Note: Aris-Taylor dispersion from single-particle point of view

Abstract

When a point Brownian particle diffuses in a straight circular tube in the presence of a laminar stationary flow of the liquid, its effective diffusion coefficient along the tube axis increases compared to its value in the absence of flow. The effective diffusion coefficient as a function of the average fluid velocity and the tube radius is given by the Aris-Taylor formula. We give a new derivation of this formula, which is based on consideration of the axial displacement of the particle that moves in the plane normal to the tube axis along a given trajectory. The result is obtained by averaging the displacement and its square over different realizations of the particle trajectory and analyzing the long-time asymptotic behavior of the two moments.

When a point Brownian particle diffuses in a straight circular tube of radius R in the presence of a laminar stationary flow of the liquid (Poiseuille flow), particle's effective diffusion coefficient along the tube axis, D_eff, is given by

$$D_{eff}=D+\frac{{\overline{v}}^2{R}^2}{48D},$$ (1)

where D is the particle diffusion coefficient in the absence of flow, and $\overline{v}$ is the average fluid velocity. The increase of the axial diffusion coefficient, D_eff > D, is due to the radial diffusion that transports the particle among layers with different axial velocities. Taylor discovered this effect in 1953.¹ Later Aris improved the Taylor theory.² In this short note we give a new derivation of the expression for D_eff in Eq. 1, which is based on consideration of the axial displacement of a single particle that moves in the plane normal to the tube axis along a given trajectory {r}_t. We derive the expression in Eq. 1 by averaging this displacement and its square over different realizations of the trajectory and analyzing the long-time asymptotic behavior of these two moments. To the best of our knowledge our derivation of D_eff is quite different from traditional derivations that one can find in the textbooks and original papers.

Consider a laminar steady flow in a straight circular tube of radius R. We use cylindrical polar coordinates (r, θ, z) with the z-axis directed towards the flow. The fluid velocity v(r) monotonically decreases from its maximum value at the center of the tube to zero at r = R,

$$v\left(r\right)=2\overline{v}\left(1-\frac{r^2}{R^2}\right),0\le r\le R,$$ (2)

where $\overline{v}$ is the average velocity given by

$$\overline{v}=2\pi {\int }_0^{R}v\left(r\right)p_{eq}\left(r\right)rdr=\frac{2}{R^2}{\int }_0^{R}v\left(r\right)rdr,$$ (3)

and p_eq(r) = 1/(πR²).

We use {r}_t as the notation for a two-dimensional particle trajectory in the plane perpendicular to the tube axis observed for time t, {r}_t = {r(t^′), 0 ⩽ t^′ ⩽ t}. The particle velocity in the axial direction at time t is

$$\dot{z}\left(t\right|\mathbf{r}\left(t\right))=v\left(r\left(t\right)\right)+f\left(t\right),$$ (4)

where f(t) is the Gaussian δ-correlated random force with zero mean, ⟨ f(t)⟩ = 0, which satisfies the fluctuation-dissipation theorem,⟨f(t)f(t^′)⟩ = 2Dδ(t − t^′). On average this force does not lead to an axial displacement, but leads to the additive contribution 2Dt to the mean squared displacement. Force f(t) is responsible for the particle diffusion along the tube axis in the absence of flow, which is represented by the first term in the right-hand side of Eq. 1. The second, non-trivial term in the right-hand side of Eq. 1 is due to the particle motion along the tube radius. With this in mind, hereafter we ignore force f(t). Then displacement of the particle in the axial direction is given by

$$\begin{array}{c}\hfill z\left(t\right|{\left\{\mathbf{r}\right\}}_t)={\int }_0^{t}v\left(r\left(t^{\prime }\right)\right|{\left\{\mathbf{r}\right\}}_t)d{t}^{\prime }.\end{array}$$ (5)

In what follows we use this displacement to derive the second term in the right-hand side of Eq. 1 by analyzing the long-time asymptotic behavior of the first two moments of the displacement averaged over different realizations of the trajectory {r}_t.

Using the identity

$$\int \delta (\mathbf{r}-\mathbf{r}(t\left)\right)d\mathbf{r}=1,$$ (6)

we can write Eq. 5 as

$$z\left(t\right|{\left\{\mathbf{r}\right\}}_t)=\int v\left(\mathbf{r}\right)\left({\int }_0^t\delta (\mathbf{r}-\mathbf{r}\left(t^{\prime }\right))d{t}^{\prime }\right)d\mathbf{r}.$$ (7)

This presentation of the displacement has a transparent physical interpretation. Indeed, $\left({\int }_0^t\delta (\mathbf{r}-\mathbf{r}\left(t^{\prime }\right))d{t}^{\prime }\right)d\mathbf{r}$ is the cumulative time spent by the trajectory in the small vicinity of point r, where the particle velocity is v(r). Thus, Eq. 7 gives the displacement as a weighted sum of the cumulative times spent by the trajectory at different points of the tube cross section, using the local velocity, v(r), as a weight factor. Such representation of the displacement is the key step in our approach to the problem.

