1st law, 2nd law, entropy by Gyftopoulos / Beretta is confusing

In summary, the authors attempt to present classical thermodynamics without the lack of logical consistency and completeness in the many presentations of the foundations of thermodynamics I have experienced in my education. Although I was looking forward to "understanding" the subject once and for all, I have to admit that this type of presentation raises more questions than it should solve. I obviously do not really understand their formulations of the 1st and 2nd law, because I can think of very simple counterexamples where they seem to fail.
  • #1
Sebi123
14
1
I'm reading Thermodynamics: Foundations and Applications by Gyftoploulos and Beretta, because the authors claim to give a presentation of classical thermodynamics without "... the lack of logical consistency and completeness in the many presentations of the foundations of thermodynamics" [from the preface], a problem I have also experienced in my education by reading some of the "standard" textbooks. Although I was looking forward to "understand" the subject once and for all I have to admit that this type of presentation raises more questions than it should solve. I obviously do not really understand their formulations of the 1st and 2nd law, because I can think of very simple counterexamples where they seem to fail:

1st law
Their 1st law is [section 3.2]:

"Any two states of a system may always be the end states of a weight process, that is, the initial and final states of a change of state that involves no net effects external to the system except the change in elevation between z1 and z2 of a weight. Moreover, for a given weight, the value of the quantity Mg(z1 — z2) is fixed by the end states of the system, and independent of the details of the weight process, where M is the mass of the weight and g the gravitational acceleration."


From this statement they directly derive a property called "energy", which can be defined as
E = E0 - Mg(z - z0) where E0 is an arbitrary reference energy and z0 the reference height of the weight.
In my opinion this statement can only be true for closed systems, because I can think of no weight process that is equivalent to a process in which the system receives or loses some amount of matter. However, they don't explicitly confine their 1st law to closed systems. And even if they rephrased the first sentence to "Any two states of a closed system ...", what about the following example:
A closed composite system consists of two subsystems, A and B, which are connected by a movable piston allowing the exchange of particles [section 6.1]:
upload_2016-9-7_20-20-27.png

The only parameter is volume V = VA + VB (that's not the point here) and the number of particles is restricted by the equation n = nA + nB.
Assuming open systems are excluded by the 1st law, I see no justification to assign a property called "energy" to each of the subsystems, because both are open. I also have a problem saying "the composite (A+B) has a property called 'energy', so each subsystem also has such a property". I mean I also can't say: "(A+B) is divisible by 2, so A and B also have the property beeing divisible 2". Not beeing able to assign "energy" to open systems seems very weird to me, so what do I misunderstand here, using only the information given in the book and no previous knowledge from other sources?

2st law

Their 2nd law is [section 4.4]:
"Among all the states of a system that have a given value E of the energy and are compatible with a given set of values n of the amounts of constituents and β of the parameters, there exists one and only one stable equilibrium state. Moreover, starting from any state of a system it is always possible to reach a stable equilibrium state with arbitrarily specified values of amounts of constituents and parameters by means of a reversible weight process."

Consider the following mechanical system of a point mass in a constant gravitational field. According to my understanding of the definition of a well defined system (chapter 1 in the book) this system can also be regarded as a valid thermodynamic system. (Indeed, the authors use a point mass system in example 3.9, p. 37 to make clearer the concept of energy.) According to the authors' formulation of the 2nd law there must a stable equilibrium state according to the energy E = mgh (where I set the reference energy E0 to zero) this system currently has.
upload_2016-9-7_20-21-3.png

However, as the authors explain in section 4.3, in a mechanical system there is only one stable equilibrium state: the state of minimum energy (point mass on the ground with speed= 0). Doesn't this contradict their formulation of the 2nd law, because the system shown with E = mgh is obviously not in a stable equilibrium state: One can easily extract energy (mgh) from the point mass while the mass is falling.

Entropy

The authors introduce entropy by the definition S = S0 + 1/cR[(E - E0) - (ΩR - Ω0R))]. ΩR is the "available energy", that is "the largest amount of energy that can be transferred to a weight out of the composite of a system A and a given reservoir R in a weight process" [section 6.6.2]. So, the definition of entropy relies on the definition of available energy which in turn relies on the definition of reservoir. In section 6.3 the term "reservoir" is defined, but the authors admit that the the statements defining a reservoir "... are so restrictive that they must be regarded as limits that a system obeying the laws of physics can approach but cannot actually reach" [section 6.3]. How can an exact property like entropy be defined in terms of an unphysical system like a reservoir?
 
