible states to evolve into determines the number of possible evolutions.

Let's take the smallest possible change of something like the coin system. Take one pixel, and flip its color. Suppose you start with an all-black state. Then, flipping one pixel's color will make the system more grey with certainty: every other possible state is one with higher 'greyness'.

Then, take the resulting state, and again, flip one pixel: if you flip any pixel other than the one you've flipped before, you will again move towards a state of higher greyness. And so on: there will be more states of higher greyness to flip to, until you've reached a 50/50 distribution of black vs. white pixels. If you flip a pixel there, you will get to a state that's ever so slightly (but undetectably) less grey. The next flip will return you to the equilibrium with a probability of 50% (actually, very slightly more than that, since there is now either one more black or white pixel).

Now, the important thing to realize is that this doesn't depend on the microdynamics; in particular, it doesn't assume that this dynamics is random. Rather, this is a conclusion that applies to generic deterministic and reversible dynamics.

Think again back to the original, all-black system: any possible microscopic evolution will lead to a more grey state. After that initial chance, still almost every possible evolution will lead to a more grey state. And so on, up until you reach equilibrium. There, half of all evolution laws will lead to a less grey state; but, even for that half that lead away from equilibrium, most will return quickly to it, with some holding out longer, and only one making it all the way to the all-white state, before going back.

In conclusion, as long as you don't take care to set up a very special evolution of the system (a point I had stressed from the beginning, you will recall), any generic microdynamics will, at any point in the evolution of the system, tend to increase the overall greyness with overwhelming probability.
  #176  
Old 05-09-2019, 04:47 PM
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Huh? Sure it is! If the initial state can be either A or B with 50% probability (equiprobability of initial states, remember!), A always yields heads, and B always yields tails, then each throw yields heads with 50% probability.


Again, that's just going from what you said: initial states can be chosen equiprobably.
If you want to consider the first throw part of the initial state, that's fine by me. But I still don't understand how it follows that the second throw has 50% probability. I assume the deterministic process that decides whether to throw A or B, the process we don't know about, depends on the currently showing side of the coin. Otherwise this whole example of throwing a single coin has no relevance to the point you were trying to make - proving a law that says "the [theoretical] system will always become more grey over time".

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Theories are build on observation, so of course, the latter must be what we start with.
Very well.

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I'm really having trouble believing you're sincere here. I am not restricting the observer's vision, or anything. Rather, that he only sees shades of grey, so to speak, is for the same reason as that you only see shades of grey in pictures like the ones here. If the density of black and white pixels were slightly different (say, a black pixel added here and there), you would not observe any difference. If there are some regions with higher density of black pixels, the gray there is darker, if there are regions with whiter pixels, they are brighter.

That's what our experimenter sees: exactly what you would see.
But we don't know that downward fluctuations in entropy are small and fast. They could just as well be large and slow. And besides, I can observe the flickering of an LCD display with any number of optical devices. On older monitors I can pick up the flickering with the naked eye. I have at my disposal a number of tools which can distinguish very slight differences between colors in photographs.

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Of course. The number of possible states to evolve into determines the number of possible evolutions.

Let's take the smallest possible change of something like the coin system. Take one pixel, and flip its color. Suppose you start with an all-black state. Then, flipping one pixel's color will make the system more grey with certainty: every other possible state is one with higher 'greyness'.

Then, take the resulting state, and again, flip one pixel: if you flip any pixel other than the one you've flipped before, you will again move towards a state of higher greyness. And so on: there will be more states of higher greyness to flip to, until you've reached a 50/50 distribution of black vs. white pixels. If you flip a pixel there, you will get to a state that's ever so slightly (but undetectably) less grey. The next flip will return you to the equilibrium with a probability of 50% (actually, very slightly more than that, since there is now either one more black or white pixel).

Now, the important thing to realize is that this doesn't depend on the microdynamics; in particular, it doesn't assume that this dynamics is random. Rather, this is a conclusion that applies to generic deterministic and reversible dynamics.
This absolutely depends on microdynamics. Is picking "any pixel other than the one you've flipped before" "until you've reached a 50/50 distribution of black vs. white pixels" a rule? How do you determine which "other" pixel to flip, is it actually random? After you reached 50/50 distribution, how do you determine which pixel to flip then? Are you implying the flips after that are picked at random? If not, how can you assign a 50% probability of evolving towards equilibrium?

