Transcript Two topics will be discussed

INTRODUCTION
The aim of computational structural biology is to
understand (and predict) the structure and function of
biological macromolecules, such as proteins and nucleic acids
based on their microscopic (atomic) interactions.
These are thermodynamic systems, which are also
affected by environmental factors such as temperature,
pressure, and solvent conditions.
Classical thermodynamics is a general methodology that
enables one to derive relations among macroscopic
properties such as volume (V), pressure (P), temperature
(T), energy (E) and entropy (S) without the need to consider
the atomic properties of the system, e.g., the equation of state,
1
PV=NRT.
In statistical mechanics, on the other hand, the macroscopic
thermodynamic behavior is obtained from the specific
microscopic description of a system (atomic interactions,
atomic masses, etc.). Therefore, statistical mechanics also
provides information that is beyond the reach of classical
thermodynamics - what is the native structure of a protein ?
It is impossible to know the exact dynamic behavior of a large
system (i.e., the atomic coordinates at time t). Appearance of a
3D configuration is only known with a certain probability.
Thus, statistical mechanics is a probabilistic theory, and
computational methods, such as Monte Carlo and molecular
dynamics are based on these probabilistic properties. Part of
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the course will be devoted to basic probability theory.
This is a short course. Therefore, the theory of statistical
mechanics will not be derived rigorously. The emphasis will
be on solving problems in statistical mechanics within the
framework of the canonical ensemble, treating polymer
systems and proteins. The probabilistic nature of the theory
will be emphasized as reflected in computer simulation
techniques.
3
Mechanical Systems are Deterministic
Refreshing some basic physics
Examples of forces F (F=|F|):
1) Stretching a spring by a distance x: F=-kx, Hook’s Law
k- spring constant.
2) Gravitation force: F= kMm/r2 - m and M masses with
distance r; k - constant. On earth (R,M large), g=kM/R2
F=mg
3) Coulomb law: F=kq1q2/r2 q1,q2 charges.
4
Newton’s second law:
F=ma - a acceleration
Mechanical work W: if a constant force is applied along
distance d, W=Fd (F=|F|). More general, W=! F. dx.
.
Potential energy: If mass m is raised to height, h
negative work is done, W = –mgh and the mass gains
potential energy,Ep= -W = +mgh - the ability to do
mechanical work: when m falls dawn, Ep is converted into
kinetic energy, Ek = mv2/2, where v2/2=gh (at floor).
A spring stretched by d: Ep= -W = k! xdx = kd2/2
In a closed system the total energy, Et = Ep+ Ek is constant but
Ep/Ek can change; e.g., oscillation of a mass hung on a spring
5
and distorted from its equilibrium position.
The dynamics of a mechanical macroscopic system in
principle is deterministic in the sense that if the forces are
known, and the positions and velocities of the masses at time
t=0 are known as well, their values at time t can in principle be
determined by solving Newton’s equations.
Simple examples: harmonic oscillator (a spring), a trajectory
of a projectile, movement of spaceship, etc. In some cases the
solution is difficult and requires strong computers.
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Stability - a system of interacting masses (by some forces)
tends to arrange itself in the lowest potential energy structure
(which might be degenerate) also called the ground state. The
system will stay in the ground state if the kinetic energy is
very small - this situation defines maximum order.
The larger the kinetic energy the larger is the disorder - in the
sense that at each moment a different arrangement of the
masses will occur (no ground state any more). Still, in
principle, the trajectories of the masses can be calculated.
Two argon atoms at rest
positioned at the lowest
energy distance e interacting through
Lennard-Jones potential.
Microscopic system.
  
 
( r)  4e   -   
r 
 r 
12
6
e
7
Thermodynamic systems and Statistical
Mechanics
A typical system is described below; examined from a
microscopic point of view – non-rigorous treatment
R
TR
TC
C
A system C of N molecules in a constant volume V is in
thermal contact with a large reservoir (R) (also called heat
bath) with a well defined temperature TR. At equilibrium (after
a long time) energy is still exchanged between R and C but the
average kinetic (and potential) energy of a molecule of C is
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constant, leading to Tc=TR.
However, in contrast to a macroscopic mechanical system,
there is no way to know the trajectories of the particles that
are changed constantly due to the energy exchange between
C&R and quantum mechanics limitations.
Relating kinetic energy to temperature, at low T, Ek is low,
the effect of the molecular forces significant - the system
arrange itself in a low potential energy state – relatively high
order. At high T, Ek is high and dominant, Ep is high – high
disorder that includes the uncertainty related to trajectories.
Therefore, a thermodynamic system at equilibrium cannot be
characterized by the positions & velocities of its 1023 particles
but only by the average values of several macroscopic
parameters such as P, T, E (internal energy) and entropy, S.9
For example, in measuring T the thermometer feels the
average effect of many molecular configurations of the tested
system and for long measurement all the microscopic states
are realized and affect the result of T.
Hence to obtain the average values of macroscopic parameters
from microscopic considerations a probability density
P(xN,vN) should be assigned to each system state (xN,vN) where
(xN,vN) = (x1,y1,z1,x2,y2,z2, ….xN,yN,zN,vx1,vy1vz1…. vxN,vyNvzN)
thus assuming that all states contribute.
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Then a macroscopic parameter M is a statistical average,
<M> = ! P (xN,vN) M(xN,vN) d(xNvN).
The entropy for a discrete and continuous system is (kB is the
Boltzmann constant),
S = <S> = -kB S Pi ln Pi and
S= -kB! P(xNvN) lnP(xN,vN) d(xNvN)+
+ ln{const.[dimension (xNvN)]}
Notice, P is a probability density with dimension 1/ (xNvN) .
The constant is added to make S independent of (xNvN). 11
The problem – how to determine P. In thermodynamics an
N,V,T system is described by the Helmholtz free energy A,
A(T,V,N)=E –TS,
which from the second law of thermodynamics should be
minimum for a given set of constraints.
We shall determine P by minimizing the statistical free energy
with respect to P.
12
A can be expressed as the average
A = <A(P)>= ! P(X)[E(X)+kBTlnP(X)]dX +const.
X=(xN,vN). We derive A with respect to P and
equate to zero; the const. is omitted.
A’=! [E(X)+kBTlnP(X) + kBT P(X) /P(X)]dX=0
E(X)+kBT[lnP(X) + 1]=0
lnP(X) = - [E(X) + kBT]/kBT= = - [E(X)/kBT] +1
P(X) =const.exp[-E(X)/kBT]
The normalization is Q=! exp[-E(X)/kBT)]
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i.e.,
PB(X)=exp[-E(X)/kBT]/Q
PB – the Boltzmann probability (density).
Q – the canonical partition function.
The intgrand defining A is E(X)+kBTlnPB(X)
Substituting PB and taking the ln gives,
E(X)+kBT[- E(X)/kBT –lnQ]= -kBTlnQ
-kBTlnQ is constant for any X and can be taken out of the
integral of A. Thus,
A=E-TS= - kBTlnQ
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The relation A= - kBTlnQ is very important. It enables
calculating A by Q that is based on the details of the system.
In classical thermodynamics all the quantities are obtained as
derivatives of A. Hence, with statistical mechanics all these
quantities can be obtained by taking derivatives of lnQ.
Also, the probabilistic character of statistical mechanics
enables calculating averages even without calculating Q.
These averages can be even geometrical properties that are
beyond the reach of thermodynamics, such as the end-to-end
distance of a polymer, the radius of gyration of a protein and
many other geometrical or dynamic properties. This is in
particular useful in computer simulation.
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(1) Probability and Statistics, M.R. Spiegel, Schaum’s Outline Series, McGRAW-Hill
ISBN 0-07-060220-4.
(2) An Introduction to Statistical Thermodynamics, T.L. Hill, Dover,
ISBN 0-486-652- 42-4. (Also Eddison-Wesley).
(3) Statistical Mechanics, R. Kubo. North Holland (ISBN 0-7204-0090-2) and
Elsevier (0-444-10637-5).
(4) Statistical Mechanics of Chain Molecules. P. J. Flory. Hanser (ISBN 3-44615205-9) and Oxford (ISBN 0-19-520756-4).
(5) Phase Transitions and Critical Phenomena. H.E Stanley (Oxford).
6) Introduction to Modern Statistical Mechanics. David Chandler (Oxford).
ISBN 0-19-504277-8.
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Lecture 2: Calculation of the partition function Q
TR
Systems at equilibrium
N particles with velocities vN and coordinates xN are moving in
a container in contact with a reservoir of temperature T.
We have seen last time that the Helmholtz free energy, A is
A=E-TS= - kBTlnQ
where
Q=! exp[-E(xN,vN)/kBT) d(xNvN)]
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E(xN,vN)= Ek(vN) + Ep(xN)
E(xN,vN) is the Hamiltonian of the system.
If the forces do not depend on the velocities (most cases) Ek is
independent of Ep and the integrations can be separated. Also,
the integrations over the velocities of different particles are
independent. Moreover, the integrations over the components
vx ,vy ,and vz are independent. Therefore, we treat first the
integration over vx (denoted v) of a single particle.
To recall: the linear momentum vector p=mv; therefore, for
one component:
Ek= mv2/2 = p2/2m
18
A useful integral (from table):

1 
2
 exp(-ax )dx 
2
a
0
Therefore, our integral is
2

p
)dp  2mk BT
 exp( 2
mk
T
B
-
because
1
a
2mkBT
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The integral over the 3N components of the momentum
(velocities) is the following product:
3
N
( 2mk BT )
Q is (h is the Planck constant – see Hill p.74; =VN),
3
N
N
( 2mk BT )
E (x )
N
Q
exp( )d x

3
N
k
T
B
N !h

The problem is to calculate the configurational integral
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E (x N ) 1 1 1 2 2 2
)dx dy dz dx dy dz ....
 exp( k BT

The origin of the division by N factorial, N!=12  3…  N
is that each configuration of the N particles in positions x1 , x2,
x3,….,xN can be obtained N! times by exchanging the particles
in these positions (introduced by Gibbs). For example, for 3
particles there are 3!=6 possible permutations.
sites
Particles 1,2,3
1
3
3
1
2
2
2
2
1
3
1
3
3
1
2
2
3
1
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Stirling approximate formula:
ln N! N ln (N/e)
The Helmholtz free energy A(N,V,T) is:
 2mkBT 

