# Can statistical mechanics explain the second law completely? [duplicate]

Statistical mechanics is restricted to the postulate of the equal a priori probability, but this postulate does not need to be considered for thermodynamics, so the valid ranges of statistical mechanics and the second law of thermodynamics are different.

For example, consider a freely falling body, the potential energy of the body is converted into the kinetic energy of random motion in the process, the driving force of the process does not satisfy the condition of the postulate of the equal a priori probability, it is easy to calculate the change of the entropy of the process by thermodynamics.

The question is: Can anyone calculate the change of the entropy of the process by statistical mechanics?

## marked as duplicate by Ben Crowell, ACuriousMind♦, Bernhard, Qmechanic♦Jul 26 '14 at 20:00

• Equilibrium statistical mechanics can tell you which one is incredibly more likely than the other in the situation where you have an ideal gas in a box and you ask yourself if it is possible that all the particles are occupying only half of the box. You easily find that starting from such a state and assuming equilibrium, it is like infinitely less likely than all the particles occupying equally the two halves of the box which is a consequence of the (additional) second principle in thermodynamics. – gatsu Jul 26 '14 at 15:03
• "the potential energy of the body is converted into the kinetic energy of random motion in the process" How could that happen? Freely falling body in gravity field increases its translational energy, but not its internal energy. Translational motion of body has no effect on entropy. – Ján Lalinský Jul 26 '14 at 15:07
• How is this different from other questions such as this one? physics.stackexchange.com/q/81465 – Ben Crowell Jul 26 '14 at 15:55
• A difference is that the answer provided in that alternative thread is 100% wrong, Ben. – Luboš Motl Jul 27 '14 at 5:33

Yes, even though some answers to nearly equivalent cousins of this question claim that the answer is No.

We really understand the right "theory of nearly everything" which means that using the microscopic theory – it's the "statistical mechanics" part that is most important here – we may calculate everything that happens in the world of everyday phenomena.

The general proof based on statistical mechanics that the entropy is increasing with time is known as Boltzmann's H-theorem

https://en.wikipedia.org/wiki/H-theorem

and there exist many versions and adjustments of this theorem and its proof – for example, in quantum field theory.

The fact that one may prove that entropy increases with time looks counterintuitive to many people because the underlying microscopic laws are time-reversal-symmetric (or at least, CPT-symmetric which makes the result pretty much equally counterintuitive). The flawed reasoning that suggests that time-reversal-breaking laws cannot emerge from time-symmetric microscopic laws was originally known as Loschmidt's paradox.

Like always in physics, there is no paradox. The reason why totally time-reversal-asymmetric – irreversible – phenomena emerge from a fundamentally T-symmetric theory is that the right laws of probability (in statistical mechanics) are inherently T-breaking even if the dynamical laws preserve T. The probabilistic reasoning breaks the symmetry, it respects the logical arrow of time (or psychological arrow of time, different terms are used).

The true mathematical origin of the asymmetry is that when we compute the probability that the ensemble of microstates A evolves into the ensemble B, we are summing over the microstates in the final state B (because the probability of "B1 or B2" is the sum of the probabilities for B1 and B2 separately) but we are averaging, perhaps with some weights (priors) over the initial microstates A1, A2. We need to average and not sum because the total probability of all initial microstates has to be kept fixed, at 100 percent, so they have to share the probability.

For this reason, we may rewrite the probability as the sum over all microstates in A as well as B as long as the result is divided by the number $N$ of microstates in A. But it is not divided by the number of microstates in B. This asymmetry encourages the evolution from lower-entropy initial states to higher-entropy final states. It is easy to see that the evolution of the T-reverted (or CPT-reverted) process is $\exp((S_B-S_A)/k)$ times less likely because the $\exp(S/k)$ factors arise from the $N$ I mentioned previously. If $S_B-S_A$ is macroscopic (of order one or higher in SI units), the process with the decreasing entropy is infinitely less likely than the increasing-entropy counterpart which means that it is impossible in the $k\to 0$ limit.

So yes, although the calculations are difficult and sometimes only doable in principle, we may not only prove that the entropy goes up but also to calculate how quickly the heat spreads from the warmer object to the cooler one, how quickly diffusion takes place in a specific dirt pumped into a specific liquid – how quickly any process involving the increase of entropy occurs. All these data are completely dictated by the microscopic theory. They have to be – otherwise the microscopic theory would be heavily incomplete, of course.

The reason why the entropy goes up has nothing whatever to do with cosmology or some assumptions about the Big Bang. The increasing entropy is a derivable result for any system, whether it's the whole Universe observed for billions of years or just its small part observed for picoseconds.

Your particular problem is easily solvable. The freely falling object behaves indistinguishably from an object outside a gravitational field. So if its internal degrees of freedom were in equilibrium at the beginning, they will stay in equilibrium and the entropy doesn't increase at all.

• Since the question is a duplicate, it might make more sense to move this and make it an answer to the earlier question. – Ben Crowell Jul 26 '14 at 18:24
• This all hangs on the claim that "the right laws of probability [...] are inherently T-breaking," because we average over the initial states but sum over the final states. But you then have to show that this asymmetry is more fundamental than the 2nd law of thermodynamics. You're begging the question -- if the 2nd law hasn't been established, we can't apply your probability rule, since we can't define "initial" and "final." Has this argument been published in more detail? I don't find it in this recent review: plato.stanford.edu/archives/fall2011/entries/time-thermo – Ben Crowell Jul 26 '14 at 18:34
• @BenCrowell: arXiv:0809.1304 cites Uffink, "Compendium of the foundations of statistical physics" (sections 7.6, 7.7) and Bacciagaluppi, "Probability and time symmetry in classical Markov processe", which both essentially conclude that there is no such T-breaking without further assumptions (which was also Boltzmann's position) – Christoph Jul 26 '14 at 19:02
• Boltzmann and I made very explicit what these assumptions are and they are nothing else than mathematical logic - they are certainly satisfied in every realistic and even un-realistic physics model ever considered that has many degrees of freedom. The claims that physics doesn't calculate something about the growth of entropy are just opinions of uninformed laymen. – Luboš Motl Jul 27 '14 at 5:33
• @LubošMotl: note that Botzmann's position on entropy evolved over the years in reaction to cricism by others like Poincaré and Zermelo; in Zu Hrn. Zermelo’s Abhandlung "Ueber die mechaniche Erklärung irreversibler Vorgänge", he mentions the idea that the universe started from an improbable low-entropy state and that there might be regions of the universe with opposite directions of the thermodynamic arrows of time (to which the subjective arrow of time would assumably align); Schrödinger (and probably others) argued that this latter case is impossible under certain (mild) assumptions – Christoph Jul 27 '14 at 8:03