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November 7, 2007, 06:31 
Dissipation local Isotropy Kolmogorov Idea

#1 
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Hi
According to Kolmogorov idea(1941) of local isotropy , since dissipation occurs at smallest scale we have: epsi (ij) =2/3 epsi * Kronokor delta (ij) dissipation occures at small scales and it is logical to accept that isotoropy is valid at that scales and e(11)=e(22) =e(33) but the thing that is not clear for me is that why the other terms are zero , for instance why we don't have disspiation for uv (e12) . what does it have to do with isotropy and the concep of invariancy. I would be thankful if anyone can help me to clear this doubt. 

November 7, 2007, 13:40 
Re: Dissipation local Isotropy Kolmogorov Idea

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Kronecker


November 7, 2007, 15:41 
Re: Dissipation local Isotropy Kolmogorov Idea

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The Kronecker symbol d_ij is, as I recall, the only isotropic secondorder tensor (other than the zero tensor). More precisely, d_ij is the rectangular cartesian component representation of the unique (to within a scalar multiplier) isotropic secondorder tensor. The isotropy is manifested in its action on vectors, i.e. its dot product with vectors. Its action preserves the direction of the vector upon which it acts.
If, v_i is the cartesian component representation of an arbitrary vector v, then the isotropy can be seen from either the component representation d_ij*v_i = v_j, or from the matrixvector representation I*V = V. In the latter, I is the identity matrix, V is the column vector representation of v, and the "*" is matrix multiplication (which implements the dot product). The component representation of the isotropic secondorder tensor in general coordinates is g_ij or g^ij or d^i_j, i.e., the metric tensor. As far as I recall, there are no nonzero isotropic tensors of odd order (at least this is true for vectors, which are order one), and there is more than one isotropic tensor of order four and higher (even) order, but my memory could be faulty. In addition to the Kronecker deltas, the permutation symbols are also involved somehow. Isotropic tensors are important in expressing tensorial invariants. The book by Heinbockel, which used to be available as a free download, probably has more details. 

November 9, 2007, 16:52 
Re: Dissipation local Isotropy Kolmogorov Idea

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I guess I forgot to explicitly answer your question as to why the offdiagonal terms are zero. The reason is that you happen to be dealing with rectangular Cartesian coordinates. Notice that in general oblique coordinates the covariant components of the metric tensor, namely the g_ij, are nonzero for all i and j. In rectangular Cartesians, g_ij reduces to d_ij with offdiagonal terms being zero. Also, in general oblique coordinates, the mixed contravariantcovariant representation of the metric tensor, namely d^i_j, again has zero for its offdiagonal terms, owing to the orthonormality of the coordinate (covariant) basis vectors relative to the dual (contravariant) basis vectors. g^ij, g_ij, d^i_j, and I are all representations of one and the same isotropic secondorder tensor, if I did not make myself clear earlier.


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