Is entropy low much of ðe time?

[Tristan L]

Hello guys,

It’s I again, Tristan Laguz. In my þread Will entropy be low much of the time?, I talk about a point boþ Sabine Hossenfelder and I have made: ðat ðe entropy of a system isn’t defined absolutely, but raðer wið regard to a choice of which subsets of state space count as macrostates. In ðe later part of ðe topic, I use ðe English (and Norse) letters Ðat (‘Ð’/‘ð’) and Þorn (‘Þ’/‘þ’) because of ðeir aesþetic appeal and because using ðem is þrice as precise and twice as fast as using “th”. Oddly, swansont took issue wið my use of ðese letters. First, he said I should use English in ðe forum, to which I replied ðat I did, in fact, write in English. Moreover, I had already justified my use of Þorn and Ðat. Swansont didn’t engage wið my arguments. Neiðer did he nicely ask me to use “th” instead of ‘Þ’ and ‘Ð’ because it’s currently more widespread even ðough it’s less precise and less efficient. Instead, he simply locked my topic, accusing me of making bad-faiþ arguments. Making such a false accusation and creating such a lie about me is unacceptable, ðough not overly surprising given swansont’s self-description above his profile picture 😉. Since my arguments in my þread are made in good faiþ and, I believe, valid and sound on top of ðat, I herewið open my topic anew to talk wið you about wheðer my aforesaid belief (ðat my arguments be valid and sound) be true.

❕ Before you read on, please mark ðat I use “partition of S” in ðe set-þeoretic sense, ðat is, to mean a set of subsets of S which are non-empty, pairwise disjoint, and togeðer cover S. Ðus, “partition of S” refers to a subset of ðe power set of S which fulfills certain conditions. For instance, {{1, 2, 3}, {4, 5}, {6}} is called “a partition of {1, 2, 3, 4, 5, 6}”. Choosing which subsets of a state space count as macrostates amounts to choosing a partition of ðe state space.

 

I add someþing in ðe beginning which I wanted to write ðe last time but forgot:

Sethoflagos wrote:

Quote

“It is not 'still the same microstate': it is the ergodic progression through various microstates of different macrostates(!)”

You’re right. IMHO, Hossenfelder is wrong to claim ðat ðe system stay in ðe same microstate.

 

Now follows ðe last part of ðe discussion in ðe closed topic and a part shortly before it:

 

On 12/15/2023 at 2:31 PM, studiot said:

Any chance you could rewrite this in English for us plebs please ?

Ðe letter Ðat (uppercase: 'Ð', lowercase: 'ð') stands for ðe 'th'-sound in "that", "then", "there" aso. whereas ðe letter Þorn (big: 'Þ', small: 'þ') represents ðe 'th'-sound in "thorn", "think", "thank" asf. Wielding 'ð' and 'þ' is þrice as precise as using 'th', since it distinguishes ðe /ð/ sound from ðe /þ/ sound and furðermore avoids mixing ðem up wið a /t/ sound followed by a /h/ sound. It is also double as efficient, because it uses only half as many letters. Ðis makes it six times as good (i.e. five times better).

On 12/15/2023 at 8:04 PM, sethoflagos said:

And it is not 'all possible microstates of ðe system', it is 'all possible microstates of ðe applicable macrostate'.

I wrote: "Ðe entropy of a system [...] has to be defined wið regard to a partition (mark ðat I use ðe linked-to set-þeoretic definition of “partition”) of [...] ðe set of all possible microstates of ðe system."

On 12/15/2023 at 8:04 PM, sethoflagos said:

Don't you mean “microstates wið respect to P"?

Nope, I really do mean macrostates wið respect to P. A macrostate wið respect to P is simply an element of P, where P is a partition (again, please mark ðat use I ðe set-þeoretic definition of "partition") of state space. Ðe elements of P are non-empty pairwise disjoint subsets of state space which togeðer cover state space. So a macrostate is a subset of state space. By contrast, a microstate is an element of state space.

