# Vector calculus Notation

1. Mar 4, 2007

### tim_lou

what does
$$\left (\vec{A}\cdot \vec\nabla \right ) \vec B$$ mean?

2. Mar 5, 2007

### neutrino

There've been a few a threads recently on "A.del". You may want to search the forum.

In short, A.del is an operator

$$\vec{A}.\vec{\nabla} \equiv A_x\frac{\partial}{\partial x}+ A_y\frac{\partial}{\partial y}+A_z\frac{\partial}{\partial z}$$

Last edited: Mar 5, 2007
3. Mar 5, 2007

### HallsofIvy

Staff Emeritus
Specifically
$$(\vec{A}\cdot \nabla)\vec{B}= \vec{A}\cdot (\nabla\vec{B})= A_x\frac{\partial B_x}{\partial x}+ A_y\frac{\partial B_y}{\partial y}+A_z\frac{\partial B_z}{\partial z}$$

4. Mar 5, 2007

### Semo727

How can vector multiplyed by scalar give scalar??

5. Mar 5, 2007

### ZioX

Do you mean HallsofIvy's? If so A_x, A_y, and A_z are vectors. Hence the whole thing is a vector.

6. Mar 5, 2007

### arildno

HallsofIvy made a mistake. The correct expression is:
$$(\vec{A}\cdot\nabla)\vec{B}=A_{x}\frac{\partial\vec{B}}{\partial{x}}+A_{y}\frac{\partial\vec{B}}{\partial{y}}+A_{z}\frac{\partial\vec{B}}{\partial{z}}$$

7. Mar 5, 2007

### Semo727

And I also guess that $(\vec{A}\cdot\vec\nabla)\, \vec{B}\neq\vec{A}\,(\vec\nabla\cdot\vec{B})$ since

$$\vec{A}\,(\vec\nabla\cdot\vec{B})=\vec{A}\,\frac{\partial B_x}{\partial x}+ \vec{A}\,\frac{\partial B_y}{\partial y}+\vec{A}\,\frac{\partial B_z}{\partial z}$$

Last edited: Mar 5, 2007
8. Mar 5, 2007

### tim_lou

wait, hold on, so the correct result is arildno's? the operation gives you a vector..?

9. Mar 5, 2007

### arunma

I see we've turned mathematics into a democracy today. :rofl:

Yes, Arildno's explanation was the correct one.

10. Mar 6, 2007

### arildno

I WON!!!!!!

11. Mar 6, 2007

### HallsofIvy

Staff Emeritus
Actually, I just screwed up the whole thing!

12. May 15, 2007

### ~Bee

interpretation

Sorry for butting in so late , but I needed to expand on this subject. I’ll try and get to Tim-Lou’s question:

but first, a few notes:

arildno said that

This can be seen as a definition of $$\nabla \vec{B}$$ since HallsofIvy pointed out that $$(\vec{A}\cdot \nabla)\vec{B}= \vec{A}\cdot (\nabla\vec{B})$$ (even though my fluids mechanics teacher gave an explicit warning in class, after deriving the expression of the convective derivative, that $$(\vec{A}\cdot \nabla)\vec{B}= \vec{A}\cdot (\nabla\vec{B})$$ only for Cartesian Systems, and then in the following class happened to mention that this is true for any curvilinear orthogonal coordinate system (it's in my class notes, so I know he said it) . . . but that doesn’t make any sense to me, because an expression written in coordinate free notation cannot depend on a coordinate system, no? I'll assume that my teacher went through a moment of mental fogginess, though when I've tried to approach him on the subject he just refers me to the bibliography. In fact, I did do some research, and most books I've seen write $$(\vec{A}\cdot \nabla)\vec{B}$$ instead of $$\vec{A}\cdot (\nabla\vec{B})$$ though they don't make any refernce to the latter not being valid - except one book, which stated flatly that $$(\vec{A}\cdot \nabla)\vec{B} \neq \vec{A}\cdot (\nabla\vec{B})$$ but didn't offer any explanation, so I'm not going to believe that. So I was happy to see HallsofIvy's post).

So let $$(\vec{A}\cdot \nabla)\vec{B}= \vec{A}\cdot (\nabla\vec{B})$$ (valid, of course, in any coordinate system, as the coordinate-free notation suggests) - as I said, this can be viewed as a definition of $$\nabla \vec{B}$$ since we know how to compute $$(\vec{A}\cdot \nabla)\vec{B}$$ from vector calculus, resulting in arildno's expression in Cartesian Coordinates (note that arildo's expression is ONLY valid for Cartsian Coordinates).

Now, if we restrict ourselves to Cartesian Coordinates, the derivative of a vector $$\vec{v}$$ with respect to the coordinates is given by

$$\frac{\partial\vec{v}}{\partial{x_i}} = \frac{\partial{v_j}}{\partial{x_i}} \vec{e_j}$$​

where $${\vec{e_i}}$$ are the standard Cartesian Basis Vectors.

Let $$\vec{v}(x,y,z)$$ represent a vector field. Remember that the differential of $$\vec{v}$$ is map that for a given $$(\Delta{x},\Delta{y},\Delta{z}) \equiv \Delta{\vec{r}}$$ returns the linear part (in terms of $$\Delta{\vec{r}}$$) of the increment $$\Delta{\vec{v}} \equiv \vec{v}(\vec{r}+\Delta\vec{r} ) - \vec{v}(\vec{r})$$, where $$\vec{r}$$ is the position vector. By definition, the differential of $$\vec{v}$$ is:

$$d\vec{v}= \frac{\partial\vec{v}}{\partial{x_i}} d{x_i} = \frac{\partial{v_j}}{\partial{x_i}} \vec{e_j} d{x_i} = d{x_i} \frac{\partial}{\partial{x_i}} ( v_j \vec{e_j}) = (d \vec{r} \cdot \nabla) \vec{v} = d \vec{r} \cdot \nabla \vec{v}$$

That is, $$d \vec{r} \cdot \nabla \vec{v} = \Delta \vec{r} \cdot \nabla \vec{v}$$ (note that the differential of the independent variable is equal to its increment) is the linear part of $$\Delta \vec{v}$$, or, in other words, the derivative of $$\vec{v}$$ along $$\Delta \vec{r}$$.

Thus, $$\vec{a} \cdot \nabla \vec{b}$$ is the directional derivative of $$\vec{b}$$ along $$\vec{a}$$, that is, the linear part (in terms of $$a_i = \Delta x_i$$ ), of $$\Delta{\vec{b}} \equiv \vec{b}(\vec{r}+\Delta\vec{a} ) - \vec{b}(\vec{r})$$.

Note that some authors define the vector gradient $$\nabla \vec{v}$$ as the transpose of how we have defined it here - no biggie, only in that case the differential of $$\vec{v}$$ would be $$d\vec{v}= \nabla \vec{v} \cdot d \vec{r}$$, and the directional derivative of $$\vec{b}$$ would be $$\nabla \vec{b} \cdot \vec{a}$$ instead.

I'll be expanding on this topic in a separate thread, as I think the questions I have to ask would no longer fall under the object of the present thread.

~Bee

Last edited: May 15, 2007
13. May 15, 2007

### neutrino

Hi Bee, welcome to PF. Are the you the same Bee that runs 'Backreaction'?

14. May 15, 2007

### ~Bee

Hi neutrino,
thank you! No, I'm a newbie bee to this and all other forums/blogs - I don't run anything. Just your average bee - please bear with me as I get lost sometimes.
~Bee