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Think again back to the original, all-black system: any possible microscopic evolution will lead to a more grey state. After that initial chance, still almost every possible evolution will lead to a more grey state. And so on, up until you reach equilibrium. There, half of all evolution laws will lead to a less grey state; but, even for that half that lead away from equilibrium, most will return quickly to it, with some holding out longer, and only one making it all the way to the all-white state, before going back.

In conclusion, as long as you don't take care to set up a very special evolution of the system (a point I had stressed from the beginning, you will recall), any generic microdynamics will, at any point in the evolution of the system, tend to increase the overall greyness with overwhelming probability.
It does not follow that the system will evolve towards a more grey state simply because there are more ways to do so than to do otherwise. There could be 999 ways to evolve to a more grey state and one way to do otherwise, but it does not follow that the system will do so or is even likely to do so. You are missing a premise.

~Max

Last edited by Max S.; 05-09-2019 at 04:49 PM.
  #177  
Old 05-09-2019, 04:57 PM
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What is an "ergodic Markov chain"? The internet definitions I found assume some sort of stochastic behavior, whereas neither Half Man Half Wit or I have backed down from the premise that microscopic dynamics are deterministic.

~Max
I was under the impression you were discussing processes that randomly mix up marbles and wanted to understand why they inevitably get scrambled rather than unscrambled.
  #178  
Old 05-09-2019, 05:02 PM
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I was under the impression you were discussing processes that randomly mix up marbles and wanted to understand why they inevitably get scrambled rather than unscrambled.
The only randomness I am aware of in the marbles hypothetical is the initial arrangement of marbles in each box. That's why I am so confused as to how Half Man Half Wit gets his probabilities.

~Max
  #179  
Old 05-09-2019, 11:51 PM
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If you want to consider the first throw part of the initial state, that's fine by me.
It's not part of it, as such, but it determines it completely.

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But I still don't understand how it follows that the second throw has 50% probability.
There's really no sense to considering a second throw in this example. As the system only has two available states, it has completed a full cycle after the first throw. But consider a coin with four possible initial states (n. B.: those don't correspond to 'heads up', that's the macrostate; rather, you can consider these states as 'ways of making the initial throw'), A and B, which yield heads and tails respectively on the first throw, and C and D, which do likewise. However, they're distinguished by the fact that both A and B will yield heads on the second throw, while C and D yield tails on the second throw. This upholds the stated determinism, but the second throw has, no matter the outcome of the first, a 50% probability of coming up heads.

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And besides, I can observe the flickering of an LCD display with any number of optical devices. On older monitors I can pick up the flickering with the naked eye. I have at my disposal a number of tools which can distinguish very slight differences between colors in photographs.
Sure, but I would ask you to take the hypothetical in the spirit it was proposed, and not try to pointlessly fight it---you're again trying to use where you think this will end up to fight the argument you think I'm making. But you're just getting sidetracked, arguing irrelevancies, so I would ask you to just take what I initially stated---an experimenter observing the system in such a way as to only be able to distinguish gross greyness, in the same way you're only able to distinguish gross greyness in the pixel images with only the naked eye and a monitor of sufficiently high resolution and quality.

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It does not follow that the system will evolve towards a more grey state simply because there are more ways to do so than to do otherwise. There could be 999 ways to evolve to a more grey state and one way to do otherwise, but it does not follow that the system will do so or is even likely to do so. You are missing a premise.

~Max
No. You only need to count.

Take my original system. It has 27 microstates. We can represent the microstate by means of a 27-dimensional vector, that has a '1' in the row corresponding to the microstate, and a '0' everywhere else, like this:
Code:
          (0)
          (0)
          (.)
          (.)
          (.)
     Si = (1)
          (.)
          (.)
          (.)
          (0)
Where the dots represent ellipses, and the index i numbers the states from 1 to 27, such that it always just gives the row where the '1' appears. We can order the states as before, with the corresponding microstates to each macrostate listed:
  • (A1B1C1): S1 - S6
  • (A2B1C0): S7 - S9
  • (A2B0C1): S10 - S12
  • (A1B2C0): S13 - S15
  • (A0B2C1): S16 - S18
  • (A1B0C2): S19 - S21
  • (A0B1C2): S22 - S24
  • (A3B0C0): S25
  • (A0B3C0): S26
  • (A0B0C3): S27

Any given evolution law of this system will be a 27 x 27 - matrix M, which takes one 27-dim vector to a different 27-dim vector by ordinary matrix multiplication:

Sj = M*Si

(That is, the system starts out in Si, is evolved for one timestep, and ends up in Sj.)

