Polymers Physics

Minne Paul Lettinga

Forschungszentrum J¨ulich Institute for Complex Systems - 3

J¨ulich, July 2011 2 Contents

1 Ideal polymer chain 3 1.1 Deﬁnition of an ideal polymer: ...... 3 1.2 End-to-end vector ...... 3 1.3 Radius of Gyration ...... 5 1.4 Measuring polymer size ...... 6 1.5 Probability distribution ...... 6 1.6 Entropic spring ...... 8

2 Non-ideal polymers 11 2.1 Fixed valency angle polymers ...... 11 2.2 Persistence length of a polymer chain ...... 13 2.3 Kuhn segments ...... 15 2.4 Occupied volume ...... 15 2.5 Obtaining persistence length experimentally ...... 17 2.6 Gaussian Chain Model ...... 19 2.6.1 Appendix A ...... 21

3 Polymer Dynamics 23 3.1 Rouse model ...... 23

Literature i

1 2 Chapter 1

Ideal polymer chain

1.1 Deﬁnition of an ideal polymer:

• no volume per joint

• no interactions

• Freely jointed chain

The consequences of these assumptions are manyfold. For example will be no interaction also between polymer coils, so that for example with ideal polymers no rubber can be formed, see later. We start with ideal polymers because it is easy to derive a few basic features of polymers.

1.2 End-to-end vector

The ﬁrst feature we will calculate is the average end-to-end distance R~ between the start and ~ ~ the end of the polymer, which consists of N segments b with orientation ~ui and length b. R is a measure of the size of the polymer, but not something one could readily measure. Because R~ can point in any direction, when R~ is averaged over all these directions it will give hR~i = 0, which is the same issue as we had with the displacement of colloidal particles. As with colloidal particles the ﬁrst non-zero parameter is

N N N N 2 X X X X R = ( ~ui)( ~uj) = ~ui~uj, (1.1) i=1 j=1 i=1 j=1

so the average of this quantity gives

3 R j

Rij

rj b i a i b ri

Figure 1.1: (a) Sketch of a freely jointed chain polymer. R~ shows the end-to-end vector of this polymer. (b) Illustration of a subfragment (in red) for which the end-to-end distance needs to be calculated.

4 N N N N N ~ 2 X X X 2 X X 2 hR i ≡ h~ui~uji = h~ui i + h~ui~uji = Nb = Lb, (1.2) i=1 j=1 i=1 i=1 j=1,6=i PN PN where L = Nb is the contour length of the polymer. We used here that i=1 j=1,6=ih~ui~uji = 0 because the angles of segments i and j are not correlated. The end to vector is thus given by R = phR2i = N 1/2b. From this we see that ideal chains form entangled coils.

1.3 Radius of Gyration

The radius of gyration Rg is a quantity that can be more readily measured, as we will see later on. Rg is deﬁned as the average distance to the center of mass of the object, in this case the polymer coil. As with the end-to-end vector R a straightforward average of the vector ~ PN Rg = i=1(~ri − ~rcm) would give zero, so again we have to calculate the square of the distance 1 PN of segment i with position ~r to the center of mass ~rcm = N j=1 ~rj.

N N N N N 1 X 1 X X 1 X X R2 = h (~r − ~r )2i = h(~r − ~r )2i == hR~ 2 i. (1.3) g N i cm 2N 2 i j 2N 2 ij i=1 i=1 j=1 i=1 j=1

Since vecRij = ~ri − ~rj is again an end-to-end distance but now from the full polymer but from a subfragment of the polymer, see Fig. 1.1b, we can use the result EQ. 1.2 to calculate vecRij. Here we assume that the number of segments subfragment is large, i.e. |i − j| 1, which will be the case for big polymers:

2 2 h(~ri − ~rj) i = |i − j|b . (1.4) Therefore

N N b2 X X R2 = |i − j|. (1.5) g 2N 2 i=1 j=1 For large N this sum can be replaced by an integral. We leave it as an exercise to show now that

