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Simple vector notation help. uiuj in non index notation

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Old   October 4, 2012, 13:39
Default Simple vector notation help. uiuj in non index notation
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Casey
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Hey Everyone,

For my paper I have about 50 equations all in the "standard" vector notation
\vec{u}, but there are 2 or 3 turbulence equations with the Reynolds stress therm in index notation that I don't know how to put in the "standard form". The term is below:

u_iu_j = ??? in standard form

I think \vec{u}^2 is not correct or at least will make the reader think I am talking about the dot product of u with u.
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Old   October 4, 2012, 13:48
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My friend just said he thinks it is

u_iu_j=\vec{u}\vec{u}
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Old   October 4, 2012, 14:01
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agd
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If you are referring to the terms in the Reynolds stress tensor, your friend is correct. The term for such a representation is dyadic.
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Old   October 4, 2012, 14:11
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I just had another friend say it is the outer product, so


u_iu_j=\vec{u}\otimes\vec{u}

is it basically the same thing?

\vec{u}\vec{u}=\vec{u}\otimes\vec{u} ?
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Old   October 4, 2012, 17:34
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They are not the same thing. The outer product of two vectors is another vector. The inner product of two vectors is, of course, a scalar. The dyadic product of two vectors is a tensor of order 2 (since the vectors are each tensors of order 1). When you write the index notation (ui)(uj) without any caveats, it generally implies the dyadic product.
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Old   October 5, 2012, 11:23
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I just looked and according to wiki (i know, i know) the outer product does produce a tensor. Are you sure about your answer?
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Old   October 7, 2012, 12:05
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Quote:
Originally Posted by meangreen View Post
I just had another friend say it is the outer product, so


u_iu_j=\vec{u}\otimes\vec{u}

is it basically the same thing?

\vec{u}\vec{u}=\vec{u}\otimes\vec{u} ?
First equation is correct. Second one should be \vec{u}\vec{u}^T=\vec{u}\otimes\vec{u} assuming column vectors.
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Old   October 7, 2012, 12:07
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Sorry - I saw the symbol and my brain interpreted it as the cross product. Yes, the outer product results in the same thing as the dyadic product. It is not the same thing as the cross product, which produces a tensor of the same order as the original entries.
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