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Mathematic expression of grad(U)

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Old   November 24, 2015, 07:35
Default Mathematic expression of grad(U)
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Tobias Holzmann
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Hi all,

I have just one question. In mostly all literature we find the gradient of a scalar \vec \nabla \phi or vector \vec \nabla \textbf{U} in that way. But in a mathematic point of view this is not correct for vectors.

So we get for any scalar:
\mathrm{grad}(\phi) = \vec \nabla \phi = \sum_{i=1}^{3} \frac{\partial \phi}{\partial x_i} \vec e_i
Of course I assume \vec \nabla as the partial differentials of the three directions and this is correct but

For any vector:
\mathrm{grad}(\textbf{U}) = (\vec \nabla \otimes \textbf{U})^T = (\vec \nabla \textbf{U})^T

My question now. Is it just a definition that we suppose that everybody know that \vec \nabla \textbf{U} is the transposed dyadic product of these two vectors? Because normally, \vec \nabla and \textbf{U} are two vectors with the matrix notation (3x1) and (3x1). Therefore we have to use the dyadic product of this operation and after that we really have to transpose that matrix to get the correct form of the gradient matrix.

Example:
\vec \nabla \textbf{U} = \vec\nabla \otimes \textbf{U}  \neq \mathrm{grad}(\textbf{U})

(\vec \nabla \textbf{U})^T = (\vec\nabla \otimes \textbf{U})^T = \vec\nabla \textbf{U}^T \overset{!}{=} \mathrm{grad}(\textbf{U})

Again. Is it a definition that in CFD numerics we say:
\vec \nabla \textbf{U} \overset{!}{=} (\vec \nabla \textbf{U})^T

I would kindly appreciate some feedback.
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Last edited by Tobi; November 24, 2015 at 11:36.
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