Eulerian transverse velocity increment

TurbJet · May 6, 2020, 23:36

Hi,

I am reading paper [1], and I run into this formula for the Eulerian transverse velocity increment

$\delta u_\perp = \|\mathcal{P}\cdot\delta u\|\cos\theta$

where $\mathcal{P} = \delta_{ij} - \hat{r}_i\hat{r}_j$ is the projection tensor projecting the vector onto the plane perpendicular to the vector

, and $\forall\theta\in[0\,,2\pi]$ , and $\hat{r}$ simply being the normalized vector of

.

I understand the projection, but I don't understand why multiplied by a $\cos\theta$ ? Can anyone shed some light?

Appreciate it.

Reference:
[1]. E. Leveque, A. Naso. Introduction of longitudinal and transverse Lagrangian velocity increments in homogeneous and isotropic turbulence. EPL, 108, 54004, 2014.

agd · May 7, 2020, 09:45

The projection involves the dot product. The cos term results from that.

May 6, 2020, 23:36	Eulerian transverse velocity increment	#1
TurbJet Senior Member Join Date: Oct 2017 Location: United States Posts: 233 Blog Entries: 1 Rep Power: 10	Hi, I am reading paper [1], and I run into this formula for the Eulerian transverse velocity increment $\delta u_\perp = \\|\mathcal{P}\cdot\delta u\\|\cos\theta$ where $\mathcal{P} = \delta_{ij} - \hat{r}_i\hat{r}_j$ is the projection tensor projecting the vector onto the plane perpendicular to the vector , and $\forall\theta\in[0\,,2\pi]$ , and $\hat{r}$ simply being the normalized vector of . I understand the projection, but I don't understand why multiplied by a $\cos\theta$ ? Can anyone shed some light? Appreciate it. Reference: [1]. E. Leveque, A. Naso. Introduction of longitudinal and transverse Lagrangian velocity increments in homogeneous and isotropic turbulence. EPL, 108, 54004, 2014.

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May 7, 2020, 09:45		#2
agd Senior Member Join Date: Jul 2009 Posts: 358 Rep Power: 19	The projection involves the dot product. The cos term results from that.