Averaging Eq. 7 over all trajectories that start from the same point, r(0) = r₀, we obtain

$$\begin{array}{ccc}\hfill {⟨z\left(t\right)⟩}_{{\mathbf{r}}_0}& =& {⟨z\left(t\right|{\left\{\mathbf{r}\right\}}_t)⟩}_{{\mathbf{r}}_0}\hfill \\ & =& \int v\left(\mathbf{r}\right)\left({\int }_0^t{⟨\delta (\mathbf{r}-\mathbf{r}\left(t^{\prime }\right))⟩}_{{\mathbf{r}}_0}d{t}^{\prime }\right)d\mathbf{r}.\hfill \end{array}$$ (8)

The averaged δ-function in Eq. 8 is the particle propagator (the Green's function) (Ref. ³) in the plane perpendicular to the tube axis,

$${⟨\delta (\mathbf{r}-\mathbf{r}(t\left)\right)⟩}_{{\mathbf{r}}_0}=g(\mathbf{r},t|{\mathbf{r}}_0).$$ (9)

Therefore, the averaged displacement is

$${⟨z\left(t\right)⟩}_{{\mathbf{r}}_0}=\int v\left(\mathbf{r}\right)\left({\int }_0^{t}g(\mathbf{r},t^{\prime }|{\mathbf{r}}_0)d{t}^{\prime }\right)d\mathbf{r}.$$ (10)

Here $\left({\int }_0^{t}g(\mathbf{r},t^{\prime }|{\mathbf{r}}_0)d{t}^{\prime }\right)d\mathbf{r}$ is the mean cumulative time spent by the trajectory, {r}_t, in the small vicinity of point r.

The propagator g(r, t|r₀) tends to p_eq(r) = 1/(πR²) as t → ∞. Denoting the difference between the propagator and p_eq(r) by u(r, t|r₀), we can write the propagator as

$$g({\mathbf{r}}_2,t|{\mathbf{r}}_1)=p_{eq}\left({\mathbf{r}}_2\right)+u({\mathbf{r}}_2,t|{\mathbf{r}}_1).$$ (11)

Substituting this into Eq. 10 we obtain

$${⟨z\left(t\right)⟩}_{{\mathbf{r}}_0}=\overline{v}t+\int v\left(\mathbf{r}\right)\left({\int }_0^{t}u(\mathbf{r},t^{\prime }|{\mathbf{r}}_0)d{t}^{\prime }\right)d\mathbf{r}.$$ (12)

Since u(r, t|r₀) vanishes as t → ∞, at asymptotically long times we have

$${⟨z\left(t\right)⟩}_{{\mathbf{r}}_0}=\overline{v}t+l\left({\mathbf{r}}_0\right),t\to \infty ,$$ (13)

where l(r₀) is given by

$$l\left({\mathbf{r}}_0\right)=\int v\left(\mathbf{r}\right)\left({\int }_0^{\infty }u(\mathbf{r},t|{\mathbf{r}}_0)dt\right)d\mathbf{r}.$$ (14)

As might be expected, the long-time behavior of ${⟨z\left(t\right)⟩}_{{\mathbf{r}}_0}$ is the sum of the product $\overline{v}t$ and a constant term that depends on r₀.

Averaging the square of the displacement in Eq. 7 over the trajectories that start from r₀ we obtain

$$\begin{array}{ccc}\hfill {⟨z^2\left(t\right)⟩}_{{\mathbf{r}}_0}& =& {⟨{\left[z\left(t\right|{\left\{\mathbf{r}\right\}}_t)\right]}^2⟩}_{{\mathbf{r}}_0}\hfill \\ & =& 2\int \int v\left({\mathbf{r}}_1\right)v\left({\mathbf{r}}_2\right)\left({\int }_0^{t}d{t}_2{\int }_0^{t_2}⟨\delta ({\mathbf{r}}_2-\mathbf{r}\left(t_2\right))\right.\hfill \\ & & \left.\times {\delta ({\mathbf{r}}_1-\mathbf{r}\left(t_1\right))⟩}_{{\mathbf{r}}_0}d{t}_1\right)d{\mathbf{r}}_{1}d{\mathbf{r}}_2.\hfill \end{array}$$ (15)

