Torque tensor

To account for the torques that the particles exert on each other, we define the elements of the "torque tensor" related to the torsions analogous to Accordingly, the torque per unit area is ... 

The potential of mean torque, parameterized by the local interaction tensors Xa and Xc (assume to be cylindrical) for the aromatic core and the C C segment, respectively, can be mapped out at each temperature by fitting the observed quadrupolar splittings in the mesophase. Furthermore, the order matrix of an average conformer of the molecule can also be calculated.17 Now pn is needed to describe the internal dynamics. 

Particles subject to Brownian motion tend to adopt random orientations, and hence do not follow these rules. A particle without these symmetry properties may follow a spiral trajectory, and may also rotate or wobble. In general, the drag and torque on an arbitrary particle translating and rotating in an unbounded quiescent fluid are determined by three second-order tensors which depend on the shape of the body ... 

A kinetic argument shows that ay - oy always. Any imbalance between these two would lead to an angular acceleration of a volume element. If this volume element were shrunk, then the torque would reduce in proportion to the linear dimension cubed, but the moment of inertia would reduce in proportion to the fifth power of the linear dimension, so that the angular acceleration would increase as the reciprocal of the square of the size of the volume element, becoming infinite in the limit. Thus reductio ad absur-dum, ay = cry. Hence there are only six independent components of the stress tensor. 

Surface tractions or contact forces produce a stress field in the fluid element characterized by a stress tensor T. Its negative is interpreted as the diffusive flux of momentum, and x x (—T) is the diffusive flux of angular momentum or torque distribution. If stresses and torques are presumed to be in local equilibrium, the tensor T is easily shown to be symmetric. 

In this case, the stress tensor is non-symmetric, if there is an external force torque. The law of conservation of angular momentum follows from the law of conservation of momentum. 

So, we shall further assume, that the internal and external rotation are balanced in polymer solutions and the stress tensor is symmetric, when there is no external force torque. 

Force equilibrium considerations show that, in the absence of an applied torque, the strain and stress tensors are symmetric o-y = o, and Sy = s, . Consequently, there are really only six independent stresses for three directions that can be applied to strain a body. The mathematical representation of stress is thus ... 

Note that A with two subscripts denotes the thermal conductivity which is a second rank tensor and A with one subscript denotes the director constraint torque which is a pseudo vector.) Heat conduction is a dissipative process. The entropy production per unit time and unit volume, a, caused by the heat flow... 

The torque density X. and the antisymmetric pressure tensor also remain unchanged, XyCy = A2C2 and PyCy = P2 2- strain is inserted into... 

We can identify four pairs of thermodynamic forces and fluxes, the symmetric traceless strain rate (Vu) and the symmetric traceless pressure tensor, the director angular velocity relative to the background, (l/2)Vxu-I2 and the torque density X, the streaming angular velocity relative to the background (l/2)Vxu- and the torque density and the trace of the strain rate V-u and difference between the trace of the pressure tensor and the equilibrium... 

The coefficients rj, fj[ and 773 are shear viscosities. The twist viscosity is denoted by 7[. The symmetric traceless pressure tensor cross couples with the trace of the strain rate and the two angular velocities (l/2)Vxu-Q and (l/2)Vxu- . The corresponding cross coupling coefficients are 772 According to the Onsager reciprocity relations, they must be equal to 72/2 and 74/2. They couple the symmetric traceless strain rate to ((l/3)7 r(P)-Pg and to the two torque densities (1) and (4)- The coeffi-... 

These results may be used to rationalize the experimental additivity schemes for the molecular tensors. Owing to the — factor of force and torque operators, the molecular wavefunction is essentially weighted in the environment of nucleus /, which could imply transferability of atomic terms from molecule to molecule in a series of structurally and chemically related homologues. 

So far only the minimal multipole moments and their contribution to the j, a., and (t2) values have been considered. However, they are not the only manifestations of the degeneracy in the characteristics of the absorption and Raman spectra under consideration. For instance, an important additional contribution to the torque can arise because of the anisotropy of the dispersion interaction between the molecules owing to the nonzero matrix elements of the anisotropic components of the polarizability tensor, i.e., the anisotropy of the cubic symmetry molecules in degenerate states. 

