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

Condiff, D. W.; Dahler, J. S.

Phys. Fluids 7, 842-854 (1964).

Keywords:

fluid mechanics

References:

[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).