Transport Properties

Abstract

This chapter deals with the classical hydrodynamic theory of steady-state transport properties, such as the translational friction and diffusion coefficients and intrinsic viscosity, of the unperturbed HW chain, including the KP wormlike chain as a special case, on the basis of the cylinder and touched-bead models. An analysis of experimental data is made from various points of view, which are based on the present theory, especially for flexible polymers. In the same spirit as that in Chap. 5, use is then made of experimental data obtained for several flexible polymers in the $\varTheta$ state over a wide range of molecular weight, including the oligomer region, and also for typical semiflexible polymers. As a result, it is pointed out that there still remain several unsolved problems for flexible polymers even in the unperturbed state. It is convenient to begin by giving a general consideration of some aspects of polymer hydrodynamics which leads to the adoption of the present hydrodynamic models.

Appendices

Appendix 1: Transport Coefficients of Spheroid-Cylinders

In this appendix we evaluate the translational and rotatory friction (or diffusion) coefficients and intrinsic viscosity of the spheroid-cylinder defined in Sect. 6.2.1 and depicted in Fig. 6.2 [27]. We introduce external (e _x , e _y , e _z ) and molecular (e ₁, e ₂, e ₃) Cartesian coordinate systems, choosing the center of mass of the body as the origin of the latter. The superscript (e) is used to indicate vectors and tensors expressed in the external system, and no superscript is used for those in the molecular system. The spatial configuration of the body may be determined by the vector position $\mathbf{R}_{\mathrm{c}}^{\mathrm{(e)}} \equiv \mathbf{R}_{\mathrm{c}} = (x_{\mathrm{c}}$, y _c, z _c) of the center of mass in the external system and the Euler angles $\varTheta = (\alpha$, β, γ) defining the orientation of the molecular system with respect to the external system. The matrix transforming the external coordinates to the molecular coordinates, which we denote by A, is then identical to the matrix Q given by Eq. (4.96) with (α, β, γ) in place of ($\tilde{\theta }$, $\tilde{\phi }$, $\tilde{\psi }$).

Now let U _c be the instantaneous velocity of the center of mass of the body, let $\boldsymbol{\Omega }$ be its instantaneous angular velocity, and let v ⁰ be the unperturbed flow field of a solvent. Under the nonslip boundary condition, the frictional force f(r) exerted by the unit area at r of the surface of the body satisfies the integral equation

$$\displaystyle{ 8\pi \eta _{0}{\bigl [\mathbf{U}_{\mathrm{c}} +\boldsymbol{ \Omega } \times \mathbf{r}_{1} -\mathbf{v}^{0}(\mathbf{r}_{ 1})\bigr ]} =\int _{\mathrm{S}}\mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{2}) \cdot \mathbf{f}(\mathbf{r}_{2})d\mathbf{r}_{2}\,, }$$

(6.154)

where r ₁ and r ₂ are the vector positions of two arbitrary points on the surface of the body, $\mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{2})$ is defined by

$$\displaystyle{ \mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{2}) = 8\pi \eta _{0}\mathbf{T}(\mathbf{r}_{1} -\mathbf{r}_{2}) }$$

(6.155)

as in Eq. (6.58), and the integration in Eq. (6.154) is carried out over the surface of the body. If we define the inverse $\mathbf{K}^{-1}(\mathbf{r}_{1},\mathbf{r}_{2})$ by

$$\displaystyle\begin{array}{rcl} \delta ^{(2)}(\mathbf{r}_{ 1} -\mathbf{r}_{2})\mathbf{I}& =& \int _{\mathrm{S}}\mathbf{K}^{-1}(\mathbf{r}_{ 1},\mathbf{r}_{3}) \cdot \mathbf{K}(\mathbf{r}_{3},\mathbf{r}_{2})d\mathbf{r}_{3} \\ & =& \int _{\mathrm{S}}\mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{3}) \cdot \mathbf{K}^{-1}(\mathbf{r}_{ 3},\mathbf{r}_{2})d\mathbf{r}_{3}{}\end{array}$$

(6.156)

with δ ⁽²⁾(r) a two-dimensional Dirac delta function and with I the unit tensor as in Eqs. (6.59), then the formal solution of Eq. (6.154) is obtained as

$$\displaystyle{ \mathbf{f}(\mathbf{r}_{1}) = 8\pi \eta _{0}\int _{\mathrm{S}}\mathbf{K}^{-1}(\mathbf{r}_{ 1},\mathbf{r}_{2}) \cdot {\bigl [\mathbf{U}_{\mathrm{c}} +\boldsymbol{ \Omega } \times \mathbf{r}_{2} -\mathbf{v}^{0}(\mathbf{r}_{ 2})\bigr ]}d\mathbf{r}_{2}\,. }$$

(6.157)

We first put v ⁰ = 0 to consider the translational and rotatory friction tensors of the body, which we denote by $\boldsymbol{\Xi }$ and $\boldsymbol{\Xi }_{\mathrm{c,r}}$, respectively. The total frictional force F and torque T _c (about the center of mass) exerted by the body on the solvent are then given by

$$\displaystyle\begin{array}{rcl} \mathbf{F} =\boldsymbol{ \Xi } \cdot \mathbf{U}_{\mathrm{c}} =\int _{\mathrm{S}}\mathbf{f}(\mathbf{r})d\mathbf{r}\,,& &{}\end{array}$$

(6.158)

$$\displaystyle\begin{array}{rcl} \mathbf{T}_{\mathrm{c}} =\boldsymbol{ \Xi }_{\mathrm{c,r}} \cdot \boldsymbol{ \Omega } =\int _{\mathrm{S}}\mathbf{r} \times \mathbf{f}(\mathbf{r})d\mathbf{r} =\int _{\mathrm{S}}\mathbf{B}_{\mathrm{c}}(\mathbf{r})^{T} \cdot \mathbf{f}(\mathbf{r})d\mathbf{r}\,,& &{}\end{array}$$

