Polymer Chapter

32Polymer Elasticity& Collapse

Polymeric Materials Are Often Elastic

Rubber is elastic, and gooey liquids such as raw eggs are viscoelastic, becausepolymer molecules have many conformations of nearly equal energy. Stretch-ing the polymer chains lowers their conformational entropy. The chain retractto regain entropy. The entropy is also lowered when polymers become com-pact, as when proteins fold or when DNA becomes encapsulated within virusheads. The simplest molecular description of polymer elasticity is the random-flight model.

Polymer Chains Can Be Modelled as Random Flights

We consider a polymer chain to be a connected sequence of N rigid vectors,each of length b. For now, each vector represents a chemical bond, but startingon page 612 we consider other situations in which each vector can represent avirtual bond, a collection of more than one chemical bond.

There are different ways to characterize the size of a polymer chain. Thecontour length L is the total stretched-out length of the chain,

L = Nb (32.1)

(see Figure 32.1(a)). The contour length has a fixed value, no matter what thechain conformation.

609

r �N

�3�4

�2

�1

(a) Contour Length L (b) End-to-end Length r

L = Nb

b

bb

bbb

Figure 32.1 Two measures of the size of a polymer molecule. If there are N bondsand each bond has length b, the contour length (a) is the sum of the lengths of allbonds, L = Nb. The end-to-end length (b) is the vector sum, r =∑N

i=1 �i, where �i isthe vector representing bond i.

Another measure of polymer chain size is the end-to-end length, the lengthof the vector pointing through space from one end of the chain to the other,when the chain is in a given conformation (see Figure 32.1(b)). The end-to-endlength varies from one chain conformation to the next. The end-to-end vectorr is the vector sum over the bonds �i for bonds i = 1,2,3, . . . , N :

r =N∑i=1

�i. (32.2)

The end-to-end length cannot be greater than the contour length. The end-to-end length changes when a polymer is subjected to applied forces. It isrelated to the radius of the molecule, and can be used to interpret the vis-cosities of polymer solutions, the scattering of light, and some of the dynamicproperties of polymer solutions.

For any vector, you can always consider the x, y , and z components indi-vidually. The x component of the end-to-end vector is

rx =N∑i=1

xi, (32.3)

where xi = b cosθi is the projection on the x-axis of the ith vector and θi isthe angle of bond i relative to the x-axis. If all the bond vectors have the samelength b, you get

rx = bN∑i=1

cosθi. (32.4)

In solution, there are many molecules, each with a different conformation.The physical properties that can be measured by experiments are averagesover all the possible conformations. To get these averages, you can use a sim-ple model, called the freely jointed chain, or random-flight model (‘flight’ inthree dimensions, or ‘walk’ in two dimensions). In this model the angle ofeach bond is independent of every other, including the nearest neighboringbonds. Assuming that all conformations have equal probabilities, the averagex-component of the end-to-end vector is

610 Chapter 32. Polymer Elasticity & Collapse

〈rx〉 =⟨b

N∑i=1

cosθi

⟩= b

N∑i=1

〈cosθi〉 = 0, (32.5)

because the average over randomly oriented vectors is 〈cosθi〉 = 0 (see Exam-ple 1.23). This means that monomer N of a random flight chain is in aboutthe same location as monomer 1, because there are as many positive steps asnegative steps, on average. Because the x, y , and z components are indepen-dent of each other, the y and z components can be treated identically and〈ry〉 = 〈rz〉 = 0.

Because the mean value of the end-to-end vector is zero (〈rx〉 = 0), 〈rx〉doesn’t contain useful information about the size of the molecule. The meansquare end-to-end length 〈r 2〉 or 〈r 2

x〉 is a more useful measure of molecularsize. It is related to the radius of gyration Rg by R2

g = 〈r 2〉/6 for a linear chain[1]. We won’t prove that relationship here. The radius of gyration, which isalways positive, is a measure of the radial dispersion of the monomers. We use〈r 2〉 instead of Rg because the math is simpler.

The square of the magnitude of the end-to-end vector can be expressed asa matrix of terms:

r · r = N∑i=1

�i

2

=

�1�1 + �1�2 + �1�3 +· · ·+ �1�N+�2�1 + �2�2 + �2�3 +· · ·+ �2�N+

......

�N�1 + �N�2 + �N�3 +· · ·+ �N�N.

(32.6)

This sum involves two kinds of terms, ‘self terms’ �i�i (along the maindiagonal) and ‘cross terms’ �i�j, i �= j. The self terms give 〈�i · �i〉 = b2,because the product of a vector with itself is the square of its length. The crossterms give 〈�i · �j〉 = b2〈cosθ〉 = 0 because there is no correlation betweenthe angles of any two bond vectors in the random-flight model. Therefore theonly terms that contribute to the sum are the self terms along the diagonal,and there are N of those, so

〈r 2〉 = Nb2. (32.7)

The root-mean-square (rms) end-to-end distance is 〈r 2〉1/2 = N1/2b. Becausethe molecular weight of a polymer is proportional to the number of monomersit contains, one important prediction of the random-flight theory is that themean radius (and the rms end-to-end length) of a polymer chain increases inproportion to the square root of the molecular weight. This result is the cen-terpiece of much of polymer theory.

How do real chains differ from random flights? An important difference ischain stiffness.

Chain Stiffness

Contrast the mathematical description of a random flight with that of a rod,for which all vectors point in the same direction. For a rod, all the terms inEquation (32.6) contribute b2 to the sum, and there are N2 terms in the matrix,so the end-to-end length equals the contour length, and 〈r 2〉 = (Nb)2 = L2.

