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


Does anyone have any clues where this comes from?


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

  • $\begingroup$ 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

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?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.