Soft Matter Physics - Solutions to Exercises Part 3

Hence the energy dissipation function Φ is obtained as Φ = πηh ∫(from 0 to a) dr r ( (4r-h²)/σ² )² = 4πηh² (7.5) (c) The free energy is given by A = −2πaħγ_s + 1/2 π a² ρgh² (7.6) Hence Ȧ = −2πaγ_s ḣ + π a² ρgh ḣ (7.7) The equation of motion is 8πηḣ = 2πaγ_s − πa²ρgh (7.8) or ḣ = (a²γ_s)/(4ηh) (1 - h/h_c) (7.9) with h_c = (2γ_s)/(ρga) (7.10) 7.2 (a) The flux of the particles is given by j = -D_c(n) ∂n/∂x + vn (7.11) Since the particles are confined in the box, their flux must be zero in the steady state. Hence ∂n/∂x = vn / (D_c(n)) (7.12) (b) The change of the particle density Δn over the distance l is estimated by Δn/n = 1/n ∫(from 0 to l) ∂n/∂x dx = ∫(from 0 to l) v / (D_c(n)) l ≈ (6πηa vl)/(k_B T) (7.13) For a = 0.1[μm] and v = 1[μm/s], and η = 1[mPas], the right hand side is equal to 400 for l = 1[mm]. Therefore, the particles are all accumulated near the wall of the box. This is also true even for the smallest particle of a = 0.01[μm]. 7.3 The particle concentration in the two boxes are n₁ = h n₀ / (h-x), n₂ = h n₀ / (h+x) (7.14) The free energy of the system is F_tot = A(h − x) f(n₁) + A(h + x) f(n₂) − Wx (7.15) The time derivative of F_tot is calculated as Ḟ_tot = A[Π(n₁) − Π(n₂)] − w]ẋ (7.16) where Π = n f'(n) - f is the osmotic pressure and w = W/A. The Rayleighian is therefore R = 1/2 ζ_m A ẋ² + A[Π(n₁) − Π(n₂)] − w]ẋ (7.17) (e) The equilibrium structure factor S_q(k) is given by S_q(k, t) at t=0. S_q(k) = (k_B T)/(N_c) * 1 / (a(T - T_c) + κ_s k²) = (k_B T)/(N_c) * 1 / (1 + ξ² k²) (8.47) where N_c = (k_B T)/(a(T - T_c)), ξ_c = (κ_s / (a(T - T_c)))¹/² (8.48) Hence N_c ∝ 1/(T - T_c) and ξ_c ∝ 1/(T - T_c)¹/². 8.6 (a) From eq.(8.82) (o(T) being set to zero), the free energy density of the system is obtained as f_el = (K/2)(∂u/∂x)² + G/2(∂u/∂y)² = (K/2)(1 + (4/3)(∂u/∂x)²) (8.49) (b) Since v_z = −[φ/(1 − φ)]u̇, the energy dissipation function becomes Φ = 1/2 ∫(from 0 to h) dx ξ(u̇ - v_z)² = 1/2 ∫(from 0 to h) dx ξ (u̇ / (1-φ))² (8.50) With the definition of κ (see eq.(8.97)), this can be written in the form of eq.(8.146). (c) The Rayleighian is written as R = 1/2 ∫(from 0 to h) u̇² dx/κ + K_e ∫(from 0 to h) [ (∂²u/∂x²) (∂u̇/∂x) + w(u̇(h) - u̇(0)) ] = 1/2 ∫(from 0 to h) u̇² dx/κ - K_e ∫(from 0 to h) [ (∂³u/∂x³)u̇ + ... ] = 1/2 ∫(from 0 to h) u̇²/κ dx + K_e ∫(from 0 to h) (∂²u/∂x²)u̇ - [∂u/∂x u̇]|(from x=h to x=0) + w(u̇(h) - u̇(0)) (8.51) Minimization of this functional with respect to u̇(x) (including u̇(h) and u̇(0) ) gives the partial differential equation (8.147) and the boundary condition (8.148). (d) The steady state solution of eq.(8.147) under the boundary condition (8.148) is u_∞(x) = -(w/K_e)x. To obtain the time dependent solution, we define f(x,t) by f(x,t) = u(x,t) + w/K_e x (8.52) then f satisfies eq.