(6.159)

where the superscript T indicates the transpose and the tensor B _c is given by

$$\displaystyle{ \mathbf{B}_{\mathrm{c}}(\mathbf{r}) = \left (\begin{array}{*{10}c} 0 & r_{3} & -r_{2} \\ -r_{3} & 0 & r_{1} \\ r_{2} & -r_{1} & 0\\ \end{array} \right ) }$$

(6.160)

with $\mathbf{r} = r_{1}\mathbf{e}_{1} + r_{2}\mathbf{e}_{2} + r_{3}\mathbf{e}_{3}$. If we substitute Eq. (6.157) into Eqs. (6.158) and (6.159), we find

$$\displaystyle{ \boldsymbol{\Xi } = 8\pi \eta _{0}\int _{\mathrm{S}}\int _{\mathrm{S}}\mathbf{K}^{-1}(\mathbf{r}_{ 1},\mathbf{r}_{2})d\mathbf{r}_{1}d\mathbf{r}_{2} = 8\pi \eta _{0}\int _{\mathrm{S}}\boldsymbol{\Psi }_{1}(\mathbf{r})d\mathbf{r}\,, }$$

(6.161)

$$\displaystyle\begin{array}{rcl} \boldsymbol{\Xi }_{\mathrm{c,r}}& =& 8\pi \eta _{0}\int _{\mathrm{S}}\int _{\mathrm{S}}\mathbf{B}_{\mathrm{c}}(\mathbf{r}_{1})^{T} \cdot \mathbf{K}^{-1}(\mathbf{r}_{ 1},\mathbf{r}_{2}) \cdot \mathbf{B}_{\mathrm{c}}(\mathbf{r}_{2})d\mathbf{r}_{1}d\mathbf{r}_{2} \\ & =& 8\pi \eta _{0}\int _{\mathrm{S}}\mathbf{B}_{\mathrm{c}}(\mathbf{r})^{T} \cdot \boldsymbol{\Psi }_{ 2}(\mathbf{r})d\mathbf{r}\,, {}\end{array}$$

(6.162)

where the tensors $\boldsymbol{\Psi }_{1}$ and $\boldsymbol{\Psi }_{2}$ are the solutions of the integral equations

$$\displaystyle\begin{array}{rcl} \int _{\mathrm{S}}\mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{2}) \cdot \boldsymbol{\Psi }_{1}(\mathbf{r}_{2})d\mathbf{r}_{2} = \mathbf{I}\,,& &{}\end{array}$$

(6.163)

$$\displaystyle\begin{array}{rcl} \int _{\mathrm{S}}\mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{2}) \cdot \boldsymbol{\Psi }_{2}(\mathbf{r}_{2})d\mathbf{r}_{2} = \mathbf{B}_{\mathrm{c}}(\mathbf{r}_{1})\,.& &{}\end{array}$$

(6.164)

Now, if we take e ₃ along the axis of revolution of the body, then $\boldsymbol{\Xi }$ and $\boldsymbol{\Xi }_{\mathrm{c,r}}$ and hence the translational diffusion tensor $\mathbf{D}_{\mathrm{c}} = k_{\mathrm{B}}T\boldsymbol{\Xi }^{-1}$ (of the center of mass) and rotatory diffusion tensor $\mathbf{D}_{\mathrm{r}} = k_{\mathrm{B}}T\boldsymbol{\Xi }_{\mathrm{c,r}}^{\ -1}$ are diagonalized. We denote their principal values by $\Xi _{j}$, $\Xi _{\mathrm{r},j}$, D _j , and D _r, j , respectively, so that

$$\displaystyle\begin{array}{rcl} D_{j} = \frac{k_{\mathrm{B}}T} {\Xi _{j}} \,,& &{}\end{array}$$

(6.165)

$$\displaystyle\begin{array}{rcl} D_{\mathrm{r},j} = \frac{k_{\mathrm{B}}T} {\Xi _{\mathrm{r},j}} & &{}\end{array}$$

(6.166)

with D ₁ = D ₂ and $D_{\mathrm{r},1} = D_{\mathrm{r},2}$ (and $\Xi _{1} = \Xi _{2}$ and $\Xi _{\mathrm{r},1} = \Xi _{\mathrm{r},2}$). The mean translational diffusion coefficient D averaged over the orientation of the body is given by

$$\displaystyle{ D = \frac{1} {3}\mathrm{Tr}\,\mathbf{D}_{\mathrm{c}}^{(\mathrm{e})} = \frac{1} {3}\mathrm{Tr}\,\mathbf{D}_{\mathrm{c}} = \frac{1} {3}k_{\mathrm{B}}T(2\Xi _{1}^{\ -1} + \Xi _{ 3}^{\ -1})\,. }$$

(6.167)

Next we consider the intrinsic viscosity [η]. The unperturbed flow field v ⁰ given by Eq. (6.65) may be expressed in the molecular coordinate system as follows,

$$\displaystyle{ \mathbf{v}^{0}(\mathbf{r}) =\epsilon _{ 0}\mathbf{A} \cdot \mathbf{e}_{x}\mathbf{e}_{y} \cdot \mathbf{A}^{T} \cdot \mathbf{r}\,. }$$

(6.168)

We may then put $\mathbf{U}_{\mathrm{c}} = \mathbf{0}$, so that

$$\displaystyle{ \boldsymbol{\Omega } \times \mathbf{r} -\mathbf{v}^{0}(\mathbf{r}) = -\epsilon _{ 0}\mathbf{m} \cdot \mathbf{r} }$$

