Maxwell's Demon ?

[ravingdave]

http://nextbigfuture.com/2009/06/nanosc ... cient.html


Interesting.


David

[TallDave]

The Second Law isn't as inviolable as is sometimes asserted. It's only true the vast majority of the time.

[IntLibber]

The Second Law isn't as inviolable as is sometimes asserted. It's only true the vast majority of the time.

Quite so, life works as a great maxwells deamon, as does water itself. We wouldnt see rich ore deposits to the point of condensation of gold nuggets and other minerals if the second law were inviolate.

[kcdodd]

The vast majority of time is all that counts when you talk about thermodynamics. It's the rule of averages. Of course you can reduce entropy in one location but it will still never decrease overall. The most likely result is to overlook an assumption of how their device would actually work. But if it does work then we can can pretty much do anything now, and we don't even need fusion power anymore.

[Helius]

The Second Law isn't as inviolable as is sometimes asserted. It's only true the vast majority of the time.

Quite so, life works as a great maxwells deamon, as does water itself. We wouldnt see rich ore deposits to the point of condensation of gold nuggets and other minerals if the second law were inviolate.

None of that is a closed system, None of that decreases entropy when all effects are considered. Assuming the Universe is a closed system, then, Entropy of it can only increase; It never decreases.

[Torulf2]

If the universe is a closed system is an open question.

[TallDave]

Assuming the Universe is a closed system, then, Entropy of it can only increase; It never decreases.

Not so. I agree those are not closed systems, but even in a closed system, it only increases the vast majority of the time.

Consider two chambers, separated by a wall, with different gases. You remove the wall, the gasses mix, entropy increases.

However, if you wait long enough (a very very very very long time, so long the age of the universe isn't even a rounding error even if the number of atoms is very small), the random motion of the gasses will eventually result in a state where the two are separate again, just by chance. Entropy has decreased.

Of course, the amount of time they are mixed exceeds the time they aren't by such a vast proportion that for all practical purposes we can generally say it doesn't matter. Even the amount of time entropy decreases such that it's a 49.9999999999/50.000000001 mix is so tiny as to be of only thereotical interest.

So the chance of observing a decrease in entropy approaches zero over any reasonable timeframe. But it does happen, given enough time.

It's easy to see why this has to be so. Imagine 50 gas X atoms in one side, 50 of Y in the other. You open and they mix. At equilibirum, there are 25 of X in either side, and 25 of Y on either side. But now every time a single atom drifts across the boundary, entropy has slightly decreased (though it will promptly increase back to equilibrium again rather than further decreasing the vast majority of the time).

[passenger66]

Assuming the Universe is a closed system, then, Entropy of it can only increase; It never decreases.

However, if you wait long enough (a very very very very long time, so long the age of the universe isn't even a rounding error even if the number of atoms is very small), the random motion of the gasses will eventually result in a state where the two are separate again, just by chance. Entropy has decreased.

Somewhere in
http://www.youtube.com/watch?v=SMcYZVylbhc
1of5 -- The Universe - Cosmic Apocalypse
actually I think part 5 of 5

they talk about how a new universe could be created by a weird entropy bubble - so over really long timescales entropy can reverse.
I think that's what they said.

[kcdodd]

Assuming the Universe is a closed system, then, Entropy of it can only increase; It never decreases.

Not so. I agree those are not closed systems, but even in a closed system, it only increases the vast majority of the time.

Consider two chambers, separated by a wall, with different gases. You remove the wall, the gasses mix, entropy increases.

However, if you wait long enough (a very very very very long time, so long the age of the universe isn't even a rounding error even if the number of atoms is very small), the random motion of the gasses will eventually result in a state where the two are separate again, just by chance. Entropy has decreased.

Of course, the amount of time they are mixed exceeds the time they aren't by such a vast proportion that for all practical purposes we can generally say it doesn't matter. Even the amount of time entropy decreases such that it's a 49.9999999999/50.000000001 mix is so tiny as to be of only thereotical interest.

So the chance of observing a decrease in entropy approaches zero over any reasonable timeframe. But it does happen, given enough time.

