High Elasticity and Stretching of an Individual Polymer

We have said that high elasticity is a common property of polymer networks.

However, this sounds a bit too general. Let’s zoom in, and examine what

116 Giant Molecules: Here, There, and Everywhere

it implies for particular molecules. Figure 7.2 b portrays a typical polymer network. You can see a set of long molecular chains, bridged together with cross-linkers (covalent chemical bonds). It looks like a kind of framework in three-dimension. What would you regard as an elementary “brick” of such a structure; what is the smallest piece we have to consider if we want to understand network deformation? The answer is clear from Figures 7.2 b and c: It is a strand of the chain between two neighboring cross-links (bridges). We shall introduce a new word, subchain, for such strands.

Because polymer chains are so flexible and randomly coiled, it is usually sufficient to uncoil slightly some of the chains to achieve a rather significant deformation of the network. Thus, if a polymer network is stretched, many subchains will be somewhat uncoiled, while some of them will be actually somewhat compressed (Figure 7.2 c). Therefore, the elasticity of the whole polymer is a sum of the elasticities of all the individual subchains. This is why it makes sense to explore elastic properties of a single subchain first, before looking at the whole network.

Thus, let’s see what happens if a single chain is pulled upon by an external force f , as shown in (Figure 7.3). Actually, such a single chain stretching experiment can be done — which is a marvelous experimental achievement. It was first done by C. Bustamante and his co-workers in the University of Oregon in 1992, and then repeated in many laboratories around the globe. There are several versions of the apparatus by which experimenters can manipulate single molecules, such as optical tweezers, magnetic tweezers, and some others; we cannot talk about it here, but encourage the reader to read on this subject, for instance, in the article [52]. It is actually somewhat ironic that the most convenient polymer for performing such single chain stretching experiment is nothing lesser than double helical DNA. Needless saying, experiments with DNA are exciting because they shed light on the properties of this most important of all molecules — and we will return to this point (see Section 7.12). But for now let’s just use the example of DNA to sort out the basics of polymer physics.We imagine that one chain end is attached to an immobile support at the origin (we always can choose origin as we like), then the position of the other end, where the force is applied, is described by the familiar end-to-end vector R (see Figure 7.3). We want to find the average value of R as it depends on the applied force f .

It turns out useful to think first about the opposite problem. Suppose we want to maintain the chain’s dangling end at the given position R; what kind of force f , on average, should we apply in order to retain the desired

R

f R

D b f

a ^{Fig. 7.3} In order to keep a given

end-to-end vector R for a single polymer chain, we need to apply an external force ff. Similar figure in the first edition of this book was considered a rather abstract theoretical concept; right now this very experiment is performed rather routinely in many laboratories using DNA as a poly-mer to test, and optical or mag-netic tweezers to apply the force.

Our cartoon illustration is unrealis-tic only in one aspect, namely, the size of the bead (to which the force is applied) in these experiments is an order of magnitude larger than the DNA coil size. Panel (a) shows the situation with weak force, when chain makes numerous loops and remains randomly coiled. Panel (b) depicts strong force regime, when chain makes only limited excursions in perpendicular direction, characterized by the length scale D; this situation is more fully considered below in Section 7.12.

value of R? You may be a little surprised by this last question. Is a force really needed to keep R unchanged? As we learned in Section 6.7, even with no force at all the end-to-end vector may have any possible value, including the value R we wish. There is no doubt about that. However, without the force, if f = 0, the dangling end will not stay at the desired point R for any length of time. It will go on fluctuating. All directions of R will be equally likely. This is why on average the end-to-end vector will be equal to zero, just as you expect from the symmetry of the distribution PN(R) (6.16). Hence, in order to keep R fixed, you need to use a force f . You can guess, based on a simple symmetry argument, that f should point in the same direction as R (Figure 7.3); indeed, what other direction might it point to? It is an external force, i.e. a force from some external object, acting on the chain. And what about the force that the chain in its turn exerts on the external object? Newton’s third law tells us that it should point in the opposite direction, i.e. towards the origin of coordinates. So it is a restoring, elastic force. Thus, we have naturally come to the conclusion that a polymer chain resists being stretched. In other words, it exhibits elasticity.

We encourage the reader to re-think the above paragraph to realize that all our arguments are perfectly applicable even to the simplest model of a chain — a freely-jointed polymer chain made of a large number N 1 of

118 Giant Molecules: Here, There, and Everywhere

elementary segments, each of the same length ` and, moreover, with all vol-ume interactions between segments completely neglected (see Section 6.3).

(The latter approximation is known as an “ideal polymer chain”; we will come back to it in Section 8.1, but for now we can think of it as just sticks of practically zero thickness).

