2.4 Two-fluid equations 412.4.1 Entropy of a distribution function
Collisions cause the distribution function to tend towards a simple final state characterized
by having the maximum entropy for the given constraints (e.g., fixed total energy). To
see this, we provide a brief discussion of entropy and show how it relates toa distribution
function.
Suppose we throw two dice, labeledAandB, and letRdenote the result of a throw.
ThusRranges from 2 through 12. The complete set of(A,B)combinations that give these
R’s are listed below:
R= 2 ⇐⇒(1,1)
R= 3 ⇐⇒(1,2),(2,1)
R= 4 ⇐⇒(1,3),(3,1),(2,2)
R= 5 ⇐⇒(1,4),(4,1),(2,3),(3,2)
R= 6 ⇐⇒(1,5),(5,1),(2,4),(4,2),(3,3)
R= 7 ⇐⇒(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)
R= 8 ⇐⇒(2,6),(6,2),(3,5),(5,3),(4,4)
R= 9 ⇐⇒(3,6),(6,3),(4,5),(5,4)
R= 10 ⇐⇒(4,6),(6,4),(5,5)
R= 11 ⇐⇒(5,6),(6,5)
R= 12 ⇐⇒(6,6)
There are six(A,B)pairs that giveR=7, but only one pair forR=2 and only one pair
forR=12. Thus, there are six microscopic states [distinct(A,B)pairs] corresponding to
R=7 but only one microscopic state corresponding to each ofR=2 orR=12. Thus,
weknow moreabout the microscopic state of the system ifR=2 or 12 than ifR=7.
We define the entropySto be thenatural logarithm of the number of microscopic states
corresponding to a given macroscopic state.Thus for the dice, the entropy would be the
natural logarithm of the number of(A,B)pairs that correspond to a giveR.The entropy
forR= 2orR= 12would be zero sinceS= ln(1) = 0, while the entropy forR= 7
would beS= ln(6)since there were six different ways of obtainingR= 7.
If the dice were to be thrown a statistically large number of times themost likely result
for any throw isR= 7;this is the macroscopic state with the most number of microscopic
states. Since any of the possible microscopic states is an equally likely outcome, the most
likely macroscopic state after a large number of dice throws is the macroscopic state with
the highest entropy.
Now consider a situation more closely related to the concept of a distribution function.
In order to do this we first pose the following simple problem: Suppose we have a pegboard
withNholes, labeledh 1 ,h 2 ,...hN and we also haveNpegs labeled byp 1 ,p 2 ,...,pN.
What are the number of ways of putting all the pegs in all the holes? Starting with hole
h 1 ,we have a choice ofNdifferent pegs, but when we get to holeh 2 there are now only
N− 1 pegs remaining so that there are now onlyN− 1 choices. Using this argument for