- k BTN ln{

2
 h

3/ 2
e
E (x N ) N
} - k BT ln  expdx
N
k
T
B

The velocities (momenta) part is completely solved; it
contains m and h - beyond classical thermodynamics!
The problem of statistical mechanics is thus to solve the
integral. For an ideal gas, E = 0 (no interactions) hence =VN
3/ 2
trivial!!
 2mk BT 

A( N ,V , T )  -k BTN ln{
 h2 
eV
}
N
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Thermodynamic derivatives of A of an ideal gas (properties
~N or V are called extensive)
Pressure:
Intensive
variable ~N/V
 A 
P  - 
 Nk BT / V  PV  Nk BT
 V T , N
Internal energy:
Extensive variable
~N
 A 
 ( ) 
3
2
T

E  -T 
 Nk BT
2
 T 

V , N
E is the average kinetic energy, proportional to T, independent
of V. Each degree of freedom contributes (1/2)kBT. If the
forces (interactions) do not depend on the velocities, T is
determined by the kinetic energy only (see first lecture). 23
Specific heat CV is independent of T and V.
3
 E 
CV   
 Nk B
 T V , N 2
The entropy is:
3/ 2 5/ 2 


E-A
2mk BT
Ve
 A 




S  

 Nk B ln 
T
N 
 h 2 
 T V , N


S increases with increasing T, V and N- extensive variable.
S is not defined at T=0 – should be 0 according
the third low of thermodynamics; the ideal gas picture
holds only for high T. Both S and E increase with T. 24
In the case of a real gas E(xN)0 and the problem is to
calculate the configurational partition function denoted Z,
where the momentum part is ignored,
Z= ! exp[- E(xN)/kBT]dxN
Notice: While Q is dimensionless, Z has the dimension of xN.
Also,
Z= ! (E)exp[- E/kBT] dE
(E) – the density of states around E; (E)dE – the volume
in configurational space with energy between E and E+ dE.
For a discrete system (n(Ei) is the degeneracy of E)
Z= exp [- Ei/kBT] =  n(Ei) exp [- Ei/kBT]
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The contribution of Z to the thermodynamic functions is
obtained from derivatives of the configurational Helmholtz
free energy, F
F=-kBTln Z
Calculating Z for realistic models of fluids, polymers, proteins,
etc. by analytical techniques is unfeasible. Powerful numerical
methods such as Monte Carlo and molecular dynamics are
very successful.
A simple example: N classical harmonic oscillators are at
equilibrium with a heat bath of temperature T - a good model
for a crystal at high temperature (the Einstein model), where
each atom feels the forces exerted by its neighbor atoms and
can approximately be treated as an independent oscillator that
does not interact with the neighbor ones.
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Therefore, QN=qN, where q is the partition function of
a single oscillator. Moreover, the components (x, y, z),
are independent as well; therefore, one can calculate qx
and obtain QN=qx3N.
The energy of a macroscopic oscillator (e.g., a mass hung on a
spring) is determined by its amplitude (the stretching distance).
The amplitude of a microscopic oscillator is caused by the
energy provided by the heat bath. This energy changes all the
time and the amplitude changes as well but has an average
value that increases as T increases. Unlike a macroscopic
mechanical oscillator, the position of the mass is unknown as
a function of t. We only know PB(x).
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The kinetic and potential energy (Hamiltonian) of an
oscillator are
p2/2m+fx2/2
f is the force constant and q=qkqp where qk was calculated
before;  is the frequency of the oscillator.
q 
k
2mk BT
h
m k BT k BT
q  2

f h
h
2k BT
qp 
f
1 f

2 m
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  A 
   
T 
2


E  -T
 k BT
 T 




 CV = kB
The average energy of one component (e.g., x direction) of
an oscillator is twice as that of an ideal gas – effect of
interaction. For N 3D oscillators, E=3NkBT – extensive (~N)
The entropy is (also extensive),
S = E/T- A/T=3NkB(1+ln [kBT/h] )
E and S increase with T.
In mechanics the total energy of an oscillator is constant,
fd2/2 where d is the amplitude of the motion and at time t
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the position of the mass is known exactly.
In statistical mechanics a classical oscillator changes its
positions due to the random energy delivered by the heat bath.
The amplitude is not constant, but the average energy is
proportional to T.
The positions of the mass are only known with their
Boltzmann probability. When T increases the energy increases
meaning that the average amplitude increases and the position
of the mass is less defined; therefore, the entropy is enhanced.
Notice: A classical oscillator is a valid system only at high T.
At low T one has to use the quantum mechanical oscillator.
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Lecture 3
Thus far we have obtained the macroscopic thermodynamic
quantities from a microscopic picture by calculating the
partition function Q and taking (thermodynamic)derivatives of
the free energy –kBTln Q. We have discussed two simple
examples, ideal gas and classical oscillators.
We have not yet discussed the probabilistic significance of
statistical mechanics. To do that, we shall first devote two
lectures to basic probability theory.
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Experimental probability
Rolling a die n times.
What is the chance to get an odd number?
n 10 50
m 7 29
100 400
46 207
Relative frequency
1000 10,000 ….
504 5,036 .…
f(n)=m/n
0.7 0.58 0.46 0.517 0.5040 0.5036 ….
f(n) → P = 0.5; P = experimental probability
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In many problems there is an interest in P and other properties.
To predict them, it is useful to define for each problem an
idealized model - a probability space.
Sample space
Elementary event:
Tossing a coin – A happened, or B happened.
Rolling a die - 1, 2, 3, 4, 5, or 6 happened.
Event: Any combination of elementary events.
An even number happened (2,4,6); a number larger than 3
happened (4,5,6).
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The empty event: impossible event – 
(2 < a number < 3).
The certain event – Ω (Coin: A or B happened).
Complementary event - Ā=Ω-A (1,2,3,4,5,6) -(2,4,6) =
(1,3,5).
Union – (1,2,3)  (2,3,5) = (1,2,3,5).
Intersection – (1,2,4)  (2,5,6) = (2)
34
A
B AB = 
AB = intersection
red
+green
AB = A and B
AB = whole
35
Elementary probability space

(
The sample space consists of a finite number n of points
(elementary events) B.

Every partial set is an event.

A probability P(A) is defined for each event A.

The probability of an elementary event B is P(B)=1/n.

P(A)=m/n; m- # of points in event A.
36
Properties of P



0  P(A)  1
P(AB)  P(A) + P(B)
i P(Ai) = 1 (Ai, elementary events)
Examples: a symmetric coin; an exact die.
However, in the experimental world a die is not exact and its
rolling is not random; thus, the probabilities of the elementary
events are not equal.
On the other hand, the probability space constitutes an ideal
model with equal P’s.
Comment: In general, elementary events can have different
probabilities (e.g., a non-symmetric coin).
37
Example:
A box contains 20 marbles, 9 white, 11 black. A marble is
drawn at random.
What is the probability that it is white?
Elementary event (EE): selection of one of the 20 marbles.
Probability of EE: 1/20
The event A – a white marble was chosen contains 9 EE,
P=9/20 This consideration involves the ideal
probability space; the real world significance: P is a
result of many experiments, P=f(n), n.
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More complicated examples:
What is the number of ways to arrange r different balls in n
cells?
Every cell can contain any number of balls.
Each ball can be put in n cells 
# ways= nn
n….n=nr
##of
nr= the number of words of length r (with repetitions)
based on n letters. A,B,C  AA, BB, AB, BA, CC, CA, AC, BC, CB
= 32 = 9
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Permutations:
# of samples of r objects out of n objects without repetitions
(the order is considered) (0!=1):
(n)r= n(n-1)…(n-r+1) = 12 ... (n-r)(n-r+1)…n = n!
12 …(n-r)
(n-r)!
(3)2  (1,2), (1,3), (2,1), (2,3), (3,1), (3,2)
(1,3)  (3,1)
# of r (2) letter words from n (3) letters:AB, BA, CA, AC, BC, CB
=6
40
Problem:
Find the probability that r people (r  365) selected at
random will have different birthdays?
Sample space: all arrangements of r objects (people) in 365
cells (days) – their number = 365r  p(EE) = 1/365r
Event A: not two birthdays fall on the same day-
# of points in A = 365364 ……. (365-r+1)=(n)r
P(A) = (n)r = 365!
365r (365-r)!365r
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Combinations
# of ways to select r out of n objects if the order is not
considered.
# of combinations = # of permutations / r!
 n  ( n )r
n!

 
 r  r! ( n - r )! r!
n=3; r=2
Permutations/2
(1,2) (2,1) /2
(1,3) (3,1) /2
(2,3) (3,2) /2
42
Problem: In how many ways can n objects be divided into k
groups of r1, r2,…..rk;  rk= n without considering the order
in each group but considering the order between the groups?
 n   n - r1   n - r1 - r2 
n
!
 



