Statistical Mechanical Expressions of Slip Length

Abstract

We provide general derivations of the partial slip boundary condition from microscopic dynamics and linearized fluctuating hydrodynamics. The derivations are based on the assumption of separation of scales between microscopic behavior, such as collision of particles, and macroscopic behavior, such as relaxation of fluid to global equilibrium. These derivations lead to several statistical mechanical expressions of the slip length, which are classified into two types. The expression in the first type is given as a local transport coefficient, which is related to the linear response theory that describes the relaxation process of the fluid. The second type is related to the linear response theory that describes the non-equilibrium steady state and the slip length is given as combination of global transport coefficients, which are dependent on macroscopic lengths such as a system size. Our derivations clarify that the separation of scales must be seriously considered in order to distinguish the expressions belonging to two types. Based on these linear response theories, we organize the relationship among the statistical mechanical expressions of the slip length suggested in previous studies.

Appendices

Appendix A: Technical Details of the Calculation of $C_{AA}(\omega )$

In the calculation of (171), we detail how to proceed from the first to the second line. We perform an integration by parts

$$\begin{aligned}&\int _{0}^{\mathcal {L}} dz_2 \int _{0}^{\mathcal {L}} dz'_2 G(z,z_2,\omega ) G(z',z'_2,-\omega ) \partial _{z_2} \partial _{z'_2} \delta (z_2-z'_2) \nonumber \\&\quad = \Big [\int _{0}^{\mathcal {L}} dz_2 G(z,z_2,\omega ) G(z',z'_2,-\omega ) \partial _{z_2} \delta (z_2-z'_2) \Big ]_{z'_2=0}^{z'_2=\mathcal {L}}\nonumber \\&\qquad -\int _{0}^{\mathcal {L}} dz_2 \int _{0}^{\mathcal {L}} dz'_2 G(z,z_2,\omega ) \partial _{z'_2} G(z',z'_2,-\omega ) \partial _{z_2} \delta (z_2-z'_2) \nonumber \\&\quad =\Big [G(z,z_2,\omega ) G(z',\mathcal {L},-\omega ) \delta (z_2-\mathcal {L}) \Big ]_{z_2=0}^{z_2=\mathcal {L}}\nonumber \\&\qquad -\int _{0}^{\mathcal {L}} dz_2 \partial _{z_2}G(z,z_2,\omega ) G(z',\mathcal {L},-\omega ) \delta (z_2-\mathcal {L}) \nonumber \\&\qquad - \Big [G(z,z_2,\omega ) G(z',0,-\omega ) \delta (z_2-0) \Big ]_{z_2=0}^{z_2=\mathcal {L}} \nonumber \\&\qquad + \int _{0}^{\mathcal {L}} dz_2 \partial _{z_2}G(z,z_2,\omega ) G(z',0,-\omega ) \delta (z_2-0)\nonumber \\&\qquad -\int _{0}^{\mathcal {L}} dz'_2 G(z,\mathcal {L},\omega ) \partial _{z'_2} G(z',z'_2,-\omega ) \delta (\mathcal {L}-z'_2)\nonumber \\&\qquad +\int _{0}^{\mathcal {L}} dz'_2 G(z,0,\omega ) \partial _{z'_2} G(z',z'_2,-\omega ) \delta (0-z'_2) \nonumber \\&\qquad +\int _{0}^{\mathcal {L}} dz_2 \int _{0}^{\mathcal {L}} dz'_2 \partial _{z_2} G(z,z_2,\omega ) \partial _{z'_2} G(z',z'_2,-\omega ) \delta (z_2-z'_2). \end{aligned}$$

(A.1)

The last expression contains integrals of the delta function weighted only at the edge of the integration range. These integrals are calculated using the property:

$$\begin{aligned} \int _0^{\mathcal {L}} dz \delta (z) = \frac{1}{2}. \end{aligned}$$

(A.2)

The Integration of the delta functions in (A.2) yields the second line of (171).

Appendix B: Sketch of the Derivation of (195) and (196)

We sketch the derivation of (195) and (196). We introduce the time correlation functions $C_{v\alpha }(z,t,t')$ as

$$\begin{aligned} C_{v\alpha }(z,t,t') \equiv \langle \tilde{\mathcal {V}}^{x}(z,t) \tilde{\mathcal {F}}_{\alpha }(t') \rangle , \end{aligned}$$

(B.1)

where $\alpha =A,B$. Equations (195) and (196) are written in terms of $C_{v\alpha }(z,t,t')$ as

$$\begin{aligned} \frac{1}{k_B T A_{xy}} \int _{0}^{\infty } dt C_{vA}(z,t,0) =\left( 1-\frac{z}{\mathcal {L}} +\frac{\mathcal {B}_B}{\mathcal {L}}\right) +\left( 1-\frac{z}{\mathcal {L}}\right) \left( \frac{\mathcal {B}_A}{\mathcal {L}} +\frac{\mathcal {B}_B}{\mathcal {L}}\right) + o(\epsilon ),\nonumber \\ \end{aligned}$$