1 R2 = Nb2. (1.6) g 6 So the ratio of the radius of gyration and the average end-to-end vector is 1/6 for ideal chains.

5 1.4 Measuring polymer size

In the past sections we have always considered average distances between the diﬀerent segments of the polymer. Suppose now that light can be scattered from each node between the segments. This would then be very similar to scattering from an ensemble of point sources as has been treated for colloids earlier. There it was shown how the structure factor of the ensemble can be obtained from light scattering:

N N 1 X X S(~k) = hexp(i~k · (~r − ~r ))i, (1.7) N i j i=1 j=1

~ ~ where k is the scattering vector with |k| = 2π/λ and ~ri are the positions of the colloids. We can now use exactly the same expression to calculated the intra-polymer structure factor taking now for ~ri the locations of the nodes, wich is also called the form factor. Doing so we immediately recognize

~ 2 2 hexp(ik · (~ri − ~rj))i = exp(−k |i − j|b /6) (1.8)

This gives

Z N Z N ~ 1 2 2 S(k) = di dj exp(−k |i − j|b /6 = Nf(kRg) (1.9) N 0 0

where f(x) = 2(exp −x2) − 1 + x2)/x4. For small wave vectors it can be shown that

~ ~ 2 2 S(k) = S(k)(1 − k Rg/3) (1.10)

1.5 Probability distribution

The probability P (R,N) to ﬁnd a chain with end-to-end vector R for a chain with N segments is given by a summation of the z possible position the chain can have before the last step from N − q to N. Of course each of these positions have there own probability P (R − bi,N − 1):

z 1 X P (R,N) = P (R − b ,N − 1) (1.11) z i i=1

6 Since the chains are very long , i.e. N 1 and |R| |bi| we can use the Taylor expansion to two variables, one of which is a vector:

∂f(r, a) f(r + dr, a + da) = f(r) + dr · f(r, a) + da (1.12) Or ∂a 1 1 ∂2f(r, a) + drdr : f(r, a) + da2 + O(3) 2 OrOr 2 ∂a2

We use this expansion to ﬁnd an expression for P (R − bi,N − 1) in terms of P (R,N). In the case of the ideal chain da = −1 and dr = −bi so that we have now

z 1 X ∂P (R,N) P (R,N) = {P (R,N) − b · P (R,N) − (1.13) z i Or ∂N i=1 1 1 ∂2P (R,N) + b b : P (R,N) + ...} 2 i i OrOr 2 ∂N 2 Let us look at each of the terms. First note that all the terms independent of i can be taken out of the summation, i.e. the ﬁrst, third and ﬁfth term. The last term can be neglected, assuming that N 1. Next, note that the second term in the equation disappears because the last step i towards the end of the polymer can come from every direction, so

z X bi = 0. (1.14) i=1

1 Pz The fourth term is less straightforward to deal with. Apparently for z i=1 bibi we have

z z 3 1 X 1 X X ∂ ∂ bibi : OrOr = bi,αbi,β z z ∂rα ∂rβ i=1 i=1 α,β=1 3 2 2 2 X ∂ ∂ b ∂ = b δαβ = 2 (1.15) ∂rα ∂rβ 3 ∂R α,β=1 In making the second step in the derivation we assumed that the polymer is located on a rectangular lattice. Using Eq. 1.14 and 1.15 in Eq. 1.13 results in the diﬀerential equation

∂P (R,N) b2 ∂2P (R,N) = , (1.16) ∂N 6 ∂R2 7 Which has the solution,

3 −3R2 P (R,N) = ( )3/2 exp 2πNb2 2Nb2 3 −3R2 = ( )3/2 exp (1.17) 2πhR2i 2hR2i

As you see, this is a Gaussian distribution of the probabilities. It is important to note that any model can be used to calculate hR2i, which we will use later when we discuss non-ideal polymers.