......




r 
r
r1! r2 !....rk !
r



3 
2
 1
Problem: How many events are defined in a sample space of n
elementary events?
 n  n  n
 n
n
         .....     2
 0  1  3
 n
n

r

 n  n
 n
 n
(1  x)n       x  ...  x r  .....    x n
 0  1
r
 n

 Binomial
 coefficient

43
Problem:
23 chess players are divided into 3 groups of 8, 8, and 7
players. What is the probability that players A, B, and C
are in the same group (event A)?
EE – an arrangement of the players in 3 groups.
# EE = 23!/(8!8!7!)
If A, B, and C in the first group the # of arrangs. 20!/(5!8!7!)
etc. 
20!2 20!
23!
21
P( A ) 


5!8!7! 8!8!4! 8!8!7! 253
44
Problem:
What is the number of ways to arrange n objects
that r1 , r2 , r3 , … rk of them are identical,  ri= n?
A permutation of the n objects can be obtained in r1 ! r2 ! … rn !
times  the n! permutations should be divided by this factor
n!
# ways 
r1! r2 !...rk !
6 6 24
5551112222
5115251222
45
Problem: 52 cards are divided among 4 players. What is the
probability that every player will have a king?
EE – a possible division of the cards to 4 groups of 13.
# EE 52!/(13!)4
If every player has a king, only 48 cards remained to be
distributed into 4 groups of 12  # of EE(A) = 48!/(12!)4
P(A)=[48!/(12!)4]/[52!/(13!)4]
46
Product Spaces
So far the probability space modeled a single experiment –
tossing a coin, rolling a die, etc. In the case of n experiments
we define a product space:
Coin; two EE: 0 ,1
P=1/2
3
n
1
2
 1    1    1   .... 1 
     
 
 0  0  0
 0
1 2 ……. n
1 ………. n
EE: (1,0,0,1,…,1); (0,1,1,1,…,0); (…….);
# (EE)=2n
If the experiments are independent, P(EE)=(1/2)n
47
Die: EE are: 1, 2 , 3 , 4 , 5 ,6
3
n
1
2
1 1 1
1
     
 
 2  2  2
 2
 3    3    3   ....   3 
 4  4  4
 4
     
 
 5  5  5
 5
     
 
6
6
6
     
 6
1…………..n 1…………...n
EE= (1,5,2,4….,3,6); (2,3,2,5,….,1,1); (4,…….)…; #(EE)=6n
In the cases of independent experiments, P(EE) = (1/6)n
48
Problem:
15 dice are rolled. Find the probability to obtain three times
the numbers 1,2, and 3 and twice, 4, 5, and 6?
EE: all possible outcomes of 15 experiments, #(EE)= 615
#(A): according to formula on p. 45: 15!/[(3!)3(2!)3]

15!
P( A )  15 3
3
6 (3! ) ( 2! )
Also: 1 can be chosen in (15 14 13)/3! ways.
2 in (12 11 10) )/3! ways etc.
49
Dependent and independent events
Event A is independent of B if P(A) is not affected if B
occurred.
P(A/B)=P(A)
P(A/B) – conditional probability. For example, die:
Independent:
P(even)=1/2; P(even/square)=1/2 [ P({2,4,6}/{1,4})=1/2]
Dependent:
P(2)=1/6; P(2/even)=1/3
P(even/odd)=0, while P(even)=1/2 (disjoint)
50
Bayes Formula
P(A/B) = P(AB)/P(B);
P(A)>0;
P(B/A) = P(AB)/P(A);
P(B)>0
P( B / A )  P( A )
P ( A / B) 
P ( B)
A=(2) B=even (2,4,6)  P(A/B)=1/3.
Using formula: P(A)=1/6; P(B)=1/2; P(AB)=1/6 
1/ 6
P ( A / B) 
 1/ 3
1/ 2
Independency:
P(AB)= P(A)P(B)
51
If an event A must result in a mutually exclusive events,
A1,….An , i.e., A=AA1 + AA2 +… AAn then
P(A)=P(A1)P(A/A1) +……+P(An)P(A/An)
Problem: Two cards are drawn successively from a deck.
What is the probability that both are red?
EE- product space: (r,r), (r,no), (no,r), (no,no)
A first card is red
{(r,r); (r,no)}
B  second card is red {(no,r); (r,r)}
P(AB)=P(r,r)=P(A)P(B/A)=1/2·(25/51)
52
Problems:
• Calculate the energy E of an oscillator by a free energy
derivative, where q=kBT/h
2) Prove the equation on p.45.
3) A die is rolled four times. What is the probability to obtain 6
exactly one time.
4) A box contains 6 red balls, 4 white balls, and 5 blue
balls. 3 balls are drawn successively. Find the probability
that they are drawn in the order red, white, and blue if
the ball is (a) replaced, (b) not replaced.
53
What is the expectation value of m in the random
variable of Poisson:
P(X = m)=mexp(-)/m! (m = 0,1,2,…..).
.
54
Summary
We have defined experimental probability as a limit of relative
frequency, and then defined an elementary probability space,
where P is known exactly. This space enables addressing and
solving complicated problems without the need to carry out
experiments.
We have defined permutations, combinations, product spaces,
and conditional probability and described a systematic
approach for solving problems.
However, in many cases solving problems analytically in the
framework of a probability space is not possible and one has
to use experiments or computer experiments, such as Monte
55
Carlo methods.
Random variables
For a given probability space with a set of elementary
events{}=, a random variable is a function X=X() on the
real line - < X() < .
Examples:
Coin: p - head; q- tail. One can define X(p)=1; X(q) =0.
However, any other definition is acceptable - X(p)=15 X(q) =2,
etc., where the choice is dictated by the problem.
Tossing a coin n times, the sample space is vectors (1,0,0,1…)
with P(1,0,0,1…). One can define X=m, where m is the
number of successes (heads).
56
Distribution Function (DF) or Cumulative DF
For a random variable X (- < X() < )
Fx(X) = P[X()]  x
Fx(X) is a monotonically increasing function
1
5/6
4/6
3/6
2/6
1/6
Die: for x <1, Fx(X) = 0
for  6, Fx(X) = 1
1
2
3
4
5
6
57
Random variable of Poisson

P( x  m)  exp(- )
m!
m
m=1,2,3,…

F ( x) 
exp(- )
m x m!
m
So far we have discussed discrete random variables.
Continuous random variable – if F(x) continuous and its
derivative f(x) = F’(x) is also continuous. f(x) is called the
probability density function; f(x)dx is the probability between
x and x+dx.
x

F ( x)   f (t )dt;  f (t )dt  F ()  1
-
-
58
The normal random variable
2
x
1
t
P( X  x )  Fx ( X ) 
 exp(- )dt
2 -
2
1
x2
f ( x) 
exp(- )
2
2
The uniform random variable
const . a  x  b
f ( x)  
otherwise
 0

f(x) is constant
between a and b
b
1
1   f ( x )  c  dx  c(b - a )  f ( x ) 
(b - a )
-
a
 0
x
x - a
Fx ( X )   f (t )dt  
-
b - a
 1
x0
a xb
xb
f(x)
1/(b-a)
a
b
59
Expectation Value
X is a discrete random variable with n values, x1, x2 ,… xn.
and P(x1), P(x2), … P(xn), the expectation value E(X) is:
n
E ( X )   P( xi ) xi
i 1
Other names are: mean, statistical average.
Coin:
X 1 ; 0 with P and 1-P.
E(X) = P1 + (1-P)0=P
Die:
X 1, 2, 3, 4, 5, 6 with P =1/6 for all.
E(X) = (1/6)(1+2+3+4+5+6)=21/6=3.5
60
Continuous random variable with f(x)

E ( X )   xf ( x )dx
-
Provided that the integral converges.
Uniform random variable
2
2
b
1
x
b -a ba
E( X )   x
dx 


2(b - a ) a 2(b - a ) 2
a b-a
b
2
Properties of E(X): If X and Y are defined on the same space
E(X+Y) =E(X)+E(Y) ; E(CX)=CE(X)
C= const.
[X()+Y()]P() =  X()P() +  Y()]P() 
61
Arithmetic average
X1 , X2 , …… Xn are n random variables defined on the same
sample space with the same expectation value =E(Xi), then
the arithmetic average:
X1  X 2  ....X n
X
is also a random variable with
n
E( X )  
X1  X 2  ....X n 1
E( X )  E(
)  E ( X1 , X 2  ....X n ) 
n
n
1
n
 [ E ( X1 )  E ( X 2 )...  E ( X n )] 

n
n
Notice: X is defined over the product space (X1, X2,…… Xn)
with P(X1, X2,…… Xn). X is important in simulations.
62
Variance

V ( X )   ( X )  E[( X - ) ]   ( x - ) f ( x)dx
2
2
2
-
V ( X )  E[( X - ) ]  E[ X   - 2 X] 
2
2
2
 E ( X )   - 2E ( X )  E ( X ) -   E ( X ) - E ( X )
2
2
2
2
2

Standard deviation:
( X )  V ( X )
63
2
Example: Random variable with an expectation value but
without variance.
P( x  n ) 
c
c
n
3
;
;
;
E

n


3
n
n
 c
n
1
n

converges
3
1
E ( X )   n  3  c   diverges
n
n
nn
2
2
c
Independence: random variables X and Y defined on the same
space are called independent if
P(X,Y)=P(X)P(Y)
Tossing of a coin twice: (1,1), (0,1), (0,0), (1,0) – the
probability of the second toss is independent of the first.
In a product space: P(X1, X2 …., Xn)=P(X1)P(X2)…….P(Xn)
64
P(X1, X2 …., Xn) _ Joint probability
Uncorrelated random variables
If X and Y are independent random variables defined on the
same sample space  they are uncorrelated, i.e.,
E(XY)=E(X)E(Y)
Proof:
E(XY) = ij xiyjP(xiyj) = ij xiyj P(xi)P(yj)=
=i xiP(xi)j yj P(yj) = E(X)E(Y)
X,Y independent  X,Y uncorrelated. The opposite is not
always true. X,Y uncorrelated defined on the same sample
space then
V(X+Y)=V(X) +V(Y)
V(CX) = E(C2X2) - E2(CX) = C2E(X2) - [CE(X)]2 = C2V(X)
65
 V is not a linear operator.
Variance of the arithmetic average
X1, X2, ….. , Xn are uncorrelated random variables with the
same  and 2 
X1  X 2  ... X n
 (X ) V(X ) V(
)
n
2
V 
V
(
X

X

...
X
)


nV


1
2
n
2
2
n n
n
n
1
1
2
While the expectation value of the arithmetic average is also
, the variance decreases with increasing n!