On 12/15/2023 at 8:04 PM, sethoflagos said:

the entropy of a system may be defined absolutely by reference to the 3rd Law.

Wiðout reference to an arbitrarily chosen partition of state space into macrostates? 🤨

 

 

On 12/15/2023 at 8:04 PM, sethoflagos said:

macrostates refer to the properties of that system observable in a volume of space-time.

Observable by whom? Ðat's ðe key point which Hossenfelder and I have made, contrary to:

On 12/15/2023 at 8:04 PM, sethoflagos said:

She doesn't. At no point does she state or even imply that:

On 12/15/2023 at 2:00 PM, Tristan L said:

Ðe entropy of a system can’t be defined absolutely. It has to be defined wið regard to a partition (mark ðat I use ðe linked-to set-þeoretic definition of “partition”) of ðe state space of ðe system, ðat is, of ðe set of all possible microstates of ðe system.

We've said ðat currently, ðe entropy of ðe Universe wið respect to a certain partition, call it "P_human", of state space is low. Accordingly, ðe Universe is currently teeming wið living beings (at least on Earþ) who divvy state space up into ðe elements of P_human. In accordance wið ðe Second Law, entropy wið regard to P_human will very likely rise until living beings, such as humans, for which ðe macrostates wi.re.to P_human matter can no longer live. It will ðen very probably take a very, very long time until entropy wi.re.to P_human is low again. However, when entropy wi.re.to P_human is high, entropy will be low wi.re.to some oðer partition, e.g. P_Chubachaba, so living beings (like Chubachabas 😉) who split state space up into ðe members of P_Chubachaba will be able to live.

Let's say we have a very simple system wið just six states. Number ðem 1 to 6. Ðe state space of ðe system is ðe set {1, 2, 3, 4, 5, 6}. If ðe system is currently in state 4, what is its current entropy? Ðe question doesn't make sense. First, we have to break state space down into macrostates, e.g. into {1}, {2, 3, 5}, and {4, 6}. By choosing {{1}, {2, 3, 5}, {4, 6}} as our partition of {1, 2, 3, 4, 5, 6}, we've just categorized ðe microstates according to primality: {1} is ðe macrostate holding all microstates which are neiðer prime nor composite, {2, 3, 5} is ðe macrostate holding ðe prime microstates, and {4, 6} is ðe macrostate containing ðe composite ones. Now, we can ask: What is ðe current entropy of ðe system wið respect to ðe partition {{1}, {2, 3, 5}, {4, 6}}? Ðe answer is -SUM_{z in current macrostate} P({z}|current macrostate)*log₂P({z}|current macrostate) = -SUM_{z in {4, 6}} P({z}|{4, 6})*log₂P({z}|{4, 6}), which = -log₂(1/2) = 1 given ðat all microstates be equally likely. Likewise, if ðe system was in ðe microstate 3 a second ago, its entropy wið regard to {{1}, {2, 3, 5}, {4, 6}} a second ago was = -log₂(1/3) ≈ 1.58, and if it was in microstate 1 two seconds ago, its entropy wið respect to {{1}, {2, 3, 5}, {4, 6}} back ðen was = -log₂(1/1) = 0.

But who says we have to choose {{1}, {2, 3, 5}, {4, 6}} as our partition? Nobody. It's just ðat (in ðis example), we happen to be living beings who care about primeness. However, living beings who care about being a power of 2 would divvy ðe state space up into {2, 4} and {1, 3, 5, 6}, ðat is, choose {{2, 4}, {1, 3, 5, 6}} as ðe partition wið regard to which ðey define entropy. Ðe key point Hossenfelder and I have made is ðat at each time t, ðe system is in a microstate, and for each z, ðere's a partition, P, of state space such ðat for ðe member (a macrostate), M, of P which contains z, ðe likelihood of {z} given M is high, so if ðe system has microstate z, its entropy wið regard to P is low.