Our arguments may not satisfy you completely. Indeed, what is the physics of polymer chain elasticity if it exists even for inextensible seg-ments with no interactions? To answer this, let’s think of an ordinary solid crystal. What makes it elastic? When a crystal is stretched, the atoms are pulled further apart (Figure 7.4). Thus, the elastic force in this case would be a result of many interatomic interactions. Sometimes it helps to describe the same thing in terms of energy. The undeformed crystal is in equilibrium; that is, the potential energy of interatomic interactions is a minimum (Figure 7.4). An external deforming force pulls the atoms up the slope from the bottom of the potential well. Suppose an external force f causes an elongation of the crystal, x. The work it does, f x, is used to increase the internal potential energy U of the crystal, which is the total energy of interatomic interactions: f x = U , or:

f = U

x . (7.1)

Equation (7.1) gives you a recipe for finding the stretching force f (and the elastic force of the crystal, which is opposite to it). All you need to know

a u

b

r r₀ r₀

r₀+$r

F F

Fig. 7.4 An illustration of the elasticity of a crystal. (a): Initially, in an non-deformed crystal, the distance, r0, between any two neighboring atoms corresponds to the minimum of the interactional potential energy U(r ). (b): To stretch the sample, we have to increase distances between atoms, to make each of them r0+ r , and the displacement r determines the shift away from the minimum of potential energy: we have to increase the potential energy, which is why the crystal develops the force of elastic response.

is how the crystal is constructed. Then you should be able to work out its internal energy U .

Now let’s go back to a polymer chain. This time, we are not talking about a stretched array of atoms, as in the case of a crystal. What we really see when a polymer chain is stretched is an increase in the end-to-end distance (Figure 7.3). The only way this can happen is, of course, if some wiggled bits of the chain straighten up and disentangle. This is particularly striking and easy to follow if we think of a freely-jointed (Figure 2.5 b) ideal chain. (Once again, ideal chain approximation assumes that the only interaction between the neighboring monomers comes from their being joined together into a chain. All other monomers do not interact at all, just like the molecules in an ideal gas. You will be getting used to this approximation.)

Suppose we apply a force f , and change the end-to-end vector R by some value R. Hence, we did an amount of work f R. Where did the energy go? Previously, when we looked at crystals, we managed to find the answer quite easily. Unfortunately, the way we did it would not work in this case. In an ideal system, interactional potential energy is zero, both before and after the deformation. Thus, Equation (7.1) is of no use. As for kinetic energy of molecules, it is determined by the temperature. If the temperature is constant during the deformation, the kinetic energy will not change either. . . Are we lost?

Before giving up, let’s think of another analogy. Strange as it may seem, help comes from an ideal gas. In a sense, an ideal gas has “elasticity the other way round”. Suppose you wanted to hold the gas under a piston (Figure 7.5) in a vessel of a certain volume (it is just like holding a polymer chain to retain a non-zero R). You would have to apply a squeezing force f = pA, where p is the gas pressure, and A is the area of the piston.

In order to reduce the volume, you would always need to do some work.

$x x

Fig. 7.5 Ideal gas in a vessel with a piston before (left) and after (right) the compression by the amount x.

120 Giant Molecules: Here, There, and Everywhere

Nevertheless, both the potential and kinetic energy of the gas molecules stay the same, given that the compression is isothermal. So where on earth does the work go? (As you see, we end up with the same question once again.)

Of course, in the long run, all the work transforms into heat which is dissipated in the surroundings. How do we know? Well, if there were no sur-rounding medium to take the heat away, both a gas when compressed and a polymer when stretched would get warmer (see also below Section 7.11).

Does that mean elasticity of a polymer chain depends on the environment which absorbs the heat? Well, we know that the pressure of an ideal gas does not depend on the type of the environment, so maybe there is some-thing similar for a polymer?

Also, is there any hope we can still manage with simple energy argu-ments? Or do we need to tackle the problem by means of mechanics, tracing all the molecules? It would not be hard for an ideal gas. Its pressure is just an overall result of all the individual hits by the molecules on the piston. This concept immediately leads to the ideal gas equation of state (we encourage the reader to reproduce this derivation, it is very beautiful).

However, there is no such simple picture for a polymer. Even in the case of an ideal chain, the motion of segments is extremely complicated, due to knots and entanglements.

But let’s think. We know that the surrounding medium plays no other role but to maintain the constant temperature T . Our proof of polymer elasticity was based on a very general idea. Indeed, we showed that when a polymer chain is stretched, it is pulled from a more probable to a less probable state. Hang on a minute! Is there perhaps some universal way of finding the energy cost of lowering the probabilities at a constant tempera-ture, without getting bogged down in the mechanics of molecular collisions?

There is indeed a very general rule, known as the Boltzmann princi-ple. It states the following. Suppose there are ⌦ ways in which molecules can occupy a certain state. (In our case, this number is proportional to the probability P (R) — see Equation (6.16)). Then we need to find the quantity

S = kBln ⌦ , (7.2)

where kB is Boltzmann’s constant. The energy equivalent of probability we are seeking is the change in the value:

Ue↵ = T S , (7.3)

where T is the absolute temperature. In the case of a polymer chain, according to (6.16),

S(R) = kB

3R²

2N `²+ const , (7.4)

or

Ue↵(R) = kBT 3R²

2N `² + const , (7.5)

where const is a quantity independent of R (which arises because ⌦ is proportional to P (R), not equal to it). Using Equation (7.5), we can easily find the elastic force from (7.1). We shall do so a little later, in Section 7.9.

Thus, the Boltzmann equation S = kBln ⌦ has rescued us when we had nearly lost hope. It is not surprising that this formula was engraved on the tombstone of the author, Ludwig Boltzmann (1844–1906)! But what does it mean, and where does it come from? It is an interesting question in its own right. We shall devote the next two sections to it, and then come back to the discussion of high elasticity.