Entropy：A concept that is not Physical Quantity

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You cannot expect to be taken seriously if you pretend that spontaneous endothermic reactions are irrelevant to a discussion of entropy.

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You cannot expect to be taken seriously if you pretend that spontaneous endothermic reactions are irrelevant to a discussion of entropy.

I have said "what you said is remote from the subject !"，if you talk about THIS PAPER, maybe I have intereste, but you said is beside the mark.

so much for this, just do as you please.

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would you like to read this paper ( the newest version ) carefully? if you are interested in this theme.

I think only you are Back To Original, you may understand ! Why? Because you have been filled with the pre-existing content, and you are firmly convinced. sometimes, theory is not only science but also belief. In fact, to a wrong theory, there is also many evidences and applications, otherwise they couldn't exist long time, e.g., caloric theory,phlogiston theory,and so on. I hope you can read this paper thoroughly, cast aside preconceived ideas, I think you can distinguish right from wrong.

The present study has demonstrated the non-existence of Clausius entropy, which simultaneously denies the Boltzmann entropy.

In statistical physics, the attempt to directly deduce entropy is untenable. On one hand, it involves a key step to translate infinitesimal into differential, which doesn’t hold. On the other hand, the unit (J/K) of entropy (Boltzmann entropy) in statistical physics is transformed from Clausius entropy. So, if Clausius entropy does not exist, there will be no transformation source for the unit (J/K) of Clausius entropy. As a result, the entropy in statistical physics is only a pure digital, with no physical meaning.

In addition, even if we do not consider the issue regarding the unit, from a pure probability point of view, in the equation S = klnΩ, Ω is the so-called thermodynamic probability, and the calculation of Ω involves the phase cell division in surpassing space μ. The phase cell is 2i-dimensional and i the total freedom degree of the molecules within the system. The essence of Ω calculation is the discretization of the continuous μ space and the generation of objective meaning. In fact, this approach does not work, and there will be no objective conclusion regardless of the amount of previous work people have done. This is due to the lack of objective, physically meaningful criteria for phase cell division, that is, Ω has no objective meaning in physics. Together the Liouville theorem and our conclusions indicate that Boltzmann entropy can be taken as a technique for displaying the irreversibility from a purely probabilistic point of view.

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would you like to read this paper ( the newest version ) carefully? if you are interested in this theme.

I think only you are Back To Original, you may understand ! Why? Because you have been filled with the pre-existing content, and you are firmly convinced. sometimes, theory is not only science but also belief. In fact, to a wrong theory, there is also many evidences and applications, otherwise they couldn't exist long time, e.g., caloric theory,phlogiston theory,and so on. I hope you can read this paper thoroughly, cast aside preconceived ideas, I think you can distinguish right from wrong.

The present study has demonstrated the non-existence of Clausius entropy, which simultaneously denies the Boltzmann entropy.

In statistical physics, the attempt to directly deduce entropy is untenable. On one hand, it involves a key step to translate infinitesimal into differential, which doesn't hold. On the other hand, the unit (J/K) of entropy (Boltzmann entropy) in statistical physics is transformed from Clausius entropy. So, if Clausius entropy does not exist, there will be no transformation source for the unit (J/K) of Clausius entropy. As a result, the entropy in statistical physics is only a pure digital, with no physical meaning.

In addition, even if we do not consider the issue regarding the unit, from a pure probability point of view, in the equation S = klnΩ, Ω is the so-called thermodynamic probability, and the calculation of Ω involves the phase cell division in surpassing space μ. The phase cell is 2i-dimensional and i the total freedom degree of the molecules within the system. The essence of Ω calculation is the discretization of the continuous μ space and the generation of objective meaning. In fact, this approach does not work, and there will be no objective conclusion regardless of the amount of previous work people have done. This is due to the lack of objective, physically meaningful criteria for phase cell division, that is, Ω has no objective meaning in physics. Together the Liouville theorem and our conclusions indicate that Boltzmann entropy can be taken as a technique for displaying the irreversibility from a purely probabilistic point of view.

While certainly effective at making use of words, simply repeating what you have said before isn't likely to convince anyone who wasn't already convinced. Neither will hand-waving and declaring questions you don't want to answer as off topic.