Last edited:
Science news on Phys.org
  • #2
Sebi123 said:
I'm reading Thermodynamics: Foundations and Applications by Gyftoploulos and Beretta, because the authors claim to give a presentation of classical thermodynamics without "... the lack of logical consistency and completeness in the many presentations of the foundations of thermodynamics" [from the preface], a problem I have also experienced in my education by reading some of the "standard" textbooks. Although I was looking forward to "understand" the subject once and for all I have to admit that this type of presentation raises more questions than it should solve. I obviously do not really understand their formulations of the 1st and 2nd law, because I can think of very simple counterexamples where they seem to fail:

1st law
Their 1st law is [section 3.2]:

"Any two states of a system may always be the end states of a weight process, that is, the initial and final states of a change of state that involves no net effects external to the system except the change in elevation between z1 and z2 of a weight. Moreover, for a given weight, the value of the quantity Mg(z1 — z2) is fixed by the end states of the system, and independent of the details of the weight process, where M is the mass of the weight and g the gravitational acceleration."


From this statement they directly derive a property called "energy", which can be defined as
E = E0 - Mg(z - z0) where E0 is an arbitrary reference energy and z0 the reference height of the weight.
In my opinion this statement can only be true for closed systems, because I can think of no weight process that is equivalent to a process in which the system receives or loses some amount of matter. However, they don't explicitly confine their 1st law to closed systems. And even if they rephrased the first sentence to "Any two states of a closed system ...", what about the following example:
A closed composite system consists of two subsystems, A and B, which are connected by a movable piston allowing the exchange of particles [section 6.1]:
View attachment 105581
The only parameter is volume V = VA + VB (that's not the point here) and the number of particles is restricted by the equation n = nA + nB.
Assuming open systems are excluded by the 1st law, I see no justification to assign a property called "energy" to each of the subsystems, because both are open. I also have a problem saying "the composite (A+B) has a property called 'energy', so each subsystem also has such a property". I mean I also can't say: "(A+B) is divisible by 2, so A and B also have the property beeing divisible 2". Not beeing able to assign "energy" to open systems seems very weird to me, so what do I misunderstand here, using only the information given in the book and no previous knowledge from other sources?

2st law

Their 2nd law is [section 4.4]:
"Among all the states of a system that have a given value E of the energy and are compatible with a given set of values n of the amounts of constituents and β of the parameters, there exists one and only one stable equilibrium state. Moreover, starting from any state of a system it is always possible to reach a stable equilibrium state with arbitrarily specified values of amounts of constituents and parameters by means of a reversible weight process."

Consider the following mechanical system of a point mass in a constant gravitational field. According to my understanding of the definition of a well defined system (chapter 1 in the book) this system can also be regarded as a valid thermodynamic system. (Indeed, the authors use a point mass system in example 3.9, p. 37 to make clearer the concept of energy.) According to the authors' formulation of the 2nd law there must a stable equilibrium state according to the energy E = mgh (where I set the reference energy E0 to zero) this system currently has.
View attachment 105583
However, as the authors explain in section 4.3, in a mechanical system there is only one stable equilibrium state: the state of minimum energy (point mass on the ground with speed= 0). Doesn't this contradict their formulation of the 2nd law, because the system shown with E = mgh is obviously not in a stable equilibrium state: One can easily extract energy (mgh) from the point mass while the mass is falling.

Entropy

The authors introduce entropy by the definition S = S0 + 1/cR[(E - E0) - (ΩR - Ω0R))]. ΩR is the "available energy", that is "the largest amount of energy that can be transferred to a weight out of the composite of a system A and a given reservoir R in a weight process" [section 6.6.2]. So, the definition of entropy relies on the definition of available energy which in turn relies on the definition of reservoir. In section 6.3 the term "reservoir" is defined, but the authors admit that the the statements defining a reservoir "... are so restrictive that they must be regarded as limits that a system obeying the laws of physics can approach but cannot actually reach" [section 6.3]. How can an exact property like entropy be defined in terms of an unphysical system like a reservoir?
Maybe a more physically and intuitively oriented discussion of the 1st and 2nd laws would work better for you. Here is a Physics Forums Insights article I wrote a while ago that may be helpful:
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
Hope it helps.
 
