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Chemistry 220A Problem Set 1(due September 4, 2014)

1. Consider a polymer chain with N units pictured in Figure 1. The position of theith unit is ri , i = 1, 2,...,N , and the separation between adjacent units is given by

ri+1 ri = ni , 1 i N 1 ,

where is a constant length and the components of ni change from one congurationof the chain to another.

(a) Suppose that each Cartesian component of ni can take one of three possiblevalues. For the x-component these are

n (x )i = 1, 0, 1,

and similarly for the other components. Suppose further that each of these

values are equally likely at equilibrium. Determine

r 2 = | ri+1 ri |2 ,

where the pointed brackets denote equilibrium average. Express your result asa function of dimension d and . Similarly, determine

| ri+1 ri |4 .

Figure 1: Conguration of a random chain with N segments.

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ith sub-volume is ni . As such, for a given micro-state, the number of molecules inthe volume is N v = n1 + n2 + ... + n N , and

P v (N ) = N N

i=1

n i , (1)

where the pointed brackets indicate an averaging operation, and

( x) = 1 , for x = 0= 0 , for x = 0 .

Assuming the different n i s are uncorrelated, the joint probability for all the n i s isthen a product of separate probabilities for each of n i , i.e.,

P (n1, n 2,...,n N ) = N

i=1

p(n i ) (2)

(a) Show that Equation (1) is indeed a general formula for the probability for N vby showing that N P v (N )g(N ) gives the average value of any function g(N v ).

(b) Use Equations (1) and (2) plus the identity for integer values of m

( m) = 12

d eim

to show thatP v (N ) =

12

d eiN [f ()] N , (3)

where

f () = exp( i n 1) =n

p(n) e in

(c) Note that f () 1, where the equality holds for = 0 only. Therefore, when N 1, only small values of contribute to the integral of Eq.(3). Hence, anasymptotic expression for f () is sufficient to evaluate the integral. Derive thefollowing asymptotic expression:

f () exp[ i n 2 (n)2 / 2], small , (4)

where n =n p(n)n and (n)2 =

n p(n)(n n )2.

(d) Use Eq.(4) to evaluate the integral in Eq.(3), noting that because only small contribute, the upper and lower limits of the integration can be extended to and , respectively. Show that the result gives the normalized Gaussiandistribution

P v (N ) = 1

2 (N v )2exp[ (N v N v )2/ 2 (N v )2 ] (5)

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The derivation of Equation (5) is thus a derivation of the central limit theoremfor this particular system. Extending this derivation to apply to more generalcases is left to you to consider on your own.

4. Exercise 3.17 in Introduction to Modern Statistical Mechanics (IMSM).

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