The above result, ( X )  n is extremely important playing a
central role in statistics &
analysis of simulation data.66
Sampling
So far we have dealt with probability spaces (ideal world),
where the probability of an elementary event is known exactly
and probabilities of events A could be calculated.
Then, we defined the notion of a random variable (X) which is
a function from the objects of a sample space to the real line,
where the function can be defined according to the problem of
interest. Accumulative distribution function and probability
density function (for a continuous random variable) were
defined.
This enables, at least in principle, calculating expectation
values E(X) and variances V(X).
67
It is important to calculate E ( ) and V of the normal
distribution (also called the Gaussian distribution), see p. 61.
2
1
x
f ( x) 
exp- [ 2 ]
 2
2
It can be shown (see integral on p.19) that f(x) is normalized,


2
1
x
exp- [ 2 ]dx  1
 f ( x )dx  
2
-
-   2
and its expectation value is 0, because x is a symmetric odd
function

E ( X )   xf ( x )dx  0
-
68
The variance is therefore:

V ( X )   x f ( x )dx 
2
-

2
1
x
2

 x exp- [ 2 ]dx 
 2 - 
2
3
2 2
 ( 2 )
2
1
2

 2
Here we used the integral:

 x exp- [ax ]dx 
-
2
2

3/ 2
2a
Thus, a Gaussian is defined by only two parameters, E(X) and
V(X) - in the above case, 0 and 2, respectively. In the general
case, f(x) ~ exp-[(x- )2/(2 2)].  defines the width of the
69
distribution.
f(x)
 = V
x
E==0
The integral of f(x) from  -   x  +  provides ~ 68% of
the total probability = 1; the integral over  - 2  x  +
2 covers ~95% of the total area.
Unlike the ideal case (i.e., known probabilities) sometimes a
distribution is known to be Gaussian but  and  are unknown.
To estimate  one can sample an x value from the distribution
70 .
– the smaller is  the larger the chance that x is closer to 
We shall discuss later how to sample from a distribution
without knowing its values. One example is a coin with
unknown p (1) and 1-p (0); tossing this coin will produce a
sample of relative frequencies (1) p, (0) 1-p.
Also, E(X) = p and V(X)=p(1-p) are unknown a-priori. If p
0 or p 1, V(X)  0 and even a single tossing experiment
would lead (with high chance) to 0 and 1, the values of E(X),
respectively.
To estimate an unknown E(X) in a rigorous and systematic
way it is useful to use the structure of probability spaces
developed thus far, in particular the product spaces and the
properties of the arithmetic average, X  X 1  X 2  ... X n .
n
71
Thus, if X1, X2, ….. , Xn are random variables with the same 
and 2 ( E ( X )  ) and if these random variables are also
uncorrelated then
2

V(X ) 
n
For example: one can toss a coin [p (1), 1-p (0), p unknown] n
times independently. The result of this experiment is one term
(vector) denoted (x1, x2, ….. , xn), [e.g., (1,0,0,1….1,0)] out of
the 2n vectors in the product space. Estimation of E ( X )  
by (x1+ x2+ ….. + xn )/n is improved in the statistical sense as
the variance V ( X ) is decreased by increasing n, the number of
experiments; for n   the estimation becomes exact because
72
V ( X )  0.
Thus, to estimate  (and other properties) one has to move
to the experimental world using the theoretical structure of the
probability spaces. Notice again that while the value of P [or
f(x)] is unknown, one should be able to sample according to P!
(see the above example for the coin).
This is the basic theory of sampling that is used in Monte
Carlo techniques and molecular dynamics. However, notice
that with these methods the random variables in most
cases are
2
correlated; therefore, to use the equation V ( X )   , the #
n
of samples generated, n’ should be larger, sometimes
significantly larger than n, the number of uncorrelated samples
used in the above equation. This topic will be discussed in
more detail later.
73
The probabilistic character of statistical mechanics
Thus far, the thermodynamic properties were obtained from
the relation between the free energy, A and the partition
function Q, A=-kBTlnQ using known thermodynamics
derivatives. However, the theory is based on the assumption
that each configuration in phase space has a certain probability
(or probability density) to occur - the Boltzmann probability,
PB(X)=exp[-E(X)/kBT]/Q
where X  xNvN is a 6N vector of coordinates and velocities
Therefore, any thermodynamic property such as the energy is
an expectation value defined with PB(X).
74
For example, the statistical average (denoted by <>) of the
energy is:
<E>= ! PB(xN,vN) E(xN,vN) d (xNvN),
where E(xN,vN) is a random variable. <E> is equal to E
calculated by a thermodynamic derivative of the free energy,
A. For an ideal gas (pp.19-20) <E> is obtained by,
QIG 
( 2mk BT )3 N
N ! h3 N
VN
2
p2
p
exp- [
] / N ! h3 N
exp- [
]
2mk BT
2mk BT
B
P 

QIG
( 2mk BT )3 N V N

 E 
p2
p2
exp- [
]d p d x

2mk BT
-  2m
( 2mk BT )3 N V N
3
 Nk BT
2
75
This is the same result obtained from thermodynamics (p. 23).
For two degrees of freedom the integral is:


-

2m
2
( p1
(
[ ( 2mk BT )

2
 p2
]dp1dp2 dx1dx2
2mk BT
2 2
2mk BT ) V
2
2
p2 )
p1
exp- [
( 2mk BT )]  2
3
2m  2


2
k
T
B
2
2
( 2mk BT )
3/ 2
76

The entropy can also be expressed as a statistical average. For
a discrete system,
S=<S>= -kB Si Pi ln Pi
If the system populates a single state k, Pk=1 and S=0  there
is no uncertainty. This never occurs at a finite temperature. It
occurs only for a quantum system at T=0 K.
On the other hand, if all states have the same probability,
Pi=1/, where  is the total number of states, the uncertainty
about the state of the system is maximal (any state can be
populated) and the entropy is maximal as well,
S= -kB ln 
This occurs at very high temperatures where the kinetic energy
is large and the majority of the system’s configurations can be
visited with the same random probability.
77
It has already been pointed out that the velocities (momenta)
part of the partition function is completely solved (pp. 19-26).
On the other hand, unlike an ideal gas, in practical cases the
potential energy, E(xN)0 and the problem of statistical
mechanics is to calculate the configurational partition function
denoted Z, where the momentum part is ignored (see p. 20)
Z= ! exp[- E(xN)/kBT]dxN
where Z has the dimension of xN. Also,
Z= ! (E)exp[- E/kBT] dE
(E) – the density of states; (E)dE – the volume in
configurational space with energy between E and E+ dE. For a
discrete system n(Ei) is the degeneracy and Z is
Z= exp [- Ei/kBT] =  n(Ei) exp [- Ei/kBT] 78
The thermodynamic functions can be obtained from
derivatives of the configurational Helmholtz free energy, F
F=-kBTln Z.
From now on we ignore the velocities part and mainly treat Z.
Thus,the configurational space is viewed as a 3N dimensional
sample space , where to each “point” xN (random variable)
corresponds the Boltzmann probability density,
PB(xN)=exp-[E(xN )]/Z
where PB(xN)dxN is the probability to find the system between
xN and xN + dxN. The potential energy E(xN) defined for xN is
also a random variable with an expectation value <E>
79
<E> = ! E(xN ) PB(xN)dxN
While <E> is identical to the energy obtained by deriving F/T
with respect to T (see p. 23 ), calculation of the latter is
significantly more difficult than calculating <E> because of
the difficulty to evaluate Z hence F. In simulations calculation
of <E> is straightforward.
Again, notice the difference between E(xN ) -the potential
energy of the system in configuration xN, and <E> - the
average potential energy of all configurations weighed by the
Boltzmann probability density.
80
The power of the probabilistic approach is that it enables
calculating not only macroscopic thermodynamic properties
such as the average energy, pressure etc. of the whole system,
but also microscopic quantities, such as the average end-toend distance of a polymer. This approach is extremely useful
in computer simulation, where every part of the system can be
treated, hence almost any microscopic average can be
calculated (distances between the atoms of a protein, its radius
of gyration, etc.).
The entropy can also be viewed as an expectation value, where
ln PB(xN) is a random variable,
S = <S> = -kB! PB(xN)ln PB(xN) dxN
81
Likewise, the free energy F can formally be expressed as an
average of the random variable, E(xN ) + kBT ln PB(xN),
F = - kBT lnZ = ! PB(xN)[E(xN ) + kBT ln PB(xN)] dxN
Fluctuations (variances)
The variance (fluctuation) of the energy (see p. 63) is:
2(E) = ! PB(xN)[E(xN ) - <E>]2 dxN =
= <E(xN )2 > - <E(xN ) >2
Notice that the expectation value is denoted by <> and E is the
energy. It can be shown (Hill p. 35) that the specific heat at
constant volume is:
Cv = (dE/dT)V = 2(E) /kBT2
82
In regular conditions Cv = 2(E) /kBT2 is an extensive variable,
i.e., it increases ~N as the number of particles N increases;
therefore, (E) ~ N1/2 and the relative fluctuation of E
decreases with increasing N,
( E )
N 1
~

E N
N
Thus, in macroscopic systems (N ~ 1023) the fluctuation (i.e.,
the standard deviation) of E can be ignored because it is ~1011
times smaller than the energy itself  these fluctuations are
not observed in macroscopic objects. Like Z,<E> can be
expressed,
<E> = ! E(E)PB(E)dE
where (E) is the density of states and PB(E) is the
83
Boltzmann probability of a configuration with E.
The fact that (E) is so small means that the contribution to <E
> comes from a very narrow range of energies around a typical
energy E*(T) that depends on the temperature
PB
E*
Therefore, the partition function can be approximated by
taking into account only the contribution related to E*(T).