This theory has two notable features. The nonlocality of molecular interaction is reflected by the ellipticity of Eq. (19) [cf Eq. (15)]. Thus, the LCP configuration is globally coupled by distortional elasticity. In addition, the elastic stress tensor is asymmetric. The mean-field torque on LCP molecules amounts to a volume torque on the material, which modifies the usual conservation of angular momentum. The antisymmetric part of the stress tensor precisely balances the volume torque computed by averaging the molecular torque. ... 

The stress tensor may be represented as a 3 x 3 matrix, with components Oy, where i and j both go from 1 to 3. The diagonal elements represent normal stresses, whereas the off-diagonal ones represent shear stresses. Positive normal stresses are tensile, while negative ones are compressive (but an opposite sign convention is sometimes used, most notably in the soil mechanics literature). Finally, from the balance of angular momentum (or torque in the static case), it follows that the stress tensor and its matrix representation are symmetric (ay = aji), meaning that only six out of the nine components are in fact independent. 

Regarding the use of Eq. (6.26) in practice we note that the same comments made earlier apply here as well [see discussion after Eq. (6.17)]. A detailed discussion of optimal choices for the Ewald parameters a and for dipolar systems can bo found in Rc fs. 243 and 244. Finally, readers who are interested in performing MD simulations of dipolar fluids are referred to Appendix F.2.2 where we present explicit expressions for forces and torques associated with the three-dimensional Ewald sum [see Eq. (6.26)]. Moreover, explicit expressions for various components of the stress tensor can be found in Appendix F.2.3. 

In a rheological experiment, one of the two cylinders is typically rotated with a known angular velocity, and the torque required to produce this motion is measured. Let us suppose that the torque is measured on the inner cylinder. Now, if we ignore the finite length of the Couette device, we have seen that there is a single nonzero component of the velocity ug(r). Hence, if we examine the various components of the rate-of-strain tensor E,... 

To actually use these results, it is of course necessary to actually calculate the components of the resistance tensors. We have seen that it is necessary to solve only three problems for translation and three problems for rotation in the coordinate directions to specify all of the components of A, B, C, and D. It probably does not need to be said that the orientation of the coordinate axes should be chosen to take advantage of any geometric symmetries that can simplify the fluid mechanics problems that must be solved. For example, if we wish to determine the force and/or torque on an ellipsoid for motions of arbitrary magnitude and direction (with respect to the body geometry), we should specify the components of the resistance tensors with respect to axes that are coincident with the principal axes of the ellipsoid, as this choice will simplify the fluid mechanics problems that are necessary to determine these components. If arbitrary velocities U and ft are then specified with respect to these same coordinate axes, the Eqs. (7-22) will yield force and torque components in this coordinate system. 

Problem 7-9. Motion of a Force- and Torque-Free Axisymmetric Particle in a General Linear Flow. We consider a force- and torque-free axisymmetric particle whose geometry can be characterized by a single vector d immersed in a general linear flow, which takes the form far from the particle y°°(r) = U00 + r A fl00 + r E00, where U°°, il, and Ex are constants. Note that E00 is the symmetric rate-of-strain tensor and il is the vorticity vector, both defined in terms of the undisturbed flow. The Reynolds number for the particle motion is small so that the creeping-motion approximation can be applied. 

Common examples of pseudo-vectors that will be relevant later include the angular velocity vector f2, the torque T, the vorticity vector co (or the curl of any true vector), and the cross product of two vectors. The inner scaler product of a vector and a pseudo-tensor or a pseudo-vector and a regular tensor will both produce a pseudo-vector. It will also be useful to extend the notion of a pseudo-vector to scalers that are formed as the product of a vector and a pseudo-vector. The third-order, alternating tensor e is a pseudo-tensor of third order as may be verified by reviewing its definition... 

Problem 8-10. Symmetry of the Grand Resistance Tensor. Use the reciprocal theorem to show that the grand resistance tensor is symmetric. The grand resistance tensor relates the hydrodynamic force/torque on a particle to its velocity/angular velocity ... 

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