(B.2)

and

$$\begin{aligned} \frac{1}{k_B T A_{xy}} \int _{0}^{\infty } dt C_{vB}(z,t,0) =\left( \frac{z}{\mathcal {L}}+\frac{\mathcal {B}_A}{\mathcal {L}}\right) -\frac{z}{\mathcal {L}}\left( \frac{\mathcal {B}_A}{\mathcal {L}} + \frac{\mathcal {B}_B}{\mathcal {L}}\right) + o(\epsilon ). \end{aligned}$$

(B.3)

Since $C_{v\alpha }(z,t,0)$ is the correlation function between the odd and even quantities under time reversal, the equality

$$\begin{aligned} C_{v\alpha }(z,t,0) = -C_{v\alpha }(z,-t,0) \end{aligned}$$

(B.4)

holds. This is in contrast to $C_{\alpha \beta }(t,0) \ (\alpha ,\beta =A, B)$, which satisfies

$$\begin{aligned} C_{\alpha \beta }(t,0) = C_{\alpha \beta }(-t,0). \end{aligned}$$

(B.5)

Because of this property, we have

$$\begin{aligned} \lim _{\omega \rightarrow 0} C_{v\alpha }(z,\omega ) = \lim _{\omega \rightarrow 0} \int _{-\infty }^{\infty } dt C_{v\alpha }(z,t,0)e^{i\omega t} = 0. \end{aligned}$$

(B.6)

Therefore, we cannot repeat the calculation of (191) in deriving (B.2) and (B.3). We calculate (B.2) and (B.3) using the following steps. First, we calculate the Fourier transform $C_{v\alpha }(z,\omega )$ and obtain the real-time correlation function in the form

$$\begin{aligned} C_{v\alpha }(z,t,0) = \int \frac{d\omega }{2\pi } C_{v \alpha }(z,\omega )e^{-i\omega t}. \end{aligned}$$

(B.7)

Next, we perform the integral in (B.7). Here, we assume that main contribution of $C_{v\alpha }(z,t,0)$ in (B.2) and (B.3) comes from the region $\omega \simeq 0$. Finally, using this result, we calculate the time integral on the left-hand side of (B.2) and (B.3).

We focus on (B.2). $C_{vA}(z,\omega )$ is written in terms of the Green function as

$$\begin{aligned} \frac{C_{vA}(z;\omega )}{2k_B TA_{xy}}&= \eta \partial _{z_2} G(z,+0;\omega ) - \eta ^2 \int _0^{\mathcal {L}} dz_2 \partial _{z_2} G(z,z_2;\omega ) \partial _{z} \partial _{z_2} G(+0,z_2;-\omega )\nonumber \\&\quad - \eta ^2 G(z,0;\omega ) \partial _{z}\partial _{z_2} G(+0,0;-\omega ) +\eta ^2 G(z,\mathcal {L};\omega ) \partial _{z} \partial _{z_2} G(+0,\mathcal {L};-\omega ) ,\nonumber \\ \end{aligned}$$

(B.8)

where we ignore terms proportional to $\delta (0)$, such as the fourth line of (174). The leading contribution of the Green function in $\omega$ is expressed as

$$\begin{aligned} G(z,z_2;\omega ) \simeq \frac{(\mathcal {L}+\mathcal {B}_B-z) (\mathcal {B}_A+z_2)}{\eta (\mathcal {B}_A+\mathcal {B}_B+\mathcal {L})-i\omega \rho c} \end{aligned}$$

(B.9)

for $z>z_2$, and

$$\begin{aligned} G(z,z_2;\omega ) \simeq \frac{(1+\mathcal {B}_B-z_2)(\mathcal {B}_A+z)}{\eta (\mathcal {B}_A+\mathcal {B}_B+\mathcal {L})-i\omega \rho c} \end{aligned}$$

(B.10)

for $z<z_2$, with

$$\begin{aligned} c \equiv \mathcal {B}_A \mathcal {B}_B \mathcal {L} + \frac{1}{2} (\mathcal {B}_A+\mathcal {B}_B) \mathcal {L}^2 + \frac{1}{6} \mathcal {L}^3. \end{aligned}$$

(B.11)

We consider the first term of (B.8). Substituting (B.9) and (B.10) into the first term of (B.8) produces

$$\begin{aligned} \eta \partial _{z_2} G(z,+0;\omega ) \simeq - \frac{\eta }{i\rho c} \frac{\mathcal {L}+\mathcal {B}_B-z}{\omega -\frac{\eta (\mathcal {B}_A +\mathcal {B}_B+\mathcal {L})}{i\rho c}}. \end{aligned}$$

(B.12)