(6.169)

with

$$\displaystyle{ \mathbf{m} = \frac{1} {2}\mathbf{A} \cdot (\mathbf{e}_{x}\mathbf{e}_{y} + \mathbf{e}_{y}\mathbf{e}_{x}) \cdot \mathbf{A}^{T}\,, }$$

(6.170)

since the body rotates about the z axis with the angular velocity $-\epsilon _{0}/2$ in the limit of ε ₀ = 0 [27]. Thus Eq. (6.157) becomes

$$\displaystyle{ \mathbf{f}(\mathbf{r}_{1}) = -8\pi \eta _{0}\epsilon _{0}\int _{\mathrm{S}}\mathbf{K}^{-1}(\mathbf{r}_{ 1},\mathbf{r}_{2}) \cdot \mathbf{m} \cdot \mathbf{r}_{2}d\mathbf{r}_{2}\,. }$$

(6.171)

As shown in Appendix 2, [η] of the body may be expressed in terms of the surface integral as

$$\displaystyle{ [\eta ] = -\frac{N_{\mathrm{A}}} {M\eta _{0}\epsilon _{0}}\int _{\mathrm{S}}\mathbf{e}_{x} \cdot {\bigl \langle\mathbf{A}^{T} \cdot \mathbf{f}(\mathbf{r})\mathbf{r} \cdot \mathbf{A}\bigr \rangle} \cdot \mathbf{e}_{ y}d\mathbf{r} = \frac{8\pi N_{\mathrm{A}}} {M} \int _{\mathrm{S}}\Psi (\mathbf{r})d\mathbf{r} }$$

(6.172)

with

$$\displaystyle{ \Psi (\mathbf{r}_{1}) =\int _{\mathrm{S}}\mathbf{e}_{x} \cdot {\bigl \langle\mathbf{A}^{T} \cdot \mathbf{K}^{-1}(\mathbf{r}_{ 1},\mathbf{r}_{2}) \cdot \mathbf{m} \cdot \mathbf{r}_{2}\mathbf{r}_{1} \cdot \mathbf{A}\bigr \rangle} \cdot \mathbf{e}_{y}d\mathbf{r}_{2}\,. }$$

(6.173)

The orientational average in Eq. (6.173) may be evaluated by expanding the matrices A and m in terms of the Wigner functions $\mathcal{D}_{l}^{mj}(\alpha,\beta,\gamma )$ [89]. The function $\Psi$ may then be expressed as

$$\displaystyle\begin{array}{rcl} \Psi (\mathbf{r})& =& \frac{1} {3}(-\psi _{11}r_{1} -\psi _{12}r_{2} + 2\psi _{13}r_{3}) + (\psi _{22}r_{3} +\psi _{23}r_{2}) + (\psi _{31}r_{3} +\psi _{33}r_{1}) \\ & & +(\psi _{41}r_{2} +\psi _{42}r_{1}) + (\psi _{51}r_{1} -\psi _{52}r_{2})\,, {}\end{array}$$

(6.174)

where r = (r ₁, r ₂, $r_{3}$), and the vectors $\boldsymbol{\psi }_{j} = (\psi _{j1}$, ψ _j2, ψ _j3) (j = 1–5) are the solutions of the integral equations

$$\displaystyle{ \int _{\mathrm{S}}\mathbf{K}(\mathbf{r}_{1},\mathbf{r}_{2}) \cdot \boldsymbol{\psi }_{j}(\mathbf{r}_{2})d\mathbf{r}_{2} = \frac{1} {5}\mathbf{m}_{j} \cdot \mathbf{r}_{1} }$$

(6.175)

with

$$\displaystyle\begin{array}{rcl} & & \qquad \mathbf{m}_{1} = -\mathbf{e}_{1}\mathbf{e}_{1} -\mathbf{e}_{2}\mathbf{e}_{2} + 2\mathbf{e}_{3}\mathbf{e}_{3}\,, \\ & & \mathbf{m}_{2} = \mathbf{e}_{2}\mathbf{e}_{3} + \mathbf{e}_{3}\mathbf{e}_{2}\,,\ \ \ \mathbf{m}_{3} = \mathbf{e}_{3}\mathbf{e}_{1} + \mathbf{e}_{1}\mathbf{e}_{3}\,, \\ & & \mathbf{m}_{4} = \mathbf{e}_{1}\mathbf{e}_{2} + \mathbf{e}_{2}\mathbf{e}_{1}\,,\ \ \ \mathbf{m}_{5} = \mathbf{e}_{1}\mathbf{e}_{1} -\mathbf{e}_{2}\mathbf{e}_{2}\,,{}\end{array}$$

(6.176)

We find exact numerical solutions of all the integral equations above for small $p = L/d$ and also solutions in the OB approximation (with the non-preaveraged Oseen tensor) for large p. We first consider the latter. General expressions for the transport coefficients of the spheroid-cylinder in the OB approximation may be derived by replacing the integrals over the surface by those over the contour distance s ($-L/2 \leq s \leq L/2$), where $\boldsymbol{\Psi }_{1}$, $\boldsymbol{\Psi }_{2}$, and $\boldsymbol{\psi }_{j}$ (j = 1–5) are then functions of s with $r_{1} = r_{2} = 0$ and r ₃ = s.

