# Is the second law of thermodynamics routinely violated?
**URL:** https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151
**Category:** Factual Questions
**Created:** [April 28, 2019, 12:15am UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151 "2019-04-28T00:15:23Z")
**Posts on this page:** 20
**Page:** 10
### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 7:58am UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/181 "2019-05-10T07:58:57Z")
> [@Max\_S](#):
>
> 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.
> [@](#):
>
> 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.
---
### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 2:38pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/182 "2019-05-10T14:38:26Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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%.
> [@](#):
>
> 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 S[SUB]i[/SUB] eventually loops back to S[SUB]i[/SUB], 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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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 3:52pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/183 "2019-05-10T15:52:08Z")
> [@Max\_S](#):
>
> 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?
> [@](#):
>
> 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 S[sub]1[/sub] could evolve to S[sub]2[/sub], and S[sub]2[/sub] to S[sub]1[/sub] (for example).
> [@](#):
>
> 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.
---
### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 4:34pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/184 "2019-05-10T16:34:27Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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 S[sub]1[/sub] could evolve to S[sub]2[/sub], and S[sub]2[/sub] to S[sub]1[/sub] (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.
> [@](#):
>
> 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:
[ul][li]A on the left, B in the middle[/li][li]A on the left, B on the right[/li][li]A in the middle, B on the left[/li][li]A in the middle, B on the right[/li][li]A on the right, B on the left[/li][li]A on the right, B in the middle[/ul][/li]
The evolution matrix would therefore be 6x6 and the state vector index six-dimensional, and is represented as such (I am representing S[SUB]i[/SUB] as a decimal):
```auto
S[SUB]j[/SUB]= 123456
S[SUB]i[/SUB]=1|000100
S[SUB]i[/SUB]=2|010000
S[SUB]i[/SUB]=3|000001
S[SUB]i[/SUB]=4|100000
S[SUB]i[/SUB]=5|000010
S[SUB]i[/SUB]=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
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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 5:13pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/185 "2019-05-10T17:13:20Z")
> [@Max\_S](#):
>
> 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.
> [@](#):
>
> 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.
> [@](#):
>
> 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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### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 5:43pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/186 "2019-05-10T17:43:16Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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 S[SUB]1-6[/SUB] could form a cycle, S[SUB]7-24[/SUB] could form another cycle, and S[SUB]25-27[/SUB] could form a third cycle. In this matrix entropy never changes in direct contradiction to your assertion.
~Max
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### Author:  [@DPRK](https://boards.straightdope.com/u/DPRK)
#### Post date: [May 10, 2019, 6:05pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/187 "2019-05-10T18:05:31Z")
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.
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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 6:14pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/188 "2019-05-10T18:14:05Z")
> [@Max\_S](#):
>
> 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 S[SUB]1-6[/SUB] could form a cycle, S[SUB]7-24[/SUB] could form another cycle, and S[SUB]25-27[/SUB] 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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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 6:28pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/189 "2019-05-10T18:28:17Z")
> [@DPRK](#):
>
> 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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### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 6:44pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/190 "2019-05-10T18:44:06Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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 S[SUB]1-2[/SUB] make one cycle, S[SUB]3-24[/SUB] make a second cycle, and S[SUB]25-27[/SUB] make a third cycle. In S[SUB]3-24[/SUB] 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 S[SUB]1-2[/SUB] or S[SUB]25-27[/SUB]. All fine so far.
You say the probability of observing a reduction in entropy during a random step in cycle S[SUB]3-24[/SUB] 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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### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 6:49pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/191 "2019-05-10T18:49:12Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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=21633256]post #167[/POST]:
> [@Half\_Man\_Half\_Wit](#):
>
> 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
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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 6:59pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/192 "2019-05-10T18:59:49Z")
> [@Max\_S](#):
>
> If I understand you correctly you are arranging the evolution matrix such that S[SUB]1-2[/SUB] make one cycle, S[SUB]3-24[/SUB] make a second cycle, and S[SUB]25-27[/SUB] 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.
---
### Author:  [@DPRK](https://boards.straightdope.com/u/DPRK)
#### Post date: [May 10, 2019, 7:05pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/193 "2019-05-10T19:05:17Z")
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.
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### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 7:05pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/194 "2019-05-10T19:05:25Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 7:19pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/195 "2019-05-10T19:19:28Z")
> [@DPRK](#):
>
> 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.
> [@](#):
>
> 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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### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 7:22pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/196 "2019-05-10T19:22:17Z")
> [@Max\_S](#):
>
> 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.
---
### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 7:25pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/197 "2019-05-10T19:25:01Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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.
> [@Half\_Man\_Half\_Wit](#):
>
> 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
---
### Author:  [@Half\_Man\_Half\_Wit](https://boards.straightdope.com/u/Half_Man_Half_Wit)
#### Post date: [May 10, 2019, 7:31pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/198 "2019-05-10T19:31:45Z")
> [@Max\_S](#):
>
> 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.
---
### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 7:44pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/199 "2019-05-10T19:44:37Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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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### Author:  [@Max\_S](https://boards.straightdope.com/u/Max_S)
#### Post date: [May 10, 2019, 7:48pm UTC](https://boards.straightdope.com/t/is-the-second-law-of-thermodynamics-routinely-violated/833151/200 "2019-05-10T19:48:58Z")
> [@Half\_Man\_Half\_Wit](#):
>
> 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.
> [ol]
> [li]Suppose you have two boxes, A and B.[/li][li]Box A is filled with white marbles, and box B is filled with black ones.[/li][li]Both boxes are placed on a vibrating plate, such that the marbles in them bounce around.[/li][li]The walls of the boxes are removable.[/li][li]Suppose you put both boxes next to one another, and remove the now adjacent walls, creating one big box.[/li][li]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.[/li][li]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.[/li][li]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.[/li][li]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.[/li][li]That observer might formulate a law, stating that black and white, if brought into contact, eventually even out to a uniform gray.[/li][li]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.[/li][li]Given a long enough timescale, the even grey will, eventually, separate into black and white patches.[/li][/ol]
>
> 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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