1.6 Entropic spring

The probability density to ﬁnd a polymer coil with an end-to-end vector length R can be directly related to the entropy of the coil, which is given by the number of ways WN (R) to obtain a coil with an end-to-end vector R:

S(R) = k ln WN (R), (1.18)

where k is the Boltzmann factor. WN (R) is directly related to the probability density P (R,N) to ﬁnd a polymer with N monomers with a size R via some constant, so that

S(R) = k ln P (R,N) + const. (1.19) 3 = − kR2/Lb + const. (1.20) 2 where we used the general result for a Gaussian chain, Eq. 1.17. Knowing the entropy of the chain, the free energy F = E − TS can be readily calculated. Since we are dealing with an ideal polymer, there are no interactions between the polymer segments so that E = 0 and

F = ln P (R,N) + const. (1.21) 3kT R2 = + const. (1.22) 2Lb

Now suppose that the chain is submitted to a force f~, resulting in small changedR~ of R. The required work dF is given by dF = fd~ R~. So to calculate the force needed to extend the polymer we have

8 PN (RN)

1/2 N l Rx

Figure 1.2: The probability density of ﬁnding an end-to-end vector of length R.

9 dF 3kT f~ = = R.~ (1.23) dR~ Lb The Hookian law for solid materials says that the force needed for deformation is propor- tional with the deformation times a modulus G, i.e. f~ = GdR~. With Eq. 3.1 we ﬁnd that this entropic modulus is

3kT G = (1.24) Lb There are a few important notices to make. First, Eq. 1.24 shows that the modulus decreases with increasing contour length so that long polymer chains are very susceptible to external actions. Second, the factor kT is typical for the entropic nature of the elasticity soft matter systems. Of course when the L is small the derivations is faulty, because at many points we used N 1. Also we didn’t consider the eﬀect of stiﬀness of the polymer, which we will treated in the next chapter.

10 Chapter 2

Non-ideal polymers

2.1 Fixed valency angle polymers

The conclusion RN1/2 is valid for ideal chain with any ﬂexibility mechanism. Many polymers have, however, a ﬁxed valency angle γ between the segments of length b and free internal rotation φ, see Fig. 2.2. 2 The direct result of such a model assumption is that h~ui~uji = b hcos θiji, and does not equal zero as is the case for ideal polymers. Here θij is the angle between segment i and j. Following Eq. 1.2 we have for our non-ideal polymer

N N ~ 2 2 2 X X hR i = Nb + b h~ui~uji (2.1) i=1 j=1,6=i

To advance we need to evaluate hcos θiji. Between two neighboring segments per deﬁnition hcos θi, i + 1i = cos γ after averaging over all angles. For j=i+2 we have hcos θi, i + 2i = cos γ2. Continuing this argument, we ﬁnd

k hcos θi,i+ki = (cos γ) (2.2)

Using this in Eq. 2.1 we have

11 φ γ

Figure 2.1: Geometry of the angles for a ﬁxed valency angle polymer. 12 N N−i ~ 2 2 2 X X hR i = Nb + 2b hcos θi,i+ki (2.3) i=1 k=1 N N−i X X = Nb2 + 2b2 (cos γ)k i=1 k=1 N X cos γ = Nb2 + 2b2 1 − cos γ i=1 cos γ = Nb2 + 2Nb2 1 − cos γ 1 + cos γ = Nb2 1 − cos γ (2.4)

For the 2nd step, see appendix. Obviously if γ = 90 the result for the ideal polymer is recovered, thus a posteriori justifying the 90 lattice we assumed in deriving the gaussian chain.

