Chapter 3, Stowe-1

8/13/2019 Chapter 3, Stowe-1

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Part 2 : Small Systems large systems & small systems

size of a system determined by the number of individual elements rather

than by the physical dimensions

Statistically speaking ex. Herd of elephants small systemConduction electrons in a pin large system

thermodynamic properties

why study small systems first?

many small systems are important understand better the need and the underlying justification for the statistical

large systems = extremely predictable

= statistical tools used are elegant and streamlined small systems = erratic and unpredictable

= statistical tools used are detailed and cumbersome

tools used for larger systems

develop an intuition for the basic causes of the thermodynamic behaviors oflarger systems and an understanding of why the statistical methods work


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CHAPTER 3 : STATISTICS FOR SMALL SYSTEMS

A. Mean Values individual elements have many possible distinct behaviors or configurations

averaging over possible individual behaviors is required

mean value of a functionf,

6

1654321 ====== PPPPPP

2

1

2

1== tailsheads PP ( ) ( )

2

10

2

11

2

1=+=+= tailstailsheadsheads fPfPf

=s

ss fPf

Ps = probability of the system being in state s

the sum is over all states s accessible to the system

Example : system = 1 coin f= number of heads showing =1 coin is heads up0 tails

Example : system = 1 rolled dice fn = n = # of dots showing upward

( ) ( ) ( ) ( ) ( ) ( ) ( )2

136

6

15

6

14

6

13

6

12

6

11

6

16

1

=+++++===n

n nPf

wherefs = value offwhen the system is in the state s

upshowdots2

1

3average,on the million2

13scoretotal =rolled 1 million dice


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Example : system = 1 rolled dice ( )21= nfn

( ) ( ) ( ) ( ) ( ) ( ) ( ) 6195614613612611610611222222

6

1

2 =+++++ ===n

n nPf

( ) fccfgfgf =+=+

iffand g are functions of the state of a system and c is a constant

B. Binomial distribution probabilities for systems of more than one element

calculate the probability for a system to be in each of its various possible

states knowing the probabilities for the behavior of a single element

p indicates the probability that the criterion for

the behavior of a single element is satisfied. q

indicates the probability that it is not.pqqp 11

Example :

1. Criterion :a given air molecule is in the front third of anempty room

2. Criterion : a flipped coin lands heads up

3. Criterion : a spin elementary particle in no external

fields has spin up

3

2

3

1

=qp2

12

1 =qp

2

1

2

1

=qp


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4. Criterion : a rolled dice lands with 2 dots up

5. Criterion :a swaggering drunk, whose next step is equallyprobable in all directions, takes his next step westwardly.

That is, his next step is somewhere in the westward half

of the area around him

6

56

1 =qp

21

21 =qp

if the system has 2 identical elements (labeled 1 and 2) 212121212211 111 qqpqqpppxqpqp

roomtheofthirdfronttheinNOTiseachthatyprobabilit

3

221 =qq

p1p2 = probability that both elements satisfy the criterion

p1 q2 = 1 does & 2 doesnt

q1p2 = 1 doesnt & 2 does

q1 q2 = both dont satisfy the criterion

Example : two air molecules (1 & 2) in an empty room

probability of being in the front third & the rear 2/3 of the room

roomtheofthirdfronttheinismoleculeeitherthatyprobabilit3

121 =pp


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BACKinBOTH9

4

3

2

3

2

BACKin1FRONT,in2

9

2

3

1

3

2

BACKin2FRONT,in19

2

3

2

3

1

FRONTinBOTH9

1

3

1

3

1

21

21

21

21

=

=

=

=

qq

pq

qp

pp

6

121 =pp

updot1withlandNOTdoBOTH

36

25

6

5

6

5

"1"""""236

5

6

1

6

5

tdoesn'2butupdot1w/lands136

5

6

5

6

1

updot1withlandBOTH36

1

6

1

6

1

21

21

21

21

=

=

=

=

qq

pq

qp

pp

Example : 2 rolled dice CRITERION : 1 dot showing UPWARD

6

521 =qq


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12 2222211 =qpqpqpqpqp

1133 332233332211 =qpqqppqpqpqpqp

2

1

2

1 =headsheads qp

p2 = probability that both elements satisfy the criterion

2pq =p1q2 + q1p2 = probability that 1 element satisfies the cri terion & 1 doesnt

q2

= both DONT satisfy the criterion

ifp1 = p2 =p AND q1 = q2 = q

for a system of 3 elements

p3 = ALL satisfy

3p2q = 2 satisfy & 1 doesnt = p1p2q3 + p1q2p3 + q1p2p3

3pq2 = 1 satisfies & 2 dont = p1q2q3 + q1p2q3 + q1q2p3

q3

= NONE satisfy

Example : system = 3 coins

8

3

2

1

2

133

2

2 =

=qpprobability of 2 heads

and 1 tail


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system ofNelements withn elements satisfying the criterion while (N-n) do

