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Rubber elasticity
Marc R. Roussel
Department of Chemistry and Biochemistry
University of Lethbridge
February 21, 2009
A rubber is a material that can undergo large deformations (e.g. stretching to five or ten times its original length) and then return to its original shape and size. Many polymeric materials, and not just rubber itself, display rubber elasticity in some range of temperatures. When we characterize these materials, we find that they are made of long chains, with occasional cross-links between the chains. In the resting state, these chains will adopt random coil configurations. As we apply a stress by pulling on a sample of rubber, the chains will tend to align along the stress, which allows the material to stretch. The cross-links pull the material back to its original shape once the stress is removed. In this note, we will look at a statistical treatment of rubber elasticity. This will allow us to relate the elastic behavior to properties of the polymer.
1 Cross-linking statistics
Figure 1 shows two primary polymer molecules which have formed some cross links. There are both internal cross-links (within a single molecule) and external cross-links (between molecules). Both types can be important to rubber elasticity. The important statistic for rubber elasticity is in fact the number of chains bounded by two cross-link junctions, a quantity called the number of active chains, denoted νe. Free ends are uninteresting because they can be displaced more-or-less freely. If we have N primary polymers, there will be 2N free ends. Let ν be the number of monomers involved in cross-links (two per cross-link). Each cross-link takes two chains and divides them into four, so the number of chains added by making ν/2 cross-links is equal to ν. From this number, we have to subtract the number of chains that end freely. Thus,
νe = ν − 2 N = ν (1 − 2 N/ν).
Now let M be the average molar mass of the primary polymers, and Mc be the molar mass per cross-linked monomer, i.e. Mc = N M/ν. Another way to look at Mc is that it’s the average molar mass of the active chains. We can then rewrite the last equation
νe = ν (1 − 2 Mc/M ). (1)
Figure 1: Two primary polymer molecules (i.e. two polymer chains) which have formed some cross-links, shown as bold dots. The chains are colored differently for ease of visualization.
2 Thermodynamics of elasticity
When we write down the differential of U in courses in chemical thermodynamics, we nor- mally only consider pressure-volume work. Here, we need to also consider the extension work. Suppose that f is the externally imposed (equilibrium, i.e. reversible) tension on an elastic body. Then dw = −P dV + f dL,
where dL represents a change in the length of the body. You will recognize the new term as a special case of the general definition of mechanical work as force times distance. Then we have dU = dw + dq = −P dV + f dL + T dS.
In this form, we have a differential of U in terms of the variables (S, V, L). Physically, these are not the most convenient variables. We would like to rewrite U as a function of (T, V, L). To do this, we need to rewrite the differential of S in terms of those variables:
dS =
∂S
∂T
V,L
dT +
∂S
∂V
T,L
dV +
∂S
∂L
T,V
dL.
If we now substitute this equation into the differential of U , we get
dU =
T
∂S
∂V
T,L
− P
dV +
T
∂S
∂L
T,V
dL + T
∂S
∂T
V,L
dT.
∂U
∂L
T,V
= T
∂S
∂L
T,V
or f =
∂U
∂L
T,V
− T
∂S
∂L
T,V
Materials scientists thus usually discuss deformation using the stress-strain relationship. (The term stress-strain relationship is used even when the strain is described in terms of the stretch ratio.) From the definition of λ, we have L = λL 0. Therefore
∂ ∂L
L 0
∂λ
Equation 3 therefore becomes
f = −
T
L 0
∂S
∂λ
T,V
4 Statistical theory of rubber elasticity
To calculate the entropy of a rubber, we will use the Boltzmann equation:
S = k ln W,
where W is the number of microstates. The active chains may have different lengths, which affects W. We can compute W by
W =
n
(Wn)νn^ ,
where νn is the number of chains containing n bonds. This gives
S = k
n
νn ln Wn.
We need to reinterpret this equation a little: In our case, we can’t count the microstates. Suppose that I give you the probability density pn(λ) that a given n-mer has end-to-end is stretched by a factor λ from its resting length. The number of microstates between λ and λ+dλ is proportional to pn(λ) dλ. Since we are going to take a logarithm, the proportionality constant, as well as the factor of dλ, end up in an additive constants:
S = k
n
(νn ln pn(λ) + constant).
Our next step will be to take the derivative in equation 4, so the constant will vanish:
f = −
kT L 0
n
νn
∂ ln pn(λ) ∂λ
T,V
Cross-links occur where two primary polymers have come sufficiently close together in the right orientation. Cross-linking should therefore have only a small effect on the locations of the monomers which became cross-linked. The distribution of relative positions of the two
cross-link junctions defining an active chain should therefore obey the Gaussian probability density derived in the last lecture:
pn(x, y, z) = (q/
π)^3 e−q (^2) (x (^2) +y (^2) +z (^2) ) , (6)
where q^2 = 3/(2n^2 ). We also had 〈r^20 〉 = n^2 = 3/(2q^2 ), where 〈r^20 〉 is the mean squared end-to-end distance of the relaxed polymer. We now introduce an affine transformation: We assume that the deformation of a sample of rubber stretches all the coordinates homogeneously by factors λx, λy and λz in each of the three Cartesian directions. Thus, (x, y, z) = (λxx 0 , λyy 0 , λz z 0 ), where (x 0 , y 0 , z 0 ) are the initial relative coordinates of the end of a chain. Taking a logarithm of equation 6 and introducing the affine transformation, we get
ln pn = −q^2
[
(λxx 0 )^2 + (λyy 0 )^2 + (λz z 0 )^2
]
where again we won’t be too concerned with the constants since they won’t contribute to the force. Rubbers are essentially incompressible, so we must have λxλyλz = 1. Suppose that we stretch a sample of rubber along the x axis. Let λx = λ. We will typically find that λy = λz.^1 Substituting this relationship into the incompressibility condition, we get λy = λz = 1/
λ. This gives ln pn(λ) = −q^2
[
λ^2 x^20 + (y^20 + z^20 )/λ
]
There is just one more thing: If we have νn polymers of length n, then we should average this quantity over all the initial end-to-end distances:
ln pn(λ) = −q^2
[
λ^2 〈x^20 〉 + (〈y^20 〉 + 〈z 02 〉)/λ
]
For the relaxed polymer, there is nothing special about the x axis, i.e. 〈x^20 〉 = 〈y^20 〉 = 〈z^20 〉. Since 〈r 02 〉 = 〈x^20 〉 + 〈y 02 〉 + 〈z^20 〉, this gives 〈x^20 〉 = 〈y 02 〉 = 〈z^20 〉 = 〈r 02 〉/3. We therefore obtain
ln pn(λ) = −
q^2 〈r^20 〉 3
λ^2 + 2/λ
λ^2 + 2/λ
Note that q and 〈r^20 〉 have both dropped out of the non-constant part of this expression. The derivative in equation 5 therefore doesn’t depend on n:
∂ ln pn(λ) ∂L
T,V
λ − 1 /λ^2
It can therefore be pulled out of the sum, leaving
n νn^ =^ νe, the number of active chains. Putting it all together, we get the elastic force
f =
kT νe L 0
λ − 1 /λ^2
(^1) Some polymers do deform differently in different directions, but then we would have to orient the sample correctly with respect to any special axes to observe this phenomenon.
