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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

Thermal & Kinetic Lecture 14 Permutations, Combinations, and Entropy Overview Distribution of energy at thermal equilibrium Permutations and combinations

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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!

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THANK YOU!