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I am doing some work for my fluid dynamics class. I was trying to do a simple proof, but hit a road block. I have an expression where a 3-vector say $\vec{u}$ dotted with a 3x3 identity matrix $\bf{I}$, and I am not sure what the result would be. I guess that it should just be $\vec{u}$ however I was taught (perhaps incorrectly) that the dot product results in a scalar thus meaning this operation maybe invalid altogether, I could be wrong but if I am right I would like to perhaps see a simple proof of why this works. Thanks for all your time!

question:

$ \vec{u} \cdot \bf{I} \stackrel{?}{=} \vec{u}$

Context:

I am doing a short proof in fluid dynamics. Show that the Eulerian operator $\frac{D}{Dt} \equiv \frac{\partial}{\partial{t}} + \vec{u} \cdot \nabla$ acting on a position vector $\vec{r}$ (where $\vec{r} = (x,y,z)$) gives the fluid velocity $\vec{u}$. That is:

$\frac{D \vec{r}}{Dt} = \frac{\partial \vec{r}}{\partial{t}} + \vec{u} \cdot \nabla \vec{r} \stackrel{?}{=} \vec{u}$ , for $\vec{r} = (x,y,z)$

So first step I simply apply the Eulerian operator to the position vector:

$\rightarrow \frac{D \vec{r}}{Dt} = \frac{\partial \vec{r}}{\partial{t}} + \vec{u} \cdot \nabla \vec{r}$

since $\vec{r}$ is a position vector its partial derivative with respect to time vanishes, and we are left with:

$\rightarrow\frac{D \vec{r}}{Dt} = \ \vec{u} \cdot \nabla \vec{r}$

doing matrix multiplication with $\nabla$ and $\vec{r}$ I get the 3x3 identity matrix $ \bf{I} $:

$\rightarrow\frac{D \vec{r}}{Dt} = \vec{u} \cdot \bf{I}$

This is the point I get stuck on, I have no familiarity with the subject of fluid dynamics prior so I am unsure of how to proceed. Since the proof requests me to show that the Eulerian operator of a position vector gives the fluid velocity, I can take a pretty good guess that $\vec{u} \cdot \bf{I} = \vec{u}$ but someone please show and confirm. Thanks!

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  • $\begingroup$ Is the dot here just matrix multiplication? $\endgroup$ Commented Sep 29, 2018 at 4:26
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    $\begingroup$ There is not enough context here to determine what the notation even means. $\endgroup$ Commented Sep 29, 2018 at 8:54
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    $\begingroup$ I have added context @HansLundmark $\endgroup$ Commented Sep 29, 2018 at 15:50
  • $\begingroup$ How did you end up with a matrix for $\nabla\vec r$? $\endgroup$
    – amd
    Commented Sep 29, 2018 at 20:50
  • $\begingroup$ Matrix multiplcation, 3x1 vector times a 1x3 vector gives a matrix. Also known as the outer product or diadic product as my professor has told me $\endgroup$ Commented Sep 29, 2018 at 22:45

1 Answer 1

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I believe everything will be clear if you think of $\vec{u} \cdot \nabla$ as single operator $$ \vec{u} \cdot \nabla = u_1 \frac{\partial}{\partial x_1} + u_2 \frac{\partial}{\partial x_2} + u_3 \frac{\partial}{\partial x_3} $$ and then apply the whole thing to $\vec{r} = (x_1,x_2,x_3)$ instead of doing the gradient first and then the dot product. Like this: $$ \left( u_1 \frac{\partial}{\partial x_1} + u_2 \frac{\partial}{\partial x_2} + u_3 \frac{\partial}{\partial x_3} \right) (x_1,x_2,x_3) = u_1 (1,0,0) + u_2 (0,1,0) + u_3 (0,0,1) . $$

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