2.2 Persistence length of a polymer chain

Though the angle between two segments is highly correlated, this correlation disappears when making more steps. We now calculate how many steps need to be made in order to loose the correlation.

k hcos θi,i+ki = (cos γ) = exp(k ln cos γ) (2.5) = exp(−k| ln cos γ|) kb = exp(− ) b/| ln cos γ| = exp(−s/˜l)

Here we use the deﬁnitions of the persistence length ˜l = b/| ln cos γ|, which is a measure of the part of the contour of the polymer, given by kb, over which the correlation with the tangent of the contour is lost. This is a principle that is also valid for non-ﬁxed valency angle an can be illustrated for a bending polymer segment, see Fig. , and is more generally expressed by

˜ hcos θuˆ0,uˆs i = exp(−s/l). (2.6) If s ˜l then we eﬀectively are dealing with a rod, if s ˜l then we are dealing with a polymer coil.

13 ^ ^ u u0 s

Figure 2.2: Tangent of the polymer decorrelates when over a distance s of the polymer contour.

14 Since eﬀectively after distance ˜l the angles are uncorrelated, just as is the case for the ideal free joint polymer. Hence we to calculate the end-to-end vector for a stiﬀ polymer we can use the result for ideal polymers Eq. 1.2 and replace b by ˜l:

√ rL p R = phR2i = Nb2 ∝ ˜l2 = N˜˜l. (2.7) ˜l Here we used that N˜ = L is eﬀectively the number of segments. Note that still R ∝ L1/2. ˜l * In the second step of Eq. 2.5 we used ln(cosγ)k = k ln cos γ, so (cosγ)k = exp k ln cos γ

2.3 Kuhn segments

Because one can, as an experimentalist, not walk over the contour, the persistence length in all its meaningful beauty is not measurable. Therefor the Kuhn segment l is often used to quantify the stiﬀness of a polymer, since it is given by the ratio of the gyration and the contour length, which can both be experimentally determined:

hR2i 6R l ≡ = g . (2.8) L L The relation between the persistence and the contour length can be calculated in case of the ﬁxed valency polymer:

1 + cos γ 1 + cos γ hR2i = Nb2 = Lb . (2.9) 1 − cos γ 1 − cos γ

1+cos γ from which, with Eq. 2.9, it follows that l = b 1−cos γ and with the result for the persistence length ˜l = b/| ln cos γ|:

l 1 + cos γ = | ln cos γ| (2.10) ˜l 1 − cos γ In Fig. 2.4 it can be seen that for a broad range of γ this ratio is 0.5, so that the stiﬀness can indeed directly be determined experimentally.

2.4 Occupied volume

It is very instructive to understand the character of a polymer to consider the volume a polymer Vpol actually occupies. Using the result of the end-to-end vector of an ideal polymer we have

15 0 . 5

0 . 4

0 . 3 p l /

l 0 . 2

0 . 1

0 . 0

- 0 . 2 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 1 . 2 1 . 4 1 . 6 γ [ r a d ]

Figure 2.3: Relation between the persistence length and the Kuhn length for varying valency angle.

16 4π 4π V = R3 = (Lb)3/2 (2.11) pol 3 3 If we assume that the backbone has a certain thickness d we can ﬁnd the fraction of the volume that is occupied by the polymer

d2L b d φ = ∝ ( )1/2( )2 1 (2.12) pol 4π 3/2 3 (Lb) L b

From Eq. 2.12 we see that polymers have very open structures. Replacing b with ˜l the structure is even more open.

2.5 Obtaining persistence length experimentally

Bending a segment means that the tangent of that segment is changed by an angle dθ over a certain distance ds. Since the energy needed to bend is independent of the direction in which dθ 2 the segment is bent the ﬁrst important term ( ds ) . The force needed to bend depends on the material via a material constant κ. To calculate the Hamiltonian of the total chain we need to integrate the contributions of all segments over the full contour

κ Z L dθ H = ds( )2 (2.13) 2 0 ds Let’s assume that the angle as a function of position along the contour can be described by a Fourier series:

X θ(s) = aq sin(s · q) (2.14) q

Implementing this in the Hamiltonian gives

κ Z L X X H = ds a a0 qq0 cos(s · q) cos(s · q0) (2.15) 2 q q 0 q q0 κ X X 0 0 0 = δ 0 a a qq cos(s · q) cos(s · q ) 2 qq q q q q0 κ X = a2q2 2 q q

17 We will use this hamiltonian to obtain the average bending in the ﬁlament. For this ﬁrst the partition sum Z = exp{−βH} needs to be calculated:

κ Z dθ Z = exp{−β ds( )2} (2.16) 2 ds Z Z κ X = da da ... exp{−β a2q2} 1 2 2 q q Y Z κ = da exp{−β a2q2} q 2 q q Y = zq, (2.17) q

where zq is the contribution of the wave with wave vector q. To calculate the average square 2 amplitude we have to multiply Eq. 2.16 with aq, average over aq and correct for the total partition sum:

R 2 κ 2 2 daqa exp{−β a q } ha2i = q 2 q q R κ 2 2 daq exp{−β 2 aqq } 2 ∂ = − ln z (2.18) κq2 ∂β q

ln zq can actually be calculated:

Z Z κ 2 2 1 −x2 ln( daq exp{−β a q }) = ln( dxe ) 2 q pβq2κ/2

1 1 Z 2 = − ln β − ln κq2/2 + ln dxe−x ) (2.19) 2 2 Inserting eq. 2.19 into eq. 2.18 gives

k T 1 ha2i = b = , (2.20) q κq2 ˜lq2 where ˜l = κ is again the persistence length. kbT Note that in deriving eq. 2.20 we have actually proven the equipartition theorem, but now for semi-ﬂexible polymers. It is (very) important to realize that exactly the same approach can be used to obtain the bending rigidity of membranes and interfaces. Especially membranes are very relevant for biological systems, i.e. cells. Eq. 2.20 supplies the experimentalist with a tool to determine the persistence length, given that the full contour can be observed.

18 In an alternative approach, also to motivate our new deﬁnition of the persistence length, we stay in real space and regard a small sequence s of the total contour. Also we consider now the tangents at both points under consideration:

Z s 0 2 2 κ 0 du(s ) 2 κ (us − u0) κ (us − u0) H = ds ( 0 ) ' 2 )s = (2.21) 2 0 ds 2 s 2 s With this hamiltonian we now calculate the average angle deviation, similar to Eq. 2.18 and 2.19:

R d(u − u )(u − u )2 exp{−β κ (u − u )2q2} h(u − u )2i = s 0 s 0 2 s 0 s 0 R κ 2 2 d(us − u0) exp{−β 2 (us − u0) q } s = , (2.22) ˜l where we have the same definition of the persistence length ˜l. What remains is to connect this definition with the decay in the correlation between the tangent at different points along the contour, similar to the result obtained for the fixed valency angle (see Kokhlov script). First note that the assumption in the derivation of Eq. 2.22 was that the range over which was integrated was very small. Hence one can consider the result in Eq. 2.22 as the first order in an expansion of an exponential. Next consider that

2 2 2 h(us − u0) i = 2(1 − h(us · u0) i) = 2(1 − hcos(θ0s) i). (2.23)

In can be shown that from this follows

2 ˜ hcos(θ0s) i = exp{−s/l}. (2.24)

Note: you don’t need to derive Eq. 2.24 that this is the case, but by now have the intuition that indeed this angle de-correlates exponentially.

2.6 Gaussian Chain Model

We have shown now that the ideal polymer chain can be replaced by a non-ideal chain that posses the same features. We want to extend this principle to calculate the elastic energy of a polymer. To do so, let us ﬁrst assume that an polymer coil of M segments consists of a sequence of subchains which are each long enough to obey gaussian statistics, i.e. M0 1 (where M0 is the number of segments of length b in this subchain), similar to the approach used to derive

19 Figure 2.4: Course graining a polymer in subchains, each of which are big enough to have Gaussian statistics so that its elasticity is well described.

