Combinatorics
Probability & Statistics
© The scientific sentence. 2010
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Probability & statistics
Macrostates and Microstates
Let's start with an example:
If we toss up two coins, the possible outcomes are:
coin 1 | coin 2 |
H | H |
H | T |
T | H |
T | T |
With: H = heads and T = tails.
The all four possible outcomes of the set
of two coins with 2 states H and T are the
microstates: H H, H T, T H, T T. But the two coins
are exactly the same; that is indistiguishable.
The two microstates 1 Head HT and TH constitutes one macrostate.
We have:
1 Head : 1 microstate:
2 Heads: 2 microstates
0 Heads: 1 microstate
The number of microstates corresponding to a given macrostate
is the multiplicity Ω.
It follows:
Ω(1) = 1, Ω(2) = 2, and Ω(0) = 1. We assume that the coins
are fair; then the all the 4 microstates are equally likely.
The probability for the four microstates are: P(HH) = P(HT) = P(TH) = P(TT) = 1/4.
The probability for the the three macrostates : P(2H) = P(0H) = 1/4, and
P(1H) = 2/4 = 1/2 ( the most probable). Generally, the probability of
n heads is equal to Ω(n)/Ω. Ω is the total number of microstates.
If we toss up 20 coins, the total number of microstates is
220 = 1,048,576 and the number of macrostates (0 H, 1 H, 2 H, ..., 20 H)
is (20 + 2 - 1)!/20! (2 - 1)! = 21!/20!1! = 21.
The number of ways to place N particles distinguishables within n places
is nN. If they are indistiguishable , the numer of ways, in this
case is: (N + n - 1)!/n!(n - 1)!
The related multiplicities for 1 Head and 2 Heads are:
Ω(0) = 1, Ω(1) = 20!/1!0! = 20
What is he multiplicity for n Heads: Ω(n)? It is just the number of
combinations of n among N. In other words, how many ways of getting n heads
among the existent N heads?. It is then equal to: C(n,N) = N!/n!(N - n)!
For n = 17, we have Ω(11) = 20!/17! 3! = 18 x 19 x 20/6 = 1140
The number of ways to arrange n objects among N, that is
the combination of n objects among N is C(n,N) = N!/n!(N - n)!
if the objects are indistinguishable; otherwise, if
they are distinguishable the number of ways is equal
to N!/(N - n)!, that is the number of arrangements.
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