The matrix evolves Si into Sj, if and only if the i-th entry in its j-th row is '1'.

This matrix must obey some constraints. First, any given state can only evolve into one other state (this is determinism). Consequently, the matrix can only have one entry different from 0 in each column.

Likewise, only one state can evolve into any given state (this is reversibility: each state must have a unique precursor). Consequently, there can be only one non-zero entry per row, as well.

Thus, all the valid laws of evolution for the system take the form of a matrix M such that each row and column have exactly one entry equal to 1, and the rest 0 (there are thus 27 nonzero entries in the matrix).

Now, we can count how often each of these evolutions will increase vs. decrease entropy. Take first the upper left 6 x 6 - submatrix: any entry here will only 'mix' the maximum-entropy states among themselves, thus yielding to constant entropy.

Then, take the 18 x 18 - submatrix 'in the middle'. This mixes the intermediate-entropy states, thus likewise not leading to entropy increase. Same for the 3 x 3 - submatrix in the lower right corner, which mixes the min-entropy states.

Now take the lower left 3 x 24 - submatrix (that is, everything left over if you take the lower right 3 x 3 away). These are more interesting: they are entries that take any of the 24 states of nonminimal entropy, and transform them into stats of minimal entropy. These are, thus, entropy decreasing. Because of the properties of the matrix, there can be three such entries.

Then, take the left 18 x 6 - strip: They are the entries that take any of the 6 maximum entropy states to one of the 18 intermediate ones. These are, likewise, entropy decreasing; again, there can be maximally 6 entries here (but, if there are any entries here, we have to be careful not to double count: if there is an entry in the first three columns, there can't be any entries in the lower left 3 x 3 - submatrix, since this would entail a maximum entropy state both transitioning to a lower entropy and a minimum entropy one, in violation of determinism).

These are all the entries that may lower entropy: transitioning from a maximum entropy state to either an intermediate or minimum entropy state, or transitioning from an intermediate entropy state to a minimum entropy state.

Consequently---and here's the kicker---any evolution law we can write down for the system can have at most nine entries that lead to a lowering of entropy, and must, consequently, have 18 entries that keep entropy constant or increase it. Hence, no matter what the microdynamics are---no matter which particular M we choose---it follows, just from the count of microstates, that it takes any given state to a lower-entropy one at most one out of three (9/27) times. Consequently, entropy increases or stays constant more often than it decreases.

And that's all I've been saying.
  #180  
Old 05-10-2019, 01:53 AM
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There's really no sense to considering a second throw in this example. As the system only has two available states, it has completed a full cycle after the first throw. But consider a coin with four possible initial states (n. B.: those don't correspond to 'heads up', that's the macrostate; rather, you can consider these states as 'ways of making the initial throw'), A and B, which yield heads and tails respectively on the first throw, and C and D, which do likewise. However, they're distinguished by the fact that both A and B will yield heads on the second throw, while C and D yield tails on the second throw. This upholds the stated determinism, but the second throw has, no matter the outcome of the first, a 50% probability of coming up heads.
The second throw only has a 50% probability of coming up heads if we assume the probability of each state is equal for subsequent throws.


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Originally Posted by Half Man Half Wit View Post
Sure, but I would ask you to take the hypothetical in the spirit it was proposed, and not try to pointlessly fight it---you're again trying to use where you think this will end up to fight the argument you think I'm making. But you're just getting sidetracked, arguing irrelevancies, so I would ask you to just take what I initially stated---an experimenter observing the system in such a way as to only be able to distinguish gross greyness, in the same way you're only able to distinguish gross greyness in the pixel images with only the naked eye and a monitor of sufficiently high resolution and quality.
Conceded.

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Originally Posted by Half Man Half Wit View Post
No. You only need to count.

Take my original system. It has 27 microstates. We can represent the microstate by means of a 27-dimensional vector, that has a '1' in the row corresponding to the microstate, and a '0' everywhere else, like this:
Code:
          (0)
          (0)
          (.)
          (.)
          (.)
     Si = (1)
          (.)
          (.)
          (.)
          (0)
Where the dots represent ellipses, and the index i numbers the states from 1 to 27, such that it always just gives the row where the '1' appears. We can order the states as before, with the corresponding microstates to each macrostate listed:
  • (A1B1C1): S1 - S6
  • (A2B1C0): S7 - S9
  • (A2B0C1): S10 - S12
  • (A1B2C0): S13 - S15
  • (A0B2C1): S16 - S18
  • (A1B0C2): S19 - S21
  • (A0B1C2): S22 - S24
  • (A3B0C0): S25
  • (A0B3C0): S26
  • (A0B0C3): S27

Any given evolution law of this system will be a 27 x 27 - matrix M, which takes one 27-dim vector to a different 27-dim vector by ordinary matrix multiplication:

Sj = M*Si

(That is, the system starts out in Si, is evolved for one timestep, and ends up in Sj.)