It's easy to see why this has to be so. Imagine 50 gas X atoms in one side, 50 of Y in the other. You open and they mix. At equilibirum, there are 25 of X in either side, and 25 of Y on either side. But now every time a single atom drifts across the boundary, entropy has slightly decreased (though it will promptly increase back to equilibrium again rather than further decreasing the vast majority of the time).

I think this argument is flawed because of how you are defining entropy. As soon as you remove the wall separating the gas you are allowing the gas to "relax". This means you can not say anything about the particular state about where either gas particles are, because ANY arrangement is possible at that point. Not just those that uniformly distribute the two gases around the chamber. To know what arrangement you have requires an observation, and of course we know from QM that doing that requires disturbing the system.

So, take your simply example where you have 50 cells, 25 particle of type X and 25 of type Y. You "relax" the distro, which basically you assign the particles to random cells. Now, by your definition of entropy, the new distribution increases entropy only if it intermingles the two species, and if it randomly assigns the particles so that you have 25 X on one side and 25 Y on the other, entropy decreased. But you have to observe the cells to know what the new distribution is, when that should not be required to say what the entropy is, only that the cells now hold random values, whatever they are.

To replace the wall, and to actually decrease entropy, you have to test each new random distribution. And again, QM says this action is not free. So if we are performing this test at all, must less over billions of years, then it is obviously not a closed system any longer.

[TallDave]

because ANY arrangement is possible at that point.

Yes, but some have more entropy than others. The ones that tend to maximize entropy are by far the more likely to evolve over time once the chambers are mixed. In the example, 25/25 would be most likely, 26/24 would be less likely, 27/23 yet more unlikely, and so forth.

As for observation, you don't have to observe the initial increase and eventual momentary decrease in entropy to know that they will happen (in fact, it would be very hard to observe a significant decrease, because it's so unlikely in any human-relevant timeframe). They just follow from the probabilities.

This is practically a truism, because saying the entropy has increased is the same as saying the particles are arranged in a way that is more likely. It naturally follows from this that there will be instances in which their arrangement is less likely, whose rarity could be expressed as a probability over a period of time, which would be a function of how much entropy needed to decrease to produce that state.

(Also, if you both demand we be able to see the closed system and at the same time forbid it, then all you've done is forbid us to draw any conclusions about any closed systems.)

Of course, none of this is very relevant in a practical sense; a glass of boiling water always cools in a room at normal temperature, and we never see a glass of water spontaneously boil because the heat in a room randomly shifted into it. From our perspective, the 2LOT is inviolate for all practical purposes.

And of course, the bigger and less entropic the system the more likely overall entropy is increasing in it. It's probably safe to say the Universe as a whole has never seen a decrease (measuring from the Big Bang, or at least post-inflation), and won't until long after the last black holes expire from Hawking radiation.

[kcdodd]

No. If you stick with the discrete case for a moment. Any single arrangement of the two species is equally likely. Drop the number down to 8 for a second and change X and Y to 0 and 1. Say we start out like this: 0000 | 1111. We remove the wall, 00001111. Ok, there is only ONE way for this distribution to be, because 00001111 is the same as 00001111? Now take the state 01010101, there is only ONE way for that state to be. But what you are saying is this state has more entropy then 00001111. How can that be if they are EQUALLY likely? If you know the exact locations of all the particles then entropy is the same as you started with, because what you basically have done is this: 0|1|0|1|0|1|0|1. And if we assumed that the entropy increased as soon as we released the original wall, then we have had to decrease the entropy in order to make this determination. That is the whole deal with the Maxwell demon. You can't do that (ie, decrease the entropy) for free.

[MSimon]

No. If you stick with the discrete case for a moment. Any single arrangement of the two species is equally likely. Drop the number down to 8 for a second and change X and Y to 0 and 1. Say we start out like this: 0000 | 1111. We remove the wall, 00001111. Ok, there is only ONE way for this distribution to be, because 00001111 is the same as 00001111? Now take the state 01010101, there is only ONE way for that state to be. But what you are saying is this state has more entropy then 00001111. How can that be if they are EQUALLY likely? If you know the exact locations of all the particles then entropy is the same as you started with, because what you basically have done is this: 0|1|0|1|0|1|0|1. And if we assumed that the entropy increased as soon as we released the original wall, then we have had to decrease the entropy in order to make this determination. That is the whole deal with the Maxwell demon. You can't do that (ie, decrease the entropy) for free.