NOT satisfy it, probability is

nNnN qpnNn

NnP !!

! 12...111! nnnnn1!0

binomial coefficient, is the number of

different configurations of the individualelements, for which n satisfy the criterion

and (N-n) do NOT.

!!!

nNn

N

Example : 5 are molecules in an empty room

Probability of 2 being in the front third of the room and 3 in back = ?

3

1, =pfrontinbeingoneanyofyprobabilit

3

2,NOT =qfrontinbeingoneanyofyprobabilit

( N = 5 AND n=2 )

243

80

3

2

3

1

!3!2

!52

32

5 =

=PExample : What is the number of different possible arrangements of the 5 moleculesthat leave 2 in FRONT and 3 in BACK?

10!3!2 !5!! ! =nNn NABCDE BCADE CDABE DEABC

ACBDE BDACE CEABD

ADBCE BEACD

AEBCD


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Stirlings formula allows us to calculate the factorial of large numbers

mmmmm 2ln21ln!ln

3

23

111 =qp

C. Statistically independent behaviors

System ofNelements with n1 elements satisfy 1st criterion & n2 elements satisfy the

2nd criterion

statistically independent MEANS the elements behavior W.R.T. 1st criteria

does NOT affect its behavior W.R.T. the other

Example : Consider an air molecule in an empty room with the 2 criteria

1. Whether it is in the front third of the room

2. Whether it is in the top half of the room

2

12

122 =qp

Whether or not the molecule is in the front third of the room has no

bearing on whether it is in the top half. The 2 behaviors are statistically

independent.


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Probability that the system is in state i W.R.T the 1st criterion AND also in statej

W.R.T. the 2nd

Pij = Pi Pj Pijkl = Pi Pj PkPl

Example : single air molecule

halfBOTTOMinyprobabilit2

1

halfTOPinyprobabilit2

1

thirds-twoREARinyprobabilit3

2

thirdFRONTinyprobabilit3

1

2

2

1

1

q

p

q

p

halfBOTTOMthirds,-twoREAR6

2

2

1

3

2

halfTOPthirds,-twoREAR6

2

2

1

3

2

halfBOTTOMthird,FRONT6

1

2

1

3

1

halfTOPthird,FRONT6

1

2

1

3

1

21

21

21

21

=

=

=

=

qq

pq

qp

pp

12423121'555 !1!4 !5!3!2 !5424,2 qpqpPPPExample : five air molecules, p=? Of 2 in FRONT third and 4 in TOP half

051.0

32

5

243

80 =

=


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Iff is a function of the configuration of the system W.R.T. one criteria

i j i ij i ijjiji ij ifPifPPifPPifPf 1, gfjgPifPjgifPPjgifPfg

i j j

j

i

iji

ji

ij ,

Iff = f(i) W.R.T. 1st criterion & g = g(j) W.R.T. 2nd criterion

ASSIGNMENT (yellow pad, due next week)

3-2 (Stowe, 1984) The energy of a spin particle in an external magnetic field

along the z-axis is E = B if it is spin UP and E = + B if it is spin DOWN.

Suppose the probability of the particle being in the lower energy state is andthat of being in the higher energy state is . That is, Pup = , Pdown = 1/4., Eup= B and Edown = + B. What would be the average value of the energy of

such a particle, expressed in terms of B?3-5 (Stowe, 1984) Consider a system of four flipped coins

(a) What is the probability of two landing heads and the other two tails?(b) How many different configurations of the individual coins are possible that have

two heads and two tails?

(c) Label the four coins 1,2,3, and 4. Make a chart that lists the various possible

configurations of these that have two heads and two tails.

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Chapter 3, Stowe-1