Why is thermodynamic system always taken as macroscopic ? What will be the changes if one considers the same as microscopic ?


1 Answer 1


For "macroscopic" systems, the law of large numbers then prevails, thanks to the atomic composition of thermodynamic materials. At thermodynamic equilibrium, this means that the probability of finding a system in a state whose particles have a distribution of energies much different from the maximum likelihood one is vanishingly small. This also answers how things differ for microscopic systems, i.e. ones with a small number of particles.

To illustrate these points, let's think of the binomial distribution, which describes, say, the probability of drawing a sample of size $N$ containing $m\leq N$ red balls when the overall population is proportion $p$ red. Let's say that $p=0.43$. If we take a small sample, say 10, then mostly you'll draw 4 or 5 balls, but the likelihood of drawing 10 or 0 is not vanishingly small. The distribution is broad and the observed sample (system) is quite likely to be found a long way from its maximum likelihood sample.

Now draw a million balls. The probability of drawing 420 000 or fewer, or 440 000 or greater red balls out of that million works out to be of the order of $10^{-90}$. Just to make sure you understand this is not a typo, that's ten to the power of minus NINETY. The probability of getting a sample with a proportion of red balls significantly different from 0.43 is vanishingly small. This effect is even stronger when you begin dealing with systems comprising Avagadro sized numbers of particles. The probability of observing a system with energy distribution of particles significantly different from the maximum likelihood one of $p(E) \propto \exp\left(-\frac{E}{k\,T}\right)$ is miniscule.

So when you deal with microscopic systems, the maximum likelihood distribution of energies doesn't automatically prevail at high probability. Another way of saying this is that there is a non-negligible for a system's entropy to spontaneously decrease significantly. The various Fluctuation Theorems allow one to estimate the probability of a spontaneous entropy drop of a given size as a function of system size. Naturally, the size of likely spontaneous drops decreases swiftly with increasing system size.

See my answer to Why are the laws of thermodynamics “supreme among the laws of Nature”? for further discussion of these ideas.

  • $\begingroup$ Many thanks for this answer, I m bit struggling to understand the 2nd paragraph with the balls, i think is very crucial part of the answer, would you mind clarifiying what p is and how you came up with those numbers? Much appreciated in advance. BTW it would be fantastic to have your opinion on this recent post: physics.stackexchange.com/questions/219391/… if time allows $\endgroup$
    – user929304
    Nov 19, 2015 at 17:53
  • $\begingroup$ @user929304 This is using the binomial distribution $P(x,n,p) = \left(\begin{array}{c} n\\x\end{array}\right) p^x (1-p)^{(n-x)};x,\,n\in\mathbb{Z};\;p\in[0,1]$ which is the probability of x successes in $n$ independent trials, each with probability $p$. I have calculated the above using the normal (Gaussian) approximation to the binomial distribution for $n=10^6$ and $p=0.43$. The $0.43$ is a figure I plucked out of the air, being something roughly a half with two significant figures to show that the actual sample will almost certainly have the same proportion of successes as $p$ to within ... $\endgroup$ Nov 19, 2015 at 21:35
  • $\begingroup$ ... two decimal places, in contrast with the $n=10$ example where the number of successes varies wildly. By the time we reach $n=10^{23}$, the proportional variation is of the order of $10^{-12}$. $\endgroup$ Nov 19, 2015 at 21:37

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