Local Anisotropy
The situation is getting considerably complicated, if one takes local anisotropy into account. The mobility of a particle along the axis of a macromolecule is considered to be bigger than that in the perpendicular direction, so that the entire macromolecule can move easier along its contour. Introduction of the local anisotropy of mobility in Chapter 3 (equations (3.12) and (3.13)) allows us to specify the extra forces of external and internal resistance and define the quantities (7.5) as
where ae and ai are measures of local anisotropy. Every internal particle of the chain can be ascribed by the direction vector
eα= rα+1− rα−1
|rα+1− rα−1|, α = 1, . . . , N− 1,
while the zeroth and the last particles having no direction, so that there are N− 1 vectors for a chain. It is convenient formally to consider the product of components of vectors e0 and eN to be defined as
e0ie0j = eNi eNj =1 3δij.
While introducing of the global anisotropy, the equation for the macro-molecular dynamics remains linear in co-ordinates and velocities, the intro-duction of the local anisotropy makes it non-linear in co-ordinates. Both global and local anisotropy are needed to describe the non-linear effects of the relax-ation phenomena in the mesoscopic approximrelax-ation.
7.2 Equations for the Non-Equilibrium Moments
The dynamics of the macromolecule in the form of a set of differential equa-tions of the first order is convenient for derivation of relaxation equaequa-tions (Volkov and Vinogradov 1984, 1985; Volkov 1990). As a starting point, we use equations (7.6) and consider m = 0 in this system, so that the second equation allows us to define
140 7 Equations of Relaxation
ϕαk = ζ(ψαk − νkjραj) + 2μT λαραk + ¯ξkα (7.13) and to rewrite the system of equations (7.6) as
d
dtραi = ψiα, d
dtψαi = 0, τ d
dtϕαi = τ ωilϕαl − ζ(ψiα− νijραj)− 2μT λαραi
− ζBανij (ψjν− νjlρνl)− ζGijαν(ψjν− ωjlρνl) + ¯ξαi + σαi. These equations allow one to find relaxation equation for different mo-ments, if the quantities Bijαν and Gijαν are given. We consider here that the quantities are independent on the co-ordinates; the more complicated case, when these quantities depend on the co-ordinates of particles, is left for other researchers. It is convenient in this section to omit the mode label and write down the above equations in the form
dρi dt = ψi, d
dtψi= 0, (7.14)
dϕi
dt = κijρj+ λijψj+ ωijϕj+1
τ( ¯ξiα+ σαi), where
κij = ζ τ
− 1
2τRδij+ νij+ Bilνlj+ Eilωlj
, λij=−ζ
τ (δij+ Bij+ Eij) , (7.15)
Bij = Bβij, Eij = Eij, τR= ζ 4μT λ.
Further on we shall use the following symbols for the moments of the considered variables
rik=ρiρk, nik=ρiϕk, yik=ρiψk, mik=ψiϕk, zik=ψiψk, lik=ϕiϕk.
The moments can be found as solutions of the set of equations, which are followed by set (7.14)
7.2 Equations for the Non-Equilibrium Moments 141
drik
dt = yik+ yki, dyik
dt = zik, dzik
dt = 0, dnik
dt = mik+ κkjrji+ λkjyij+ ωkjnij+1
τρi( ¯ξiα+ σiα), dmik
dt = κkjyji+ λkjzji+ ωkjmij+1
τψi( ¯ξiα+ σαi), dlik
dt = κijnjk+ κkjnji+ λijmjk+ λkjmji+ ωijljk+ ωkjlji
+1
τϕi( ¯ξiα+ σiα(t)) + ϕk( ¯ξαi + σαi).
To determine the average quantities, which contain the random forces in the above-written equations, one can consider the equilibrium situation. The unknown terms can be evaluated through the equilibrium values of moments, which can be used to rewrite the equations for the moments. Finally, the set of relaxation equations has the form of the above-written equations, where the terms with random forces and the terms containing the product of velocity gradient and an equilibrium moment are omitted. Instead of moments, the differences in the moments and their equilibrium values, such as rik− r0ik instead of rik, for instance, ought to be written, so that one has
drik
dt = yik+ yki, dyik
dt = zik, dzik
dt = 0, dnik
dt = mik− m0ik+ κkjrji− κ0kjr0ji+ λkjyij+ ωkjnij, dmik
dt = κkjyji+ ωkjmij, dlik
dt = κijnjk− κ0ijn0jk+ κkjnji− κ0kjn0ji+ λijmjk+ λkjmji
+ ωijljk+ ωkjlji
(7.16)
where κ0ij=−ζτ2τ1Rδij is the value of κij given by (7.15) at zero velocity gradients. The fact that some of the equilibrium moments are equal to zero, as shown below, has already been taken into account in equations (7.16).
The system of equations (7.16) determines the unknown quantities as func-tions of time and the parameters of the problem: ζ, B, E, and the parameters of anisotropy. To find a solution, one uses the equilibrium values of moments, three of which are found in Section 4.1.2
142 7 Equations of Relaxation
rik0 = 1
2μλδik, zik0 = T
mδik, yik0 = 0. (7.17) The others are contained in the following relations, which are consequences of relation (7.13)
nik= n0ik+ ζ(yik− νkjrji) + ζ
2τR(rik− r0ik), mik= m0ik+ ζ(zik− νkjyji) + ζ
2τRyki, (7.18)
lik= l0ik+ ζ(mki− νkjnji) + ζ
2τR(nki− n0ki).
One can note that, with help of the one of the above relations, the fourth equation in the set (7.16) can be written as
dnik
dt = ζzik− ζνkjyji+ ζ
2τRyki+ κkjrji− κ0kjrji0 + λkjyij+ ωkjnij. From the other side, after having differentiated the first of equations (7.18), one has
dnik
dt = ζ(zik− νkj(yji+ yij)) + ζ
2τR(yik+ yki).
These equations are followed by the relation τ
2τRδkj+ δkj+ Bkj+ Ekj
yij
=− 1
2τR(rik− r0ik) + τ
2τRδkl+ δkl+ Bkl+ Ekl
ωljrji+ Bklγljrji
+ τ νkjyij+ τ ωkjyji+τ
ζωkjn0ji+ τ
2τRωkjrji0, (7.19) where terms containing velocity gradients in the second power are already excluded.
The equation (7.19) has to be considered as an equation for the quantity yik. When the velocity gradients are absent,
yik=− 1
2τ♦(rik− r0ik), τ♦ =τ
2+ τR(1 + B + E) . (7.20) The symbol♦is used here to show the place of label of relaxation times which are identical to relaxation times (4.26). Then, one can see that the last four terms in (7.19) can be neglected and the last equation allows us to find the relation
yik=− 1
2τ♦(rij− rij0)bjk+ ωkjrji+BτR
τ♦ rijγjlclk, (7.21) where the following notations are used