Polymeric Materials Are Often Elastic 611

Because 〈r 2〉 ∼ N when bonds are uncorrelated, and 〈r 2〉 ∼ N2 whenbonds are perfectly correlated, you might expect that partial correlation be-tween bonds, which is a more realistic model for polymers, would lead to adependence of 〈r 2〉 on N with some exponent between 1 and 2. But this is notthe case.

b

bK

ψ

Figure 32.2 The Kuhnmodel of a polymer. b isthe length of the chemicalbond, bK is the length of avirtual, or Kuhn, bonddirected along the chainaxis, and ψ is the anglebetween the chain axis andthe chemical bond.

To illustrate, consider a model in which bonds have a weak angular corre-lation between first neighbors, 〈�i · �i+1〉 = b2γ, where γ < 1 is a positiveconstant. There are no correlations beyond first neighbors. If N is large, Equa-tion (32.6) contains only terms along the main diagonal and the two adjacentdiagonals. So for large N , you have 〈r 2〉 ≈ Nb2 + 2Nb2γ = Nb2(1+ 2γ).

Angular correlations between neighboring bonds do not change the scaling,〈r 2〉 ∼ N . Neighbor correlations only change a multiplicative constant. Thisconclusion also holds if you include second neighbors and third neighbors,etc., provided only that the correlations decay to zero over a number of bonds|i − j| � N that is much smaller than the chain length. So the random-flightmodel is useful even for real chains, for which bonds are not fully independentof their neighbors. Chain stiffness is a term sometimes used to describe thedegree of angular correlation between neighboring bonds.

In general, the effects of bond correlations can be described by

〈r 2〉 = CNNb2, (32.8)

where the constant CN is called the characteristic ratio. The subscript indicatesthat CN can depend on the chain length N . The characteristic ratios of poly-mers are typically greater than one, indicating that there are correlations oforientations between near-neighbor bonds along the chain.

These arguments suggest a strategy for treating real chains. The idea isto represent a real chain by an equivalent freely jointed chain. The equivalentchain has virtual bonds, or Kuhn segments, each of which represents more thanone real chemical bond. The number of virtual bondsNK and the length of eachbond bK are determined by two requirements. First, the Kuhn model chainmust have the same value of 〈r 2〉 as the real chain, but it is freely jointed soits characteristic ratio equals one,

〈r 2〉 = CNNb2 = NKb2K. (32.9)

Second, the Kuhn chain has the same contour length as the real chain,

L = Nb cosψ = NKbK, (32.10)

where ψ is the angle between the real bond and the long axis of the extendedreal chain (see Figure 32.2), since that axis defines the direction of the virtualbonds. Dividing Equation (32.9) by (32.10) gives

bKb= CN

cosψ(32.11)

so

NKN

= cos2ψCN

. (32.12)

Example 32.1 gives the Kuhn length for polyethylene.


EXAMPLE 32.1 The Kuhn model of polyethylene. Polyethylene has ameasured characteristic ratio of 6.7. For tetrahedral valence angles, ψ =70.5◦/2 = 35.25◦ [1]. Equations (32.11) and (32.12) give the Kuhn lengthas bK/b = 6.7/ cos 35.25◦ ≈ 8 times the chemical bond length, and there areN/NK = 6.7/ cos2(35.25◦) ≈ 10 chemical bonds per virtual bond.

Figure 32.3 Example of arandom-walk chainconformation. Source: LRGTreloar, Physics of RubberElasticity, 2nd edition,Oxford University Press,New York, 1958.

Another model is the wormlike chain model [2]. Its measure of stiffness isthe persistence length, which is half the Kuhn length. The persistence and Kuhnlengths characterize the number of bonds over which orientational correlationsdecay to zero along the chain. These are simplified models of bond correlations.A better treatment, which we won’t explore here, is the rotational isomeric statemodel [1, 3], which accounts not only for the angles and lengths of chemicalbonds, but also for the different statistical weights of the various possible bondconformations.

To understand why polymeric materials are elastic, you need to know morethan just the first and second moments, 〈r〉 = 0 and 〈r 2〉 = Nb2; you need thefull distribution function of the end-to-end lengths.

Random-flight Chain Conformations Are Describedby the Gaussian Distribution Function

To get the distribution of end-to-end lengths of a polymer chain, we use therandom-flight model that predicted the diffusion of a particle in Chapter 4(pages 57 to 59). In diffusion, a particle moves a distance b in a random direc-tion at each time step. For a polymer, imagine laying down a bond at a time,from bond i = 1 to N , so the growing end of a random-flight chain moves adistance b in a random direction as each bond vector is added to the chain.Adding each randomly oriented bond vector to the growing chain correspondsto a particle moving one time step in the diffusion model. Figure 32.3 showshow the conformation of a random-flight polymer chain looks like the path ofa diffusing particle.

To be more quantitative, we first consider a one-dimensional random walk,then we generalize to three dimensions. Each bond vector, because it is ori-ented randomly, can have a different projection onto the x-axis. We first com-pute the distribution of the number of forward and reverse steps, then wecompute the average x-axis distance travelled per step.

