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July 4, 2010, 12:33 
How to derive this simple stress tensor calculation?

#1 
Member

Most turbulence models are written in tensor expressions so we have to write out the full formulation before we can implement them. I am not very experienced in difficult tensor analysis. So can anyone explain a little about writing this term in the standard ke model?
From other books, the LHS should equals RHS, which reads: How to get from LHS to RHS? In my understanding, for 3D domain LHS should be a 3x3 matrix as it's a product of two ranktwo tensors. But RHS is a scalar. How can this happen? Thank you very much. 

July 4, 2010, 13:46 

#2 
Administrator

Check out: http://www.cfdonline.com/Wiki/Einst...ion_convention
The LHS you have written is also a scalar and uses the Einstein summation convention 

July 4, 2010, 14:29 

#3  
Member

Quote:
The difficulty is that the index does not repeat in any individual term, we don't have any summation. http://topex.ucsd.edu/geodynamics/shearer.pdf this is about solid mechanics, but those concepts such as strain tensor and rotation tensor also apply to fluid mechanics. 

July 4, 2010, 17:03 

#4 
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Guillaume Fournier
Join Date: Jun 2010
Posts: 3
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Actually, (dui/dxj + duj/dxi)*dui/dxj = (dui/dxj)² + duj/dxi*dui/dxj
Now, you have two terms in which the indexes repeat themselves. Therefore, there is a summation on both i and j for each terms After a few step, you will get the RHS from the LHS in your first post 

July 4, 2010, 18:27 

#5 
Member

yes, Guillaume, you're right from your direction.
On the other hand, isn't that (not fully written because of Latex limit here) and They are all 3x3 matrices, how can their product be a scalar? Anything wrong with the above expressions? Last edited by bearcat; July 4, 2010 at 19:00. 

July 5, 2010, 03:45 

#6 
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Guillaume Fournier
Join Date: Jun 2010
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I think you're right in the way you wrote the expressions of and
However, you're mistaken in the fact that is not only a product of matrices when using Einstein's convention: you have to sum over i and j, which corresponds to computing the trace of the matrix. Let me show you. I won't develop the Stress tensor, and I will call each of its components S11, S12.... Let's start with the Einstein's summation convention That will eventually lead to the RHS term you wrote in your first message Without the convention, it would be a classical matrix product between two 3x3 matrices, yielding a 3x3 matrix as well. If we call it M, we have and so on for the next lines. We can notice that the RHS term only contains the sum of the diagonal terms of M. Therefore, if you prefer to use the matrix product rather than the Einstein convention to compute and develop your term, you just have to remember that this convention is equivalent to the computation of the trace of the resultant matrix Here we are. I hope it will help a little Cheers Guillaume Last edited by Guillaume_Fournier; July 5, 2010 at 04:42. 

July 5, 2010, 13:34 

#7 
Member

Thank you for your explanation. This is very helpful discussion.
Quote:


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