The matrix evolves Si into Sj, if and only if the i-th entry in its j-th row is '1'.

This matrix must obey some constraints. First, any given state can only evolve into one other state (this is determinism). Consequently, the matrix can only have one entry different from 0 in each column.

Likewise, only one state can evolve into any given state (this is reversibility: each state must have a unique precursor). Consequently, there can be only one non-zero entry per row, as well.

Thus, all the valid laws of evolution for the system take the form of a matrix M such that each row and column have exactly one entry equal to 1, and the rest 0 (there are thus 27 nonzero entries in the matrix).

Now, we can count how often each of these evolutions will increase vs. decrease entropy. Take first the upper left 6 x 6 - submatrix: any entry here will only 'mix' the maximum-entropy states among themselves, thus yielding to constant entropy.

Then, take the 18 x 18 - submatrix 'in the middle'. This mixes the intermediate-entropy states, thus likewise not leading to entropy increase. Same for the 3 x 3 - submatrix in the lower right corner, which mixes the min-entropy states.

Now take the lower left 3 x 24 - submatrix (that is, everything left over if you take the lower right 3 x 3 away). These are more interesting: they are entries that take any of the 24 states of nonminimal entropy, and transform them into stats of minimal entropy. These are, thus, entropy decreasing. Because of the properties of the matrix, there can be three such entries.

Then, take the left 18 x 6 - strip: They are the entries that take any of the 6 maximum entropy states to one of the 18 intermediate ones. These are, likewise, entropy decreasing; again, there can be maximally 6 entries here (but, if there are any entries here, we have to be careful not to double count: if there is an entry in the first three columns, there can't be any entries in the lower left 3 x 3 - submatrix, since this would entail a maximum entropy state both transitioning to a lower entropy and a minimum entropy one, in violation of determinism).

These are all the entries that may lower entropy: transitioning from a maximum entropy state to either an intermediate or minimum entropy state, or transitioning from an intermediate entropy state to a minimum entropy state.

Consequently---and here's the kicker---any evolution law we can write down for the system can have at most nine entries that lead to a lowering of entropy, and must, consequently, have 18 entries that keep entropy constant or increase it. Hence, no matter what the microdynamics are---no matter which particular M we choose---it follows, just from the count of microstates, that it takes any given state to a lower-entropy one at most one out of three (9/27) times. Consequently, entropy increases or stays constant more often than it decreases.

And that's all I've been saying.
Matrix operations are just beyond my level of mathematics education, so it will take me a couple days to fully understand what you have written here. But it seems to me that this sentence need not be true: "Thus, all the valid laws of evolution for the system take the form of a matrix M such that each row and column have exactly one entry equal to 1, and the rest 0 (there are thus 27 nonzero entries in the matrix)."

Could not the row and column both be all zeroes? For situations where some microstates are never visited. My intuition is that most systems do not visit every microstate.

~Max
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Old 05-10-2019, 02:58 AM
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The second throw only has a 50% probability of coming up heads if we assume the probability of each state is equal for subsequent throws.
There is no question of probability at this point any more. The second throw is perfectly determined by the initial state. You just choose that state, then the evolution proceeds deterministically, like clockwork.




Quote:
But it seems to me that this sentence need not be true: "Thus, all the valid laws of evolution for the system take the form of a matrix M such that each row and column have exactly one entry equal to 1, and the rest 0 (there are thus 27 nonzero entries in the matrix)."



Could not the row and column both be all zeroes? For situations where some microstates are never visited. My intuition is that most systems do not visit every microstate.



~Max
No, that's not possible. If a column were all zeroes, then a certain microstate would evolve to nothing. If a row were all zeroes, then another row must have more than one 1, since we still need to specify where all the 27 possible states evolve to, and we would violate reversibility.
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Old 05-10-2019, 09:38 AM
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There is no question of probability at this point any more. The second throw is perfectly determined by the initial state. You just choose that state, then the evolution proceeds deterministically, like clockwork.
Oh, I see. I misread your last post, and now concede that the chance on the second throw is 50%.