20 the Radius of Gyration where we also divided the polymer in subchains. We coarse grain the ideal polymer, replacing the microscopic parameters M0 and b by

2 2 a = M0b N = M/M0, (2.25)

where N and a are phenomenological parameters. It is important to note that we made a similar replacement in Eq. 2.7 when we rescaled the polymer to its stiﬀ elements. These elements can be chosen according to whatever information is available on the polymer under study. Thus the polymer can be thought of as a succession of ”beads” i located at positions ~r1, ~r2, ..., ~ri, ..., ~rN connected by springs of length a along the vectors ~a1 = ~r−~r2, ...,~ai = ~ri − ~ri+1, ...,~aN−1 = ~rN−1 − ~rN . With Eq. and Eq. 1.24 the elastic energy of a subchain i is given by

3kT (~a − ~a )2 F~ = i+1 i . (2.26) i 2 a2 The total energy of the chain is then given by

N−1 3kT X F = (~a − ~a )2. (2.27) total 2a2 i+1 i i

2.6.1 Appendix A For x > 1 one can write down

N X i 2 N x = SN = 1 + x + x + ... + x (2.28) i=0

2 N+1 xSN = x + x + ... + x (2.29)

From which follows

N+1 (1 − x)SN = 1 − x (2.30)

21 N X 1 − xN+1 1 S = xi = (2.31) N (1 − x) ≈ (1 − x) i=0 However, in the second line of Eq. 2.3 we the summation starts at i = 1. There we need the identity

N N X X x xi = x xi = (2.32) (1 − x) i=1 i=0 to make the step to the third line.

22 Chapter 3

Polymer Dynamics

3.1 Rouse model

We now know how to obtain the static properties of a single isolated polymer coil. The next step is to calculate the dynamic properties of a unperturbed polymer in equilibrium, which can be done on the base of the Langevin equation. There are three forces to consider: a randomizing force θi mediated by the surrounding medium on a polymer segment i, a friction d~ai force of a segment of the polymer being dragged through the surrounding medium ζ dt , and a deformation force of the polymer coil. In principle one should also consider the interaction between segments of the polymer, which are due to steric and hydrodynamic interactions. If we again, as in Chapter 1 and 2, assume a ideal polymer which can cross itself, then the steric interaction can be discarded. In principle a moving segment induces a ﬂow which will be sensed by surrounding segments, which makes that the contribution of the hydrodynamics is very complex. This interaction will be also discarded. In order to calculate the deformation force of the polymer we regard the polymer with the Gaussian chain model, so that it consists of N coarse grained connected beads with a spring constant set by the entropic force needed for deforming a polymer, i.e. in this case a subchain of the polymer. Each bead experiences a force from its two neighbors

3kT f~ = [(~a − ~a ) + (~a − ~a )]. (3.1) i,i+1 a2 i+1 i i−1 i

Thus the Langevin equation for a bead i can now be written by

d~a 3kT ζ i = − [2~a − ~a − ~a ] + θ . (3.2) dt a2 i i+1 i−1 i 23 In the continuous limit ~ai is replaced by a function R(s, t) with the parameter s following the chain from 0 to N. In the continuous limit the elastic force is replaced by a second derivative

∂R(s, t) 3kT ∂2R(s, t) ζ = − + θ(s, t). (3.3) ∂t a2 ∂s2 So Eq. 3.3 is the Langevin equation of motion of a chain segment where we neglected the inertia of this chain segment. Since there is also no external force applied on the chain we have the boundary conditions

∂R(s, t) | = 0. (3.4) ∂s s=0,N Because Eq. 3.3 is a linear equation, the position of the segments R(s, t) can be expressed in eigen modes which are also the called the Rouse modes