Z = ! (E)exp[- E/kBT] dE  fT(E*) = (E*)exp[- E*/kBT]
F  E*(T)+kBTln (E*) = E*(T)-TS
84
The entropy, -kBln (E*) is the logarithm of the degeneracy of
the most probable energy. For a discrete system
Z =  n(Ei) exp [- Ei/kBT]
where Ei are the set of energies of the system and n(Ei) their
degeneracies. For a macroscopic system the number of
different energies is large (~N for a discrete system) while
only the maximal term
n(E*) exp [- E*/kBT]
contributes. This product consists of two exponentials. At very
low T the product is maximal for the ground state energy,
where most of the contribution comes from exp[- EGS*/kBT]
while n(EGS*) ~ 1 (S=0). At very high T the product is
maximal for a high energy, where n(E*) is maximal
(maximum degeneracy  maximum entropy) but the
exponential of the energy is small. For intermediate T 85
n(E)e-E/kBT
fT(E)
e-E*/kBT
n(E)
E*(T)
E
86
The fact that the contribution to the integrals comes from an
extremely narrow region of energies makes it very difficult to
estimate <E>, S and other quantities by numerical integration.
This is because the 3N dimensional configurational space is
huge and the desired small region that contributes to the
integrals is unknown a-priori.
Therefore, dividing the space into small regions (grid) would
be impractical and even if done the corresponding integration
would contribute zero because the small important region
would be missed - clearly a waste of computer time.
The success of Monte Carlo methods lies in their ability to
find the contributing region very efficiently leading to precise
estimation of various averages, such as <E>.
87
Numerical integration
f(x)
x 1 x2 x3
! f(x)dx  i f(xi)xi
xn x
xi= xi-xi-1
88
What is the probability to find the system in a certain energy
(not xN)?
PB(E)=n(E)exp-[E/kBT]/Z
So, this probability depends not only on the energy but also on
the degeneracy n(E). The relative population of two energies is
therefore:
P ( E1 ) n( E1 ) exp(- E1 / k BT ) n( E1 )




exp(

E
/
k
T
)
B
B
P ( E2 ) n( E2 ) exp(- E2 / k BT ) n( E2 )
B
E  E1- E2
89
Problems:
1. Show that the number of ways n objects can be divided into
k groups of r1, r2,…..rk;  rk= n without considering the order
in each group but considering the order between the groups is
n!/(r1!r2!….rk!)
2. Two random variables X and Y are uncorrelated if
E(XY)=E(X)E(Y). Show that in this case V(X+Y)=V(X)+V(Y).
V is the variance.
3. The configurational partition function of an one
dimensional oscillator qp is defined on p.28. Calculate the
90
average potential energy <E>. Use the integral on p. 69.
Solving problems in statistical mechanics
The first step is to identify the states of the system and the
corresponding energies (e.g., the configurations of a fluid
and their energies). Then, three options are available:
1) The thermodynamic approach: Calculate the partition
function Z  the free energy F=-kBTlnZ and obtain the
properties of interest as suitable derivatives of F.
2) Calculate statistical averages of the properties of interest.
3) Calculate the most probable term of Z and the most
dominant contributions of the other properties.
91
Problem: N independent spins interact with a magnetic field H.
the interaction energy (potential) of a spin is H or - H,
depending of whether , the magnetic moment is positive or
negative. Positive  leads to energy - H. Calculate the
various thermodynamic functions (E,F,S, etc.) at a given
temperature T.
2N stats of the system because each spin is or +1(-) or 1().
Potential energy of spin configuration i: Ei = -N+ H +N- H
or
Ei = -(N-N-)H + N-  H where N+ and N- are the numbers
of +1 and –1 spins. The magnetization of i is:
92
Option 1: Thermodynamic approach
We have to calculate Z = i exp-[Ei/kBT] ; i runs over all
the 2N different states of the system!
This summation can be calculated by a trick. The spins
are independent, i.e., they do not interact with each
other  changing a spin does not affect the other
spins. Therefore, the summation over the states of N
spins can be expressed as the product Z=(z1)N where z1
is the partition function of a single spin.
z1 = exp(-H/kBT) + exp(+H/kBT)=2cosh[H/kBT]
cosh(x)=[exp(x)+exp(-x)]/2
Z= [2cosh(H/kBT)]N
93
F=-kBTlnZ =-kBTNln[2cosh(H/kBT)]
Entropy:

H
H
H 
 F 
S  -   Nk B ln{ 2 cosh(
)} tanh(
)
k BT
k BT
k BT 
 T  H

T=, S/N=ln 2 ; T=0,
S=0
e x - e - x sinh( x )
tanh( x ) 

e x  e - x cosh(x )
Energy:
F
 
H
2  T H
E  F  TS  -T
 - NH tanh(
)
T
k BT
94
Magnetization: M=N+- N-
H
 F 
M  -
)
  N tanh(
k BT
 H T
Specific heat:
 E 
C  -

 T  H
 H 
Nk B 

k BT 


2
H
cosh (
)
k BT
2
95
Option 2: Statistical approach
Again we can treat first a single spin and calculate its average
energy. z1 for a single spin is:
z1 = exp(-H/kBT) + exp(+H/kBT)= 2cosh[H/kBT]
The Boltzmann probability for  spin is: exp[ H/kBT] / z1
The average energy is
<E>1= [-Hexp(+H/kBT) + Hexp(-H/kBT)]/z1
= -H {2sinh[H/kBT]}/ {2cosh[H/kBT]}=
= -H tanh(H/kBT) ;
HNtanh(H/kBT)
<E>= 96
The entropy s1of a single spin is:
s1= -kB[P+lnP+ +P-lnP-], where P is the Boltzmann
probability.
s1= -kB{exp[+H/kBT][H/kBT -ln z1] +
exp[-H/kBT][-H/kBT -ln z1]}/z1=
=-kB{H/kBT [e+ - e-]/ z1- ln z1[e+ +e-]/ z1}=
= -kB {H/kBT tanh[H/kBT ] –ln z1}
The same result as on p. 94.
97
Option 3: Using the most probable term
E =-N+  H + N-  H  E’= E/H =-N++ NN = N++NN- = (N+ E’)/2 ; N+= (N- E’)/2
M =  (N+ - N-)
E =-MH
The # of spin configurations with E, W(E)=N!/(N+!N-!)
The terms of the partition function have the form,
N!
fT ( E  ) 
exp[- E / k BT ]
 N  E   N - E 

!
!
 2  2 
98
For a given N, T, and H we seek to find the maximal term. We
take ln fT(E), derive it with respect to E’, and equate the result
to 0, using the Stirling formula, lnN!  NlnN.
N  E   N  E   N - E   N - E  E

ln fT ( E )  N ln N - 
 ln
-
 ln
 2   2   2   2  k BT
The maximum or minimum of a function f(x) with respect to x, is
obtained at the value x* where (df/dx)|x* = f’(x*)= 0 and (df’/dx)|x*<0 or
>0, respectively.
fT ( E )
1 N  E  1 N - E  H
 - ln
 ln

0
E 
2
2
2
2
k BT
1 N - E  H
N - E
2H
ln


 exp
2 N  E  k BT
N  E
k BT
99
2H
H 
H
H 
1 - exp(
)
exp(
)  exp() - exp(
)
k BT
k BT
k BT
k BT


E  N
N
2H
H 
H
H 
1  exp(
)
exp(
) exp()  exp(
)
k BT
k BT 
k BT
k BT 
H
E   - N tanh[
]
k BT
The most probable energy, E* for given T and H is:
E*=-NH tanh(H/kBT)
and
M= N tanh(H/kBT)
As obtained before.
100
degeneracy
1 (min. S=0)
(H)
N
N!/[(N-2)!2!]
=N(N-1)/2
energy
typical T
-NH (min.)
T0 = 0
…
-(N-1)H+H very low (T0)
= -NH+2H
 …
-NH+4H
 …
k

T1>T 0
.
N!/[(N-k)!k!]
…
.
.
N! / [(N/2)!(N/2)!]

(max. degeneracy
S = ln 2)
spin configurations
-NH +2kH
Tk >Tk-1
N/2
H (N/2-N/2)
=0
high T
N/2
. .
101
102
Several points:
The entropy can be defined in two ways:
1) As a statistical average:
S = -kBi PilnPi (Pi – Boltzmann) and
2) as
S  kB ln n(E*)
n(E*) - degeneracy of the most probable energy. For large
systems the two definitions are identical.
As a mechanical system the spins “would like” to stay in the
ground state (all spins are up; lowest potential energy  most
stable state), where no uncertainty exists (maximum order) 
the entropy is 0.
103
However, the spins interact with a heat bath at a finite T,
where random energy flows in and out the spin system. Thus,
spins parallel to H (spin up) might absorb energy and “jump”
to their higher energy level (spin down), then some of them
will release their energy back to the bath by returning to the
lower energy state () and vice versa.
For a given T the average number of excited spins () is
constant and this number increases (i.e., the average energy
increases) as T is increased.
The statistical mechanics treatment of this model describes
this physical picture. As T increases the average energy (=
E*(T) - the most probable energy) increases correspondingly.
104
As E*(T) increases the number of states n(E*) with energy
E*(T) increases as well, i.e., the system can populate more
states with the same probability  the uncertainty about its
location increases  S increases.
So, the increased energy and its randomness provided by the
heat bath as T increases, is expressed in the spin system by
higher E*(T) and enhanced disorder, i.e., larger ln n(E*) 
larger S(T).
The stability of a thermodynamic system is a “compromise”
between two opposing tendencies: to be in the lowest potential
energy and to be in a maximal disorder. At T=0 the potential
energy wins; it is minimal  complete order S=0 (minimal).
At T= the disorder wins S and E* are both maximal. 105
At finite T the stability becomes a compromise between the
tendencies for order and disorder; it is determined by finding
the most probable (maximal) term of the partition function [at
E*(T)]:
n(E*) exp-[E*/kBT]
or equivalently the minimal term of the free energy,
E* + kBTln n(E*) = E*-TS*
Notice that while the (macroscopic) energy is known very
accurately due to the small fluctuations, the configuration
(state) is unknown. We only know that the system can be
located with equal probability in any of the n(E*) states.
106
Simplest polymer model – “ideal chain”
This model only satisfies the connectivity of the chain’s
monomers; the excluded volume interaction is neglected, i.e.,
two monomers can occupy the same place in space. In spite of
this unrealistic feature the model is important as discussed
later. We study this model on a d-dimensional lattice; the chain
starts from the origin ;no kinetic or potential energy.
a4
end–to-end
distance
a2
a1
N=8 bonds (steps) or
N+1=9 monomers
Probabilistic approach: Sample space – ensemble of all chain
configurations for a given N. No interactions, E=0, exp-E/kBT =1
107
for all chains  they are equally probable.
This ensemble can be built step-by-step. For a square lattice:
first bond – 2d=4 possible directions, and the same for the rest.
 the partition function Z is the total # of chain configurations
Z= i1=4N= (2d)N ; PB(i)=1/Z= (1/4)N = (1/2d)N
S = -kB i PB(i)ln PB(i) = NkB ln4 = NkB ln2d
There is interest in global geometrical properties of a
polymer such as the root mean square end-to-end
distance (ETED) R. Denoting by V the ETED vector,
V= iai
and R2 = <VV> = ij <ai  aj> = i <ai  ai>= Na2
a=|a|
R=N½a
108
This is an important result: even though the chain can intersect
itself and go on itself many times, the global dimension of the
chain increases as N1/2 as N is increased; this means that the
number of open structures is larger than that of the compact
ones dominating thereby the statistical averages.
j
1
4
2
About the calculation:
i
3
For any direction of i :
i  1=1 ; i 2 = 0 ; i  3 = -1 ; i  4 = 0
 if i j
<aiaj> = 0
A similar proof applies to a continuum chain.
109
Problem:
One dimensional (d=1) ideal chain of n bonds each of length
a starts from the origin; the distance between the chain ends is
denoted by x (ETED). Find the entropy as a function of x and
the relation between the temperature and the force required to
hold the chain ends at x.
Comments: 1) The force is defined as =(F/x)T parallel
to the definition of the pressure of gas, P=-(F/V)T
2) We have defined the entropy in terms of the degeneracy of
the most probable energy. Here it is defined as the degeneracy
of the most probable x.
Thus, the entropy is maximal for x=0 and minimal when
110
completely stretched, x=na (without force <x2>=an).
We shall use the most probable term method
+
x
can be defined by n+ bonds in direction + and n- bonds in
rection _
x= (n+ - n- )a
n= n+ + n- 
n+= (na+x)/2a=(n+x/a)/2 ; n-=(na-x)/2a = (n-x/a)/2
he number of chain combinations for given n+ and n- (i.e., a
ven x) is
w(x)=n!/ (n+!n-!)
111
Using Stirling’s formula,
S(x)=kBlnw(x)= kB (nln n - n+ln n+ - n-ln n-)
1
x
x 1
x
x
S  nk B {ln 2 - (1  ) ln(1  ) - (1 - ) ln(1 - )}
2
na
na 2
na
na
F = -TS

