S10.G.1. Fluid Flow Around the Brownian Particle

Transcription

1 Rea Reichl s introuction. Tables & proofs for vector calculus formulas can be foun in the stanar textbooks G.Arfken s Mathematical Methos for Physicists an J.D.Jackson s Classical Electroynamics. S0.G.. Flui Flow Aroun the Brownian Particle Consier the linearize hyroynamic equations [see (0.36-7) of 0.C. with F = 0] Δρ + ρ 0 r v = 0 t v ρ 0 t = - r ΔP + ζ + 3 η r r v + η r v Taking the Fourier transform on t gives (0.353) -i ω ρ ω (r) +ρ 0 r v ω (r) = 0 -i ω ρ 0 v ω (r) = - r P ω (r) + ζ + 3 η r r v ω (r) + η r v ω (r) (0.354a) r ( r v) = ϵ i j k j (ϵ k l m l v m ) = ( δ i l δ j m - δ i m δ j l ) j l v m = j i v j - j j v i = r ( r v) - r v (0.354b) to eliminate r v ω (r), (0.354a) becomes -i ω ρ 0 v ω (r) = - r P ω (r) + ζ η r r v ω (r) - η r r v ω (r) (0.354) Consier a spherical Brownian particle of raius R moving through the flui with velocity u(t) = u(t) z. Let the surface of the Brownian particle be highly sticky so that the velocity v(r, t) of the flui is equal to that of the Brownian particle everywhere on its surface. The total force on the Brownian particle is F = S P S (0.355) = S P r S where S is the surface of the Brownian particle an P = P + Π is the pressure tensor of the flui [see 0.B.]. For an incompressible flui, r v(r, t) = 0 r v ω (r) = 0 an (0.353) & (0.354) simplify to (0.356a) -i ω ρ ω (r) = 0 -i ω ρ 0 v ω (r) = - r P ω (r) -η r r v ω (r) (0.356)

2 S0.G.._Flui lowarounthebrownianparticle.nb (0.356b) r r f = 0 we can write (0.356) as f -i ω ρ 0 r v ω (r) = -η r r r v ω (r) = -η r r r v ω (r) - r v ω (r) [ (0.354b) use. ] = η r r v ω (r) [ (0.356b) use. ] = η ϵ i j k j ( m m v ω k ) = η m m ϵ i j k j v ω k = η r [ r v ω (r) ] (0.358) Now, the bounary conition at the surface of the Brownian particle means that the flow is symmetric about u = z, i.e., it is inepenent of ϕ. Setting where = r - r c = r c (t) = position of the center of the Brownian particle at time t so that, θ, ϕ form a spherical coorinate basis, we can use to write u = θ cos θ v(r, t) = u(t) A(, θ) + u(t) B(, θ) v ω (r) = u ω A(, θ) + u ω B(, θ) (0.357a) Thus, v ω is always in the z, plane & hence inepenent of ϕ, both in magnitue & irection. Alternatively, consier r [ r (u ω g )] = [ (u ω g )] If we set g = g(), we get u ω g( ) = where we have use ψ(, θ, ϕ) = ψ = [ ( u ω g )] - (u ω g ) = ( u ω g ) - u ω g + θ u ω cosθ = -θ u ω sinθ ψ θ + ϕ sinθ -u ω g + u ω cosθ g -u ω g (0.357b) Since z,, θ are in the same plane, we can replace (0.357a) with the more efficient form (0.357) v ω (r) = v ω (r c + ) = u ω g( ) ψ ϕ