1 Z N pπs Xp(t) = cos( )R(s, t). (3.5) N s=0 N This is a similar trick as was used earlier in the Kahn-Hilliard approach for spinodal de- composition, where the goal was to obtain density ﬂuctuations, whereas here we are looking for ﬂuctuations in the density of bead positions, i.e. the breathing of the polymer. The relation with the function R(s, t) is given by a Fourier transformation of these modes

X pπs R s, t) = X (t) + X (t) cos( ). (3.6) ( 0 p N p>0

To equation of motion for the Rouse modes is obtained by inserting Eq. 3.5 in Eq. 3.3, giving

∂X (t) Nζ 0 = θ (t) for P = 0 (3.7) ∂t 0 (3.8)

∂X (t) p2π3kT 2N p = −2Nζ X (t) + θ (t) for P > 0. (3.9) ∂t Na2 p p We will now use, without proving it, the ﬂuctuation-dissipation theorem

24 0 hθ0,α(t)θ0,α(t )i = (Nζ)(2kT )δαβ (3.10)

0 hθp,α(t)θp,α(t )i = (2Nζ)(2kT )δαβ (3.11)

Note that here the index p indicates the mode of the wave and α, β are indices for the individual beads. The effect of the fluctuation terms is thus to ensure that correlation functions only have a value when α = β. For p=0 In Eq. 3.5 we recognize for p = 0 the definition of the center of mass ~rcm as we used it in section 1.3, but now in the continuous limit. So Eq. 3.7 describes the center of mass motion of a polymer coil. The friction on the center of mass is the sum of all frictions of the individual segments and the corresponding diffusion constant D

kT D = , (3.12) Nζ scales as the inverse of the molecular weight. For p > 0 each Rouse mode performs a motion in a potential well. At equilibrium this motion is characterized by the correlation function

Na2 t hXpα(t)Xqβ(0)i = ζ 2 2 exp , (3.13) π p τr where

N 2ζa2 R2 τ = ∼ g , (3.14) r 3kT π2 D is the Rouse time. The Rouse time is the slowest mode of relaxation of the polymer and scales 2 like the molecular weight squared. Again it can be expressed in the very general parameters Rg and D and is thus applicable for any coarse grain model. In general in experiments the full spectrum of Rouse modes will be detected, obscuring the experiments. This result is called the Rouse model, after P. E. Rouse how published a paper with this model in 1953. The model suﬀers of two main deﬁciencies: it does not correctly incorporate hydrodynamic interactions between the beads and it allows chain segments to cross.

25 i

• M. Doi and S. F. Edwards , The Theory of Polymer Dynamics , Clarendon Press, Oxford

• P. G. de Gennes, Introduction to Polymer Physics, Cambridge University Press

• P. G. de Gennes, Scaling Concepts in Polymer Physics, Cornell University Press

• J.-L. Barrat and J.-P. Hansen, Basic Concepts for Simple and Complex Liquids Experimental Rouse: Moduli (exact) Experimental Rouse: Moduli (exact)

Experimental Rouse: Mean Square Displacement (NMR)

D ∼M-1

But data taken in polymer Melt!!! Add hydrodynamic interactions:

NEXT PROBLEM: Rouse valid in polymer melts BUT only till a critical molecular weight! Molecular weight dependence of the diffusion Add that crossing is not allowed:

Direct observation of reptation

J. Kas: et al., Nature (1994); Biophysical J. (1996).

Fluorescently labeled a small fraction of ACTIN filaments in an entangled solution. Overlay chain contours from 60 micrographs taken at 0.1 sec intervals

Lm ≈= 4.0/1 μρm

• Conditions: 0.1 mg Actin / ml water Lmm =≈1/ρ 0.51 μ • The polymer rapidly explores a tubelike region of topologically accessible contours. • Over much longer times, reptation can be observed

Time sequences (At = 8 s) of the reptation motion of actin filaments of 40.0 ,um (right side) and 6.9 ,um length (left side) in a matrix of F-actin of 0.1 mg/ml or 2.0 mg/ml, respectively.