 = (F/x)T = -T(S/x) =
x

1 
2

k BT
x
x
x
na
  ln(1  )(1  

ln

....)
2a  1 - x 
na
na n 2 a 2


 na 
Using 1/(1-q)= 1+q+q2+q3+ …… for q < 1 q = x/na 1
112
i.e., x is much smaller than the stretched chain.
Because x/2a 1 it is justified to take only the leading terms
in the expansion, 1 and 2x/na.
k BT
2x
k BT 2 x
k BT

ln(1   ...) 
  ...  2  x  ...
2a
na
2a na
na
We also used here the expansion, ln(1+y)  y for y  1.
Therefore, for x  na we obtain Hook’s law where kBT/na2 is
the force constant. This derivation applies to rubber elasticity
which is mainly caused by entropic effects.
113
Solution of set problems 1:
1) Calculate the energy E of an oscillator by a free energy
derivative, where q=kBT/h
F=-kBTlnq= =-kBTln(kBT/h)
F
k BT
k BT
 
[-k B ln
]
[ln
]
2 T 
2
2
2 h k B
h

h

E  -T
 -T
 k BT
T
 k BT
T
T
T
k BT h
For N oscillators:
E=3NkBT
114
2) Prove the equation on p.45.
3) A die is rolled four times. What is the probability to obtain
6 exactly one time.
This experiment is defined in the product space, where
elementary events are vectors (i,j,k,l), 0 i,j,k,l 1.
The total number of EE: 64 (P=1/6)
The event “6 occurred exactly one time” consists of the
following events:
A) i=6, j,k,l remain between 1-5  53 EE.
B) j=6 the rest are between 1-5  53 EE, etc. 
P = 453/64 = 0.3858
115
4) A box contains 6 red balls, 4 white balls, and 5 blue
balls. 3 balls are drawn successively. Find the probability
that they are drawn in the order red, white, and blue if
the ball is (a) replaced, (b) not replaced.
(a) The problem is defined in the product space:
(15)×(15)×(15), where the experiments are independent.
P(red)=6/15 ; P(white)=4/15 ; P(blue)=5/15
P(red,white,blue) =(645)/153= 120/3375=0.0356
(b)
P(red,white,blue)=(6/15)(4/14)(5/13)=0.0440
116
About force and entropy
In the mechanical systems the force is obtained from a
derivative of the potential energy; in gravitation E=mgh; force
= -dE/dh=-mg. For a spring, E=kx2/2; force = -dE/dx =-kx.
If the spring is stretched and released the mass will oscillate.
The potential energy is converted into kinetic energy and vice
versa.
TS has a dimension of energy. Can a force be obtained from
pure entropy? Answer – yes.
To show that we examine the 1d ideal chain studied recently.
117
We have shown on p. 108 that the entropy of an1d ideal chain
(i.e., ln of total # of chains) is, S = NkB ln2d = NkB ln2.
On p. 113 we obtained, -/T = dS/dx=-const.ln(1+2x/na+..)
for x  na. Equating this derivative to 0 and solving for x
leads to x=0, which is the ETED value for which S is maximal
(the second derivative of S with respect to x is negative at
x=0).
For x=0, n+= n-= n/2;  the # of chain configurs. for x=0 is:
w(x=0)=n!/[(n/2)!(n/2)!]
and
ln(w) = nlnn - nln(n/2) = nln2 = S/kB of this system! 
118
using Stirling approximation the # of configurations at x=0
 x=0 is the most probable ETED, i.e., it has the maximal
entropy; thus, to “hold” the chain ends at a distance x>0 one
has to apply force against the “will” of the chain to remain in
its most probable state (x=0). In other words, the force is
required for decreasing the chain entropy.
This is an example for what is known as the “potential of
mean force” (PMF), which is the contribution to the free
energy of configurations that satisfy a certain geometrical
restriction. In our case this restriction is a certain value of the
ETED x.
Deriving the PMF (free energy) with respect to the
corresponding restriction (e.g., x), leads to the average force
required to hold the system in the restricted geometry. 119
Rubber elasticity stems from such entropic effects. Rubber
consists of polymer chains that are connected by cross-links.
When the rubber is stretched bond angles and lengths are not
changed (or broken). The main effect is that the polymers
become more ordered, i.e., their entropy is decreased. When
the rubber returns to its normal length the energy invested is
released as heat which can be felt. Thus, this “thermo-dynamic
spring” differs from the mechanical one.
Calculation of PMF values and the corresponding forces is
also carried out for proteins. An example is the huge muscle
protein titin (~33,000 residues), which is stretched by atomic
force microscopy and the forces are compared to those
obtained from derivatives of PMF calculated by MD.
(see, Lu & Schulten, Biophysical Journal vol. 79, 51-65,1202000.
For x/2a 1 we have obtained Hook’s law for the force  :
k BT
2x
k BT 2 x
k BT

ln(1   ...) 
  ...  2  x  ...
2a
na
2a na
na
Thus, the force required to hold the chain at x increases
linearly with increasing T, because the free energy (-TS) at x=0
decreases, i.e., the system becomes more stable at higher T and
it is difficult to stretch it to a larger x.
Also, the force is proportional to 1/(na2 ), meaning that the
longer the chain the easier is to stretch it.
121
Phase transitions
A phase transition occurs when a solid become liquid and the
latter gas. A Phase transition also occurs when a magnet loses
its magnetization or a liquid crystal becomes disordered.
While phase transitions are very common they involve a
discontinuity in thermodynamic properties and it was not clear
until 1944 whether this phenomenon can be described within
the framework of equilibrium statistical mechanics.
In 1944 Onsager solved exactly the Ising model for magnetism
where the properties of phase transition appeared in the
solution. This field was developed considerably during the last
40 years. It was also established that polymers correspond
122 to
magnetic systems and show a phase transition behavior.
First order phase transition is a discontinuity in a first
derivative of the free energy F. Examples of first derivatives
are the energy, <E> = T2 [(F/T)/T]N,V and the
magnetization M of a spin system in a magnetic field, <M> =
(F/ H)T.
A known example for a system undergoing first order
transition is a nematic liquid crystal. The molecules are elliptic
or elongated. At low T they are ordered in some direction in
space and <E> is low. A random arrangement (and high <E>)
occurs at high T. At the critical temperature Tc the system can
co-exist in two states random and ordered.
123
There is finite difference in the energy, E (latent heat) and
In other words, the two states are equally stable at Tc.
T< Tc ordered state
low energy & entropy
T >Tc disordered state
high energy & entropy
E
-<E>
Fordered=Fdisordered
ordered
disordered
T
124
For a regular system the function fT(E)=n(E)exp[-E/kBT] has a
single sharp peak at E*(T). In a first order phase transition
two peaks exist at Tc, around Eorder and Edisorder .
Notice that in a small system the peaks will be smeared, while
in a macroscopic system they become very sharp.
In the case of a peptide or a protein it is possible that several
energies contribute significantly to Z and the molecule will
populate all of them significantly in thermodynamic
equilibrium. Because the system is finite and relatively small
the peaks will be smeared.
fT(E)
125
E1 E2
E3
E4
Question: The ``jump” in energy at a first order transition is
E (latent heat). What is S in terms of E and Tc?
F1=F2  E1- TcS1=E1- TcS1  E2-E1 = Tc (S2-S1) 
E = TcS

S = E/Tc
In a second order phase transition the first derivatives of the
free energy F are continuous but there is a discontinuity (or
divergence to ) in a second derivative. Examples: the
specific heat Cv & the magnetic susceptibility T,
Cv= (E/T)V - E itself is a first derivative of F
T = -(M/H)T ;
M= -(F/H)T
126
A second order phase transition occurs at the critical point of
a magnetic system or a liquid, where the susceptibility & the
compressibility, respectively diverge to . Polymers in some
respect behave as a system undergoing such a transition.
To understand this phenomenon we define on an LL=N
square lattice a simple spin model called the Ising model. On
lattice site i a spin i is defined with two states, ``up” and
``down”, I = +1 or –1, respectively;  the total # of spin
configurations is 2N. Unlike the model studied on p. 92, spin i
interacts with a nearest neighbor spin j with energy ij=-Jij,
where the strength of the interaction is J >0 (ferromagnetic),
but the magnetic field, H=0.
ij(++)= ij(--)= -J
ij(+-)= ij(-+)= J
127
The Ising model on a square lattice was solved analytically by
Onsager in 1944; on a cubic lattice only approximate solutions
exist. In some respects the behavior of this model is similar to
the model of independent spins in a magnetic field studied
earlier.
In the ground state (typical for T=0) all spins are up (i=+1 ),
while the random spin configuration are the most probable at
high T. However, unlike the previous model, a critical T exists
at which all the first derivatives of F are continuous but the
second derivatives diverge with typical critical exponents.
++++++
++++++
++++++
T=0, M=1
-+++-+
+-++++
+++-++
T <Tc , M=0.6
++++-+++--++----
+-++-+
+++-++---++
T =Tc , M=0
T  Tc , M=0
128
large droplets of + & -
At Tc, CV/N and /N diverge as:
Tc
T