  • #3
Chestermiller said:
Maybe a more physically and intuitively oriented discussion of the 1st and 2nd laws would work better for you. Here is a Physics Forums Insights article I wrote a while ago that may be helpful:
https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
Hope it helps.

I find the basic idea behind the treatment of thermodynamics in the book so appealing, because everything there relies on lifting and lowering weights in a constant gravitational field. That is very intuitive and leads to an operational defintion of the terms "energy" and "entropy". Additionally, their definition of entropy is related to the difference in energy and available energy, which make me think about changes in entropy as a measure of energy that must be extracted from or added to a reservoir, which in turn is a measure of "wasted" energy - according to my understanding - because a reservoir is always in an equilibrium state, so energy transferred to it cannot be regained. So, their approach gives more insight into the concept of entropy, e.g., than I ever got from other sources.

I just don't want to give up on this approach, although your tutorial is nice to read. I'd like to start from the very beginning and question my understanding to the concepts of system, temperature, heat, etc. I know no presentation of thermodynamics that introduces such concepts in an intuitive and operational way while not relying on a statistical approach. In the end I always have the impression that temperature, e.g., is just what I experience when I touch a hot rod of iron.
On the other hand, my lack of understanding of the basics presented in the book makes it difficult to trust it.
 
Last edited:
  • #4
Well, I find these definitions very confusing. Isn't it much simpler to use statistical mechanics to define the quantities?
 
  • #5
vanhees71 said:
Well, I find these definitions very confusing. Isn't it much simpler to use statistical mechanics to define the quantities?

The statistical approach might be simpler.

I wonder why the book seems not to have received that attention I thought it would have after reading the preface, where the authors say they "... have composed an exposition of the foundations and the applications of thermodynamics that many enthusiastic M.I.T. students have found clarifying, rewarding, and inspiring." At least there are only very few posts in the forum that can be found with the authors' names and the book title as key words.
Sounded so promising.
 
  • #6
Well, it's completely subjective whether you like a textbook or not. Perhaps it's a very good book, but from the cited passage, I'd think, I wouldn't like it, but that doesn't mean that it isn't a very good source for somebody else who likes "many words" instead of "many formulae" :-).
 
  • #7
Gyftoploulos and his advisor at MIT promoted the school of Quantum thermodynamics which did not find its way to fame among physicists. As such this text is rarely used as a standard textbook today. It was used by Gyftoploulos at MIT, however. Google Quantum Thermodynamics to learn more about their ideas and thoughts. I think some of the papers from this school eventually found their way to respected journals such as Phys. Rev. E and others. In the past many of these papers accumulated in arXiv only without eventual peer review publication.

I would read this text with another "mainstream" one. I strongly recommend the first 8 chapters in Herbet Callen's thermodynamics and an introduction to thermostatics. It is always insightful to read the same subject from different points of view.
 
  • Like
Likes Sebi123 and dextercioby
  • #8
Callen is indeed a great book on thermodynamics and statistics.
 

What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another.

What is the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process.

What is entropy?

Entropy is a measure of the disorder or randomness in a system. It is a thermodynamic property that describes the amount of energy that is no longer available for work in a system.

Why is the book "1st law, 2nd law, entropy by Gyftopoulos / Beretta" confusing?

The book is considered confusing by some because it presents the first and second laws of thermodynamics in a different way than the traditional approach. This can be challenging for readers who are familiar with the traditional approach and may find it difficult to understand the concepts in a new framework.

What is the best way to understand the concepts of the first and second laws of thermodynamics?

The best way to understand these concepts is to approach them from different perspectives and to use a variety of resources. It is also helpful to practice solving problems and to seek clarification from experts in the field.

Similar threads

  • Thermodynamics
Replies
2
Views
776
  • Thermodynamics
Replies
33
Views
2K
Replies
1
Views
661
Replies
4
Views
978
Replies
12
Views
1K
Replies
20
Views
1K
Replies
15
Views
1K
Replies
3
Views
730
Replies
5
Views
2K
Replies
5
Views
2K
Back
Top