We want the probability, P(m,N) that the chain takes m steps in the +xdirection out of N total steps, giving N−m steps in the −x direction. Just likethe probability of getting m heads out of N coin flips, this probability is givenapproximately by the Gaussian distribution function, Equation (4.34),

P(m,N) = P∗e−2(m−m∗)2/N,

where m∗ is the most probable number of steps in the +x direction.Now we convert from the number of steps m to coordinate position x. The

net forward progress x is a product of two factors: (1) the number of forwardminus reverse steps, which equalsm−(N−m) = 2m−N ; (2) the average x-axisdistance travelled per step. The average x-axis projection of the bond vectorsis 〈rx〉 = 0 (see Equation (32.5)) because of the symmetry between forward and

Random-flight Chain 613

reverse steps. However, because we want the average step length, without itssign, we use the rms x-axis projection: 〈b2 cos2 θ〉1/2 = b〈cos2 θ〉1/2 = b/√3(see Equation (1.45)).

The product of these two factors is x = (2m −N)b/√3. Since the averagenumber of forward steps ism∗ = N/2, you havex = 2(m−m∗)b/

√3. Squaring

both sides and rearranging gives

(m−m∗)2 = 3x2/4b2.

Substituting this expression into Equation (4.34) gives

P(x,N) = P∗e−3x2/2Nb2. (32.13)

You can find the normalization constant P∗ by integrating:∫∞−∞P∗e−3x2/2Nb2

dx = 1.

This is solved by using the integral (Appendix D, Equation (D.1))∫∞−∞e−βx

2dx =

√πβ, where β = 3/(2Nb2). (32.14)

Equation (32.13) becomes

P(x,N) =[βπ

]1/2e−βx

2 =[

32πNb2

]1/2e−3x2/2Nb2

. (32.15)

Equation (32.15) gives you the relative numbers of all N-mer chain conforma-tions that begin at the origin of the x-axis and end at x. The Gaussian distribu-tion has a peak at x = 0, implying that most of the chains have about as many+ steps as − steps, as we noted in Chapter 4.

Relatively few chains are highly stretched. If you want to convert fromthe fractions of chain conformations to total numbers of chain conformations,you can multiply by the approximate total number of chain conformations zN ,where z is the number of rotational isomers per bond, to get zNP(x,N). Nowwe show that the Gaussian model predicts that rubber and other polymericmaterials are highly deformable and elastic.

Polymer Elasticity Follows Hooke’s Law

When a polymer is stretched and then released, it retracts like a Hooke’s lawspring. The retractive force is proportional to the extension. The retractiveforce is entropic.

Suppose you pull the two ends of a polymer chain apart along thex-direction.Because all the conformations of random-flight chains have the same energy,the free energy is purely entropic, F = −TS. Substituting Equation (32.15)into the expression S/k = lnP(x,N) gives, for the entropy and free energy ofstretching,

FkT

= −Sk= − lnP(x,N) = βx2 + constant = 3x2

2Nb2+ constant. (32.16)


20

10

0

0.4

0.2

0

Force (pN)

Force (pN)

0 10 20 26 30

0 10 20 30

Extension (µm)

Extension (µm)

Figure 32.4 Stretching a single molecule of DNA leads to a retractive force that islinear in the extension at low extensions, but steeper at higher extensions. The insetis an enlargement of the y-axis, for small deformations. Source: SB Smith, L Finzi andC Bustamante, Science 258, 1122–1126 (1992).

The retractive force felastic is defined as the derivative of the free energy,

felastic = −dFdx = −2kTβx = −3kTxNb2

. (32.17)

Equation (32.17) shows that the retractive force felastic is linear in the displace-mentx, like a Hooke’s law spring. Polymeric materials can be stretched to manytimes their undeformed size. (Metal can be stretched too, but only to a muchsmaller extent, and by a different, energetic, mechanism.) Fully stretched, thesize of a chain is determined by its contour length, L = Nb. In its undeformedstate, its average size is 〈r 2〉1/2 = N1/2b, so a polymer chain can be stretchedby nearly a factor of

√N . N

f1

Figure 32.5 Chainconformational elasticity is avector force tending tocause the chain ends(labeled N and 1) to be neareach other, on average.

Figure 32.4 shows how the retractive force depends on the stretched lengthsof single molecules of DNA. You can see the linearity between force and exten-sion for small stretching. At large deformations, the retractive forces becomemuch stronger and the Gaussian distribution no longer applies.

Now we generalize to three dimensions to describe elasticity as a vectorforce that drives the chain ends together (see Figure 32.5).

Elasticity in Two and Three Dimensions

Let r represent the vector from the origin, where the chain begins, to (x,y, z),where the chain ends (see Figure 32.6(a)). The length of the end-to-end vector

Polymer Elasticity Follows Hooke’s Law 615

P ( r,N ) = P (x,y, z,N )

r

(a)

x

z

y

r

dxdydz

z

P ( r,N ) = 4πr P ( r,N )

r

(b)

x

y

r

dr

Figure 32.6 Two different probability densities for the termini of chains that beginat the origin (0,0,0). (a) The probability density P(r, N) = P(x,y, z,N) that the chainend is between (x,y, z) and (x + dx, y + dy, z + dz). The most probabletermination point is the origin. (b) The probability density P(r ,N) = 4πr 2P(r, N) thatthe chain end is anywhere in a radial shell between r and r + dr . The peak of thisdistribution is predicted by Equation (32.23). Source: CR Cantor and PR Schimmel,Biophysical Chemistry, Part II, WH Freeman, San Francisco, 1980.

is r , where r 2 = x2 +y2 + z2. Because each bond is oriented randomly, the x,y , and z projections of a random-flight chain are independent of each other.So the probability density of finding the chain end in a small volume elementbetween (x,y, z) and (x + dx,y + dy,z + dz) is the product of independentfactors:

P(x,y, z,N)dx dy dz = P(x,N)P(y,N)P(z,N)dx dy dz

=[βπ

]3/2e−β(x

2+y2+z2) dx dy dz. (32.18)

In terms of the vector r, you have

P(r, N) = P(x,y, z,N) =[βπ

]3/2e−βr

2. (32.19)

Equations (32.18) and (32.19) give the probability that the chain terminus islocated at a specific location r between (x,y, z) and (x +dx,y +dy,z+dz),if the chain originates at (0,0,0) (see Figure 32.6(a)). However, sometimesthis is not exactly the quantity you want. Instead you may want to know theprobability that the chain terminates anywhere in space at a distance r fromthe origin (see Figure 32.6(b)). We denote this quantity P(r ,N), without thebold r notation. Because the number of elements having a volume dx dy dzgrows with distance r from the origin as 4πr 2, the probability of finding the


200

100

00 2 4 6

r (µm)

Number of Molecules Figure 32.7 Distribution of end-to-enddistances r of T3 DNA adsorbed oncytochrome c film (in microns, µm).The line represents the theoreticaldistribution for two-dimensionalrandom walks. Source: D Lang,H Bujard, B Wolff, and D Russell, J MolBiol 23, 163–181 (1967).

two ends separated by a distance r in any direction is P(r ,N) = 4πr 2P(r, N).The normalization is given by∫∞

−∞

∫∞−∞

∫∞−∞P(x,y, z,N)dx dy dz =

∫∞0

4πr 2P(r, N)dr = 1. (32.20)

The function P(r ,N) has a peak because it is a product of two quantities: theprobability P(r, N) diminishes monotonically with distance from the origin, butthe number of volume elements, 4πr 2, increases with distance from the origin.Figure 32.7 shows that the two-dimensional end-to-end distance distribution ofT3 DNA molecules adsorbed on surfaces and counted in electron micrographsis well predicted by two-dimensional random-walk theory.

Example 32.2 shows how the Gaussian distribution function is used for pre-dicting polymer cyclization equilibria and kinetics.

Figure 32.8 For polymercyclization, the two chainends must be closetogether.

EXAMPLE 32.2 Polymer cyclization (Jacobson–Stockmayer theory). Whatis the probability that the two ends of a polymer chain come close enoughtogether for them to react with each other? This probability is useful for cal-culating the rate and equilibrium constants for cyclization processes in whichlinear chains form circles (see Figure 32.8). Here we follow the random-flightmodel of H Jacobson and WH Stockmayer [4].

Suppose that the polymer has N bonds. To determine the probability thatthe chain end is within a bond distance b of the chain beginning, integrateP(r ,N), the probability of finding the ends a distance r apart, from 0 to b:

Pcyclization =∫ b

0P(r ,N)dr

=[

32πNb2

]3/2 ∫ b0e−3r2/2Nb2

4πr 2 dr. (32.21)

When the ends are together, r = b is small (r 2/Nb2 � 1), so e−3r2/2Nb2 ≈ 1.

Then the integral in Equation (32.21) is∫ b0 4πr 2 dr = 4πb3/3, and

Pcyclization =[

32πNb2

]3/2 (43πb3

)=(

6π

)1/2N−3/2.

The main prediction of the Jacobson–Stockmayer theory is that the cyclizationprobability diminishes with chain length as N−3/2. The longer the chain, thesmaller is the probability that the two ends are close enough to react. This

Polymer Elasticity Follows Hooke’s Law 617

10−8

10−9

10−7

102 103 104

10−8

10−7

(a) Concentration of Cyclized Long Chains (M)

(b) Higher Resolution Cyclization Concentrations (M)

DNA Length (base pairs)DNA Length (base pairs)245235 255

Figure 32.9 (a) For long chains the DNA cyclization probability diminishes aspredicted by the Jacobson–Stockmayer theory. Shortening the chain stiffens it,reducing the cyclization probability below the Jacobson–Stockmayer value. (b) Athigher resolution the cyclization probability depends on periodicities in the polymer,neglected in the random-flight treatment. Source: D Shore and RL Baldwin, J Mol Biol170, 957–981 (1983).

Figure 32.10 Experimental molarcyclization equilibrium constants K (inmol dm−3) for cyclic[O(CH2)10OCO(CH2)4CO]n inpoly(decamethylene adipate) melts at423 K ( ) versus chain length n arecompared with values calculated ( )from the Jacobson–Stockmayer theory.Source: JA Semylen, Cyclic Polymers,Elsevier, London, 1986.

−2

−3

−1

1.2 1.6

20 30 40 50 100

2.0Log N

NLog K

prediction is confirmed by experiments shown in Figures 32.9 and 32.10. Fig-ure 32.9 shows that the probability of DNA cyclization diminishes with chainlength for long chains, as predicted by the theory, but you can also see theeffect of chain stiffness, which is not treated by the theory. If chains are tooshort, their stiffness prevents them from bending enough to cyclize, so theircyclization probability is smaller than is predicted by the random-flight model.Also, chemical details can matter for short chains: the ends must be orientedcorrectly with respect to each other to cyclize.

Example 32.3 shows how to find the most probable radius R0.

EXAMPLE 32.3 Most probable radius R0. What is the most probable end-to-end distance R = R0 of a Gaussian chain? Because this is related to the sizeof the molecule, this quantity is often called the most probable radius. Beginwith the entropy, S(r) = k lnP(r ,N). Because F = −TS, using Equation (32.19)gives


λzz0

λyy0 λxx0

y0

z0

x0

Figure 32.11 An affine deformation of a material fromdimensions (x0, y0, z0) to (λxx0, λyy0, λzz0).