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No, that's not possible. If a column were all zeroes, then a certain microstate would evolve to nothing. If a row were all zeroes, then another row must have more than one 1, since we still need to specify where all the 27 possible states evolve to, and we would violate reversibility.
Why do you need to specify where all 27 states evolve to? It should be enough that Si eventually loops back to Si, whether that takes two steps or 26 steps. I can conceive of a system that never reaches every possible microstate, two moving points on a line segment which cannot occupy the same spot and bounce off both each other and the ends of the segment. No matter how you arrange the two points on the line, there exists at least two microstates that will never be reached: the points cannot switch places at the ends of the line segment. If I made an evolution matrix of that system, the columns (or is it rows?) for those two microstates are undefined.

You would, however, need to add a rule that the evolution matrix has a value for the initial state vector, and that each path through the matrix comes full circle.

~Max
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Old 05-10-2019, 10:52 AM
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Why do you need to specify where all 27 states evolve to?
Because otherwise, you'll end up with some states for which you don't have any evolution law. I mean, what'll happen if you set up the system in that state? If the matrix doesn't say, what does?

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I can conceive of a system that never reaches every possible microstate, two moving points on a line segment which cannot occupy the same spot and bounce off both each other and the ends of the segment.
Cyclic motions are not a problem for this setup: state S1 could evolve to S2, and S2 to S1 (for example).

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No matter how you arrange the two points on the line, there exists at least two microstates that will never be reached: the points cannot switch places at the ends of the line segment.
If you can't set up the system in such a way that the points start out at every place, then the states you can't set up are just not possible states of the system---you've got additional constraints, such that certain configurations are just impossible. If you can set the system up in this state, then the evolution law will have to specify how it evolves.

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  #184  
Old 05-10-2019, 11:34 AM
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Because otherwise, you'll end up with some states for which you don't have any evolution law. I mean, what'll happen if you set up the system in that state? If the matrix doesn't say, what does?

Cyclic motions are not a problem for this setup: state S1 could evolve to S2, and S2 to S1 (for example).
That's right, some evolution matrices will be incompatible with any given initial state. This is a consequence of the initial state being but one instant in a periodic cycle.

The other option is to have multiple independent paths in an evolution matrix, such that perhaps M[27] points to itself (does not evolve) and all of the other 26 states form a loop: 1->2->3->...->26->1->2->...

Actually that makes more sense.


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If you can't set up the system in such a way that the points start out at every place, then the states you can't set up are just not possible states of the system---you've got additional constraints, such that certain configurations are just impossible. If you can set the system up in this state, then the evolution law will have to specify how it evolves.
The points can start out in any configuration, so long as the two points do not occupy the same position, which I think you will allow me to rule out. But they can never switch sides - if point A is initially on the left and point B is initially on the right, the evolution of that system will never allow point A to be on the right and point B on the left. If point A is initially on the right and point B is initially on the left, again the system will never evolve so as to allow the points to switch sides.

Let's say this system has exactly six distinct microstates:
  • A on the left, B in the middle
  • A on the left, B on the right
  • A in the middle, B on the left
  • A in the middle, B on the right
  • A on the right, B on the left
  • A on the right, B in the middle

The evolution matrix would therefore be 6x6 and the state vector index six-dimensional, and is represented as such (I am representing Si as a decimal):

Code:
Sj=  123456
Si=1|000100
Si=2|010000
Si=3|000001
Si=4|100000
Si=5|000010
Si=6|001000
Do I correctly understand your idea of evolution matrices? If so, it does not make sense to count all of the states in any particular sub-matrix because it is not guaranteed that any evolutionary path visits all of those states.

~Max

Last edited by Max S.; 05-10-2019 at 11:37 AM.
  #185  
Old 05-10-2019, 12:13 PM
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That's right, some evolution matrices will be incompatible with any given initial state.
Then it's just not an evolution law. As any physical law, it must apply to every state of the system to yield a valid dynamics.


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The other option is to have multiple independent paths in an evolution matrix, such that perhaps M[27] points to itself (does not evolve) and all of the other 26 states form a loop: 1->2->3->...->26->1->2->...



Actually that makes more sense.
This is perfectly well possible, of course. But it doesn't change the conclusion.




Quote:
Do I correctly understand your idea of evolution matrices? If so, it does not make sense to count all of the states in any particular sub-matrix because it is not guaranteed that any evolutionary path visits all of those states.



~Max
I can't really make sense of your notation, sorry. I will try to explain the evolution matrices more thoroughly, but I might not get to it for a couple of days. Basically, the matrix changes one vector into another via matrix multiplication; perhaps read up on that in the meantime.