Then the function f _D defined by Eq. (6.37) may be expressed as [27]

$$\displaystyle{ f_{D} = \frac{1} {4}{\bigl (2F_{1}^{\ -1} + F_{ 2}^{\ -1}\bigr )}\,, }$$

(6.177)

where

$$\displaystyle{ F_{j} =\int _{ -1}^{1}\Psi _{ 1j}(x)dx }$$

(6.178)

with $x = 2s/L$. In Eq. (6.178), $\Psi _{1j}$ (j = 1, 2) are the solutions of the integral equations

$$\displaystyle{ \int _{-1}^{1}K_{ j}(x_{1},x_{2})\Psi _{1j}(x_{2})dx_{2} = 1\,, }$$

(6.179)

where $K_{j}(x_{1},x_{2})$ (j = 1, 2) are given by

$$\displaystyle\begin{array}{rcl} K_{1}& =& \frac{2p^{2}(x_{1} - x_{2})^{2} + 3[1 - h(x_{1})]} {d[p^{2}(x_{1} - x_{2})^{2} + 1 - h(x_{1})]^{3/2}}\,,{}\end{array}$$

(6.180)

$$\displaystyle\begin{array}{rcl} K_{2}& =& \frac{4p^{2}(x_{1} - x_{2})^{2} + 2[1 - h(x_{1})]} {d[p^{2}(x_{1} - x_{2})^{2} + 1 - h(x_{1})]^{3/2}}\,,{}\end{array}$$

(6.181)

with $p = L/d$ and with

$$\displaystyle\begin{array}{rcl} h(x)& =& 0\ \ \ \mathrm{for}\ 0 \leq \vert x\vert <1 - \frac{\epsilon } {p} \\ & =& {\biggl [1 -\frac{p(1 -\vert x\vert )} {\epsilon } \biggr ]}^{2}\ \ \ \mathrm{for}\ 1 - \frac{\epsilon } {p} <\vert x\vert \leq 1\,.{}\end{array}$$

(6.182)

In the limit of $p \rightarrow \infty$ (with $h =\epsilon = 0$), Eq. (6.177) becomes the second line of Eqs. (6.49) [24] if we find the asymptotic solution of Eq. (6.179) by a Legendre polynomial expansion method [20].

The rotatory diffusion coefficient $D_{\mathrm{r},1}$ and intrinsic viscosity [η] may be expressed as [33]

$$\displaystyle\begin{array}{rcl} D_{\mathrm{r},1} = \frac{3k_{\mathrm{B}}T} {\pi \eta _{0}L^{3}F_{\mathrm{r}}}\,,& &{}\end{array}$$

(6.183)

$$\displaystyle\begin{array}{rcl} [\eta ] = \frac{\pi N_{\mathrm{A}}L^{3}} {90M} (3F_{\mathrm{r}} + F_{\eta _{\infty }})\,,& &{}\end{array}$$

(6.184)

where

$$\displaystyle\begin{array}{rcl} F_{\mathrm{r}} = 3\int _{-1}^{1}x\Psi _{ 21}(x)dx\,,& &{}\end{array}$$

(6.185)

$$\displaystyle\begin{array}{rcl} F_{\eta _{\infty }} = 6\int _{-1}^{1}x\Psi _{ 22}(x)dx& &{}\end{array}$$

(6.186)

with $\Psi _{2j}$ (j = 1, 2) the solutions of the integral equations

$$\displaystyle{ \int _{-1}^{1}K_{ j}(x_{1},x_{2})\Psi _{2j}(x_{2})dx_{2} = x_{1}\,. }$$

(6.187)

In the OB approximation we have $\Xi _{\mathrm{r},3} = 0$ from Eqs. (6.162) since $r_{1} = r_{2} = 0$, so that the rotation of the cylinder about its axis cannot be considered. In the limit of $p \rightarrow \infty$, we find [33] by the Legendre polynomial expansion method

$$\displaystyle\begin{array}{rcl} F_{\mathrm{r}}^{-1} =\ln p + 2\ln 2 -\frac{11} {6} + \mathcal{O}(p^{-1})\,,& &{}\end{array}$$

(6.188)

$$\displaystyle\begin{array}{rcl} F_{\eta _{\infty }}^{\ -1} =\ln p + 2\ln 2 -\frac{17} {6} + \mathcal{O}(p^{-1})\,,& &{}\end{array}$$

(6.189)

so that Eq. (6.184) becomes Eq. (6.86).

In the following, we complete expressions for f _D , D _r, 1, and [η].

• (a) Translational Diffusion Coefficient

As mentioned above and in Sect. 6.3.1, the OB approximation along with the preaveraging of the Oseen tensor and with the KR approximation [3] as in Eq. (6.35) can give the correct asymptotic result for f _D as given by Eq. (6.49). We therefore evaluate f _D in this way from

$$\displaystyle{ f_{D} = \frac{1} {2}L\int _{-1}^{1}\int _{ -1}^{1}K(x_{ 1},x_{2})dx_{1}dx_{2} }$$

(6.190)

with

$$\displaystyle{ K(x_{1},x_{2}) ={\bigl \{ (x_{1} - x_{2})^{2} +{\bigl [ 1 - h(x_{ 1})\bigr ]}p^{-2}\bigr \}}^{-1/2}\,. }$$

(6.191)

The result reads

$$\displaystyle\begin{array}{rcl} f_{D}& =& \sinh ^{-1}(2p-\epsilon ) - \frac{\epsilon } {p}\sinh ^{-1}\epsilon - \frac{1} {2p}{\bigl \{{\bigl [(2p-\epsilon )^{2} + 1\bigr ]}^{1/2} - (\epsilon ^{2} + 1)^{1/2}\bigr \}} \\ & & - \frac{\epsilon } {2p}\ln \left (2(p-\epsilon ){\bigl \{2p -\epsilon +{\bigl [(2p-\epsilon )^{2} + 1\bigr ]}^{1/2}\bigr \}} + 1\right ) \\ & & + \frac{\epsilon } {2p}\ln {\bigl [4p(p-\epsilon ) + 1\bigr ]} + f'_{D} {}\end{array}$$