FkT

= −Sk= − ln

(4πr 2P(r, N)

)= βr 2 − 2 ln r + constant. (32.22)

To get the equilibrium value r = R0, find the minimum free energy:

ddr

(FkT

)r=R0

= 2βR0 − 2R0

= 0 (32.23)

�⇒ R20 =

1β= 2Nb2

3.

The quantity R20 = 2Nb2/3 = (2/3)〈r 2〉 is the square of the most probable ra-

dius of the Gaussian chain. We use it starting on page 621 for treating polymercollapse and folding processes.

Now we use the theory of polymer chains to describe elastic materials.

The Elasticity of Rubbery MaterialsResults from the Sum of Chain Entropies

Rubber and other polymeric materials are elastic. Polymeric elastomers are co-valently cross-linked networks of polymer chains. Here we describe one of thesimplest and earliest models for the retractive forces of polymeric materials,the affine network model.

Suppose you have m chains, all of the same length N . The bond lengthis b. The chain ends are covalently cross-linked at junction points. Assumethere are no intermolecular interactions, and that the total elastic free energyof the material is the sum of the elastic free energies of each of the chains.For the undeformed material, indicated by subscript 0, the end of a chain is at(x0, y0, z0) and

r 20 = x2

0 +y20 + z2

0 = Nb2. (32.24)

If the material is deformed by a factor λx in the x-direction, λy in the y-direction, and λz in the z-direction (see Figure 32.11), so that x = λxx0, y =λyy0, and z = λzz0, the chain end is moved to (x,y, z),

r 2 = x2 +y2 + z2 = λ2xx

20 + λ2

yy20 + λ2

zz20. (32.25)

Elasticity of Rubbery Materials 619

Using F = −kT lnP(x,y, z,N), β = 3/(2Nb2), and Equation (32.19), you havethe free energy for deforming a single chain, F1:

∆F1 = Fdeformed − Fundeformed = −kT lnP(x,y, z,N)P(x0, y0, z0, N)

= kTβ(r 2 − r 20 ).

Summing over m independent chains gives the total free energy Fm:

∆Fm = kTβ∑m(r 2 − r 2

0 ). (32.26)

Assuming that all chains are equivalent, you have∑m(r 2 − r 2

0 ) = m[〈r 2〉 −〈r 2

0 〉], and Equation (32.26) becomes

∆Fm = kTβm(〈r 2〉 − 〈r 20 〉)

= kTβm[(λ2x − 1)〈x2

0〉 + (λ2y − 1)〈y2

0 〉 + (λ2z − 1)〈z2

0〉]. (32.27)

If the undeformed chains are isotropic (all the directions are equivalent), then〈x2

0〉 = 〈y20 〉 = 〈z2

0〉 = (Nb2)/3, and Equation (32.27) becomes

∆FmkT

= m2(λ2x + λ2

y + λ2z − 3). (32.28)

In a macroscopic material, applying a force in one direction can cause forcesand deformations in other directions. The forces per unit area in various di-rections are called stresses and the relative displacements are called strains.Stresses are derivatives of the free energy with respect to strains. If you deforma material by a factor α, where the x-axis, y- axis, and z-axis deformations de-pend on α, the free energy is a function F(λx(α), λy(α), λz(α)). The derivativecan be written in terms of λ2

x , λ2y , and λ2

z as

(∂F∂α

)=(∂F∂λ2

x

)dλ2

xdα

+(∂F∂λ2

y

)dλ2

y

dα+(∂F∂λ2

z

)dλ2

zdα

. (32.29)

Suppose that you stretch an elastomer along the x-direction by a factorα = x/x0. Then the x-direction force is

fx = −(∂∆Fm∂x

)= 1x0

(∂∆Fm∂α

)= −α

x

(∂∆Fm∂α

). (32.30)

The stress τ acting in the opposite direction equals the force given by Equation(32.30) divided by the deformed cross-sectional area yz,

τ = − fxyz

= αV

(∂∆Fm∂α

), (32.31)

where V = xyz is the final volume of the material [5, 6]. Examples 32.4 and32.5 are applications of this elasticity model.

EXAMPLE 32.4 Stretch a rubber band along the x-axis. If you stretch rub-ber, its volume remains approximately constant. Therefore, stretching alongthe x-direction by λx = α leads to

λy = λz = 1√α. (32.32)


Then Equation (32.28) becomes

∆FmkT

= m2

(α2 + 2

α− 3

).

Because ∂∆Fm/∂λ2x = ∂∆Fm/∂λ2

y = ∂∆Fm/∂λ2z = mkT/2, Equation (32.28)

gives

∂Fm∂α

=mkT(α− 1

α2

). (32.33)

Substituting Equation (32.33) into Equation (32.31) gives

τ = mkTV

(α2 − 1

α

). (32.34)

Figure 32.12 shows that the model predicts experimental data adequately atextensions below about α = 3–5, but that the chains become harder to stretchthan the model predicts at higher extensions. 6 842

0

20

40

60

α

τ (kg cm−2)

Figure 32.12 Stretching arubber string gives stress τversus elongation α, ( )experimental data; ( )curve predicted byEquation (32.34). Source:P Munk, Introduction toMacromolecular Science,Wiley, New York, 1989. Dataare from LRG Treloar, TransFaraday Soc 40, 59 (1944).