The point you're making isn't valid, however: while of course, you can still t up cyclical evolutions, this doesn't change the fact that since there's more ways for a state to evolve to a higher entropy, typically, you will observe entropy increase.

Indeed, it's not hard to see that eventually, any evolution will get back to the initial state. This doesn't change the conclusion: if you observe the system in a low entropy state, there are more cases that the next state is a higher entropy one than a lower entropy one.

I mean, if it's a minimum entropy state, thus should be clear: every step either yields constant entropy, or an increase. If every lowest-entropy state leads to another lowest-entropy state, then, in particular, none of the intermediate or high entropy states can evolve into a lowest-entropy state.

Then, for every intermediate-entropy state, each next step will either increase entropy, or will leave it constant. And so on.

The thing is, you need to consider every possible state. If you eliminate the possibility of entropy increase for one, then another must (except for the limiting case where entropy always is constant, as e. g. if every state stays the same). Since there are more ways to increase entropy, some of them must go uncompensated.
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Old 05-10-2019, 12:43 PM
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I can't really make sense of your notation, sorry. I will try to explain the evolution matrices more thoroughly, but I might not get to it for a couple of days. Basically, the matrix changes one vector into another via matrix multiplication; perhaps read up on that in the meantime.

The point you're making isn't valid, however: while of course, you can still t up cyclical evolutions, this doesn't change the fact that since there's more ways for a state to evolve to a higher entropy, typically, you will observe entropy increase.
Sorry about the notation. And right, that fact is exactly what I don't understand.

I still don't see how there are necessarily more ways to increase entropy. Those entropy-changing steps could all be part of a different evolutionary cycle within the matrix. For example S1-6 could form a cycle, S7-24 could form another cycle, and S25-27 could form a third cycle. In this matrix entropy never changes in direct contradiction to your assertion.

~Max
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Old 05-10-2019, 01:05 PM
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I don't think this example of some states (27?) evolving according to an arbitrary permutation matrix is a good illustration of mixing or ergodicity or entropy production. For instance, suppose your matrix M above is the identity matrix.
  #188  
Old 05-10-2019, 01:14 PM
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Sorry about the notation. And right, that fact is exactly what I don't understand.



I still don't see how there are necessarily more ways to increase entropy. Those entropy-changing steps could all be part of a different evolutionary cycle within the matrix. For example S1-6 could form a cycle, S7-24 could form another cycle, and S25-27 could form a third cycle. In this matrix entropy never changes in direct contradiction to your assertion.



~Max
I explicitly allowed for entropy staying constant (see my last post). After all, that's the second law: there's a greater-or-equal sign there. Also, I have stipulated from the beginning that I'm not considering such limiting cases, as in for instance where all of the gas molecules just oscillate in lockstep between one side of the room and the other.

What I said was that you can't have an uncompensated entropy decrease, and if you have the chance of a decrease, there will be a greater chance of increase. Take the case where two of the minimum entropy states just oscillate among each other, one evolves to an intermediate state, one intermediate state evolves to that low entropy state, and the high entropy states just cycle among themselves.

Then, if the system is in a high-entropy state, it will stay high. In an intermediate state, one out of 18 times, we will observe a reductuon, otherwise, entropy stays constant.

In a low entropy state, however, one out of three times, you get an entropy increase.

Hence, the chance of observing an increase (or no change) is much higher than of observing a decrease.
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Old 05-10-2019, 01:28 PM
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I don't think this example of some states (27?) evolving according to an arbitrary permutation matrix is a good illustration of mixing or ergodicity or entropy production. For instance, suppose your matrix M above is the identity matrix.
Well, if you turn into off the dynamics, then you'll indeed have constant entropy, but that's also the case with a gas. Plus, you can write every evolution law in this way (if you discretize the system), so it's sufficiently general.
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Old 05-10-2019, 01:44 PM
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I explicitly allowed for entropy staying constant (see my last post). After all, that's the second law: there's a greater-or-equal sign there. Also, I have stipulated from the beginning that I'm not considering such limiting cases, as in for instance where all of the gas molecules just oscillate in lockstep between one side of the room and the other.

What I said was that you can't have an uncompensated entropy decrease, and if you have the chance of a decrease, there will be a greater chance of increase. Take the case where two of the minimum entropy states just oscillate among each other, one evolves to an intermediate state, one intermediate state evolves to that low entropy state, and the high entropy states just cycle among themselves.

