lecture 9

# lecture 9 - The microscopic origins of irreversibility...

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Unformatted text preview: The microscopic origins of irreversibility and entropy in statistical mechanics For more on this issue, see the video of Feynman’s lecture “The Distinction Between Past and Future” from his “Character of Physical Law” series: http://research.microsoft.com/apps/tools/tuva/index.html Our world consists of molecules, which obey microscopic laws of motion that are well understood. Can we derive the laws of thermodynamics from those? The 1st law is nothing but the energy conservation, which we know from physics. What about the 2nd law, which determines the direction of the “time arrow”? Can we derive the direction of this “time arrow” starting from microscopic laws? All microscopic laws of nature that we know are reversible. For example, Newton’s laws are reversible. If, at any moment, we change the sign of the velocity of every molecule it will exactly follow its original trajectory backwards. If, for example, a gas is irreversibly expanding from V/2 to V (see previous Lecture) and, at the end of the experiment, we reverse the velocity of every molecule of gas (and also every molecule of the container, strictly speaking), we will see the gas going back into a half of the vessel, even though it would violate the 2nd law. Then why don’t we see this all the time? Is it because our microscopic laws are wrong? To understand what’s going on consider N molecules in a container of volume V. Each individual molecule obeys the simple laws of classical mechanics (actually, one should be using quantum mechanics but classical mechanics provides a reasonable approximation as long as the temperature is not too low). Because there are many other molecules inside the container and in its walls, with which that our molecule is colliding with, its trajectory will appear quite erratic. If we watch this molecule for a while we will find that it spends, on the average, the same time in the left and the right part of the container. We say it is equally likely for the molecule to be on the left or on the right. We can describe the microscopic state of the gas in the container as a string LRLL…R. The microscopic state (or microstate) is different from macroscopic states (or simply states) that we are concerned with in thermodynamics. Macroscopic state describes global properties (e.g. P, V, T) of a gas. Microscopic states involve detailed information about the molecules of the gas that is typically hard to obtain experimentally. For example if N=4, the string LRRL means that the molecule number 1 is on the left, molecule number 2 on the right, number 3 on the right and number 4 is on the left. All strings (LLRL, LLLR, LLLL etc.) are equally likely. The number of molecules in our “gas” is so small that we can easily enumerate all possible strings of our gas. In the Table below, for each such state, we tabulate the number of molecules on the left (i.e., the number of L’s in the string): “microscopic state” RRRR Number of molecules on the left 0 LLLL RLLL LRRR LRLL RLRR LLRL RRLR LLLR RRRL RRLL LLRR RLRL LRLR RLLR LRRL 4 3 1 3 1 3 1 3 1 2 2 2 2 2 2 We see that there is one microstate with all four molecules on the left, 4 microstates with 1 or 3 molecules on the left and 6 microstates with half of he molecule son the left. It is 6 times more likely to have 2 molecules on the left and 2 on the right than to have all molecules on the left. Now let N be large. There is a special notation for the number of strings, each of length N, that consist of n R’s and N ­n L’s. This number is denoted ⎛N⎞ ⎜ ⎟ ⎝n⎠ we just saw that ⎜ ⎟ = 6 . Generally, we have ⎜ It can be shown that when N is large, ⎜ ⎛ 4⎞ ⎝ 2⎠ ⎛N⎞ N! ⎟= ⎝ n ⎠ n !( N − n)! total number of strings of length N is also 2N. In other words, the overwhelming majority of ⎛N⎞ N ⎛N⎞ ⎟ ~ 2 . The ⎟ has a maximum for n = N/2 and ⎜ ⎝ N / 2⎠ ⎝n⎠ strings have approximately N/2 R’s and N/2 L’s in them. That is, the majority of configurations of the gas will have N/2 molecules on the right and N/2 on the left. Again, we emphasize that any given string is just as likely as LLL…L. It’s just there are many more different strings each containing N/2 R’s and N/2 L’s in them. Denote Ω the total number of microscopic states (i.e., strings) corresponding to the given state. For the gas in one half of the vessel Ω = 1 and for the gas evenly distributed over the container Ω ∼ 2Ν. Boltzmann postulated that S = kB ln Ω where kB = R/Na is called the Boltzmann constant. This formula is carved on Boltzmann’s tombstone (http://www.chemcollective.org/chem/MIT/images/boltzmann.jpg) and is really the foundation of statistical mechanics. Let us show that by using this definition we can recover our previous results for the entropy of an expanding ideal gas. Suppose that, as in a previous example, our gas expands from one half of the vessel of volume V (where it occupied the volume V(1) = V/2) into the entire vessel (V(2) = V). According to Boltzmann’s formula, we obtain ΔS = kB ln Ω(2) − kB ln Ω(1) = kB ln Ω(2)/Ω(1) = kB ln 2N = N kB ln 2. If N = Na then ΔS = R ln 2, exactly what we found in example 2. Processes disallowed by the 2nd law do not happen because they are very unlikely, which, in turn, is a consequence of the fact that macroscopic objects consists of large numbers of atoms (think of the gas in a vessel – if the gas consists of only 4 molecules then gathering of all of them in one half of the container is not all that unlikely). If we had enough time to wait, we would eventually see processes violating the 2nd law happen every once in a while. Less dramatic violations of the 2nd law happen all the time. They are called fluctuations. For example, the number of molecules in one half of the vessel may not be exactly equal to N/2. Thermodynamics becomes exact only in the limit of an infinite number of particles N. In single molecule experiments one sees fluctuations all the time. Note . There is an uncertainty in our definition of a microscopic state. We defined it as a string of letters RL…R… corresponding to the molecules being on the left or right. Instead we could divide the vessel into 4 parts and characterize the state of a molecules by which of the 4 parts of the vessel it is in. This would change the value of Ω and therefore S. It will not change, however, the value of the entropy difference, ΔS. In quantum mechanics there is no uncertainty in the definition of the state. Each system has discrete states and the number of such states Ω can be calculated. Therefore the absolute value of entropy is well defined. In particular, at T=0 the system is in a sate with the lowest energy called the ground state. Assuming there is only one such state, Ω=1, and using Boltzmann’s formula we find S(T=0)=0. This statement is called the 3rd law of thermodynamics or Nernst’s theorem ...
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