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Dissipation local Isotropy Kolmogorov Idea

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Old   November 7, 2007, 06:31
Default Dissipation local Isotropy Kolmogorov Idea
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Hatef
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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.

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Old   November 7, 2007, 13:40
Default Re: Dissipation local Isotropy Kolmogorov Idea
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Hatef
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Kronecker
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Old   November 7, 2007, 15:41
Default Re: Dissipation local Isotropy Kolmogorov Idea
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Ananda Himansu
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The Kronecker symbol d_ij is, as I recall, the only isotropic second-order 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 second-order 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 matrix-vector 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 second-order 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 non-zero 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.
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Old   November 9, 2007, 16:52
Default Re: Dissipation local Isotropy Kolmogorov Idea
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Ananda Himansu
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I guess I forgot to explicitly answer your question as to why the off-diagonal 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 non-zero for all i and j. In rectangular Cartesians, g_ij reduces to d_ij with off-diagonal terms being zero. Also, in general oblique coordinates, the mixed contravariant-covariant representation of the metric tensor, namely d^i_j, again has zero for its off-diagonal 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 second-order tensor, if I did not make myself clear earlier.
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