On 12/20/2023 at 7:23 PM, swansont said:

Using a common language makes it better. Using terminology that you understand but others don’t is not better.

English is the international language of science. 

Not only is ðe speech I use English, but ðe letters I'm writing are English letters.

On 12/15/2023 at 8:04 PM, sethoflagos said:

out-of-equilibrium

On 12/15/2023 at 8:04 PM, sethoflagos said:

thermodynamic equibrium

Wið regard to which partition of state space into macrostates?

On 12/15/2023 at 8:04 PM, sethoflagos said:

I suspect that she is encountering problems in attempting to square her curious views on determinism with the 2nd Law.

I agree; in a deterministic universe, each probability is eiðer 0 or 1, so ðe entropy is always 0 ... ðough ðis is in accordance wið ðe 2^nd Law, of course.

On 12/15/2023 at 8:04 PM, sethoflagos said:

Hence a macrostate has a uniquely defined entropy even if some of the constituent microstates seem to exhibit spooky patterns.

Ðat's correct and you're right. Ðe entropy of a macrostate is defined absolutely. However, when we ask about ðe entropy of ðe system, we have to ask: Which subsets of state space count as macrostates?

[Markus Hanke]

48 minutes ago, Tristan L said:

Ðis makes it six times as good (i.e. five times better).

I disagree. While I know that Icelandic distinguishes these sounds in phonology and orthography, English effectively doesn’t, so there’s no point in this at all. Furthermore the vast majority of people here on this forum presumably will use English keyboards, so these letters are not straightforwardly accessible to them, making this not at all efficient to most of us.

There are good reasons why English orthography is largely standardised, so deviating from this is unwise, really comes across as silly, and makes it hard for some to read your text. Trust me, this isn’t making a good impression.

My advice to you, if you wish to engage in a proper discussion of your ideas and be taken seriously while doing so, is to stick to standard English orthography. You don’t have to agree with it, you just need to use it.

[exchemist]

1 hour ago, Tristan L said:

Hello guys,

 

It’s I again, Tristan Laguz. In my þread Will entropy be low much of the time?, I talk about a point boþ Sabine Hossenfelder and I have made: ðat ðe entropy of a system isn’t defined absolutely, but raðer wið regard to a choice of which subsets of state space count as macrostates. In ðe later part of ðe topic, I use ðe English (and Norse) letters Ðat (‘Ð’/‘ð’) and Þorn (‘Þ’/‘þ’) because of ðeir aesþetic appeal and because using ðem is þrice as precise and twice as fast as using “th”. Oddly, swansont took issue wið my use of ðese letters. First, he said I should use English in ðe forum, to which I replied ðat I did, in fact, write in English. Moreover, I had already justified my use of Þorn and Ðat. Swansont didn’t engage wið my arguments. Neiðer did he nicely ask me to use “th” instead of ‘Þ’ and ‘Ð’ because it’s currently more widespread even ðough it’s less precise and less efficient. Instead, he simply locked my topic, accusing me of making bad-faiþ arguments. Making such a false accusation and creating such a lie about me is unacceptable, ðough not overly surprising given swansont’s self-description above his profile picture 😉. Since my arguments in my þread are made in good faiþ and, I believe, valid and sound on top of ðat, I herewið open my topic anew to talk wið you about wheðer my aforesaid belief (ðat my arguments be valid and sound) be true.

 

❕ Before you read on, please mark ðat I use “partition of S” in ðe set-þeoretic sense, ðat is, to mean a set of subsets of S which are non-empty, pairwise disjoint, and togeðer cover S. Ðus, “partition of S” refers to a subset of ðe power set of S which fulfills certain conditions. For instance, {{1, 2, 3}, {4, 5}, {6}} is called “a partition of {1, 2, 3, 4, 5, 6}”. Choosing which subsets of a state space count as macrostates amounts to choosing a partition of ðe state space.

