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This question already has an answer here:

Can you formalize statistical mechanics, like some people have done with relativity, and prove the second law of thermodynamics from more foundational axioms?

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marked as duplicate by Ben Crowell, ACuriousMind, Bernhard, Qmechanic Jul 26 '14 at 19:58

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    $\begingroup$ I don't think that a true proof exists; the controversies listed on the wikipedia page haven't been totally resolved. In particular, the Poincaré recurrence theorem says that, if you wait long enough, the second law of thermodynamics will eventually fail. However, as the article notes, this happens so infrequently that it's not worth worrying about. Still, things like that prevent a totally rigorous proof. $\endgroup$ – lnmaurer Jul 18 '14 at 12:17
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    $\begingroup$ duplicate of physics.stackexchange.com/q/81465 $\endgroup$ – Ben Crowell Jul 26 '14 at 15:56
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This is a very difficult problem, I try to explain why.

In statistical mechanics, tending to the most probable distribution is a probability event, and for Boltzmann' entropy, $dS\ge 0$ is also a probability event but not an inevitable result. So you can’t prove $dS\ge 0$ as an inevitable result from statistical mechanics.

If we want to obtain the mathematical proof of the second law from thermodynamics, we must consider the mathematical proof of the entropy first, as a state function. Clausius’ definition $dS=δQ/T$ cannot be proven in mathematics, as an exact differential, so the definition $dS=δQ/T$ can only depend on imaginary reversible cycles. On the other hand, in C.Carath´eodory or M.Planck approaches, the expressions of the entropy are the mathematical equations but not the definition of a physical concept because the equation contains the difference of functions (please see bellow, Euler’s equation), the physical image of the entropy and both the second law are not clear, that is why we cannot explain the physical meaning of entropy according to these approaches. In such case, to prove the second law in mathematics will be very difficult, it is not an isolated problem.

The following are the some steps of the link paper, the paper introduce a new approach, and the new statement on the second law can be considered as an axiom.

1) According to the fundamental equation of thermodynamics (Euler’s equation).

\begin{align} dS=\frac{dU}{T}-\frac{Ydx}{T}-\sum_j\frac{\mu_jdN_j}{T}+\frac{pdV}{T}.\end{align}

2) Define the function that

\begin{align}dq=dU-Ydx-\sum_j\mu_jdN_j.\end{align}

Here $dq$ can be proven as an exact differential in mathematics, the physical meaning of $q$ is the heat energy within the system. (but not the heat in transfer $Q$)

3) Such that we get

\begin{align}dS=\frac{dq}{T}+\frac{pdV}{T}.\end{align}

Here $dS$ can be proven as an exact differential in mathematics.

4) Consider an interaction between the two locals, then we can get the total differential of the entropy production.

\begin{align}d_iS=\nabla \left(\frac{1}{T}\right)dq+\frac{1}{T}\nabla Ydx+\sum_j\frac{1}{T}\nabla \mu_jdN_j+\nabla \left(\frac{p}{T}\right)dV.\end{align}

This is a non- equilibrium thermodynamic equation.

5) Prove the total differential of the entropy production $d_iS\ge0$.

The new statement of the second law: "irreversibility root in a fundamental principle: the gradients of the four thermodynamic forces spontaneously tend to zero".

The four gradients of thermodynamic forces are \begin{align} \nabla \left(\frac{1}{T}\right),\,\,\,\, \nabla Y,\,\,\,\, \nabla \mu_j,\,\,\,\, \nabla \left(\frac{p}{T}\right). \end{align}

The conditions of thermodynamic equilibrium are these four gradients equal to zero.

Please compare the different statements about the second law, and see which statement can be considered as an axiom. 1)-4) are quoted from http://en.wikipedia.org/wiki/Second_law_of_thermodynamics

1) Clausius statement

  • Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.

2) Kelvin statement

  • It is impossible, by means of inanimate material agency, to derive mechanical effect from any portion of matter by cooling it below the temperature of the coldest of the surrounding objects.

3) Planck's statement

  • Every process occurring in nature proceeds in the sense in which the sum of the entropies of all bodies taking part in the process is increased. In the limit, i.e. for reversible processes, the sum of the entropies remains unchanged.

4) Principle of Carathéodory

  • In every neighborhood of any state S of an adiabatically enclosed system there are states inaccessible from S.

5) The new statement.

  • The gradients of the four thermodynamic forces spontaneously tend to zero.

Note

  • Statistical mechanics is restricted to the postulate of the equal a priori probability, but this postulate does not need to be considered in thermodynamics, so the valid ranges of the two theories are different. The valid range of H-theorem is less than the second law of thermodynamics. There are some computer simulations for H theorem, the changes in H are not monotonous.

