Dear Remi,

Thanks for your response.

However, your answer is a post hoc ergo propter hoc.

No-and the description is wrong. A real particle doesn't behave the way described. If it's isolated, it conserves energy (first law of thermodynamics) and, since it is isolated, i.e. has zero temperature, has zero entropy (third law of thermodynamics). It's in its vacuum state. 

If it's not isolated, for instance is in equilibrium with a heat bath at finite temperature, then everything about it, in particular transition probabilities,  is described by its canonical partition function. 

If it's in contact with a bath of other particles, everything, once more transition probabilities,  is described by the grand canonical partition function, so the chemical potential is, also, required. 

These statements hold whether the object of interest is a classical particle, or a quantum particle. In the latter case it's subject to quantum fluctuations, also, that's all.

Dear Stam, thanks a lot for your skillful insights.

However, there's something I do not understand.  I say that some real particle falls in the vacuum.  You said that  "a real particle, if isolated, conserves energy (first law of thermodynamics) and, since it is isolated, i.e. has zero temperature, has zero entropy (third law of thermodynamics). Therefore, this real particle is in its vacuum state".  Exactly what I said...

Furthermore, if my framework is wrong, how do explain that the galaxies are increasing their distance at the speed of 40 km/sec? I suppose that it depends on an increase in vacuum volume among them.  If it's right, where could the energy that increases the vacuum volume come from, other than the former real particles' energy?

A particle doesn't ``fall in the vacuum''-that statement is meaningless, since it confuses two different notions of vacuum. 

The expansion of the Universe doesn't have anything to do with the question and, in particular with any ``vacuum volume''.  The expansion takes place in spacetime-not space.

The  expansion of the Universe doesn't come from the energy of matter. Matter tends to contract spacetime, not to expand it. The expansion of the Universe is described by general relativity-there exists a family of solutions of Einstein's equations, as found by Friedmann in the 1920s, that describe a uniformly  expanding spacetime. And the cosmological constant, introduced by Einstein for totally irrelevant reasons, was found by de Sitter to be contributing to accelerating the expansion, if its sign is positive. And it's now known that the cosmological constant is positive-the expansion of the Universe, described as a solution of Einstein's equations,  is accelerating. 

One shouldn't confuse the vacuum state, implied by the term ``quantum vacuum'' with the vacuum=absence of matter in spacetime. 

Dear Stam, the vacuum energy does exert an anti-gravitational energy, that is expected to contribute to the cosmological constant, which affects the expansion of the universe.  The pressure of the quantum vacuum is always negative, implying a repulsive gravitational field.  Therefore, if the Universe expansion is accelerating, this means that the vacuum volume might be increasing.  The first law of thermodynamics is preserved at the level of the total cosmic sum of vacuum and non-vacuum.  Why the cosmological constant is now positive? A possible explanation is that the vacuum volume is simply increasing... 

Furthermore, the QED quantum vacuum can be described exaclty as an absence of "matter", "real particles" (not of energy, of course!) in spacetime.  

Otherwise, you have to explain me why the cosmological constant is positive.  

There isn't any such notion as the pressure of the quantum vacuum. In any quantum field theory, in the absence of gravity, the vacuum state can be defined to  have zero energy: only energy differences are physical, because time translation invariance is a global symmetry. One shouldn't confuse the fluctuations of energy about its average value, with the average value itself.

In any classical field theory, coupled to gravity, the vacuum energy is equal to the cosmological constant. 

The cosmological constant is part of classical gravity. That its value, in our Universe,  is positive is an experimental fact-there's no particular reason from general relativity that would favor a positive value over a zero or a negative value. But general relativity is consistent whatever its value-the fact that it's positive doesn't mean anything particular for classical gravity. It doesn't make sense to discuss classical gravity together with quantum matter, however, because it's not known how quantum matter affects spacetime.

I would like you gentlemen to have a look at my paper: Creation of Universe and Innumerable Dimensions on ResearchGate site: https://www.researchgate.net/publication/289004971_Socio-Physicochemical_Theory_Part_-1_Creation_of_Universe_Innumerable_Dimensions 

Research Socio-Physicochemical Theory Part -1: Creation of Universe &...

Dear Remy, 

when I am a pediatrician, I am very conservative: I never dare to try a therapy, unless it is written in the most influential guidelines.  However, when I approach theoretical issues in physics and biology, I like provocative and unconventional views.   Sometimes I say bollocks, sometimes not, but who cares? I feel free to play the linguistic joke I prefer, and I can think what I want.  I do not eat thanks to Academics, I do not have students to bother, I do not care about "publish or perish". I'm not frustrated at all, I'm a privileged.    And my strategy works, because I published scattered IF papers everywhere, JUST FOR MY PLEASURE. 

A suggestion: if you want to publish something and improve your flebile CV, make a better use of RG.  RG, apart a (very) few provocative idiots, is a treasure of good, propositive, polite researchers.  The most of my papers were published with researchers met on RG, who I never met face to face in my life.  On RG, even if someone says something that is stupid, you cannot attack him in such an hasty way, otherwise you will never proceed further in your pursued academic career. People is open to (also severe) critics, but not to hasty, frustrated, injuring comments.    Take the example of Stam: he says politely that I'm wrong, but he did not use your sarcasm.   With your way to proceed, you just loose chances.  For example, when I have a good idea (it does not seems to be the case, but, I repeat, who cares?) and some commenter appreciates me, I always try to involve also him/her in my projects.     

Best wishes, and take care of your mind and emotions.  

Dear Remi, 

Ok. Peace.

I suggest to you, that are a peer of mine, to try to publish your stuff.  Try to do like me and my predatory Journals: I published my miserable stuff on Nature, NEJM, PLOS Biology, Physics of Life Review, Frontiers in Human Neuroscience... 

I also published, despite my intrinsic limits, on a small  physics Journal: https://link.springer.com/article/10.1007/s10773-016-2998-7

It's not enough, I know, but to start with it could be enough... Don't keep your secrets hidden: try!  You surely will find favourable reviewers, because your stuff seems good. Very good.

Ciao, and peace again!

Entropy: A concept that is not a physical quantity

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

Calculus is not "take for granted".

ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

It is well known that calculus has a definition.

Any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.