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I have said "what you said is remote from the subject !"

You ( or others ) can raise interminable irrelevant questions as this one, surely it doesn't mean that I ( or others )should answer it.

certainly, I can answer this question, but this relates to another paper, I don't want to release this paper now.

OK, How does "This Paper" explain spontaneous endothermic reactions?

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While certainly effective at making use of words, simply repeating what you have said before isn't likely to convince anyone who wasn't already convinced. Neither will hand-waving and declaring questions you don't want to answer as off topic.

I have answered questions, if you can not understand it, I have nothing to say.

In fact, my paper ( the newest version ) has expounded very clearly.

OK, How does "This Paper" explain spontaneous endothermic reactions?

I do don't think one should put forward question when he is Confused and don't know what is said,

Attention, I never said " "This Paper" explain spontaneous endothermic reactions? " !!!

what I said is:

" I have said "what you said is remote from the subject !"

You ( or others ) can raise interminable irrelevant questions as this one, surely it doesn't mean that I ( or others )should answer it.

certainly, I can answer this question, but this relates to another paper, I don't want to release this paper now. "

I sincerely suggest you read this three lines of words again and again, when you really think you know its meanings, then, you can continue to talk here.

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Moderator Note

shufeng-zheng,

1. The insults stop.

2. You'll answer their questions. As John Cuthber noted, simply telling members to read the paper does not constitute an answer.

If you cannot comply to the rules of this forum, this will be closed.

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Moderator Note

shufeng-zheng,

1. The insults stop.

2. You'll answer their questions. As John Cuthber noted, simply telling members to read the paper does not constitute an answer.

If you cannot comply to the rules of this forum, this will be closed.

1) I have answered their questions.

2) I didn't insult him, I only crack a joke with John Cuthber, if this caused misunderstanding, I say sorry.

3) Some questions such as "explain spontaneous endothermic reactions" is unrelated with this theme, I don't think I have to answer any question someone posed.

4) I think I have spoken clearly, but sometimes they could not understand.

5) I would like to answer anyone question RELATED to this theme, but if he could not understand, what to do next?

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Shufeng-zhang

Two responders have not argued with your paper.

Both posted questions designed to clarify what you are saying and help you with a discussion.

You have been rude and dismissive to both and completely failed (refused?) to clarify your points.

I have not come across a more tolerant scientific forum than this one so I fear you will soon loose your voice here if you carry on discussion in this way.

I have not yet sought to debate with you since all I have done has been to try to establish the basis for your paper.

I think the following sequence speaks for itself.

Eclectic post#2

Let me see if I understand what you are saying.

Shufeng-zhang post#12

From what you say as abovementioned, I think you misunderstood this paper, would you like to read the paper( The newest version ) again carefully?

Shufeng-zhang post#20

continuing such debate will be unmeaning but a waste of time

Studiot post#22

You did not answer one of them.

A discussion is two way and based on what can be agreed, not what is in doubt.

So if you wish a real discussion let us start again.

Shufeng-zhang post#23

Just do as you wish. Debate can't solve problem, to me, so much for this. You can discuss with other, if you have interest in this paper.( you really read this paper carefully?)

Shufeng-zhang post#34

1) I have answered their questions.

Edited by studiot
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I think it's clear enough that anyone who doesn't understand the relationship between entropy and endothermic reactions does not understand thermodynamics and is, by that fact, unqualified to offer a meaningful opinion about entropy.

On that basis, and also because the OP's refusal to discuss his ideas is a breach of rule 8, I suggest closing the thread.

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It is a pity to attribute to someone long dead something he did not say as the basis for a paper.

Clausius actually said

"Die Warme kann nich von selbst aus einem kalteren in einem warmeren Korper ubergangen"

I see not even the ghost of a cyclic integral in that statement.

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5) I would like to answer anyone question RELATED to this theme, but if he could not understand, what to do next?

John Cuthber and I have been trying to point out to you that if you do away with the concept of entropy then you have no explanation for why [observed] spontaneous processes occur.