N (T - Tc ) 
CV
Tc

N (T - Tc )
And the magnetization (# spins up - # of spins down)/N
approach to 0 at Tc from the cold side as:

(T - Tc )
 M 
Tc
CV
M
1
=1/8 ; =1.75 for 2d Ising
model
129
T=0
Tc
T > Tc
What is happening?
<M> is a first derivative of F and it is continuous.
<E> (not shown) is a first derivative & continuous as well.
The divergence of CV/N and the susceptibility T/N stems
from the fact that CV ~ 2(E) (as has already been pointed out)
and  ~ 2(M)N. At non-critical temperatures 2(E) & 2(M)N
are extensive (i.e., ~ N), hence CV /N and T/N are finite.
At Tc 2(M)N & 2(E) increase with N stronger, as N1+x & N1+y;
x, y >0 are critical exponents  CV/N and T/N diverge.
This increase in the fluctuations is a result of the large ( for
an  system) droplets of spin up and down that occur at130
Tc.
Another measure of this “ criticality” is the correlations
between spins as a function of distance
corr(i, k ) 
 i k  -   2
 -
2
2
In this equation  is M. Thus, two spins on the same droplet
will have the same sign  they will be correlated. The
correlation length  increases as T Tc because the droplet
size increases,

Tc
++--+-+

+++--+-(T - Tc )
---++---where  is another critical exponent
(=1for a 2d Ising model).
131
The interesting property of these exponents is that they are
universal - i.e., they do not depend on the system (in most
cases) but depend on the dimension  they are different in 1,
2 and 3 dimensions. There are exact relations between them
and inequalities.
It turns out that a polymer chain- like a magnetic or a fluid
system at Tc is also a critical system with characteristic critical
exponents,  , , and . For example, we have already
found that the ETED of an ideal chain behaves like an, where
=1/2.
132
Models for polymers and proteins
Proteins, nucleic acids, or poly-saccharides are all long
polymers; therefore, before treating such biological
macromolecules (in particular proteins) we shall study general
properties of polymers.
The most general model is of many polymers and solvent
molecules (e.g., water) contained in a volume V. The potential
energy is based on intra-polymer, polymer-polymer, polymersolvent and solvent-solvent interactions. We shall be interested
mainly in the very dilute regime where a polymer ”meets”
another one very rarely  one can treat a single polymer
surrounded by solvent molecules.
133
This model describes faithfully proteins than do not aggregate.
The accuracy of the interaction energy depends on the
questions asked. If one is interested in the specific 3D structure
of a protein the potential energy function (also called force
field) will be complicated, typically based on electrostatic, van
der Waals, hydrogen-bond and other interactions.
An analytical solution of Z and other thermodynamic and
geometrical parameters is impossible and the only available
approach is numerical, in particular Monte Carlo (MC) and
molecular dynamics (MD) simulation methods.
On the other hand, if one is interested in global properties of
polymers (e,g., shape, size), relatively simple lattice models are
sufficient.
134
Simple models for polymers – ideal chains
As we have seen, an ideal chain only satisfies the connectivity
of a polymer but not the excluded volume (EV) interaction,
i.e., the chain can intersect itself and go on itself. No potential
energy (attractive) is applied.
Other names – Gaussian chain, random walk, and the
unperturbed chain. If the chain is off-lattice (i.e., in
continuum) it is called the freely jointed chain or a random
flight chain.
When the EV interaction is considered the chain is called a
“real chain”.
135
Importance of ideal chains
• Can be treated analytically; provide the basis for
understanding real chains.
• Explain rubber elasticity.
• Describe a polymer at the Flory -point.
• An Ideal chain models a polymer in a dense multiplepolymer system.
We shall treat a single long polymer of N bonds (steps) (i.e.,
N+1 monomers) ignoring the volume. The chain interacts with
a heat bath at temperature
136
The global shape of a polymer
Besides the ETED, the spatial size of the entire chain can be
si
expressed by the radius of gyration s, R
s
2
cM
N 1
1
2

s
 i
N  1 i 1
where si is the distance of monomer i from the center of mass,
cM of the chain,
 mi xi
cM 
xi- vector coordinates
 mi
It can be shown (Flory, p.5) that
s 
2
1

( N  1) 0i  j  N
2
rij are distances between monomers i and j.
rij2
137
Notice, s2 is based on all rij whereas the ETED on a single
distance; therefore, in simulations <s2> converges faster than
<R2> (ETED).
For a sphere of radius R the radius of gyration is
5
R
R
1
3

4

3
R
3 2
2
4
4
s  
 R
  r sin dddr 
 r dr  3
3
volume 0
4R 0
R 5 5
3
2
 s   R  0.77 R
5
For a protein with a shape close to spherical <s2>1/2 will be
~ 0.8R of the radius of the protein.
138

It was shown that on a lattice and in continuum while <R>=0
<R2>=Na2
; <R2>½ = N½a=Na
and one can show that =1/2 also for the radius of gyration
<s2>= <R2>/6

<s2> ½ = N½a/6
 the shape of the chain increases with critical exponent
=1/2,  the open chains dominate these averages as N is
increased; this result is true for any dimension, d = 1,2 ,3 …
Recall, in the Ising model  describes the increase of the spin
droplets (correlation length) as TTc.
139
Adding various geometrical restriction to an ideal chain does
not change the value =1/2 as long as the EV interaction is
ignored. The only change occurs in the pre-factor.
Thus, for a chain with constant bond angles ,
R 
2 1/ 2
(1 - cos ) 1 / 2

N
a
(1  cos )
If =109.5o  <R2>½ = 2N½a ; for =90o  <R2>½ =
N½a
Like for a freely-jointed chain.
Adding restrictions to the bond angles will open the chain (the
size of the shortest possible loop increases), <R2> will grow
leading however only to a larger pre-factor (2 for
140
o
Adding long-range interactions such as EV will affect the
exponents as well.
One can define a statistical unit a’, based on several
monomers. Thus, for a chain of N monomers a, M units a’ can
be defined where,
<R2>½ = M½a’ (M < N; a’ > a)
For =109.5o  M = N/2 ; a’=2a. a’ is related to the
persistence length.
Notice, the relation, <R2>½ = N½a is satisfied, not only for
the chain ends, but for any pair of monomers inside the chain
(internal monomers) – the chain “tails” do not change this
relation; however, the pre-factor might be changed.
141
Distribution of the end-to-end distance
For an ideal chain the probability density to be at ETED, x can
be obtained exactly and an excellent approximation is a
Gaussian function (normal distribution; see pp. 59 & 68). It is
easy to derive this probability density for the 1d lattice chain
model studied on pages110-113. For simplicity we assume that
the bond length is a=1.
One has to calculate the ratio between the number of chains
with ETED=x and the total number of chains, 2N.
We write this expression exactly, simplify it by taking ln and
applying Stirling’s formula; the result is then exponentiated
back; it is a Gaussian distribution.
142
n+ , n- : # of bonds in direction +, n= n+ + nx = n+ - n- but also -x = n- - n+
n+ = (n+x)/2
; n- = (n-x)/2
;
+
x
total # of chains 2n
n! ( 2)
P( x ) 
The ratio is:
n n x n-x
2 (
)! (
)!
2
2
Using Stirling’ formula:
n x n x n-x n-x
ln P( x)  n ln n  ln 2 ln
ln
- n ln 2
2
2
2
2
143
After opening the ln terms and regrouping:
n
x n x
ln P( x )  n ln n  ln 2 - ln( n - x )( n  x ) - ln

2
2 n-x
 x
1 

2
n
x
x
2
 n ln n  ln 2 - ln n (1 - ) - ln n 
2
n 2 2  1 - x 
 n
For xn ln(1+x/n) ~ x/n 
n x2 x  x x 
x2 x2
x2
ln P( x )  ln 2 
-     ln 2  -  ln 2 2 n2 2  n n 
2n n
2n
x2
x2
P( x )  exp(- )  exp()
2n
2 R 2 

This Gaussian should be normalized (due to approx.)  ln2 is ignored
.
144
For a freely jointed chain in 3 dimensions,
 3 

P( r )dr  
2 
 2 R  
3/ 2
 3r 2 
2
exp4

r
dr

2
 2 R  
|-----------------------------------|
This is the probability to find the chain in a spherical shell
defined by the radii r and r+dr. It is zero at r=0.The first part
is the probability density per volume as obtained in the 1d
case; this part is maximal at r=0 (see figure). 4 comes from
integrating over the angles  and .
This Gaussian is approximate because it is non 0 also for
r>Na – the fully stretched chain! However, this is a very good
approximation for large N. The Gaussian distribution exists
145
also for distant enough internal monomers i,j.
n=10
r/na
146
n=4
n=10
r/na
n=20
147
Ideal chains and random walks
There is one-to-one correspondence, ideal chains  random
walks. E.g., on a lattice, the walker starts from the origin and
“jumps” to one of the nearest neighbor sites with equal prob.
& continues in the same way. Its pathway defines a chain
configuration. The decision of the walker does not depend on
previous jumps – no memory. Every pathway of the walker
has the same probability- (1/2d)N as that of an ideal chain.
Indeed, it is known from Brownian motion, which describes a
random walk, that the average distance squared <S2> is
<S2> = 2D·t = (2kBT/f)·t
D- diffusion constant ; f- viscosity constant ; t-time  n ;<S2>
 <R2>; it is also known that <S2> has Gaussian distribution.
148
Problems - third set
(1) Given: <E> = i Ei exp[-Ei/kBT]/Z
Z = i exp[-Ei/kBT]
Prove: 2(E) = <E2> - <E>2 = kBT2CV
where,
(2) A system of N0 adsorption points and N adsorbing
molecules (N N0) is given. Calculate the chemical potential:
 F 