The ones and zeros need individual identities to make sense of what is going on. ABCD1234.

ABCD1234 is not the same as BACD1234.

It is like working out the probabilities of two dice. All sevens are not the same.

[TallDave]

It is like working out the probabilities of two dice. All sevens are not the same.

Better than I could have explained it.

As for the 00001111, it's easy to see there are only 2 solutions for 4/0 or 0/4,

0000-1111
1111-0000

but many more for the others:

32 solutions (16 each) for 3/1 and 1/3

0001-0111
0010-0111
0100-0111
1000-0111
0001-1011
0010-1011
0100-1011
1000-1011
0001-1110
0010-1110
0100-1110
1000-1110
0001-1101
0010-1101
0100-1101
1000-1101
(and now reverse for the other 16)

I think there are 72 solutions for 2/2.

So you can see why the low-entropy states of (4/0,0/4) are the least probable, and the high entropy (2/2) is most probable. Given a random distribution, you're 36 times more likely to see 2/2 than (0/4,4/0).

This is why glasses of water never spontaneously boil. You'd have to outlive the Universe to see something that unlikely, given the 36000000000000000000000000 atoms involved.

[kcdodd]

The ones and zeros need individual identities to make sense of what is going on. ABCD1234.

ABCD1234 represents 8 different species. Atoms of the same element are indistinguishable. You can't "mark" particles like that and follow them around. 00001111 is the same no matter what order you put the individual 1s or 0s in. This is experimentally verified. If you use ABCD1234 you get too many possible states.

As for the 00001111, it's easy to see there are only 2 solutions for 4/0 or 0/4,

0000-1111
1111-0000

but many more for the others:

32 solutions (16 each) for 3/1 and 1/3

You are right, but missing a subtle point.

Let's start counting the number of possible states, instead of individual states. Say the number q represents a quantity for the system, which can take on values ranging from 0 to 4, marking the number of 1s on the right side. q=3 represents all the states where you have three 1s on the right side. So, 0010111, etc like you have said. q=1 represents all states with one 1 on the right side. 01110100 etc. and so on. Also lets define an operator N(q) that counts the number of possible states for any q. So, N(4) = 1, N(3) = 16, N(2) = 36, N(1) = 16, N(0) = 1.

Ok, so this is what you have already said, I'm just added a little structure. It is true, q=2 has the most possible number of states. However, now what you are doing is trying to determine the value of q, instead of the positions. Say we start out with q=4 (00001111). After you remove the wall, you can no longer say what the value of q is. There is no individual value of q any more. All we can do is say what the probability of the value of q: P(q) = N(q)/(N(0)+N(1)+N(2)+N(3)+N(4)). To actually know the value of q you have to perform a measurement, and measurements are not free, which goes right back to my previous post.

[Art Carlson]

It is like working out the probabilities of two dice. All sevens are not the same.

Better than I could have explained it.

As for the 00001111, it's easy to see there are only 2 solutions for 4/0 or 0/4,

0000-1111
1111-0000

but many more for the others:

32 solutions (16 each) for 3/1 and 1/3

0001-0111
0010-0111
0100-0111
1000-0111
0001-1011
0010-1011
0100-1011
1000-1011
0001-1110
0010-1110
0100-1110
1000-1110
0001-1101
0010-1101
0100-1101
1000-1101
(and now reverse for the other 16)

I think there are 72 solutions for 2/2.

So you can see why the low-entropy states of (4/0,0/4) are the least probable, and the high entropy (2/2) is most probable. Given a random distribution, you're 36 times more likely to see 2/2 than (0/4,4/0).

This is why glasses of water never spontaneously boil. You'd have to outlive the Universe to see something that unlikely, given the 36000000000000000000000000 atoms involved.

Entropy always has to do with ignorance (or maybe apathy). You get so many combinations for (2/2) because you don't know or don't care about anything other than the total number of each type on each side. If you were obsessed with details, you would ask, How many ways can I have exactly 1000-1101? Just one. High entropy. If I ask you if the entropy of the state 010101 is high or low, you might say low, until I tell you that is my name in Morse code. With that additional knowledge the same state suddenly has low entropy.