Allow me to clarify:

The internal energy of a system expressed in natural variables is:

$U(S,V,N) = TdS-PdV+\sum_{i} \mu _{i} dN_{i}$

By definition of the differential we have:

$dU=\frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V}dV + \sum_{i} \frac{\partial U}{\partial N_{i}}dN_{i}$

By inspection we can see that T, -P, and "mu" correspond to the partial derivatives of internal energy with respect to its natural variables. Which, by the way, how do you intend to define the internal energy without S?

Alright so now let's define the Gibb's energy which is a great measure of spontaneity near equilibrium. We can get an expression for the Gibb's energy and all the other thermodynamic state functions by Legendre transforming the internal energy any number of times. For the Gibb's energy we Legendre transform in V and S:

$G= TS-PV+\sum_{i} \mu _{i} dN_{i}-V\frac{\partial U}{\partial V}-S\frac{\partial U}{\partial S}$

differentiating with knowledge of the coefficients established above gives:

$dG(P,T,N_{i})=VdP-SdT+\sum_{i} \mu _{i} dN_{i}$

You can do some algebra and show that the enthalpy is hidden inside the expression (yes I know technically you can't integrate the expression as is because S has a T dependence but the approximation is valid near equilibrium and for a relatively small temperature change):

$dG=dH-SdT$

$\Delta G= \Delta H - T \Delta S$

So for spontaneous processes [not just chemical reactions by the way] that require a net input of energy, i.e. those with a positive change in enthalpy, there must be a change in some other quantity in order to meet the spontaneity requirement of the change in Gibb's Energy being negative. How do you explain that?

This is all from classical thermodynamics. But your argument also makes no sense on a statistical mechanics level. Entropy is a volume on the phase space of an ensemble. I don't see how entropy being not directly experimentally measurable, or not unique affects that. Do you not agree that for a larger phase space volume there are a greater number of accessible microstates?

I read your paper by the way. No matter what you justify through the Liouville Theorem or by redefining the Carnot cycle, your result must agree with observed experiment.

EDIT: LaTeX hiccup

Edited by mississippichem
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Shufeng-zhang

Two responders have not argued with your paper.

Both posted questions designed to clarify what you are saying and help you with a discussion.

You have been rude and dismissive to both and completely failed (refused?) to clarify your points.

I have not come across a more tolerant scientific forum than this one so I fear you will soon loose your voice here if you carry on discussion in this way.

I have not yet sought to debate with you since all I have done has been to try to establish the basis for your paper.

I think the following sequence speaks for itself.

Debate can't solve the problem, I have said, just do as you please.

John Cuthber and I have been trying to point out to you that if you do away with the concept of entropy then you have no explanation for why [observed] spontaneous processes occur.

Allow me to clarify:

The internal energy of a system expressed in natural variables is:

$U(S,V,N) = TdS-PdV+\sum_{i} \mu _{i} dN_{i}$

By definition of the differential we have:

$dU=\frac{\partial U}{\partial S}dS + \frac{\partial U}{\partial V}dV + \sum_{i} \frac{\partial U}{\partial N_{i}}dN_{i}$

By inspection we can see that T, -P, and "mu" correspond to the partial derivatives of internal energy with respect to its natural variables. Which, by the way, how do you intend to define the internal energy without S?

Alright so now let's define the Gibb's energy which is a great measure of spontaneity near equilibrium. We can get an expression for the Gibb's energy and all the other thermodynamic state functions by Legendre transforming the internal energy any number of times. For the Gibb's energy we Legendre transform in V and S:

$G= TS-PV+\sum_{i} \mu _{i} dN_{i}-V\frac{\partial U}{\partial V}-S\frac{\partial U}{\partial S}$

differentiating with knowledge of the coefficients established above gives:

$dG(P,T,N_{i})=VdP-SdT+\sum_{i} \mu _{i} dN_{i}$

You can do some algebra and show that the enthalpy is hidden inside the expression (yes I know technically you can't integrate the expression as is because S has a T dependence but the approximation is valid near equilibrium and for a relatively small temperature change):

$dG=dH-SdT$

$\Delta G= \Delta H - T \Delta S$

So for spontaneous processes [not just chemical reactions by the way] that require a net input of energy, i.e. those with a positive change in enthalpy, there must be a change in some other quantity in order to meet the spontaneity requirement of the change in Gibb's Energy being negative. How do you explain that?