 N T
Hint: calculate # of arrangements
149
(3) Nc monomeric units are ordered along a straight line and
create a chain. Every unit can be in state  or  where its
length is a or b with energy E or E, respectively. Find the
relation between the chain length and the force K between the
chain ends.
K
K
x
See previous model solved in class.
150
Two comments on ideal chains
ETED – Because the statistical average of the end-to-end
distance is zero, i.e., <R>=0, <R2> is the variance. This should
be compared to the energy, where <E>#0, <E> ~N and the
variance is <E2>-<E>2. However, in both cases the distribution
is Gaussian with a very good approximation,
<R> = 0, 2(R) = <R2> ~Na
<E> ~ N, 2(E) = <E2> - <E>2 ~ N
P(R)dV = C’exp[-3R2/22(R)]
P(E)dE = Cexp[-(E - <E>)2/22(R)]
151
Ideal chain and ideal gas: Both do not satisfy the EV
interaction, but because of connectivity, # of chain
configurations is smaller than those of an ideal gas. Also, the
chain has a global structure that increases with increasing N.
Chains with excluded volume – real chains
Assume again a chain of N steps (bonds), i.e., N+1 monomers
anchored to the origin on a square lattice. The chain is not
allowed to intersect itself because two monomers occupying
the same site have  interaction energy. Therefore, this model
is called self-avoiding walks (SAWs) where ideal chains are
also called random walks (RW). Due to the EV interaction,
exact analytical solution of Z is not available and the
152
information about the model is based only on
numerical calculations (mostly simulations) and approximate
analytical treatments. Two interaction energies are defined for
chain configuration i, Ei=0 if i is a SAW and Ei =  for a selfintersecting walk. The partition function is
Z =  exp[-Ei /kBT] =  1i
RW i
SAW i
Thus, Z is the total number of SAWs – they constitute a
subgroup of the random walks (ideal chains).
SAW,
Ei = 0
Self-intersecting walk,
Ei = 
153
As for random walks, each SAW contribute 1 to the partition
function (Ei=0); therefore, the Bontzmann probability is equal
for all SAWs.
PB(i)=1/Z
Z can be calculated exactly for relatively short SAWs on
lattices (N<50 on a square lattice) by enumerating all the
different SAWs with a computer. Approximate value of Z for
large N can then be obtained by a suitable extrapolation.
Because of the repulsion among the monomers of SAWs (EV
interaction) the expansion of the chain’s shape with increasing
N is larger than that found for random walks (ideal chains).
[the compact RW configurations where a chain go on itself do
not contribute to the average ETED(SAWs)]. Thus, the
increase of the ETED (<R2>) and the radius of gyration, 154
<s2>
is governed by an exponent  > ½.
<R2>1/2= BNa
<s2>1/2 = DNa
B and D are constant pre-factors that depend on the kind of
lattice, the model, and the dimension.  on the other hand
depend only on the dimensionality – all the d=2 models will
have the same  (e.g., a square or triangular lattice, etc.) –
universality. An excellent approximation for  of SAWs is
given by Flory’s formula
 = 3/(d+2)
d
1
2
3
4
(Flory)
(best)
1
0.75
0.60
0.50
1
0.75
0.588
0.5
method
exact (trivial)
almost an exact solution
simulations (lattice)
exact
155
For d=1  is maximal, =1 because the chain can go left or
right only; this  also describes the increase of a rod-like
molecule such as an -helix or a DNA. As d increases the EV
interaction becomes less effective because the chain has more
alternative pathways to circumvent the EV disturbances;
therefore,  decreases.
In 4 and higher dimensions the spatial freedom is so large that
the chain does not “feel” the EV disturbances at all and it
grows like an ideal chain, i.e.,
<R2>1/2= CN1/2a ,  =1/2.
SAW in three dimensions is sometimes called a swollen chain
(because its shape is larger than that of a RW); it is a model for
a polymer in a good solvent, where the solvent attracts the
156
monomers and opens up the chain.
Returning to Z. One can express Z(SAW) as follows:
Z = CNN-1
C - a constant pre-factor that depends on the model.
 - growth parameter - effective coordination numberdepends on the model.
 - critical exponent that depend only on d.
To understand this formula we consider a random walk on a ddimensional lattice, where immediate return is forbidden; 
2d-1 directions are available for each step besides the first one
that has 2d possible directions 
Z=2d(2d-1)N-1=[2d/(2d-1)](2d-1)N
157
In this case, C = 2d/(2d-1); =2d-1, =1 (N-1=N0=1)
For SAW on a simple cubic lattice  4.6 (as expected, it is
smaller than =2d-1=5 for RW). On a square lattice  2.6
(again, smaller than 2d-1=3 for RW). For d=4 (SAW)=2d1=7 – as for RW.
For large N, the N term dominates the N-1 term. Again, for
a given d,  depends on the model (lattice) while  depends
only on the dimensionality- i.e., it is universal.
(d=2)=1.34…
(d=3)=1.16
(d=4)=1
exact
numerical calculations
exact- like random walks
Summary: As d increases a SAW becomes more and more like
a random walk, because the EV interactions weaken. For158d4
the EV becomes ineffective  a RW behavior.
Polymer chain as a critical system
For the magnetic Ising model and a fluid we defined several
critical exponents that describe the divergence of second
derivatives of the free energy at the critical temperature Tc of a
second order phase transition.
CV
Tc

N
(T - Tc ) specific heat (variance of energy)
Tc


N (T - Tc )  Susceptibility (variance of magnetization)

Tc
(T - Tc )

correlation length (growth of droplet)
Other magnetic lattice models exist which undergo second
159
order transition.
Of special interest is the n-vector model in the limit of n 0.
We shall not define the model here but will only say that there
is an one-to-one correspondence between the exponents of this
model and the exponents we defined for SAWs (, and ).
One can consider the radius of gyration of a chain as defining
the size of a droplet. In the magnetic model Tc/(T-Tc)
while <s2>1/2 ~ N = 1/(1/N)
So, 1/N corresponds to (T-Tc)/Tc. As N increases the chain
approaches its critical temperature. What happens is that the
contribution to  of the magnet comes from the values of
<R2> calculated for chains of connected spins that do not selfintersect (SAWs). Therefore, the two models have the same .
160
This correspondence is important for several reasons:
First, sophisticated renormalization group techniques were
developed for the magnetic systems  critical exponents for
the polymer systems can be obtained by treating magnetic
models, and vice versa – certain critical exponents of magnetic
systems can be obtained very accurately by Monte Carlo
simulations of long polymers.
Also, this correspondence discovered in1972 by de Gennes (a
Nobel prize winner) has shown that a branch of analytical
approximations for polymers that had been active for 25 yearswas incorrect!!!
A more philosophical note: Critical behavior occurs in high
energy and solid state physics, liquids, and polymers – these
very different fields live now under the same roof ,i.e., all
are
161
treated by the normalization group theory.
These ideas are summarized in a book by de Gennes “scaling
concepts in polymers”. Another polymer scientist who in
many respects established this field, is Paul Flory who won the
Nober prize in 1974, before de-Gennes.
Distribution of the end-to-end distance of SAWs

First, the growth with N of the ETED
of two monomers i and j
that are not end monomers is, <R2(ij)>1/2=A|j-i|
with the same exponent  as for the end monomers
(<R2(ij)>1/2=B|j-i|) but with a different pre-factor, i.e., AB.
j
ETED
i
When i and j approach the ends
A  B (might be important for
162
analyzing fluorescence experiments)
.
A non-Gaussian distribution for R has been developed
Pij ( r )  i - j
 -1
f ( x)  x  ,
- 3
f(
r
i- j
x  1 ;

)
1
f ( x )  exp(- x1-  ),
x  1
f(x)
x
Other parameters (not exponents) appearing in this function change when
i j are internal, end- internal etc. See Valleau JCP 104, 3071 (1996).
163
As discussed later, denatured proteins belong to the same
regime as SAWs. In fluorescence experiments one can
measure the ETED distributions of pairs of certain residues
along the chain  one can use the formulas for Pij(r) and
Rij~|i-j| for analyzing the data.
The effect of solvent and temperature on chain size
A polymer in a good solvent is open due to its strong favor
interaction with the solvent molecules. In a bad solvent the
attractions among the monomers are stronger than with the
solvent; therefore, the polymer’s size decreases and in extreme
conditions the solute molecules will aggregate and precipitate.
164
A simple modeling for solvent (or temperature) effects is a
SAW on a lattice where an attraction energy  = -|| is
defined between any two nearest neighbor monomers that are
non-bonded.
Mi=5 # of attractions
 
 

Z=  exp [-Ei /kBT] =  exp [-Mi  /kBT]
SAWs
SAWs
By decreasing the temperature the most probable chains are
those with the lower energy, i.e., the more compact ones – a
bad solvent condition. Notice that two kinds of long-range
interactions exist here attractions and the EV repulsions.165
This model defines three solvent (or temperature) regimes:
T >  - high temperature regime – high energy- open chain –
good solvent behavior - exponents of SAWs( = 0).
T =  -  - Flory’s theta temperature – the polymer behaves,
to a large extent, as an ideal chain (random walk). A
counterbalance between the attractions and repulsions
(EV) exists - ideal chain exponents ( =1/2, =1).
T <  - the quality of the solvent decreases further – the
polymer collapses – compact structure – low energy –
corresponds to the state of a folded protein -  =1/d –
other exponents are not defined – no connection to the
166
n-vector model.
The -point
It is expected that at T=  the attractions compensate the EV
repulsions and the chain has an ideal behavior. The following
parameters are obtained:
d
method
2
3
4
RW

0.571…
~0.5
0.5
0.5

1.143…
1.0050.02
1
1

3.2
5.058
exact
simulation (MC)
3(5)
Conclusions: in two dimensions (d=2) at the balance point, the EV
167 is to
prevails (even though  =3.2 rather than 3. In d=3 the behavior