Then, if the system is in a high-entropy state, it will stay high. In an intermediate state, one out of 18 times, we will observe a reductuon, otherwise, entropy stays constant.

In a low entropy state, however, one out of three times, you get an entropy increase.

Hence, the chance of observing an increase (or no change) is much higher than of observing a decrease.
If I understand you correctly you are arranging the evolution matrix such that S1-2 make one cycle, S3-24 make a second cycle, and S25-27 make a third cycle. In S3-24 there is exactly one low-entropy state that evolves into an intermediate entropy state, and exactly one intermediate entropy state that evolves into a low-entropy state. Then you say the entropy does not change during the evolution cycles S1-2 or S25-27. All fine so far.

You say the probability of observing a reduction in entropy during a random step in cycle S3-24 is 1/18 and the probability of observing no change is 17/18. That doesn't follow at all - there are 21 steps in that cycle. The probability of any random step increasing entropy is 1/21, the probability of entropy remaining constant is 19/21, and the probability of entropy decreasing is 1/21.

~Max
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Old 05-10-2019, 01:49 PM
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Well, if you turn into off the dynamics, then you'll indeed have constant entropy, but that's also the case with a gas. Plus, you can write every evolution law in this way (if you discretize the system), so it's sufficiently general.
You cannot make an evolutionary matrix that uses hidden variables, such as the law you suggested in post #167:
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Now suppose that whatever deity has created the universe has made it so that the actual law is: stuff falls down, except once every sextillion times, when it just hovers in place.
But I don't see a need to assume such laws exist anyways.

~Max
  #192  
Old 05-10-2019, 01:59 PM
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If I understand you correctly you are arranging the evolution matrix such that S1-2 make one cycle, S3-24 make a second cycle, and S25-27 make a third cycle.
No, that's not quite it. 1-6 evolve among themselves (whether they make one cycle, or more than one), and one state from 7-24 evolves to one from 25-27, while one from 25-27 evolves to one from 7-24.

So if the system is in one of the states 1-6, entropy will remain constant. For one state among 7-24, entropy will decrease. For one state from 25-27, entropy will increase.

Hence, the probability of observing a reduction given that the system is in an intermediate entropy state is 1/18. The probability of observing an increase in entropy given that the system is in a low-entropy state is 1/3.

The 'given that' is what's usually called the 'past hypothesis'. We find the universe in a low entropy state; the second law concerns the probability of what happens given that we do so.

In other words, for more of the low entropy states that we could find the system in, we observe an increase (or constance) than a decrease.
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Old 05-10-2019, 02:05 PM
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OK, but aren't we getting away from dynamic considerations of entropy (and note, by the way, that considered as a discrete dynamical system the Kolmogorov-Sinai entropy of your process is always zero) and saying things that verge on the tautological, along the lines of, if we partition our space into a big subset and a little subset, then permute all the states, then a majority of points land in the big subset? Or that the system is most probably in the most probable state?

Classically (too classically?), it seems that the entropy of a body describes its average properties when at equilibrium, or at least observed over some non-infinitesimal period of time so that some probability distribution arises.

Last edited by DPRK; 05-10-2019 at 02:07 PM.
  #194  
Old 05-10-2019, 02:05 PM
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The probability of observing an increase in entropy given that the system is in a low-entropy state is 1/3.
How did you get this probability of 1/3? Shouldn't it be zero?

~Max

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  #195  
Old 05-10-2019, 02:19 PM
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OK, but aren't we getting away from dynamic considerations of entropy (and note, by the way, that considered as a discrete dynamical system the Kolmogorov-Sinai entropy of your process is always zero)
Sure, but that's not really a relevant notion here. We're considering the amount of information we may gain by discovering the precise microstate, basically.

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and saying things that verge on the tautological, along the lines of, if we partition our space into a big subset and a little subset, then permute all the states, then a majority of points land in the big subset? Or that the system is most probably in the most probable state?
Pretty much, yes. That's ultimately all the second law comes down to (as I think I said earlier).
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Old 05-10-2019, 02:22 PM
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How did you get this probability of 1/3? Shouldn't it be zero?



~Max
No. Basically, if there's a way in, there's a way out. So if one of the intermediate entropy states evolves to a low entropy one, then (at least) one of the low entropy states can't evolve to another low entropy state (since every state has a unique precursor, and the precursor of one of the low entropy states is an intermediate entropy state, and thus, can't be a low entropy state), and hence, must evolve to a higher entropy state.
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Old 05-10-2019, 02:25 PM
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No. Basically, if there's a way in, there's a way out. So if one of the intermediate entropy states evolves to a low entropy one, then (at least) one of the low entropy states can't evolve to another low entropy state (since every state has a unique precursor, and the precursor of one of the low entropy states is an intermediate entropy state, and thus, can't be a low entropy state), and hence, must evolve to a higher entropy state.
What? You just said all of the low entropy states are inaccessible from intermediate or high-entropy states.