This is all from classical thermodynamics. But your argument also makes no sense on a statistical mechanics level. Entropy is a volume on the phase space of an ensemble. I don't see how entropy being not directly experimentally measurable, or not unique affects that. Do you not agree that for a larger phase space volume there are a greater number of accessible microstates?

I read your paper by the way. No matter what you justify through the Liouville Theorem or by redefining the Carnot cycle, your result must agree with observed experiment.

EDIT: LaTeX hiccup

You are argueing in a circle.

Debate can't solve the problem.

I think it's clear enough that anyone who doesn't understand the relationship between entropy and endothermic reactions does not understand thermodynamics and is, by that fact, unqualified to offer a meaningful opinion about entropy.

On that basis, and also because the OP's refusal to discuss his ideas is a breach of rule 8, I suggest closing the thread.

Thanks for your " suggest closing the thread " , I'm tired of debate.

It is a pity to attribute to someone long dead something he did not say as the basis for a paper.

Clausius actually said

"Die Warme kann nich von selbst aus einem kalteren in einem warmeren Korper ubergangen"

I see not even the ghost of a cyclic integral in that statement.

What I want to say is: Thermodynamics and statistical physics, from this, Physical chemistry, will be rebuilt, believe it or not.

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Debate can't solve the problem, I have said, just do as you please.

Then why have you brought this to a forum where people typically come to debate!?

You are argueing in a circle.

Debate can't solve the problem.

You are arguing with poor sentence structure. You've yet to address even one argument raised in this thread. I'll accept that as your statement of defeat.

Just becuause I used entropy in my argument does not mean that I am using circular logic. Show me the circular logic in my argument. I was showing how entropy must exist in order to explain many of the phenomena we observe.

I ask again, what are the natural variables for internal energy? I find it I'll advised for you to evoke fancy stat. mech. arguments when you clearly do not understand rudimentary classical thermodynamics.

Thanks for your " suggest closing the thread " , I'm tired of debate.

You've yet to debate. I wish you would at least attempt to tackle even one of these arguments. People have put time into their posts and you effectively respond with "Nah-uh".

What I want to say is: Thermodynamics and statistical physics, from this, Physical chemistry, will be rebuilt, believe it or not.

And you expect to rebuild it without the presence of spontaneous endothermic processes, and a flawed interpretation of Liouville's theorem.

Thanks for a good laugh. Physical chemistry currently works fine and currently pays my salary. Physical chemistry was already built from thermo and stat. mech. Show where current chemical thermodynamics is inconsistent with these notions from physics.

I'm not going to let you keep making unfounded assertions and snide empty comments. Pony up or shut up. It's quite simple really.

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Then why have you brought this to a forum where people typically come to debate!?

You are arguing with poor sentence structure. You've yet to address even one argument raised in this thread. I'll accept that as your statement of defeat.

Just becuause I used entropy in my argument does not mean that I am using circular logic. Show me the circular logic in my argument. I was showing how entropy must exist in order to explain many of the phenomena we observe.

I ask again, what are the natural variables for internal energy? I find it I'll advised for you to evoke fancy stat. mech. arguments when you clearly do not understand rudimentary classical thermodynamics.

You've yet to debate. I wish you would at least attempt to tackle even one of these arguments. People have put time into their posts and you effectively respond with "Nah-uh".

And you expect to rebuild it without the presence of spontaneous endothermic processes, and a flawed interpretation of Liouville's theorem.

Thanks for a good laugh. Physical chemistry currently works fine and currently pays my salary. Physical chemistry was already built from thermo and stat. mech. Show where current chemical thermodynamics is inconsistent with these notions from physics.

I'm not going to let you keep making unfounded assertions and snide empty comments. Pony up or shut up. It's quite simple really.

I post the paper here, I want other can read it and think alone, that's all.

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Moderator Note

Debate can't solve this problem? Then why would you choose to bring it up in a discussion forum of all places if you had no intention of discussing anything? That's rather against the nature of this website, don't you think?

You don't get to cherry pick what you will and will not respond to. I have asked you to answer all of the questions asked of you in this thread and you still refuse to do so. This is against our rules.

Since you refuse to comply and refuse to even attempt to foster discussion, I am closing this thread. You are not permitted to reintroduce the topic.

Edit: cross posted with shufeng-zheng.