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?
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?
On
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?
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 $(\vec{u}\cdot\vec{n})\vec{n}=\vec{u}^{T}(\vec{n}\otimes\vec{n})=(\vec{n}\otimes\vec{n})\vec{u}$