EXAMPLE 32.5 Stretch a rubber sheet biaxially. Stretch a rubber sheetalong the x-axis by an amount λx = α1 and along the y-axis by an amountλy = α2, where α1 is independent of α2. If the volume is constant, λz =1/(α1α2). Then the stress along the x-axis is

τx = mkT2V

α1

(dλ2

xdα1

+ dλ2z

dα1

)= mkT

V

(α2

1 −1

α21α

22

). (32.35)

The retractive stress in elastomers depends not only on the deformation butalso on the cross-link density, through m/V , the density of chains. In a unitvolume there are 2m total chain ends. If each junction is an intersection of jchain ends, then there will be (1 junction/j chain ends) × (2m chain ends) =(2m/j) junctions, so the number of junctions is proportional to m. Thereforethe retractive force increases linearly with the cross-link density of the network.Bowling balls are made of a type of rubber that has a much higher cross-linkdensity than rubber bands.

The main advance embodied in theories of polymer elasticity was the recog-nition that the origin of the force is due mainly to the conformational freedomof the chains, and is mainly entropic, not energetic. The conformational en-tropies of polymers are important not only for stretching processes. They alsooppose the contraction of a polymer chain to a radius smaller than its equilib-rium value.

Polymers Expand in Good Solvents,Are Random Flights in Theta Solvents,and Collapse In Poor Solvents

According to the random-flight theory, the ‘size’ of a molecule increases inproportion to N1/2, where size means either the average end-to-end distance〈r 2〉 = Nb2 or the most probable radius R2

0 = 2Nb2/3. We showed on pages

Polymers Expand 621

611–613 that near-neighbor bond angle correlations (called local interactions)do not change this scaling relationship. But now we show that solvent (non-local) interactions can change this relationship because different solvents ortemperatures cause the chain to swell or contract.

Random-flight behavior, which is dominated by local interactions, appliesonly to a limited class of solvent and temperature conditions, called θ-condi-tions, θ-solvents, or θ-temperatures. In contrast, in a good solvent, monomer–solvent interactions are more favorable than are monomer–monomer interac-tions. Chains expand in good solvents. The radius of a polymer molecule growsmore steeply with N than is predicted by the random-flight model, mainly be-cause of the self-avoidance of the segments of the chain. In a third class of con-ditions, called poor solvents, the chain monomers are attracted to each othermore strongly than they are attracted to the solvent. Poor solvents cause anisolated chain to collapse into a compact globule, with radius ∼ N1/3. Poor sol-vents can also cause multiple chains to aggregate with each other. For example,oil chains such as polymethylene phase-separate from water because water isa poor solvent. Here is the simplest model of the expansion and collapse of asingle chain, due to PJ Flory [7, 8].

How does the conformational free energy depend on the radius? Considera chain with N monomers. As in Equation (32.23), our strategy is to find theradius r = R that maximizes the entropy, or minimizes the free energy. How-ever, now we include an additional contribution to the free energy. In additionto the elastic free energy Felastic, we now also account for the solvation freeenergy Fsolvation, using the Flory–Huggins theory of Chapter 31,

ddr

(Felastic(r)+ Fsolvation(r))r=R = 0. (32.36)

The solvation free energy depends on the radius through the mean chainsegment density, ρ = N/M . M is the number of sites of a lattice that containsthe chain. We defineM in terms of the chain radius below. Low density, ρ → 0,means the chain is expanded, while high density, ρ → 1, means the chain iscompact.

Chain collapse is opposed by the conformational entropy but is driven by thegain in monomer–monomer contacts, which are favorable under poor-solventconditions. To get the conformational entropy for Fsolvation, you can begin withEquation (31.4), which gives ν1, the number of conformations of a single chain.But we make two changes. First, we leave out the factor (z−1)N−1 because it isa constant that doesn’t change with the density ρ, and leaving it out simplifiesthe math. (The independence of the conformational entropy from the factor(z−1)N−1 in the Flory model has the important implication that chain stiffnessand local interactions do not contribute to entropies of collapse or expansion.)Second, our focus on a single chain means that its center-of-mass position inspace is irrelevant. This is equivalent to neglecting the placement of the firstmonomer, so we replace the factor M−(N−1) with M−N and the conformationalentropy of the chain is

Sk= lnν1 = ln

M !(M −N)!MN . (32.37)

Multipy the numerator and denominator of Equation (32.37) by N ! to put part


+P1 P2 P1P2 P2

Figure 32.13 The contact free energy ∆g describesdesolvating two polymer chain segments P1 and P2 andbringing them into contact. The interaction parameter χis defined for the process with the opposite sign (seeFigure 15.6).

of this expression into a more familiar form,

Sk= ln

(M !

(M −N)!N !

)+ ln

(N !MN

)

= −N ln(NM

)− (M −N) ln

(1− N

M

)+N ln

(NM

)−N ln e

= −(M −N) ln(

1− NM

)−N, (32.38)

by using Stirling’s approximation N !/MN ≈ (N/eM)N = (N/M)Ne−N . Nowdivide by N to get the entropy per monomer, and express the result in termsof the average density ρ = N/M ,

SNk

= −(

1− ρρ

)ln(1− ρ)− 1. (32.39)

You can check that S → 0 as ρ → 0 and S → −Nk as ρ → 1.

