thermal & kinetic lecture 14 permutations, combinations, and entropy
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Thermal & Kinetic Lecture 14
Permutations, Combinations, and Entropy
Overview
Distribution of energy at thermal equilibrium
Permutations and combinations
Interacting atoms and energy transfer
Last time…
Diffusion
The Einstein model of a solid
Oscillators and quanta
Permutations and combinations.
Energy distributions
Consider bringing two identical blocks together.
What is the most probable distribution of energy amongst the two blocks?
Most probable distribution is ‘intuitively’ that where total thermal energy is shared equally between the two blocks.
However, what is the probability that the first block has more energy than the second or, indeed, ends up with all the thermal energy?
Need to consider possible arrangements of energy quanta……
Energy distributions
We have 3N independent simple harmonic oscillators (where N is the total number of atoms in the solid).
Number of ways of distributing quanta of energy amongst these oscillators?
Say we have 3 quanta of energy to distribute amongst 2 oscillators:
The arrangement above is one possible distribution of 3 quanta amongst two oscillators. Sketch the remaining possibilities. How many possibilities in total are there???
Oscillator 1 Oscillator 2
03
00
3 energy quanta distributed between 2 oscillators
Oscillator 1 Oscillator 2
03
00
Oscillator 1 Oscillator 2
02 01
Oscillator 1 Oscillator 2
01 02
ANS: Total of 4 possibilities (including arrangement shown on previous slide)
Counting arrangements
Clearly, we are not going to count by hand every arrangement of energy possible for 3N oscillators in, e.g., a mole of solid (N ~ 6 x 1023).
Need to consider permutations and combinations.
A permutation is an arrangement of a collection of objects where the ordering of the arrangement is important.
A CD reviewer is asked to choose her top 3 CDs from a list of 10 CDs and rank them in order of preference. How many different lists can be formed? (ABC ≠ BAC)??ANS: There are 10 choices for the 1st CD, 9 for the 2nd,
and 8 for the 3rd. Hence, 720 different lists.
The number of permutations of r objects selected from a set of n distinct objects is denoted by nPr where nPr = n! / (n - r)!
A CD club member is asked to pick 3 CDs from a list of 10 CDs. How many different choices are possible???
Counting arrangements
ANS: We need to divide the previous 720 arrangements by the total number of different possible permutations of 3 choices (e.g. ABC = BAC = CAB). This is 3! permutations. Hence, 120 choices are possible.
The number of combinations of r objects selected from a set of n distinct objects is denoted by nCr where
)!(!
!
rnr
nCrn
Counting arrangements
Take a collection of 10 pool (billiard) balls, 6 of which are yellow and 4 of which are red. How many different arrangements of the coloured balls are possible (eg RRYYYYYYRR)?
??ANS: = 210 arrangements
!4!6
!10
Counting arrangements
Returning to the distribution of energy quanta amongst a collection of oscillators, we need to establish a formula for the number of possible arrangements.
Consider the case of 3 quanta of energy distributed between 2 oscillators as before. We’ll adopt the same representation as Chabay and Sherwood, p. 348…….
=1 2 1 2
Thus, we have 4 objects arranged in a certain sequence. We need N - 1 vertical bars to separate N oscillators.
=1 2 1 2
Counting arrangements
=1 2 1 2
This problem thus reduces to the pool ball problem except instead of red and yellow pool balls we have to arrange | and objects.
So, total number of arrangements of 3 quanta amongst 2 oscillators =
!1!3
!4Total number
of objects
Number of quanta
Number of boundaries between oscillators (= N-1)
Counting arrangements
Number of ways to arrange q quanta of energy amongst N 1D oscillators:
)!1(!
)!1(
Nq
Nq
How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? ??
How many ways can four quanta of energy be arranged amongst four oscillators?
21 42 256 35
0%
100%
0%0%
a) 21
b) 42
c) 256
d) 35
How many ways can 4 quanta of energy be arranged amongst four 1D oscillators? ??
ANS: = 35 ways!3!4
!7
Counting arrangements
Microstates and macrostates
Each of the 35 different distributions of energy is a microstate (i.e. an individual microscopic configuration of the system, as shown above).
The 35 different microstates all correspond to the same macrostate - in this case the macrostate is that the total energy of the system is 4 0
FUNDAMENTAL ASSUMPTION OF STATISTICAL MECHANICSEach microstate corresponding to a given macrostate is equally probable.