(8.147), and the following initial and boundary conditions f(x, 0) = w/K_e x (8.53) ∂f/∂x = 0 at x = 0, h (8.54) This can be solved by the method of separation of variables. The final solution of the equation is u(x,t) = -w/K_e [x - h/2 - Σ(from n=1,3,5,... to ∞) (4h)/(n²π²) cos(nπx/h) e^(-n²t/τ)] (8.55) where τ = h²/ (κ K_e π²) (8.56) Therefore Δh(t) = u(h, t) - u(0, t) = -w/K_e [h - Σ(from n=1,3,5,... to ∞) (8h)/(n²π²) (1 - e^(-n²t/τ))] (8.57) Therefore for long time,Δh(t) approaches to the equilibrium value -w/K_e with relaxation time τ. On the other hand, for short time t ≪ τ, Δh(t) decreases as −(κ_s K_e t)¹/². The equation of motion is ξ_m ẋ = w − [Π(n₁) − Π(n₂)] (7.18) If w is small, x is small, and eq.(7.18) is approximated as ξ_m ẋ = w − 2K/h ẋ (7.19) where K = n∂Π/∂n is the osmotic modulus. The solution of eq.(7.19) is x = (wh)/(2K) (1 - e^(-t/τ)) (7.20) with τ = (ξ_m h)/(2K) (7.21) (b) When the piston is moving with velocity ẋ, the fluid velocity is constant along the axis of the tube. On the other hand, the particles are not moving with the same velocity as the fluid because if all particles move with the same velocity as the fluid, a region empty of particles is created at the semipermeable membrane:) To maintain a uniform concentration, the particles must move relative to the fluid by diffusion. The relative velocity depends on the particle position: at the semi-permeable membrane, the relative velocity is -ẋ; while at the wall of the container, it is zero. Therefore, the energy dissipation due to the particle diffusion can be written as Φ = 1/2 α ξ (h + x)A ẋ². (7.22) where α is a certain number between 0 and 1. Therefore, the time evolution equation for x can be written as αξ(h+x)ẋ = w − [Π(n₁) − Π(n₂)] (7.23) For small x, this can be written as αξh ẋ = w − 2K/h x (7.24) The solution can be written in the same form as eq.(7.20), but the relaxation time is now given by τ = (αξh²)/(2K) (7.25) Notice that this is essentially equal to the relaxation time by diffusion h²/D_c. 7.4 (a) For the functional (7.52) in the text, the functional derivative is calculated as δA/δn(x) = k_B T(ln(n(x)) + 1) + U(x) (7.26) We denote the right hand side as f(x). Equation(7.131) in the text then becomes ṅ(x) = -∫ dx' μ(x,x')f(x') (7.27) = -∫ dx' ∂/∂x [n(x)/ζ δ(x - x')] ∂/∂x f(x') = ∂/∂x [n(x)/ζ ∂/∂x f(x)] = ∂/∂x [n(x)/ζ ∂/∂x [k_B T ∂n/∂x + n ∂U/∂x]] = (k_B T/ζ) ∂/∂x [n ∂/∂x [∂n/∂x + n/k_B T ∂U/∂x]] (7.28) (b) Since ∫ dx' μ(x,x') f(x') = − ∂/∂x [n(x)/ζ ∂f(x)/∂x] (7.29) we have ∫ dx ∫ dx' μ(x,x') n₁(x) n₂(x') = ∫ dx n₁(x) ∂/∂x [n(x)/ζ ∂n₂(x)/∂x] = -∫ dx (∂n₁(x)/∂x) [n(x)/ζ ∂n₂(x)/∂x] (7.30) where integration by parts has been used. Similarly. ∫ dx ∫ dx' μ(x',x) n₁(x) n₂(x') = −∫ dx' n₂(x') ∂/∂x' [n(x')/ζ ∂n₁(x')/∂x'] = ∫ dx' (∂n₂(x')/∂x') [n(x')/ζ ∂n₁(x')/∂x'] (7.31) which is equal to eq.(7.30). 7.5 (a) Since |ṙ| is constant, ṙ ⋅ r̈ = 0 (7.32) For such ṙ, ω² is equal to ṙ²/r². Hence the energy dissipation function is given by Φ = 1/2 ∫ dr ζ_r ω² ψ = 1/2 ∫ dr ζ_r (ṙ²/r²) ψ (7.33) (b) Equation (7.135) in the text is rewritten by integral by parts R = 1/2 ∫ dr ζ_r (ṙ²/r²) ψ + ∫ dr [ r̈ ⋅ ∂ψ/∂ṙ + k_B T (ln(ψ) + 1) ] + ∫ dr λ ṙ ⋅ r̈ (7.34) (c) δR/δṙ = 0 gives ζ_r (ṙ/r²) ψ + k_B T ∂ψ/∂ṙ + λr̈ = 0 (7.35) λ is determined by ṙ ⋅ r̈ = 0: λ = -(k_B T / r̈²) ṙ ⋅ ∂ψ/∂ṙ (7.36) From eqs.(7.34) and (7.35) r̈ψ = - (k_B T/ζ_r) [r²r̈ − ṙṙ²] ∂ψ/∂ṙ = −D_r [r²r̈ − ṙṙ²] ∂ψ/∂ṙ (7.37) where D_r = k_B T/ζ_r. Therefore ∂ψ/∂t = D_r ∂/∂r ⋅ (r²r̈ − ṙṙ²) ∂ψ/∂ṙ (7.38)