# 4 Microscopic dynamics

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1 4 Microscopic dynamics In this section we will look at the first model that people came up with when they started to model polymers from the microscopic level. It s called the Oldroyd B model. We will spend one double lecture to derive it and in the last lecture, look at its macroscopic (top-down properties: Linear rheology Stress in shear Stress in extensional flow 4.1 The stress tensor Remember we had the three Navier-Stokes equations (in incompressible form: u = 0 This is mass conservation: integrated over a volume of space, it tells us that fluid flow in matches fluid flow out, so fluid does not accumulate in one place. ( u ρ t + u u = σ σ = pi + η( u + ( u In the last two equations, σ is the stress tensor. The physical meaning of this is calculated from the force the force exerted across a small surface element with area δs and unit normal n: n The force acting on the top surface of the tiny pillbox is σ n δs. 4.2 Beads in a viscous fluid Probably the simplest object you could put into a flow is a solid spherical bead. Let s look briefly at the motion of a microscopic solid sphere in a Newtonian fluid. There is really only one parameter in Newtonian fluid flow (once you ve made everything dimensionless: the Reynolds number, Re, which gives the ratio of inertial to viscous forces. If the scale of the flow is very small, then the Reynolds number is very small and we can assume it is zero. Then the governing equations are the Stokes equations: u = 0 η 2 u = p. 17

3 which also depends on the typical size of the relaxed polymer coil. If the bead at x is moving with velocity U 1, and the fluid around it has velocity u(x, the drag force on the bead will be 6πηa[u(x U 1 ]. The total force on the bead (which must be zero is so we can work out the velocity of the bead: 3kT a 2 r + 6πηa[u(x U 1] = 0 U 1 = kt 2πηa 3 r + u(x. The other bead is positioned at x + r and the spring force on it is λr so its velocity is U 2 = kt r + u(x + r. 2πηa3 In the absence of Brownian motion, these two velocities tell us how the bead positions evolve: dx dt d(x + r dt = kt 2πηa 3 r + u(x = kt r + u(x + r 2πηa3 Finally we subtract the two to get the evolution of r: dr dt = kt r + u(x + r u(x πηa3 Since r is microscopically small relative to the scale over which u changes, we can use a Taylor series for u(x + r: dr dt = kt πηa r + r u(x 3 which is correct up to terms of order r. The constant 2kT/πηa 3 is called 1/τ; it represents the inverse relaxation time of the dumbbell following distortion caused by the flow. dr dt = 1 2τ r + r u. 19

4 4.3.3 Stress in the fluid Now suppose we have a suspension containing many of these dumbbells. What will the extra stress they contribute be? Of course, the fluid they are suspended in, which has its own viscosity η, will continue to contribute a Newtonian stress; we re just looking for the polymer extra stress, σ p. Again, we look at the tiny surface element with area δs and unit normal n. interested in dumbbells that cross the surface. n We are The force associated with a single dumbbell that crosses the surface is λr; how many will cross the surface? Clearly a long dumbbell is more likely to cross the surface than a short one; and a dumbbell which is lined up with n is more likely to cross the surface than one which is perpendicular to it. If there are m dumbbells per unit volume, then we expect the number crossing our surface element to be mr n δs. Then the extra stress exerted by the dumbbells is σ p n δs = λr(mr n δs so σ p = mλrr = 3πηam rr. 2τ Again, we introduce a name for the new constant: this time, G = 3πηam/2τ so that σ p = Grr Adding Brownian motion We have worked out how a dumbbell will evolve if its motion is deterministic, and the stress it will cause. But to complete the model, we need to add Brownian motion of the two beads. We simply add a standard 3D Brownian motion (i.e. each of the x, y, z components is a Brownian motion to the evolution of the vector r: ( dr = 1 2τ r + r u(x dt + τ 1/2 db t We have chosen τ 1/2 as the coefficient for two reasons: it has the correct dimension; but also because (as we will see later this leads to a dumbbell with no flow having a natural length of 1. 20

5 Because we eventually want to work out the extra stress, we then define the conformation tensor A(x, t = IE [r(x, tr(x, t] which will give us σ p = GA. Now we use the first-step method (see GM01 part 1. As we take a time-step dt, the position x of our dumbbell will move to So when we take our time-step, we get x + dx = x + u(xdt. A(x + u(xdt, t + dt = IE [(r + dr(r + dr] = IE [r r + r dr + dr r + dr dr] A(x + u(xdt, t + dt = IE [r r] + IE [r dr] + IE [dr r] + IE [dr dr] Now let s look at these four terms separately. As usual, we only keep terms up to order dt. IE [r r] = A(x, t [ (( IE [r dr] = IE r 1 [ = IE r ] 2τ r + r u(x dt + τ 1/2 db t (( 1 2τ r + r u(x ] dt = 1 Adt + A udt 2τ [(( IE [dr r] = IE 1 ] 2τ r + r u(x dt + τ 1/2 db t r [(( = IE 1 ] 2τ r + r u(x dt r = 1 2τ Adt + ( u Adt [(( IE [dr dr] = IE r (( 2τ + r u dt + τ 1/2 db t r ] 2τ + r u dt + τ 1/2 db t = IE [( ( ] τ 1/2 db t τ 1/2 db t = τ 1 Idt We also need to look at the left-hand side term. A(x + u(xdt, t + dt = A(x + u(xdt, t + dt A(x, t + dt + A(x, t + dt = (udt A + A(x, t + dt = (udt A + A(x, t + dt A(x, t + A(x, t = (u Adt + A dt + A(x, t t 21

6 Putting these all together gives: ( A + (u A dt = 1 1 Adt + A udt t 2τ 2τ Adt + ( u Adt + 1 τ Idt A t + (u A A u ( u A = 1 τ and the polymer extra stress is σ p = GA. ( A I The terms on the left-hand side of the A equation are called the co-deformational time derivative because they represent what happens to a line element that is deforming with the flow. We have now derived the Oldroyd-B equations: A u = 0 ( u ρ t + u u = σ σ = pi + η ( u + ( u + GA t + (u A A u ( u A = 1 τ ( A I In the final lecture we will see how the fluid corresponding to these equations behaves. 22

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