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No, that's not quite it. 1-6 evolve among themselves (whether they make one cycle, or more than one), and one state from 7-24 evolves to one from 25-27, while one from 25-27 evolves to one from 7-24.
Bolding mine.

~Max
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Old 05-10-2019, 02:31 PM
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What? You just said all of the low entropy states are inaccessible from intermediate or high-entropy states.
1-6 are the high-entropy states, corresponding to the (single) macrostate (A1B1C1), see above.
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Old 05-10-2019, 02:44 PM
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1-6 are the high-entropy states, corresponding to the (single) macrostate (A1B1C1), see above.
Oh, OK. My mistake.

So we have the following rules:

There are 27 possible states. The initial state is equiprobable: 1/27.

The probability of the initial state being a high-entropy state is 6/27.
The probability of the initial state being an intermediate-entropy state is 18/27.
The probability of the initial state being a low-entropy state is 3/27.
6+18+3=27, so that checks out.

The probability of any random observation of a high entropy state changing entropy is 0.
The probability of any random observation of a high entropy state keeping consistent entropy is 1.

The probability of any random observation of an intermediate entropy state changing to a low entropy state is 1/18.
The probability of any random observation of an intermediate entropy state keeping consistent entropy is 17/18.
The probability of any random observation of an intermediate entropy state changing to a high entropy state is 0.

The probability of any random observation of a low entropy state changing keeping consistent entropy is 2/3.
The probability of any random observation of a low entropy state changing to an intermediate entropy state is 1/3.
The probability of any random observation of a low entropy state changing to a high entropy state is 0.

Therefore, the probability of a random observation showing an increase in entropy is 1/3 * 3/27 = 1/27.
The probability of a random observation showing no change in entropy is 6/27 + 17/18 * 18/27 + 2/3 * 3/27 = 25/27.
The probability of a random observation showing a decrease in entropy is 1/18 * 18/27 = 1/27.

1/27 + 25/27 + 1/27 = 27/27, so that checks out.

But wasn't your assertion that the probability of observing an increase is greater than the probability of observing a decrease?

~Max
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Old 05-10-2019, 02:48 PM
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Don't worry about that for now. This is going to be a bit of a journey, and you keep running into confusions by getting ahead of yourself. But for now, the first step has been taken: we agree that it's possible to have a law, valid to all appearances in the macroscopic realm, concerning a certain quantity ('greyness') such that once that quantity has been understood on a microscopic level, we understand that the original law can only hold in an approximate sense. We can build from here.

Now, to the next step.
  1. Suppose you have two boxes, A and B.
  2. Box A is filled with white marbles, and box B is filled with black ones.
  3. Both boxes are placed on a vibrating plate, such that the marbles in them bounce around.
  4. The walls of the boxes are removable.
  5. Suppose you put both boxes next to one another, and remove the now adjacent walls, creating one big box.
  6. Marbles from the white box A will bounce into the black box, and marbles from the black box B will bounce into the white box.
  7. There are more ways of realizing a state that's pretty uniformly grey, than there are to realize a state that's (say) all white in box A, and all black in box B.
  8. Consequently, there are more ways to go from a state that's slightly inhomogeneous to one that's more homogeneous, than there are ways to go to a state that's even more inhomogeneous.
  9. To any observer who, as before, is only capable of seeing gross colors, the formerly black-and-white separated box will gradually tend to a shade of even gray.
  10. That observer might formulate a law, stating that black and white, if brought into contact, eventually even out to a uniform gray.
  11. Knowing the microscopic description, we know that this is just, again, a law of averages: there is nothing that prohibits, say, all or a sizable fraction of the white marbles from bouncing back into box A.
  12. Given a long enough timescale, the even grey will, eventually, separate into black and white patches.

I expect greater resistance with this example. But again, try not to think ahead to the rest of this discussion; just consider the above system, as I have presented it. Do you agree that the conclusion is reasonable, here? That there is once again a law that appears valid thanks to the limited observations made at the macroscale, which we can see must be violated once we know about the microscopic level?
I'm still not making the connection between 7. and 8. in the marbles example.

~Max
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