Are there any analytical proofs for the 2nd law of thermodynamics?

Or is it based entirely on empirical evidence?

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    $\begingroup$ In the presence of gravity there is no proof (now and never will). The 2nd law was stated outside of gravity environment. $\endgroup$ – Helder Velez Jun 2 '11 at 11:17
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    $\begingroup$ never proved, always loved, it is not a fundamental physical law, is actually a computer science law. $\endgroup$ – Wolphram jonny Jul 13 '16 at 23:31
  • $\begingroup$ related to (and possibly duplicated by): physics.stackexchange.com/questions/81465/… physics.stackexchange.com/questions/20401/… $\endgroup$ – user4552 Oct 16 '17 at 21:49
  • $\begingroup$ The 2nd law reconciles if it is made tighter and asserts "entropy of a closed system can never change over time" - in other words, entropy is only an interesting property of open systems, and is basically a measure of "closed-ness". $\endgroup$ – Steve Jan 25 '18 at 21:36
  • $\begingroup$ Yes, there is a classical proof. See my comment below. $\endgroup$ – ratsalad Feb 23 '19 at 20:04

It's simple to "roughly prove" the second law in the context of statistical physics. The evolution $A\to B$ of macrostate $A$, containing $\exp(S_A)$ microstates, to macrostate $B$, containing $\exp(S_B)$ microstates, is easily shown by the formula for the probability "summing over final outcomes, averaging over initial states", to be $\exp(S_B-S_A)$ higher than the probability of the inverse process (with velocities reversed). Because $S_B-S_A$ is supposed to be macroscopic, such as $10^{26}$ for a kilogram of matter, the probability in the wrong direction is the exponential of minus this large difference and is zero for all practical purposes.

The more rigorous versions of this proof are always variations of the 1872 proof of the so-called H-theorem by Ludwig Boltzmann:


This proof may be adjusted to particular or general physical systems, both classical ones and quantum ones. Please ignore the invasive comments on the Wikipedia about Loschmidt's paradoxes and similar stuff which is based on a misunderstanding. The H-theorem is a proof that the thermodynamic arrow of time - the direction of time in which the entropy increases - is inevitably aligned with the logical arrow of time - the direction in which one is allowed to make assumptions (the past) in order to evolve or predict other phenomena (in the future).

Every Universe of our type has to have a globally well-defined logical arrow of time: it has to know that the future is being directly evolving (although probabilistically, but with objectively calculable probabilities) from the past. So any universe has to distinguish the future and the past logically, it has to have a logical arrow of time, which is also imprinted to our asymmetric reasoning about the past and the future. Given these qualitative assumptions that are totally vital for the usage of logic in any setup that works with a time coordinate, the H-theorem shows that a particular quantity can't be decreasing, at least not by macroscopic amounts, for a closed system.

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    $\begingroup$ Isn't there an underlying assumption that the microstate probability density doesn't vary or do any fishy business for the derivations in statistical thermodynamics to work. I mean, as you write it is easy to visualize entropy when looking at microstates in configuration space and macrostates, but that moves the "problem" one level deeper: crudely put, the reason it works is because particles are "dumb". Am I opening another box of worms here? :) $\endgroup$ – BjornW Jun 2 '11 at 9:10
  • $\begingroup$ Dear Bjorn, it's an OK can of worms. In the particular proof Boltzmann chose, he assumed molecular chaos, a particular assumption about the lack of correlations in the initial state. There can be correlations but it's still true that it's exponentially unlikely that they will modify the qualitative answer - they would just make the quantitative calculation more ambiguous. Moreover, there may also be an assumption that particles are dumb but this assumption can be proved, cannot it? ;-) Things that are clever, like eggs and fridges, may lower their entropy but they increase $S_{environment}$. $\endgroup$ – Luboš Motl Jun 2 '11 at 12:36
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    $\begingroup$ Yes it can be and is demonstrated but even if it's trivial, it's still an underlying assumption that the reductionalist in me is left with after consuming the other derivations :) After all, it is the whole business with Maxwell's demons in another guise, right.. but I guess that a "clever" particle (for example a particle which starts with but also maintains and changes correlations) would undermine and destroy the derivation in other ways anyway! :) $\endgroup$ – BjornW Jun 2 '11 at 22:31
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    $\begingroup$ @Lubos: Lorschmidt's paradox is not as silly as you make it sound--- if you actually reverse the spins of a relaxing nuclear spin system, it will decrease in apparent entropy with time. This is a physical realization of Lorschmidt's paradox. The absence of entropy decreasing correlations in the motion of molecules is really a separate assumption, which is equivalent to the statement that the entropy is low only in the past, and the state is generic in the future. $\endgroup$ – Ron Maimon Nov 15 '11 at 20:17
  • $\begingroup$ @LubošMotl "Every Universe of our type has to have a globally well-defined logical arrow of time". Is that something you can derive from ST? If so then, yes, I can understand your reasoning - you're saying that the arrow must be one way or the other and experimentally we know which way it is. But does that invalidate the Loschmidt argument? I mean, it certainly tells us why the LP is not a problem, but it seems to me that this is done by bringing other info (your statement) and the experimental fact to the table. LP then says, "we can't decide without further info" and you say "here it is!". $\endgroup$ – Selene Routley Oct 22 '13 at 10:17

It was first found empirically, and later derived from various more theoretical assumptions.

There is a proof in Section 7.2 of Chapter 7: Phenomenological Thermodynamics of Classical and Quantum Mechanics via Lie algebras, based on a few axioms for thermodynamics, and a proof in Chapter 9 that these laws follow from the standard assumptions in statistical mechanics.