Microstates and macrostates: poker hands
What’s the probability of being dealt the hand of cards shown above in the order shown? ??
What’s the probability of being dealt the hand of cards in the order shown?
1/2
112
1/8
192
1/2
,256
,781
1/3
11,8
75,2
00
4%
84%
6%6%
a) 1/2112
b) 1/8192
c) 1/2,256,781
d) 1/311,875,200
Microstates and macrostates: poker hands
What’s the probability of being dealt the hand of cards shown above in the order shown? ??
As compared to the “junk” hand of cards, the probability of being dealt the Royal Flush is:
Hig
her
Lower
Exa
ctly
the
sam
e
Don’t
know
0% 2%
90%
8%
a) Higher
b) Lower
c) Exactly the same
d) Don’t know
What’s the probability of being dealt the hand of cards shown above in the order shown? ??ANS: This is 1 possibility out of a total of 52!/47! possibilities.
What’s the relevance of this to thermal equilibrium and entropy?! ??You’ll have to bear with me again for the answer......
Microstates and macrostates: poker hands
Two interacting atoms: six 1D oscillators
Remember, we’re still trying to find out why the total thermal energy is shared equally between the two blocks.
Let’s consider the smallest possible blocks – two interacting atoms. As each atom comprises three 1D oscillators this means we have six 1D oscillators in total.
How many ways are there of distributing 4 quanta of energy amongst six 1D oscillators???
Are many ways are there of distributing four quanta of energy amongst six 1D oscillators?
126
1260
1,02
4,25
6
None
of the
se
93%
4%0%2%
a) 126
b) 1260
c) 1,024,256
d) None of these
If four quanta of energy are given to one atom or the other, how many ways are there of distributing the energy quanta?
15 30 45 60
55%
18%
9%
18%
a) 15
b) 30
c) 45
d) 60
If all four quanta of energy are given to one atom or the other, how many ways are there of distributing the energy???
ANS: 4 quanta distributed amongst 3 oscillators = 15 ways x 2 = 30 ways
Two interacting atoms: six 1D oscillators
If three quanta are given to one atom, and one quantum to the other, how many ways are there of distributing the energy?
10 30 60 120
0% 0%0%0%
a) 10
b) 30
c) 60
d) 120
Two interacting atoms: six 1D oscillators
If three quanta are given to one atom, and one quantum to the other how many ways are there of distributing the energy???
ANS: 60 ways
3 quanta distributed amongst 3 oscillators on atom 1 : 10 ways.1 quantum distributed amongst 3 oscillators on atom 2: 3 ways.
However, could also have three quanta on atom 2, one quantum on atom 1
If the four quanta are shared equally between the atoms, how many ways are there of distributing the energy???
ANS: 36 ways. (i.e. 126 – 30 – 60, but make sure you can get the same result by counting the states as above.)
Two interacting atoms: six 1D oscillators
0 1 2 3 40
10
20
30
40
Nu
mb
er o
f d
istr
ibu
tio
ns
Number of quanta in atom 1
For two interacting atoms it is most probable that the thermal energy is shared equally.
We can look at this result in two ways: (i) if frequent observations of the two atom system are made, in 29% of
the observations - i.e. 36 out of 126 – the energy will be split evenly;
(ii) for 100 identical two atom systems, at any given instant 29% will have the thermal energy split evenly between the two atoms.
Increasing the number of atoms……
For two atoms, 29% of the time the system will adopt a state where the energy is shared equally.
So, although this is the most probable distribution, it happens < ⅓ of the time. It is almost as likely (24%) to find all the energy on one atom or the other.
What happens as we add more atoms?
How many ways are there of distributing 10 quanta of energy amongst 10 atoms??? ANS: 6.35 x 108
How many ways are there of arranging the system so that the 10 quanta of energy are on one specific oscillator??? ANS: 1
Increasing the number of atoms……
No. of arrangements increases VERY quickly for small changes in numbers of oscillators.
For 300 oscillators (100 atoms) there are ~ 1.7 x 1096 ways of distributing 100 quanta of energy.
1 mole of any material contains ~ 6 x 1023 atoms.
Is it possible that all the energy could be concentrated on 1 atom?
Are we ever likely to see this happen?
Yes!
No!
The aims and objectives were made clear
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Handing out printed lecture notes is a good idea
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Attempting questions during the lectures using the keypads aided my learning
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The printed lecture notes are easy to follow
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