## Fluid mechanical aspects of antisymmetric stress.(English)Zbl 0125.15801

fluid mechanics
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 [1] Dahler, Proc. Roy. Soc. (London) A275 pp 504– (1963) [2] Dahler, J. Chem. Phys. 30 pp 1447– (1959) [3] Research Frontiers of Fluid Dynamics, edited by G. Temple and R. J. Seeger (Interscience Publishers, Inc., New York, to be published). [4] Born, Z. Physik 1 pp 221– (1920) [5] Our notation is that of Gibbs where the tensor elements are to be considered as components of the dyadic expanded in the appropriate set of base vectors. [6] See the second part of Ref. 2. [7] J. Serrin, inHandbuch der Physik, edited by S. Flügge, (Julius Springer-Verlag, Berlin, 1959), Vol. VIII, Pt. 1, p. 230,et seq. [8] R. Aris,Vectors, Tensors, and the Basic Equations of Fluid Mechanics(Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1962), Chap. 5, p. 106,et seq. · Zbl 0123.41502 [9] This can be seen by choosing Cartesian coordinates with thezaxis in the direction of {$$\times$$}u-2{$$\omega$$}0. The matrix associated with the dyadic dual ({$$\times$$}u-2{$$\omega$$}0) will then be of the form 0M0-M00000 while that of Ta will be 0{$$\tau$$}3-{$$\tau$$}1-{$$\tau$$}30{$$\tau$$}2{$$\tau$$}2-{$$\tau$$}20. Isotrophy implies {$$\tau$$}1 = {$$\tau$$}2 = 0 since a rotation through {$$\pi$$} about thezaxis replaces {$$\tau$$}2 and {$$\tau$$}1 by -{$$\tau$$}2 and -{$$\tau$$}1 respectively. This shows that {$$\tau$$}2 = -{$$\tau$$}2 since they must be the same function of the matrix 0M0-M00000 which is unchanged by the rotation. [10] Sather, Phys. Fluids 5 pp 754– (1962) [11] P. Debye,Polar Molecules(Dover Publications, Inc., New York, 1929). · JFM 55.1179.04 [12] To see this, substitute {$$\omega$$}0 = 12{$$\times$$}u in (12) and (13). Then subject to the stated restrictions one finds that 0 = -P+{$$\zeta$$}({$$\times$$}({$$\times$$}u))+({$$\phi$$}+13{$$\eta$$}-{$$\zeta$$})()+({$$\zeta$$}+{$$\eta$$})2u+{$$\rho$$}fP+({$$\phi$$}+13{$$\eta$$})()+{$$\eta$$}2u+{$$\rho$$}f, (12) 0 = ({$$\nu$$}2+13{$$\nu$$}1)({$$\times$$}u))+({$$\nu$$}1+{$$\nu$$}2)2({$$\times$$}u){$$\nu$$}1+{$$\nu$$}2)2({$$\times$$}u). (13) Equation (13) is identical to the equation obtained by taking the curl of (12), whereas (12) is just the Navier-Stokes equation for a structureless fluid. Thus {$$\omega$$}0 = 12{$$\times$$}u satisfies the spin equation (13) wheneverusatisfies the Navier-Stokes equation for a structureless fluid. These solutions for {$$\omega$$}0 andusatisfy both (12) and (13).
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