The reversibility objections (Loschmidt's paradox) are unjustified since the Poincare recurrence theorem assumes that the system in question is bounded, which is (most likely) not the case for the real universe.

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    $\begingroup$ note this is a similar "proof" statistically (much like Boltzmann's H-theorem) and not a full proof (of course whether this is relevant or not, we will have to see) $\endgroup$ – Nikos M. Jul 3 '14 at 23:35

If we assume time evolution is unitary and hence reversible, and the total size of the phase space subject to constraints based upon the total energy and other conserved quantities is finite, then the only conclusion is Poincaré recurrences cycling ergodically through the entire phase space. Boltzmann fluctuations to states of lower entropy might occur with exponentially suppressed probabilities, but the entropy would increase both toward its past and future. This is so not the second law as Boltzmann's critics never tire of pointing out.

The H-theorem depends upon the stosszahlansatz assumption that separate events in the past are uncorrelated, but that is statistically exceedingly improbable assuming a uniform probability distribution.

If the total size of the phase space is infinite, Carroll and Chen proposed that in eternal inflation there can be some state with finite entropy with entropy increasing in both time directions.

To me, the most likely scenario is to drop the assumption of unitarity and replace that with time evolution using Kraus operators acting upon the density matrix.

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The problem when you include gravity or other long range forces, is that thermodynamics becomes non extensive. For instance, the energy of the union of two systems is not the sum of the energies of the individual systems.

To handle those cases, generalized entropies have been proposed. By generalized it means that these formalisms allow for long range forces and non-extensivity, for certain parameters of the definition of entropy, but reduces to the classical extensive entropy for certain value of the parameter. One of such extended entropies is Tsallis entropy. It depends on a parameter $q$, and for $q=1$ it reduces to the standard classical entropy.

It has been shown that this entropy works well in some gravitational systems, where it predicts the correct distribution of temperatures and densities, for instance in a polytropic model of a self-gravitating system. It has also been shown that this entropy satisfies the second law for any parameters $q$ in the classical case, and at least for $q\in(0,2]$ in the quantum case.

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In the strict sense of the question: no. Physics is science based on empirical evidence. But this applies to all laws of physics. E.g. if by tomorrow you find and confirm experimental evidence which contradict current theories, you have to expand the theories (or invent new ones), and you gain insight in the domain of applicability of your old theory (which still stays valid in its domain).

Of course you might be able to derive/prove the second law from certain assumptions, but if you were to find an experiment where the second law doesn't hold, then you start to know the limitations of your assumptions.

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There is actually a very simple derivation of the Second Law in classical thermodynamics for an ideal gas, assuming only classical mechanics and the First Law. Here is a brief sketch -- whether this constitutes a "proof" depends largely on taste, the level of rigor desired, and how comfortable you are with thermo-style derivations.

The First Law of Thermodynamics is:

\begin{align} dU = dq + dw \end{align}

where the differentials refer to changes of the system. By convention we have defined a gain of energy or heat by the system as positive, work done on the system as positive, and work done by the system on the surroundings as negative.

Without loss of generality, we consider pressure-volume work. The work done by the system is quantified by the amount of work done in the surroundings, and so the relevant pressure is the external pressure $P_{ext}$ in the surroundings that the system is pushing against. Then, the work done by the system is

\begin{align} dw = -P_{ext} dV \end{align}

If the internal pressure of the system is greater than the external pressure of the surroundings,

\begin{align} P_{int} \ge P_{ext} \end{align}

then according to classical mechanics the system will expand against the surroundings, i.e. $dV \ge 0$.

For a reversible change, the internal and external pressures are equal ($P_{int} = P_{ext}$), and so the work done by the system in a reversible process is

\begin{align} dw_{rev} = -P_{int} dV \end{align}


\begin{align} P_{int} dV &\ge P_{ext} dV \\ -P_{int} dV &\le -P_{ext} dV \\ dw_{rev} &\le dw \end{align}

which means that the magnitude of work done by the system on the surroundings is maximal during a reversible process. Combining this result with the First Law gives:

\begin{align} dq_{rev} &\ge dq \end{align}

We now define the state function entropy $S$ classically as

\begin{align} dS = \frac{dq_{rev}}{T} \end{align}

From the previous inequality for reversible heat, we see that

\begin{align} dS = \frac{dq_{rev}}{T} \ge \frac{dq}{T} \end{align}

which is the generalized Clausius inequality. This is a complete mathematical statement of the Second Law of Thermodynamics. All consequences of the Second Law can be derived from it, including the proposition that heat always spontaneously flows from hot to cold.

The one missing part is that we did not establish that entropy $S$ is a state function for an ideal gas, but this can be found in any introductory thermodynamics treatment (e.g. [1]).

[1] https://en.wikiversity.org/wiki/Physics_equations/Introduction_to_entropy

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  • $\begingroup$ This is wrong. You can't derive a time-asymmetric law from time-symmetric microscopic laws, unless you introduce some other element that breaks the time-reversal symmetry. $\endgroup$ – user4552 Jan 13 at 17:47
  • $\begingroup$ @BenCrowell If it is wrong, then please show where. $\endgroup$ – ratsalad Jan 14 at 14:50
  • $\begingroup$ @BenCrowell You downvoted my answer and claim it is wrong. I'd appreciate it if you would support your assertion and explain where it is wrong. Note, it is of course possible that my answer is both correct and implicitly breaks time-reversal symmetry. $\endgroup$ – ratsalad Feb 12 at 23:09

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