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Will entropy be low much of the time?


Tristan L
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7 hours ago, studiot said:

I was hoping to introduce this in a more measured way, but you have jumped the gun

I didn't and don't want to do that; rather, I only wanted and still want to make sure that there are no misunderstandings before I answer your other points and go on with the discussion.

 

7 hours ago, studiot said:

I am not, and never have, disagreed with your outlining of basic mathematical set theory in respect of the word 'partition'.

The important thing is that in this whole thread, I have only ever talked about partitioning the set of microstates into macrostates, not partitioning the system into subsystems. However, you seem to imply that I have done the latter by saying

21 hours ago, studiot said:

For the subsytems (partitions in your parlance)

In reality, I have only ever talked about partitioning the phase space.

 

7 hours ago, studiot said:

Mathematical partitioning of Mi (is based on) equipartition and implicitly assumes the 'equipartition theorem' of thermodynamics.

In what way does it do that?

Does that mean that without the equipartition theorem, one microstate could belong to more than one macrostate?

 

7 hours ago, studiot said:

You have also mentioned disjoint partitions, which is important in the mathematical statistics of this.

What exactly do you mean by that?

All members of a partition are pairwise disjoint by definition.

Being sets, some partitions P, Q are disjoint (share no common subsets of the ground-set), while others are not.

 

7 hours ago, studiot said:

Nor can it arise in information technology, whose partitions are disjoint.

Partitions of what?

 

7 hours ago, studiot said:

Finally, I do hope, you are not trying to disagree with Cartheodory.

Of course not, but as I said, I want to do away with any misunderstandings on my or your part before talking about your other points, including Caratheodory.

 

7 hours ago, studiot said:

And you seem to be disagreeing with my classical presentation, because it is much shorter than the same conclusion reached in the statistical paper you linked to  ?

Actually, I linked to the paper mainly because I find it interesting that there may be a way to outsmart the Second Law 😁.

Edited by Tristan L
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  • 1 year later...
On 6/4/2020 at 1:55 PM, studiot said:

I was hoping to introduce this in a more measured way, but you have jumped the gun.

I was hoping that you'd actually read and try to understand what I've written, at least the last post of mine, where I sweetle very clearly that I'm not talking about partitioning physical 3D space, but rather PHASE-ROOM. All I'm saying about disjointness is that macro-states are eachotherly disjoint sets of microstates; can we agree/forewyrd on that? Also, what exactly do you have in mind when you talk of disjointness?

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On 6/4/2020 at 7:22 PM, Tristan L said:

Being sets, some partitions P, Q are disjoint (share no common subsets of the ground-set), while others are not.

I think you misunderstand the meaning of disjoint in set theory.

This should be cleared up prior to any other consideration.

 

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17 hours ago, studiot said:
On 6/4/2020 at 9:22 PM, Tristan L said:

others are not.

I think you misunderstand the meaning of disjoint in set theory.

This should be cleared up prior to any other consideration.

We should indeed clear up any misunderstandings before we go on, so we'll do that right now:

For any sets S, T, "S is disjoint with T" means that S and T share no elements in common.

For any sets P, S, "P is a partition of S" means that all members of P are subsets of S and any two members T, B of P are disjoint, th.i. share no elements of S in common, and the union of all members of P is S.

Now to my above quote, which I'll sweetle/explain with the help of a byspel/example: {1, 2, 3, 4} is our groundset. All three of the following are partitions of {1, 2, 3, 4}:

{{1, 2}, {3, 4}}

{{1, 3}, {2, 4}}

{{1}, {2}, {3, 4}}

The first and the second are disjoint since they have no elements in common, but the first and the third one are not, for they have the member {3, 4} in common.

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51 minutes ago, Tristan L said:

We should indeed clear up any misunderstandings before we go on, so we'll do that right now:

For any sets S, T, "S is disjoint with T" means that S and T share no elements in common.

For any sets P, S, "P is a partition of S" means that all members of P are subsets of S and any two members T, B of P are disjoint, th.i. share no elements of S in common, and the union of all members of P is S.

Now to my above quote, which I'll sweetle/explain with the help of a byspel/example: {1, 2, 3, 4} is our groundset. All three of the following are partitions of {1, 2, 3, 4}:

{{1, 2}, {3, 4}}

{{1, 3}, {2, 4}}

{{1}, {2}, {3, 4}}

The first and the second are disjoint since they have no elements in common, but the first and the third one are not, for they have the member {3, 4} in common.

 

It is you attitude which impedes progress.

I said,  "I think that........"

I did not try to lay down the law.

 

Your response could have been  that you think you do understand set theory perfectly and so you wonder what I mean.

Instead you tried to lay down the law.

 

I agree with this first statement.

51 minutes ago, Tristan L said:

For any sets S, T, "S is disjoint with T" means that S and T share no elements in common.

But this second statement is inaccurate.