(6.192)

with

$$\displaystyle\begin{array}{rcl} f'_{D}& =& \frac{\epsilon } {2p(\epsilon ^{2} - 1)^{1/2}}\biggl \{\epsilon \ln \bigl [\epsilon ^{2} + (\epsilon ^{4} - 1)^{1/2}\bigr ] + (2p-\epsilon ) \\ & & \times \,\ln \biggl |\frac{\epsilon (2p-\epsilon ) - (\epsilon ^{2} - 1)^{1/2}[(2p-\epsilon )^{2} + 1]^{1/2}} {2p[(\epsilon ^{2} - 1)^{1/2}-\epsilon ] + 1} \biggr |\biggr \}\ \ \ \mathrm{for}\ \epsilon> 1 \\ & =& \frac{1} {2p}\bigl \{2p -\bigl [ (2p - 1)^{2} + 1\bigr ]^{1/2} + \sqrt{2}\bigr \}\ \ \ \mathrm{for}\ \epsilon = 1 \\ & =& \frac{\epsilon } {2p(1 -\epsilon ^{2})^{1/2}}\biggl [2\epsilon \tan ^{-1}\biggl (\frac{1 -\epsilon ^{2}} {1 +\epsilon ^{2}}\biggr )^{1/2} + (2p-\epsilon ) \\ & & \times \biggl \{\sin ^{-1} \frac{\epsilon (2p-\epsilon )} {[4p(p-\epsilon ) + 1]^{1/2}} +\sin ^{-1} \frac{1 - 2p\epsilon } {[4p(p-\epsilon ) + 1]^{1/2}}\biggr \}\biggr]\ \ \ \mathrm{for}\ \epsilon <1\,.{}\end{array}$$

(6.193)

Equation (6.192) with Eqs. (6.193) is the desired result and reduces to the result of Norisuye et al. [90] when ε = 1 (prolate spherocylinder), and to the well-known result [91, 92] for the spheroid when ε = p. The latter, which we denote by $f_{D,(\mathrm{SD})}$, is given by

$$\displaystyle{ f_{D,(\mathrm{SD})} =\epsilon F(\epsilon ) }$$

(6.194)

with

$$\displaystyle\begin{array}{rcl} F(\epsilon )& =& \frac{1} {(\epsilon ^{2} - 1)^{1/2}}\cosh ^{-1}\epsilon \ \ \ \mathrm{for}\ \epsilon> 1 \\ & =& 1\ \ \ \mathrm{for}\ \epsilon = 1 \\ & =& \frac{1} {(1 -\epsilon ^{2})^{1/2}}\cos ^{-1}\epsilon \ \ \ \mathrm{for}\ \epsilon <1\,.{}\end{array}$$

(6.195)

Figure 6.17 shows double-logarithmic plots of f _D against p. The solid and dot-dashed curves represent the values calculated from Eq. (6.192) for the spheroid-cylinders with ε = 0. 5 and 1.0 and from Eq. (6.194) for the spheroid (ε = p), respectively, and the unfilled (ε ≠ p) and filled (ε = p) circles represent the exact numerical solutions. The former values are seen to agree well with the latter. Thus we may adopt Eq. (6.192) with Eqs. (6.193) as a useful interpolation formula for f _D for the spheroid-cylinder. When ε = 0, Eq. (6.192) gives the f _D, (R) from Eq. (6.39), and the values for this limiting case are represented by the dotted curve. It is seen that the end effects may be ignored for $p \gtrsim 5$. For comparison, the values calculated from the Broersma equation [18], which are not very different from those from his new version [93], are represented by the dashed curve B. Indeed, his solution of the integral equation is not exact, although asymptotically correct in the limit of $p \rightarrow \infty$.

Fig. 6.17

Double-logarithmic plots of f _D against p. The solid, dot-dashed, and dotted curves represent the values calculated from Eq. (6.192) for the spheroid-cylinders, spheroid (ε = p), and cylinder (ε = 0), respectively, and the unfilled and filled circles represent the exact numerical solutions. The dashed curve B represents the values by Broersma [18] for cylinders

• (b) Rotatory Diffusion Coefficient

We construct an interpolation formula for D _r, 1 on the basis of the exact numerical solutions and the OB asymptotic solution above. The result for $F_{\mathrm{r}} = F_{\mathrm{r}}(p,\epsilon )$ in Eq. (6.183) reads

$$\displaystyle{ F_{\mathrm{r}}(p,\epsilon )^{-1} =\ln p + 2\ln 2 -\frac{11} {6} + \frac{a_{\mathrm{r}0}(\epsilon )} {\ln (1 + p)} +\sum _{ i=1}^{6}a_{\mathrm{ r}i}(\epsilon )\,p^{-i/4} }$$

(6.196)

with

$$\displaystyle\begin{array}{rcl} & & a_{\mathrm{r}0}(\epsilon ) ={\bigl [\ln (1+\epsilon )\bigr ]}{\biggl [f_{\mathrm{r}}(\epsilon )^{-1} -\ln \epsilon -2\ln 2 + \frac{11} {6} -\sum _{i=1}^{6}a_{\mathrm{ r}i}(\epsilon )\,\epsilon ^{-i/4}\biggr ]}\,,{}\end{array}$$

(6.197)

$$\displaystyle\begin{array}{rcl} a_{\mathrm{r}i}(\epsilon ) =\sum _{ j=0}^{2}a_{\mathrm{ r}ij}\,\epsilon ^{j}\,,& &{}\end{array}$$

