PHY306 Homework 1
Due February 3, 2004
This exercise illustrates the idea of different states that a system can take and gives a feel for the statistics involved. Consider 3 coins being flipped - the 8 possible outcomes are:
Coin 1 |
Coin 2 |
Coin 3 |
H |
H |
H |
H |
H |
T |
H |
T |
H |
T |
H |
H |
H |
T |
T |
T |
H |
T |
T |
T |
H |
T |
T |
T |
Each separate outcome is a microstate - a microstate specifies the state of each individual coin (or particle, in a general physical system). If we specify only how MANY heads or tails there are, we specify the macrostate. In this example we have 4 possible macrostates (3 heads, 2 heads, 1 head and no heads). If we know the microstate, we know the macrostate, but not the other way around.
The number of different microstates leading to the same macrostate is called the multiplicity, W. For our 4 macrostates, the multiplicities are (1, 3, 3, 1) - can you see why that is? The probability of finding any given macrostate is the ratio of the number of its microstates to the total number of possible microstates.
Probability of n heads = W (n)/ W (all)
For example, the probability of getting 2 heads is W (n)/ W (all) = 3/8.
Things get more tricky if you have many coins, like 100, for example. Many microstates are possible: 2^{100 }since each coin has two possible states (H or T). The number of macrostates, however, is only 101: 0 heads, 1 head, … up to 100 heads. Now, we investigate how many microstates each macrostate has.
The denominator is just n! We can write the numerator as (100!)/(100-n)!, and then the general formula becomes W(n) = 100! / (n! * (100-n)!)
W(N,n) = N! / (n! * (N-n)!)
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Problem 1:
Let's flip 4 coins.
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Problem 2:
Now you flip 20 coins
HTHHTTTHTHHHTHHHHTHT (in exactly that order)?
c) what is the probability of getting 12 heads and 8 tails in any order?