EXAMPLE 32.6 Polymer collapse entropy. When a protein folds, it col-lapses to a nearly maximally compact state. What is the entropic componentof the free energy opposing collapse at T = 300 K? If the chain length isN = 100monomers, the equations below Equation (32.39) give

∆Fcollapse = −T(Scompact − Sopen) = NkT≈ (100)(1.987 cal mol−1 K−1)(300 K) ≈ 60 kcal mol−1

Next, we determine how the contact energy depends on the chain radius,or density. Let wpp,wss , and wsp represent the energies of a contact betweentwo polymer segments, between two solvent molecules, and between a sol-vent molecule and a polymer segment, respectively. Figure 32.13 and Equa-tion (15.11) indicate that the free energy ∆g for desolvating two polymer seg-ments and forming a contact is

∆g = −2(wsp −

wss +wpp2

)= −2

zχkT . (32.40)

The minus sign indicates that the process defining ∆g is the reverse of theprocess defining χ (see Figure 15.6). The mean-field approximation gives anestimate of the number of contacts among chain monomers. The probabilitythat a site adjacent to a polymer segment contains another polymer segment isρ = N/M , and there are z sites that are neighbors of each of the N monomers,so the contact energy U relative to the fully solvated chain is

U = Nρz2∆g = −NkTρχ. (32.41)

Polymers Expand 623

The factor of 1/2 corrects for the double counting of interactions. Combiningthe entropy Equation (32.39) with the energy Equation (32.41) gives the confor-mational free energy of the chain as a function of its compactness,

Fsolvation

NkT= UNkT

− SNk

=(

1− ρρ

)ln(1− ρ)+ 1− ρχ. (32.42)

When the chain is relatively open and solvated, the density is small, ρ � 1.Then you can use the approximation ln(1− ρ) ≈ −ρ − (1/2)ρ2 − . . . to get(

1− ρρ

)ln(1− ρ)+ 1 ≈

(1− ρρ

)(−ρ − ρ

2

2− . . .

)+ 1

≈ 12

(ρ + ρ2

). (32.43)

Substituting Equation (32.43) into Equation (32.42), and keeping only the first-order approximation, gives the solvation free energy as a function of the aver-age segment density,

Fsolvation

kT= Nρ

(12− χ

). (32.44)

In Flory theory, a simple approximation relates the density ρ to the size r

ρ = NM= Nvr 3, (32.45)

where v is the volume per chain segment. This relationship defines the valueof M , the number of sites on the lattice that contains the polymer chain. Com-bining Equation (32.45) with Equations (32.22) and (32.44) gives

Felastic

kT+ Fsolvation

kT= βr 2 − 2 ln r + N

2vr 3

(12− χ

)+ constant. (32.46)

Taking the derivative of Equation (32.46) and finding the value r = R thatcauses the derivative to be zero (the most probable value of r ; Equation (32.36))gives

2βR − 2R− 3N2v

R4

(12− χ

)= 0. (32.47)

You can express this in terms of R20 = 2Nb2/3 = β−1, the most probable ra-

dius of the unperturbed chain (see Equation (32.23)). Multiplying both sides ofEquation (32.47) by R4/(2R3

0), and rearranging gives

(RR0

)5

−(RR0

)3

=(

3N2v2R3

0

)(12− χ

)

=(

32

)5/2 vb3

(12− χ

)√N. (32.48)

Theta Solvents Give Random-flight Behavior

For solvent and temperature conditions that cause χ = 0.5, the right-hand sideof Equation (32.48) equals zero. In that case, multiplying Equation (32.48) by


1.0

1.8

0.2

0.90 1.00 1.10T/θ

Expansion Factor Figure 32.14 Homopolymers collapse when thetemperature T is less than the temperature θ at whichχ = 0.5. Chains expand at higher temperatures (χ < 0.5,good solvents). The transition steepens with increasingchain lengths. (( ) molecular weight M = 2.9× 103,( ) M = 1× 105, ( ) M = 2.6× 107.) Source: ST Sun,I Hishio, G Swislow and T Tanaka, J Chem Phys 73,5971–5975 (1980).

(R0/R)3 gives the random-flight prediction,(RR0

)2

= 1 �⇒ R2 = R20 =

2Nb2

3, (32.49)

given by Equation (32.23). That is, when the solvation free energy is zero (themonomer–monomer attraction just balances the excluded volume), then themost probable radius of the chain is given simply by the elastic free energyalone.

Good Solvents Expand Polymers

For solvent and temperature conditions that cause χ < 0.5 (called good sol-vents), the right-hand side of Equation (32.48) is positive. In this regime, ex-cluded volume causes chain expansion. For large N , the fifth-power term ismuch larger than the third-power term, so(

RR0

)5

≈(

32

)5/2 vb3

(12− χ

)√N �⇒ R5 ∝ R5

0N1/2 ∝ N3

�⇒ R ∝ N0.6. (32.50)

In good solvents, the chain radius grows more steeply with chain length (R ∝N0.6) than in θ-solvents (R ∝ N0.5).

Polymers Collapse in Poor Solvents

Solvent and temperature conditions that cause χ > 0.5 are called poor solvents.For a poor solvent, the right-hand side of Equation (32.48) is negative, soR/R0 <1, and the polymer collapses into a compact conformation. For poor solventconditions, Equation (32.48) is no longer sufficient because of the low-densityapproximation we used to derive it. A collapsed chain has a high segmentdensity. If you were to keep the next higher term in the density expansionfor ln(1 − ρ), however, you would find that the model predicts R ∝ N1/3, asexpected for a compact chain.

Polymers Expand 625

Polymers can undergo very sharp transitions from the coil to compact statesas the solvent and temperature are changed. These are called coil-to-globuletransitions. The reason they are so sharp is found in Equation (32.48). For largeN , the right-hand side of Equation (32.48) can change abruptly from being verypositive to being very negative with only a small change in χ at about χ = 0.5.Figure 32.14 shows the collapse process in homopolymers.