3 S0.G.._Flui lowarounthebrownianparticle.nb 3 Furthermore, using we have z = cos θ - θ sin θ g() = = g v ω (r) = θ u ω sinθ + g - u ω cosθ which is the representation of v ω (r) in spherical coorinates. we have A = sin θ θ sinθ f() = sin θ θ sinθ ϕ θ ϕ A A θ sinθ A ϕ = ϕ sinθ cosθ f() = sin θ so that (0.365) gives (0.359) we have v ω (r) = ϕ u ω sinθ = ϕ u ω sinθ - = ϕ u ω sinθ = ϕ u ω sinθ ϕ = -sinϕ x + cosϕ y ψ(, θ, ϕ) = θ sinθ ϕ θ ϕ 0 sinθ f() 0 sinθ [ f()] = ϕ f + f θ sinθ ϕ θ ϕ cosθ f () g + - g ψ + sinθ ϕ = - x sinϕ + y cosϕ + g g + 3 g 3 = sin θ ϕ -sinϕ x +cosϕ y = sin θ sinϕ x -cosϕ y + g + = ϕ sinθ f - + g + 3 g 3 θ sinθ ψ θ + sin θ ψ ϕ (0.365)

4 4 S0.G.._Flui lowarounthebrownianparticle.nb = - sin θ ϕ Also, [sinθ f()] = sinθ f() + f() sin θ θ = sinθ f() cosθ f() + sin θ ϕ sinθ f() = - sin θ ϕ sinθ f() + ϕ = cosθ - sinθ sinθ sinθ θ f() + sinθ f() ϕ = sinθ - f() + f() ϕ sinθ f() cosθ f() + sin θ (0.359) then gives (0.359a) [ v ω (r)] = ϕ u ω sinθ - g + Putting (0.359-a) into (0.358) gives -i ω ρ 0 ϕ u ω sinθ g = η ϕ u ω sinθ - -i ω ρ 0 (0.360) = (0.360) becomes -i ω ρ 0 Setting g = η - + = + g = η - g + = g + = η 4 g g g g + = g g + (0.360a) k = i ω ρ 0 η k = ω ρ 0 η e i π /4 = ω ρ 0 η ( + i) (0.360b) we get

5 S0.G.._Flui lowarounthebrownianparticle.nb 5 4 g + k g = 0 (0.36) 4 g + k g = C C = const (0.36) Now, far away from the Brownian particle, the flui is unisturbe so that v(r, t) = 0 as t v ω (r) = 0 as n C = 0 n g = 0 as n Thus, (0.36) becomes (0.36a) g + k g = 0 with solution that vanishes as given by (0.363) g = c ei k (0.363) can be written as = c e i k (0.364) = c e i k +c [see Coe] = c e i k k + i k + c [see Coe] = c e i k k + i k + c To impose the bounary conition at = R, we nee to express v ω (r) in the cylinrical coorinate basis ρ, ϕ, z. Putting = ρ sin θ + z cos θ into (0.365) gives v ω (r) = ρ cos θ - z sin θ u ω sinθ = ρ u ω cosθ sinθ -z u ω sin θ = ρ u ω sinθ cosθ - + g + g θ = ρ cos θ - z sin θ + g - sinθ cosθ + cos θ - ρ sin θ + z cos θ u ω cosθ + g - z u ω sin θ - + g +

6 6 S0.G.._Flui lowarounthebrownianparticle.nb = ρ u ω sinθ cosθ g - z u ω sin θ - 3 The bounary conition + g + (0.365a) v ω (r) then requires g g = R = - 3 = z u ω θ =R = 0 =R Putting these into (0.363) gives - 3 = c R ei k R c = - 3 R (0.366) e-i k R while (0.364) gives - R = R c e i k R k + R i k + c = R = - (0.365b) (0.365c) = R - 3 R k + R i k + c [ (0.366) use. ] c = - R3 + 3 R k + R i k = R k R (0.363, 4 & 5) become g = - 3 R (0.366b) ei k (-R) = - 3 R ei k (-R) k + i k v ω (r) = θ u ω sinθ - + g = θ u ω sinθ 3 3 R ei k (-R) k + - u ω cosθ 3 (0.366a) + R k R (0.366c) - u ω cosθ i k R ei k (-R) k + - R i k k R + R k R (0.366) v ω (r) as expecte. = -θ u ω sinθ + u ω cosθ = u ω z =R

7 S0.G.._Flui lowarounthebrownianparticle.nb 7 Coe DSolvef''[] + f'[] + k f[] 0, f, f Function{}, e- -k C[] + e -k C[] -k ei k e i k k - i k