 

 

 

I add someþing in ðe beginning which I wanted to write ðe last time but forgot:

 

Sethoflagos wrote:

You’re right. IMHO, Hossenfelder is wrong to claim ðat ðe system stay in ðe same microstate.

 

 

 

Now follows ðe last part of ðe discussion in ðe closed topic and a part shortly before it:

 

Ðe letter Ðat (uppercase: 'Ð', lowercase: 'ð') stands for ðe 'th'-sound in "that", "then", "there" aso. whereas ðe letter Þorn (big: 'Þ', small: 'þ') represents ðe 'th'-sound in "thorn", "think", "thank" asf. Wielding 'ð' and 'þ' is þrice as precise as using 'th', since it distinguishes ðe /ð/ sound from ðe /þ/ sound and furðermore avoids mixing ðem up wið a /t/ sound followed by a /h/ sound. It is also double as efficient, because it uses only half as many letters. Ðis makes it six times as good (i.e. five times better).

I wrote: "Ðe entropy of a system [...] has to be defined wið regard to a partition (mark ðat I use ðe linked-to set-þeoretic definition of “partition”) of [...] ðe set of all possible microstates of ðe system."

Nope, I really do mean macrostates wið respect to P. A macrostate wið respect to P is simply an element of P, where P is a partition (again, please mark ðat use I ðe set-þeoretic definition of "partition") of state space. Ðe elements of P are non-empty pairwise disjoint subsets of state space which togeðer cover state space. So a macrostate is a subset of state space. By contrast, a microstate is an element of state space.

Wiðout reference to an arbitrarily chosen partition of state space into macrostates? 🤨

 

 

Observable by whom? Ðat's ðe key point which Hossenfelder and I have made, contrary to:

We've said ðat currently, ðe entropy of ðe Universe wið respect to a certain partition, call it "P_human", of state space is low. Accordingly, ðe Universe is currently teeming wið living beings (at least on Earþ) who divvy state space up into ðe elements of P_human. In accordance wið ðe Second Law, entropy wið regard to P_human will very likely rise until living beings, such as humans, for which ðe macrostates wi.re.to P_human matter can no longer live. It will ðen very probably take a very, very long time until entropy wi.re.to P_human is low again. However, when entropy wi.re.to P_human is high, entropy will be low wi.re.to some oðer partition, e.g. P_Chubachaba, so living beings (like Chubachabas 😉) who split state space up into ðe members of P_Chubachaba will be able to live.

Let's say we have a very simple system wið just six states. Number ðem 1 to 6. Ðe state space of ðe system is ðe set {1, 2, 3, 4, 5, 6}. If ðe system is currently in state 4, what is its current entropy? Ðe question doesn't make sense. First, we have to break state space down into macrostates, e.g. into {1}, {2, 3, 5}, and {4, 6}. By choosing {{1}, {2, 3, 5}, {4, 6}} as our partition of {1, 2, 3, 4, 5, 6}, we've just categorized ðe microstates according to primality: {1} is ðe macrostate holding all microstates which are neiðer prime nor composite, {2, 3, 5} is ðe macrostate holding ðe prime microstates, and {4, 6} is ðe macrostate containing ðe composite ones. Now, we can ask: What is ðe current entropy of ðe system wið respect to ðe partition {{1}, {2, 3, 5}, {4, 6}}? Ðe answer is -SUM_{z in current macrostate} P({z}|current macrostate)*log₂P({z}|current macrostate) = -SUM_{z in {4, 6}} P({z}|{4, 6})*log₂P({z}|{4, 6}), which = -log₂(1/2) = 1 given ðat all microstates be equally likely. Likewise, if ðe system was in ðe microstate 3 a second ago, its entropy wið regard to {{1}, {2, 3, 5}, {4, 6}} a second ago was = -log₂(1/3) ≈ 1.58, and if it was in microstate 1 two seconds ago, its entropy wið respect to {{1}, {2, 3, 5}, {4, 6}} back ðen was = -log₂(1/1) = 0.