Two natural collapse processes are the folding of proteins into their com-pact native states in water, and the compaction of DNA molecules for insertioninto virus heads and cell nuclei. While this homopolymer collapse model illus-trates the principle of coil-to-globule transitions, neither protein folding norDNA collapse follow it exactly, because both polymers also have electrostaticinteractions and specific monomer sequences.

Summary

Polymers have many conformations of nearly equal energy, so they have broaddistributions of conformations. Stretching or squeezing or otherwise perturb-ing polymers away from their equilibrium conformations leads to entropicforces that oppose the perturbations. This is the basis for rubber elasticity.One of the most important models is the random-flight theory, in which thedistribution of chain end distances is Gaussian. The rms distance betweenthe two ends of the chain increases as the square root of the chain length N ,〈r 2〉1/2 = (Nb2)1/2. The random-flight theory applies when the chain is in aθ-solvent. In that case, the steric tendency to expand is just balanced by theself-attraction energy causing the chain to contract. When the solvent is poor,the self-attractions between the chain monomers dominate and chains collapseto compact configurations. When the solvent is good, the self-attractions areweak and chains expand more than would be predicted by the random-flighttheory.


Problems1. Stretching a rubber sheet. What is the free energyfor stretching an elastomeric material uniformly along thex and y directions, at constant volume?

2. Stretching DNA. Figure 32.4 shows that it takesabout 0.1 pN of force to stretch a DNA molecule to anextension of 20µm. Use the chain elasticity theory to es-timate the undeformed size of of the molecule, 〈r 2〉1/2.

3. Stretched polymers have negative thermal expansioncoefficients. The thermal expansion coefficient of a ma-terial is α = (1/V)(∂V/∂T)p . Consider the correspondingone-dimensional quantity αp = (1/x)(∂x/∂T)f for a sin-gle polymer molecule stretched to an end-to-end length xby a stretching force f .

(a) Compute αp for the polymer chain.

(b) What are the similarities and differences between apolymer and an ideal gas?

4. An ‘ideal’ solvent expands a polymer chain. If apolymer chain is composed of the same monomer unitsas the solvent around it, the system will be ideal in thesense that the polymer–polymer interactions will be iden-tical to polymer–solvent interactions, so χ = 0.

(a) Write an expression for the most probable radius Rfor a chain in an ideal solvent.

(b) Show that such a chain is expanded relative to arandom-flight chain.

(c) Describe the difference between an ideal solvent anda θ-solvent.

5. Computing conformational averages. Using the ex-pression for the distribution P(r ,N) for the end-to-endseparation of a polymer chain of length N , compute 〈r 2〉and 〈r 4〉.

6. Contour length of DNA. The double-stranded DNAfrom bacteriophage λ has a contour length L = 17 ×10−6 m. Each base pair has bond length b = 3.5 Å.

(a) Compute the number of base pairs in the molecule.

(b) Compute the molecular weight of the DNA.

7. Using elasticity to compute chain concentrations.Figure 32.12 shows the stress–strain properties of a rub-ber band. Use the figure and chain elasticity theory to:

(a) Estimate the number of polymer chains in a cubicvolume 100 Å on each side.

(b) If each monomer occupies 100 Å3, what is the lengthof each chain between junction points?

References

[1] PJ Flory. Statistical Mechanics of Chain Molecules, Wi-ley, New York, 1969.

[2] JE Mark ed. Physical Properties of Polymers Hand-book, Chapter 5, American Institute of Physics,Woodbury, 1996.

[3] WL Mattice and UW Suter. Conformational Theory ofLarge Molecules: the Rotational Isomeric State Modelin Macromolecular Systems, Wiley, New York, 1994.

[4] H Jacobson and WH Stockmayer. J Chem Phys 18,1600 (1950).

[5] B Erman and JE Mark. Ann Rev Phys Chem 40, 351–374 (1989).

[6] JE Mark and B Erman. Rubberlike Elasticity: a Molec-ular Primer, Wiley, New York, 1988.

[7] PJ Flory. Principles of Polymer Chemistry, Series ti-tle: George Fisher Baker Nonresident Lectureship inChemistry at Cornell University, Cornell UniversityPress, Ithaca, 1953.

[8] HS Chan and KA Dill. Ann Rev Biophys and BiophysChem 20, 447–490 (1991).

Suggested Reading

The classic texts on rotational isomeric state and random-flight models:

CR Cantor and PR Schimmel, Biophysical Chemistry, VolIII, WH Freeman, San Francisco, 1980.

M Doi, Introduction to Polymer Physics, trans by H See, Ox-ford University Press, New York, 1996.

PJ Flory, Principles of Polymer Chemistry, series title:George Fisher Baker Nonresident Lectureship in Chem-istry at Cornell University. Cornell University Press,Ithaca, 1953.

WL Mattice and UW Suter, Conformational Theory ofLarge Molecules: the Rotational Isomeric State Modelin Macromolecular Systems, Wiley, New York, 1994.

Excellent summaries of rubber elasticity:

B Erman and JE Mark, Ann Rev Phys Chem 40, 351–374(1989).

JE Mark and B Erman, Rubberlike Elasticity: a MolecularPrimer, Wiley, New York, 1988.

One of the first models of a polymer collapse, applied toDNA:

CB Post and BH Zimm, Biopolymers 18, 1487–1501(1979).

Problems 627

Polymer Chapter

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