Similar to Eq. 9 the averaged product of the δ-functions in Eq. 15 is the product of the propagators,

$$\begin{array}{ccc}& & {⟨\delta ({\mathbf{r}}_2-\mathbf{r}\left(t_2\right))\delta ({\mathbf{r}}_1-\mathbf{r}\left(t_1\right))⟩}_{{\mathbf{r}}_0}\hfill \\ & & =g({\mathbf{r}}_2,t_2-t_1|{\mathbf{r}}_1)g({\mathbf{r}}_1,t_1|{\mathbf{r}}_0),t_2>t_1.\hfill \end{array}$$ (16)

This allows us to write ${⟨z^2\left(t\right)⟩}_{{\mathbf{r}}_0}$ in terms of the propagator

$$\begin{array}{ccc}\hfill {⟨z^2\left(t\right)⟩}_{{\mathbf{r}}_0}& =& 2\int \int v\left({\mathbf{r}}_1\right)v\left({\mathbf{r}}_2\right)\left({\int }_0^{t}d{t}_2{\int }_0^{t_2}g({\mathbf{r}}_2,t_2-t_1|{\mathbf{r}}_1)\right.\hfill \\ & & \left.\times g({\mathbf{r}}_1,t_1|{\mathbf{r}}_0)d{t}_1\right)d{\mathbf{r}}_{1}d{\mathbf{r}}_2.\hfill \end{array}$$ (17)

Asymptotic long-time behavior of ${⟨z^2\left(t\right)⟩}_{{\mathbf{r}}_0}$ can be found using Eq. 11,

$$\begin{array}{ccc}\hfill {⟨z^2\left(t\right)⟩}_{{\mathbf{r}}_0}& =& {\overline{v}}^2{t}^2+2t\left[\overline{v}l\left({\mathbf{r}}_0\right)+\int v\left(\mathbf{r}\right)l\left(\mathbf{r}\right)p_{eq}\left(\mathbf{r}\right)d\mathbf{r}\right],\hfill \\ & & t\to \infty .\hfill \end{array}$$ (18)

Thus we have

$$\begin{array}{ccc}\hfill \underset{t\to \infty }{lim}\frac{{⟨z^2\left(t\right)⟩}_{eq}-{⟨z\left(t\right)⟩}_{eq}^2}{2t}& =& \int v\left(\mathbf{r}\right)l\left(\mathbf{r}\right)p_{eq}\left(\mathbf{r}\right)d\mathbf{r}\hfill \\ & =& \frac{2}{R^2}{\int }_0^{R}v\left(r\right)l\left(r\right)rdr,\hfill \end{array}$$ (19)

where l(r) is function l(r) averaged over θ. We will see that the integral in the right-hand side is just the second term in the expression for D_eff, Eq. 1.

To finish the derivation we take advantage of the fact that the propagator g(r, t|r₀) considered as a function of r₀ satisfies

$$\frac{\partial g(\mathbf{r},t|{\mathbf{r}}_0)}{\partial t}=D{\Delta }_{{\mathbf{r}}_0}g(\mathbf{r},t|{\mathbf{r}}_0),$$ (20)

where ${\Delta }_{{\mathbf{r}}_0}$ is the two-dimensional Laplace operator, with reflecting boundary condition on the tube wall. As a consequence, function u(r, t|r₀) satisfies the same equation and boundary condition. Integrating both sides of this equation with respect to time from zero to infinity one obtains the equation for ${\int }_0^{\infty }u(\mathbf{r},t|{\mathbf{r}}_0)dt$. Multiplying both sides of the resulting equation by v(r) and then integrating over the tube cross section one finds that l(r₀), Eq. 14, satisfies

$$\frac{D}{r_0}\frac{d}{d{r}_0}\left(r_0\frac{dl\left(r_0\right)}{d{r}_0}\right)=\overline{v}-v\left(r_0\right)=\overline{v}\left(2\frac{r_0^2}{R^2}-1\right)$$ (21)

with the boundary condition dl(r)/dr|_{r = R} = 0. In addition, l(r) satisfies the requirement

$${\int }_0^{R}l\left(r\right)rdr=0.$$ (22)

This can be obtained using Eq. 14 and the fact that ${\int }_0^{R}u({\mathbf{r}}_2,t|{\mathbf{r}}_1)d{\mathbf{r}}_1=0$, which follows from Eq. 11. Integrating Eq. 21 one finds

$$l\left(r\right)=\frac{\overline{v}}{24D{R}^2}\left(3r^4-6R^2{r}^2+2R^4\right).$$ (23)

Substituting this into the integral in Eq. 19 and carrying out the integration one recovers the second term in the expression for D_eff given in Eq. 1.