But who says we have to choose {{1}, {2, 3, 5}, {4, 6}} as our partition? Nobody. It's just ðat (in ðis example), we happen to be living beings who care about primeness. However, living beings who care about being a power of 2 would divvy ðe state space up into {2, 4} and {1, 3, 5, 6}, ðat is, choose {{2, 4}, {1, 3, 5, 6}} as ðe partition wið regard to which ðey define entropy. Ðe key point Hossenfelder and I have made is ðat at each time t, ðe system is in a microstate, and for each z, ðere's a partition, P, of state space such ðat for ðe member (a macrostate), M, of P which contains z, ðe likelihood of {z} given M is high, so if ðe system has microstate z, its entropy wið regard to P is low.

Not only is ðe speech I use English, but ðe letters I'm writing are English letters.

Wið regard to which partition of state space into macrostates?

I agree; in a deterministic universe, each probability is eiðer 0 or 1, so ðe entropy is always 0 ... ðough ðis is in accordance wið ðe 2^nd Law, of course.

Ðat's correct and you're right. Ðe entropy of a macrostate is defined absolutely. However, when we ask about ðe entropy of ðe system, we have to ask: Which subsets of state space count as macrostates?

I'm intrigued to see that the "thorn", present in Old English, is something used in Icelandic. The other characters I don't recognise.

However I must agree with Markus that if you are interesting in discussing science, as opposed to striking an affected linguistic pose, you are better off sticking to the English alphabet. I've just tried to read your post and gave up in annoyance after only a paragraph. 

 

[KJW]

Hmmm. I don't think this is going to end well for Tristan L. Not only restarting a thread closed by a moderator, but actually persisting with the action that led to the original thread being closed.

 

 

[Genady]

3 hours ago, Tristan L said:

I use ðe English

This is not English. See English alphabet - Wikipedia.

Plus, the topic of an alphabet is not Physics.

Edited December 22, 2023 by Genady

[studiot]

Quote

Wikipedia

Today Old Norse has developed into the modern North Germanic languages Icelandic, Faroese, Norwegian, Danish, Swedish, and other North Germanic varieties of which Norwegian, Danish and Swedish retain considerable mutual intelligibility while Icelandic remains the closest to Old Norse.

 

Quote

Quora

The Old Norse language, spoken by the Norse people, was a North Germanic language, while Old English, spoken by the Anglo-Saxons, was a West Germanic language. Although there were some similarities due to their shared Germanic roots, the languages had distinct grammar, vocabulary, and pronunciation.

 

Quote

Scholar Commons

Could Old English and Old Norse understand each other?

 

 

In the past, it was the general agreement of historians that there existed little or no amount of mutual intelligibility between the English and Norse peoples of this time.

 

English is descended from AngloSaxon not Old Norse.

Out of interest I borrowed my wifes's copy of Sweet's AnglosSaxon Reader, 5th Ed,  1895.

When she was at Londondon university all students of English were required to be competent in Old English.

The interesting thing is that I could not find your letters in Sweet, although there is some similarity.

 

 

Modern English please, because it may be that you have something interesting and worthwhile to discuss.

 

Ed I have just corrected my spelling mistakes.
I think that is more than enough to lay onto other members, without adding alphabetic ones as well.

Edited December 22, 2023 by studiot

[joigus]

I hope you also notice that English doesn't identify a particular sequence of letters with a sound. Eg,

4 hours ago, Tristan L said:

but raðer wið regard to

(my emphasis)

the "th" sound in "rather" is very different from "th" sound in "with."

Edited December 22, 2023 by joigus
minor addition

[swansont]

!

Moderator Note

Your thread was locked. You don't get to open a new one.

Others have pointed out the flaws in your claim to be using English and the issues this poses. Claiming that you are using the English alphabet is, in my estimation, a bad-faith argument.