51 minutes ago, Tristan L said:

For any sets P, S, "P is a partition of S" means that all members of P are subsets of S and any two members T, B of P are disjoint, th.i. share no elements of S in common, and the union of all members of P is S.

This means that the set S must contain sets as members.

Which is not a requirement of set theory.

 

The issue of failure to distinguish between subsets and elements is further compounded by your sudden switch from elements to members.

The number 1 is a member, but not a subset of your set S = {1,2,3,4}

The partition {1} is a subset of S but not a member of S.

 

I don't take kindly to the attitude that you know everything and no one else knows anything.

Particularly as you are introducing mocking irrelevancies such as "byspel", which are no longer funny.

Edited by studiot
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Not at all, you quite misunderstood; I would never dare to lay down the law for anything and never will; I only repeat 8th-grade mathematical definitions and knowledge.

4 hours ago, studiot said:

But this second statement is inaccurate.

In truth, this statement of yours is simply incorrect.

4 hours ago, studiot said:

This means that the set S must contain sets as members.

Which is not a requirement of set theory.

Could you please enlighten me as to the sinn/sense of what you've written there? Set theory does indeed not need every set to only contain sets (though ZFC does actually rule out ur-elements for convenience), but a *partition* does indeed only contain sets as elements, namely disjoint subsets of the ground-set.

4 hours ago, studiot said:

your sudden switch from elements to members.

I brook/use the words "element" and "member" in exactly one and the same meaning.

4 hours ago, studiot said:

The number 1 is a member, but not a subset of your set S = {1,2,3,4}

Oh, I thank you for the info, but my friend's 13 year old cousin already told me today morning. 😉

4 hours ago, studiot said:

The partition {1} is a subset of S but not a member of S.

{1} is not a partition of S; it's a member/element e.g. of {{1}, {2, 3, 4}} and of {{1}, {2, 3}, {4}}, which in turn are partitions of S; accordingly, it's an underset of S.

 

It's funny that you borrowed my lines which I was about to write to you 🤣:

4 hours ago, studiot said:

It is you attitude which impedes progress.

 

4 hours ago, studiot said:

The issue of failure to distinguish between subsets and elements

 

4 hours ago, studiot said:

I don't take kindly to the attitude that you know everything and no one else knows anything.

 

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15 hours ago, Tristan L said:

In truth, this statement of yours is simply incorrect.

You are right it was my sloppy use of language.

I should have said

The set {1} is part of the partition set of S so is a subset of S but not a member of S.

This difference between the use of 'partition' in the physical sciences and in mathematics strikes again.

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That ambiguity in meaning of words can indeed be very hindering. A much better word imho is German/Theech "Zerlegung", which unmistakably means that which is meant by English "partitioning" (should we use that word from now on for the mathematical concept?) or "sectioning" or something like that.

This brings us to the minor side-issue of speech:

23 hours ago, studiot said:

Particularly as you are introducing mocking irrelevancies such as "byspel", which are no longer funny.

I don't mean to mock or anything; I just have a side-hobby of bringing back English's true potential, and that includes brooking/using truly English words, for byspel "byspel", which is the proper English word for "example" and cognate/orbeteed to German "Beispiel". I brook this proper English on purpose/ettling where it's not the object but only the tool of talking; after all, that's the ord/point of speech. But again, that's just a hobby of mine.

On 6/2/2020 at 3:35 PM, studiot said:

Here is another simple problem

 

tube1.jpg.7c7280be993ec0fe695642bddc28c851.jpg

 

Suppose you have a sealed adiabatic tube containing an adiabatic frictionless piston dividing the tube into two chambers, A and B as shown.

Let both sides of the system contain an ideal gas.

Discuss the time evolution of the system in terms of entropy.

This is a very intrysting problem indeed 🤔. I'd say that it evolves as follows:

If the pressure/thrutch is the same on both sides of the resting piston, nothing will happen. Otherwise, 1. the piston will start to go from the high-thrutch side to the low-thrutch side. 2. As it goes in that direction, internal energy of the high-pressure gas is transferred to kinetic energy of both gases and internal energy of the low-pressure gas. 3. When both thruthes become equal, the piston goes on shrithing/moving thanks to the inertia of the gases. 4. Now, the kinetic energy of the gases and internal energy of the former high-thrutch gas (now the low-pressure gas) are transferred to internal energy of the former low-thrutch gas (now the high-pressure one), slowing down the piston, until 5. the piston is at rest again. Now, the whole ongoings repeat in the other righting/direction.

Entropy doesn't change/wrixle during the whole process, so the Second Law of Thermodynamics is of no brook/use. But of course, we have another Second Law, namely that of Newton, and this one helps us further here.

Actually, I believe that we shouldn't find this too surprising 🤔, for there are other systems which wrixle/change although their entropy stays the same, e.g. a frictionless pendulum swinging. Mark that the system doesn't have to be periodic, I think; for instance, two  bodies shrithing/moving at not-zero relative speed forever in space (forget about gravitational waves) make up such a system.

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