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Page 999 In the book "Introduction to Theoretical and Computational Fluid Dynamics" by Constantine Pozrikidis mentions something called a 'tangential projection operator'

I-N(x)N

Does anyone have any clues where this comes from?

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Let $\vec{u}$ be a vector, and let $\vec{t}$ and $\vec{n}$ be the tangential and normal vectors in our system of reference. The vector $\vec{u}$ can be decomposed into: $$\vec{u} = (\vec{u}\cdot\vec{t})\vec{t}+(\vec{u}\cdot\vec{n})\vec{n}=\vec{u}_t+\vec{u}_n$$

Therefore the tangential part, noted by $\vec{u}_t=(\vec{u}\cdot\vec{t})\vec{t}$ can be rewritten as: $$\vec{u}_t = \vec{u}-\vec{u}_n=\vec{u}-(\vec{u}\cdot\vec{n})\vec{n}=(I-\vec{n}\otimes\vec{n})\vec{u}$$ Being the tensor within brackets $(I-\vec{n}\otimes\vec{n})$ your "tangential operator"

I've applied the identity $(\vec{u}\cdot\vec{n})\vec{n}=\vec{u}^{T}(\vec{n}\otimes\vec{n})=(\vec{n}\otimes\vec{n})\vec{u}$

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  • $\begingroup$ Could you explain about the identity you mentioned? $\endgroup$ – Analysis Newbie Nov 21 '17 at 14:26
  • $\begingroup$ This is a simple reordering of the dot product of two vectors times another vector. Since the result of this operation is a vector, there must me some matrix that applied to a vector gives the desired result. You can verify this componentwise. $\endgroup$ – HBR Nov 21 '17 at 14:46
  • $\begingroup$ Assume $\vec{u} = \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}$ and $\vec{n} = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}$, then $(\vec{u}\cdot\vec{n})\vec{n} = (u_1 n_1 + u_2 n_2)\begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}$. However, $\vec{n}\otimes\vec{n} = \vec{n}\vec{n}^T = \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix}$, thus $\vec{u}^T(\vec{n}\otimes\vec{n}) = \begin{bmatrix}{u_1 \\ u_2}\end{bmatrix} \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix}$ is ill-defined, similarly for $(\vec{n}\otimes\vec{n})\vec{u}$ on RHS. In which step am I wrong? $\endgroup$ – Analysis Newbie Nov 21 '17 at 15:45
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For the sake of simplicity, let us consider 2-d case, we assume row vector $\vec{u} = \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}$ and column vector $\vec{n} = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}$, then $\vec{u}\cdot\vec{n} = (u_1 n_1 + u_2 n_2)$ which is a scalar, so the column vector $(\vec{u}\cdot\vec{n})\vec{n} = (u_1 n_1 + u_2 n_2)\begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}$ is well-defined on the LHS.

However, $\vec{u}^T = \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}^T = \begin{bmatrix}{u_1 \\ u_2}\end{bmatrix}$, and $\vec{n}\otimes\vec{n} = \vec{n}\vec{n}^T = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix} \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix}^{T} = \begin{bmatrix}{n_1 \\ n_2}\end{bmatrix} \begin{bmatrix}{n_1 \; n_2}\end{bmatrix} = \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix}$, thus $\vec{u}^T(\vec{n}\otimes\vec{n}) = \begin{bmatrix}{u_1 \\ u_2}\end{bmatrix} \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix}$ is ill-defined.

In addition, $(\vec{n}\otimes\vec{n})\vec{u} = \begin{bmatrix}{n_1 n_1 \; n_1 n_2 \\ n_2 n_1 \; n_2 n_2}\end{bmatrix} \begin{bmatrix}{u_1 \; u_2}\end{bmatrix}$ is also ill-defined.

In which step am I wrong?

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  • $\begingroup$ Vectors without $T$ are always column vectors. $\endgroup$ – HBR Nov 21 '17 at 15:55
  • $\begingroup$ I've realized that $\vec{u}$ should be a column vector in order to make the dot product $\vec{u} \cdot \vec{n}$ well-defined. However, even if we assume $\vec{u}$ a column vector, the identity $\vec{u}^{T}(\vec{n}\otimes\vec{n}) = (\vec{n}\otimes\vec{n})\vec{u}$ does not hold, because LHS will be a row vector while RHS a column vector. $\endgroup$ – Analysis Newbie Nov 21 '17 at 17:36
  • $\begingroup$ Obviously... some transpose operator is missing. But you could have deduced that. $\endgroup$ – HBR Nov 21 '17 at 18:10

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