Next, after performing the integral in $\omega$, we obtain

$$\begin{aligned} \eta \int \frac{d\omega }{2\pi } \partial _{z_2} G(z,+0;\omega ) e^{-i\omega t} = \frac{\eta (\mathcal {L}+\mathcal {B}_B-z)}{\rho c} e^{-t\frac{\eta (\mathcal {L}+\mathcal {B}_A+\mathcal {B}_B)}{\rho c}}, \end{aligned}$$

(B.13)

which is a part of $C_{vA}(z,t,0)/2k_B TA_{xy}$ obtained from the first term of (B.8). Finally, integrating the right-hand side of (B.13) in t gives

$$\begin{aligned} \int _0^{\infty } dt \frac{\eta (\mathcal {L}+\mathcal {B}_B-z)}{\rho c} e^{-t\frac{\eta (\mathcal {L}+\mathcal {B}_A+\mathcal {B}_B)}{\rho c}} =\frac{\mathcal {L}+\mathcal {B}_B-z}{\mathcal {L}+\mathcal {B}_A+\mathcal {B}_B}. \end{aligned}$$

(B.14)

We calculate the remaining terms of (B.8) in the same manner. By collecting these results and extracting the first order terms in $\epsilon$, we obtain (B.2). Finally, we note that (B.3) is obtained by a similar procedure.

Appendix C: Derivation of (221) from the Linearized Fluctuating Dynamics

In this Appendix, we show a proof of (221) by employing the model in Sect. 7.

In our model, the average velocity of the particles in the slab, $\hat{u}^x_{slab}(\varGamma )$, is replaced with the fluid velocity near the wall, $\tilde{\mathcal {V}}^x(+0,t)$. Based on this, we transform (221) to the expression with respect to frequency. By recalling that

$$\begin{aligned} \lim _{\omega \rightarrow 0} A(\omega ) = \int _0^{\infty } ds A(s), \end{aligned}$$

(C.1)

where A(t) is any function and $A(\omega )$ is the Fourier transform of A(t), (221) is rewritten as

$$\begin{aligned} \frac{\eta }{b_A^{HTD}} = \frac{\lim _{\omega \rightarrow 0}D_{vA}(+0,\omega )}{\lim _{\omega \rightarrow 0}D_{vv}(+0,\omega )}, \end{aligned}$$

(C.2)

where $D_{vA}(z,\omega )$ is defined by (B.7) and $D_{vv}(z,\omega )$ is given by

$$\begin{aligned} D_{vv}(z,\omega ) = \int _0^{\infty } dt \langle \tilde{\mathcal {V}}^{x}(z,t) \tilde{\mathcal {V}}^{x}(z,0) \rangle e^{i\omega t}. \end{aligned}$$

(C.3)

In Appendix B, $\lim _{\omega \rightarrow 0}D_{vA}(+0,\omega )$ is calculated as

$$\begin{aligned} \lim _{\omega \rightarrow 0}\frac{D_{vA}(+0,\omega )}{k_B T A_{xy}} =\frac{\mathcal {L}+\mathcal {B}_B}{\mathcal {L}+\mathcal {B}_A+\mathcal {B}_B}, \end{aligned}$$

(C.4)

where we have ignored the terms proportional to $\delta (0)$. Then, we calculate $D_{vv}(z,\omega )$ in $\omega \rightarrow 0$. The leading contribution of the Green function in $\omega \rightarrow 0$ is expressed as

$$\begin{aligned} G(z,z_2;\omega ) \rightarrow \frac{(\mathcal {L}+\mathcal {B}_B-z)(\mathcal {B}_A+z_2)}{\eta (\mathcal {L}+\mathcal {B}_A+\mathcal {B}_B)} \end{aligned}$$

(C.5)

for $z>z_2$, and

$$\begin{aligned} G(z,z_2;\omega ) \rightarrow \frac{(\mathcal {L}+\mathcal {B}_B-z_2) (\mathcal {B}_A+z)}{\eta (\mathcal {L}+\mathcal {B}_A+\mathcal {B}_B)} \end{aligned}$$

(C.6)

for $z<z_2$. By recalling that $D_{vv}(z,\omega )$ is given by (171), we substitute (C.5) and (C.6) into (171). This yields

$$\begin{aligned} \lim _{\omega \rightarrow 0} \frac{D_{vv}(+0,\omega )}{k_B T A_{xy}} =\frac{\mathcal {B}_A(\mathcal {L}+\mathcal {B}_B)}{\eta (\mathcal {L} +\mathcal {B}_A+\mathcal {B}_B)}, \end{aligned}$$

(C.7)

where we have ignored the terms proportional to $\delta (0)$. Substituting (C.4) and (C.7) into (C.2) produces

$$\begin{aligned} b^{HTD}_A = \mathcal {B}_A. \end{aligned}$$

(C.8)

This is what we want to prove.