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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
=
=
=
=
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
=
=
=
=
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
=
=
=
=
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.