(6.198)

$$\displaystyle\begin{array}{rcl} f_{\mathrm{r}}(\epsilon )^{-1} = F_{\mathrm{ r}}(\epsilon,\epsilon )^{-1} = \frac{\pi \eta _{0}d^{3}\epsilon ^{3}} {3k_{\mathrm{B}}T}D_{\mathrm{r},1,\mathrm{(SD)}}\,,& &{}\end{array}$$

(6.199)

where a _rij are numerical constants and their values are given in Table 6.7; and $D_{\mathrm{r},1,(\mathrm{SD})}$ is the rotatory diffusion coefficient D _r, 1 of the spheroid and is given by [91, 92]

$$\displaystyle\begin{array}{rcl} \frac{\pi \eta _{0}d^{3}D_{\mathrm{r},1,(\mathrm{S}D)}} {k_{\mathrm{B}}T} & =& \frac{3} {2(\epsilon ^{4} - 1)}{\bigl [(2\epsilon ^{2} - 1)F -\epsilon \bigr ]}\ \ \ \mathrm{ for}\ \epsilon \neq 1 \\ & =& 1\ \ \ \mathrm{for}\ \epsilon = 1 {}\end{array}$$

(6.200)

with F being given by Eqs. (6.195). Note that at ε = p, Eq. (6.183) with Eq. (6.196) gives the exact solution given by Eqs. (6.200) for the spheroid. The range of application of Eq. (6.196) is limited to $0.6 \lesssim \epsilon \lesssim 1.3$. We note that $D_{\mathrm{r},3,(\mathrm{SD})}$ is given by [91, 92]

$$\displaystyle\begin{array}{rcl} \frac{\pi \eta _{0}d^{3}D_{\mathrm{r},3,(\mathrm{SD})}} {k_{\mathrm{B}}T} & =& \frac{3} {2(\epsilon ^{2} - 1)}(\epsilon -F)\ \ \ \ \mathrm{for}\ \epsilon \neq 1 \\ & =& 1\ \ \ \mathrm{for}\ \epsilon = 1\,, {}\end{array}$$

(6.201)

Table 6.7 Values of $a_{\mathrm{r}ij}$ in Eq. (6.198)

Figure 6.18 shows double-logarithmic plots of $D_{\mathrm{r},(\mathrm{S})}/D_{\mathrm{r},1}$ against p, where D _r, (S) is the rotatory diffusion coefficient of the Stokes sphere and is equal to $D_{\mathrm{r},1,(\mathrm{SD})}$ with ε = 1. The solid and dot-dashed curves represent the values calculated from Eq. (6.183) with Eq. (6.196) for the spheroid-cylinders with ε = 0. 63 and 1.0 and from Eqs. (6.200) for the spheroid (ε = p), respectively, and the unfilled (ε ≠ p) and filled (ε = p) circles represent the exact numerical solutions. It is seen that the former values agree well with the latter, and that the end effects are rather small even for small p. In the figure are also shown the values calculated from the Broersma equation [18], which is also correct only for large p. (Note that they are not very different from the values from his new version [93].)

• (c) Intrinsic Viscosity

Fig. 6.18

Double-logarithmic plots of $D_{\mathrm{r},(\mathrm{S})}/D_{\mathrm{r},1}$ against p (see the text). The solid and dot-dashed curves represent the values calculated from Eq. (6.183) with Eq. (6.196) for the spheroid-cylinders and spheroid (ε = p), respectively, and the unfilled and filled circles represent the exact numerical solutions. The dashed curve B represents the values by Broersma [18] for cylinders

As in the case of D _r, 1, we construct an interpolation formula for [η] on the basis of the exact numerical solutions and the OB asymptotic solution above (with the nonpreaveraged Oseen tensor). The result reads

$$\displaystyle{ [\eta ] = \frac{2\pi N_{\mathrm{A}}L^{3}} {45M} F_{\eta }(p,\epsilon ) }$$

(6.202)

with

$$\displaystyle\begin{array}{rcl} & & F_{\eta }(p,\epsilon )^{-1} =\ln p + 2\ln 2 -\frac{25} {12} + \frac{a_{\eta 0}(\epsilon )} {\ln (1 + p)} +\sum _{ i=1}^{5}a_{\eta i}(\epsilon )\,p^{-i/4}\,,{}\end{array}$$

(6.203)

$$\displaystyle\begin{array}{rcl} & & a_{\eta 0}(\epsilon ) ={\bigl [\ln (1+\epsilon )\bigr ]}{\biggl [f_{\eta }(\epsilon )^{-1} -\ln \epsilon -2\ln 2 + \frac{25} {12} -\sum _{i=1}^{5}a_{\eta i}(\epsilon )\,\epsilon ^{-i/4}\biggr ]}\,,{}\end{array}$$

(6.204)

$$\displaystyle\begin{array}{rcl} a_{\eta i}(\epsilon ) =\sum _{ j=0}^{2}a_{\eta ij}\,\epsilon ^{j}\,,& &{}\end{array}$$

(6.205)

$$\displaystyle\begin{array}{rcl} f_{\eta }(\epsilon ) = F_{\eta }(\epsilon,\epsilon ) = \frac{45M} {2\pi N_{\mathrm{A}}d^{3}\epsilon ^{3}}[\eta ]_{(\mathrm{SD})}\,,& &{}\end{array}$$