Please see

http://arxiv.org/pdf/1201.4284v5.pdf

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  • $\begingroup$ You can edit your old (already accepted) answer. It is frowned upon to post multiple answers unless they are really very different. $\endgroup$ – ACuriousMind Jul 20 '14 at 2:10
  • $\begingroup$ @Freeman "the physical image of the entropy and both the second law are not clear, ..." This is not entirely true. The physical picture generally has to do with particle states (statistics, the Boltzmann equation and variations thereof) which readily connects to quantum mechanics and can be used to derive thermodynamics equations (for example, different equations of state). Your above derivation follows from macroscopic (classical) thermodynamics. As such, it is hard to gain physical insight at that level. $\endgroup$ – user3814483 Jul 20 '14 at 3:12
  • $\begingroup$ @user3814483, I don’t think so, there are some issues that can hardly be explained only by statistical mechanics, for example, by Euler’s equation, $Yx$ and $μN$ make no contribution to entropy, so the numbers of the degrees of freedom/states in $Yx$ and both $μN$ make no sense for calculating entropy, can you distinguish these two parts from the total energy of the system $E$ and the particles energy $ε$ for a general thermodynamic systems? If not, the physical picture of the entropy and the second law will be still unclear. $\endgroup$ – Freeman Jul 20 '14 at 4:41
  • $\begingroup$ This appears to be a plug for the linked paper, which has all the hallmarks of being a kook paper: it makes "gee whiz" claims; has references only to old papers rather than current work in the field; hasn't been published in a peer-reviewed journal; and is written in unintelligible English. $\endgroup$ – Ben Crowell Jul 26 '14 at 18:14
  • $\begingroup$ @Ben Crowell, I don't mind your comments, but “Mathematical proof of the Second Law of Thermodynamics” and “How do you prove the second law of thermodynamics from statistical mechanics?” are the two different questions, the first question includes thermodynamic approach, but the second question does not. I noticed your answer for the second question, If you think that, and I agree, “you can’t” prove the second law of thermodynamics from statistical mechanics, why don’t give it a try by thermodynamic approach? $\endgroup$ – Freeman Jul 27 '14 at 2:47
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I believe it was Boltzmann who first made the connection between entropy and micro states. chapter 12 of "Classical and Statistical Thermodynamics" by Ashley H. Carter discusses Boltzmann's arguments. To summarize from that book:

Entropy ($S$) corresponds to a particular configuration of an ensemble of particles called a macro state. A macro state can be achieved using a number of different micro states ($w$). Therefore, $S = f(w)$. Micro states represent the probability of being in the macro state (they just need to be normalized). If two systems are combined, the total entropy is just $S = S_A + S_B$, or $f(w) = f(w_A) + f(w_B)$. The likelihood of being in micro state $w$ is just $w_A \cdot w_B$, since independent probabilities are multiplicative. Therefore, $S(w_A w_B) = S(w_A) + S(w_B)$. Carter then says that the only function that satisfies this property is the natural logarithm, so $S\propto ln(w)$. The constant of proportionality is $k_B$. A lot more details in the text, and examples of micro states (discussion in the context of quantum as well, leading up to density of states).

The book has a good general treatment of statistical mechanics. It starts off with classical thermodynamics and then moves onto statistical mechanics and makes the connection to quantum mechanics.

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There is a new statement on the second law: "irreversibility root in a fundamental principle: the gradients of the four thermodynamic forces spontaneously tend to zero". Please see

http://arxiv.org/abs/1201.4284

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    $\begingroup$ Not sure why this answer was accepted. It doesn't formalize statistical mechanics or really discuss it at length in the context of the second law. $\endgroup$ – user3814483 Jul 19 '14 at 21:47
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I think that the second law of thermodynamics is a direct consequence of the symmetry of spacetime. Imagine you have two identical cubes of metal A and B with one common side, the first one warmer than the other. If you wait long enough, A and B will end up being at the same temperature. The same phenomenon would occur if B were warmer than A. Now imagine that you reverse time, so that from two cubes at the same temperature, you end up with one warmer than the other one. But why would nature choose arbitrarily between A and B? This would imply that the laws of nature depend on the direction of space that you consider, which would be contradictory with relativity.

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  • $\begingroup$ It's easy to make up similar examples that violate the second law but don't violate spatial symmetry. E.g., a sphere initially at uniform temperature could develop a radial temperature gradient. $\endgroup$ – Ben Crowell Jul 26 '14 at 18:18
  • $\begingroup$ What I mean here is that if the second law of thermodynamics were false, there would be one instance of violation of this law that would be contradictory with relativity. $\endgroup$ – Sylvain JULIEN Jul 26 '14 at 19:03
  • $\begingroup$ "the laws of nature depend on the direction of space that you consider, which would be contradictory with relativity." does not make sense. What is the direction of space? $\endgroup$ – Hans Oct 7 '18 at 19:47

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