(6.206)

where a _{η i j} are numerical constants and their values are given in Table 6.8; and [η]_(SD) is the intrinsic viscosity of the spheroid and is given by [94, 95]

$$\displaystyle\begin{array}{rcl} [\eta ]_{\mathrm{(SD)}}& =& \frac{\pi N_{\mathrm{A}}d^{3}} {30M} \epsilon (\epsilon ^{2} - 1)^{2}\biggl \{ \frac{2[-(4\epsilon ^{2} - 1)F + 2\epsilon ^{3}+\epsilon ]} {3\epsilon (3F + 2\epsilon ^{3} - 5\epsilon )[(2\epsilon ^{2} + 1)F - 3\epsilon ]} \\ & & + \frac{28} {3\epsilon (3F + 2\epsilon ^{3} - 5\epsilon )} + \frac{4} {(\epsilon ^{2} + 1)(-3\epsilon F +\epsilon ^{2} + 2)} \\ & & + \frac{2(\epsilon ^{2} - 1)} {\epsilon (\epsilon ^{2} + 1)[(2\epsilon ^{2} - 1)F-\epsilon ]}\biggr \}\ \ \ \mathrm{for}\ \epsilon \neq 1 \\ & =& \frac{5\pi N_{\mathrm{A}}d^{3}} {12M} \ \ \ \mathrm{for}\ \epsilon = 1 {}\end{array}$$

(6.207)

with F being given by Eqs. (6.195). Note that at ε = p, Eq. (6.202) gives the exact solution given by Eqs. (6.207) for the spheroid. We also note that the derivation of the first of Eqs. (6.207) by Simha [94] is not correct, although his result happens to be correct, as shown by Saito [95], the second being originally due to Einstein [1, 96]. The range of application of Eq. (6.202) is limited to $0.6 \lesssim \epsilon \lesssim 1.3$.

Table 6.8 Values of a _{η i j} in Eq. (6.A.52)

Figure 6.19 shows double-logarithmic plots of M[η]∕M ₀[η]_E p against p, where [η]_E is given by Eq. (6.99). The solid and dot-dashed curves represent the values calculated from Eq. (6.202) for the spheroid-cylinders with the indicated values of ε and from Eqs. (6.207) for the spheroid (ε = p), respectively, and the unfilled (ε ≠ p) and filled (ε = p) circles represent the exact numerical solutions. It is seen that the end effects on [η] are more remarkable than those on D and D _r, 1.

Fig. 6.19

Double-logarithmic plots of $M[\eta ]/M_{0}[\eta ]_{\mathrm{E}}p$ against p (see the text). The solid and dot-dashed curves represent the values calculated from Eq. (6.202) for the spheroid-cylinders and spheroid (ε = p), respectively, and the unfilled and filled circles represent the exact numerical solutions

Finally, we note that in the case of cylinders we have found that the values of [η] with the nonpreaveraged and preaveraged Oseen tensors agree with each other to within 1 % for $60 \lesssim p \lesssim 150$ [27]. This fact has been used in the construction of the interpolation formula for $[\eta ]_{\mathrm{R}}$ given by Eq. (6.92).

Appendix 2: Excess Stress Tensor for the Touched-Bead Model

In this appendix we derive an expression for the excess stress tensor due to an addition of a single touched-bead (or generally subbody) model chain to an incompressible fluid with viscosity coefficient η ₀ [38]. In the unperturbed flow field v ⁰ given by Eq. (6.65) with ε ₀ the velocity gradient, the intrinsic viscosity [η] may be written in the form

$$\displaystyle{ [\eta ] = \frac{N_{\mathrm{A}}} {M\eta _{0}\epsilon _{0}}\langle \sigma '_{xy}\rangle = \frac{N_{\mathrm{A}}} {M\eta _{0}\epsilon _{0}}\mathbf{m}:\langle \boldsymbol{\sigma } '\rangle \,, }$$

(6.208)

where m is given by Eq. (6.67), and $\sigma '_{xy}$ is the xy component of the excess stress tensor $\boldsymbol{\sigma }'$ for the single chain.

Now the equation of motion for the (incompressible) fluid in steady flow may be written in the form [1]

$$\displaystyle{ \nabla \cdot \boldsymbol{\sigma } (\mathbf{r}) + \mathbf{f}(\mathbf{r}) = \mathbf{0}\,, }$$

(6.209)

where $\mathbf{f}(\mathbf{r})$ is the force density due to the external (frictional) force exerted on the fluid (per unit volume) at a point r, and $\boldsymbol{\sigma }$ is the stress tensor given by

$$\displaystyle{ \boldsymbol{\sigma }(\mathbf{r}) = -p(\mathbf{r})\mathbf{I} +\eta _{0}{\bigl \{\nabla \mathbf{v}(\mathbf{r}) +{\bigl [ \nabla \mathbf{v}(\mathbf{r})\bigr ]}^{T}\bigr \}} }$$

(6.210)

with $\mathbf{v}(\mathbf{r})$ the fluid velocity, p the pressure, I the unit tensor, and the superscript T indicating the transpose. Note that substitution of Eq. (6.210) with Eq. (6.2) into Eq. (6.209) leads to Eq. (6.1). In the present case of the single chain composed of N beads (subbodies), f(r) is given by

$$\displaystyle{ \mathbf{f}(\mathbf{r}) =\sum _{ j=1}^{N}\int _{ \mathrm{S}_{j}}\delta (\mathbf{r} -\mathbf{r}_{j} -\hat{\mathbf{r}}_{j})\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})d\hat{\mathbf{r}}_{j}\,, }$$

(6.211)

where r _j is the vector position of the center of the jth bead, $\hat{\mathbf{r}}_{j}$ is the radius vector from its center to an arbitrary point on its surface, $\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})$ is the frictional force exerted by the unit area at $\hat{\mathbf{r}}_{j}$ on the fluid, and the integration is carried out over its surface (S _j ) (see Fig. 6.3).

The stress tensor $\boldsymbol{\sigma }$ may be written as a sum of the stress tensor $\boldsymbol{\sigma }_{0}$ of the pure fluid and the excess stress tensor $\boldsymbol{\sigma }'$ due to the force density f; that is, $\boldsymbol{\sigma }=\boldsymbol{\sigma } _{0} +\boldsymbol{\sigma } '$. Equation (6.209) may therefore be rewritten as

$$\displaystyle\begin{array}{rcl} \nabla \cdot \boldsymbol{\sigma }_{0} = \mathbf{0}\,,& &{}\end{array}$$

(6.212)

$$\displaystyle\begin{array}{rcl} \nabla \cdot \boldsymbol{\sigma }' + \mathbf{f} = \mathbf{0}\,.& &{}\end{array}$$

(6.213)

We take the Fourier transform of both sides of Eq. (6.213),

$$\displaystyle{ i\mathbf{k} \cdot \tilde{\boldsymbol{\sigma }}'(\mathbf{k}) +\tilde{ \mathbf{f}}(\mathbf{k}) = \mathbf{0}\,, }$$

(6.214)

where

$$\displaystyle{ \tilde{\boldsymbol{\sigma }}'(\mathbf{k}) =\int \boldsymbol{\sigma } '(\mathbf{r})\exp (i\mathbf{k} \cdot \mathbf{r})d\mathbf{r}\,, }$$

(6.215)

$$\displaystyle\begin{array}{rcl} \tilde{\mathbf{f}}(\mathbf{k})& =& \int \mathbf{f}(\mathbf{r})\exp (i\mathbf{k} \cdot \mathbf{r})d\mathbf{r} \\ & =& \sum _{j=1}^{N}\int _{ \mathrm{S}_{j}}\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})\exp {\bigl [i\mathbf{k} \cdot (\mathbf{r}_{j} +\hat{ \mathbf{r}}_{j})\bigr ]}d\hat{\mathbf{r}}_{j}\,.{}\end{array}$$

(6.216)

Let $\mathbf{R}_{\mathrm{c}}$ be the vector position of the center of mass of the chain, and let S _j be the vector distance from it to the center of the jth bead. We have $\mathbf{r}_{j} = \mathbf{R}_{\mathrm{c}} + \mathbf{S}_{j}$, and the second line of Eqs. (6.216) may be rewritten as

$$\displaystyle\begin{array}{rcl} & & \tilde{\mathbf{f}}(\mathbf{k}) =\exp (i\mathbf{k} \cdot \mathbf{R}_{\mathrm{c}})\sum _{j=1}^{N}\mathbf{F}_{ j} + i\mathbf{k} \cdot \exp (i\mathbf{k} \cdot \mathbf{R}_{\mathrm{(C)}}) \\ & & \qquad \times \sum _{j=1}^{N}\int _{ \mathrm{S}_{j}}{\biggl \{\int _{0}^{1}\exp {\bigl [i\xi \mathbf{k} \cdot (\mathbf{S}_{ j} +\hat{ \mathbf{r}}_{j})\bigr ]}d\xi \biggr \}}(\mathbf{S}_{j} +\hat{ \mathbf{r}}_{j})\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})d\hat{\mathbf{r}}_{j}\,,{}\end{array}$$

(6.217)

where F _j is the total frictional force exerted by the jth bead and is given by Eq. (6.27). Under the condition of ordinary viscosity measurements, there is not any external force other than shear flow field, so that the total frictional force (sum of F _j ) must vanish. We then obtain from Eqs. (6.214) and (6.217)

$$\displaystyle\begin{array}{rcl} \tilde{\boldsymbol{\sigma }}'(\mathbf{k})& =& -\exp (i\mathbf{k} \cdot \mathbf{R}_{\mathrm{c}})\sum _{j=1}^{N}\int _{ \mathrm{S}_{j}}{\biggl \{\int _{0}^{1}\exp {\bigl [i\xi \mathbf{k} \cdot (\mathbf{S}_{ j} +\hat{ \mathbf{r}}_{j})\bigr ]}d\xi \biggr \}} \\ & & \times (\mathbf{S}_{j} +\hat{ \mathbf{r}}_{j})\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})d\hat{\mathbf{r}}_{j}\,. {}\end{array}$$

(6.218)

Finally, we take the configurational average of both sides of Eq. (6.218), noting that R _c is distributed uniformly in the fluid and that the average over R _c may be taken independently of the other variables. We then obtain

$$\displaystyle{ \langle \tilde{\boldsymbol{\sigma }}'(\mathbf{k})\rangle = -(2\pi )^{3}\delta (\mathbf{k})\sum _{ j=1}^{N}{\biggl [\langle \mathbf{S}_{ j}\mathbf{F}_{j}\rangle +{\biggl \langle\int _{\mathrm{S}_{j}}\hat{\mathbf{r}}_{j}\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})d\hat{\mathbf{r}}_{j}\biggr \rangle}\biggr ]}\,. }$$

(6.219)

Thus, by Fourier inversion of Eq. (6.219), we obtain

$$\displaystyle{ \langle \boldsymbol{\sigma }'\rangle = -\sum _{j=1}^{N}{\biggl [\langle \mathbf{F}_{ j}\mathbf{S}_{j}\rangle +{\biggl \langle\int _{\mathrm{S}_{i}}\mathbf{f}_{j}(\hat{\mathbf{r}}_{j})\hat{\mathbf{r}}_{j}d\hat{\mathbf{r}}_{j}\biggr \rangle}\biggr ]}\,. }$$

(6.220)

Substitution of Eq. (6.220) into the second of Eqs. (6.208) leads to Eq. (6.102). In the case of a single rigid body, it also reduces to the first of Eqs. (6.172).