# (A * del)

1. Jun 29, 2006

### StatusX

$$(\vec A \cdot \nabla)$$

Is this operator well defined? It appears in many vector calculs identites, and it has an easy enough explicit formula in cartesian coordinates. But I've heard it cannot be written generally in the curvilinear coordinates. I assume this is because this operator can be applied to vectors or scalars, and has different forms depending on which. Is this the case? Why is this operator always glossed over in vector calculus classes? Why are explicit curvilinear formulas for it never given? Does it have a name?

For example:

$$( (\vec v \cdot \nabla) \vec v)_r \neq (\vec v \cdot \nabla) v_r$$

although I was never told this, and still don't completely understand it.

Last edited: Jun 29, 2006
2. Jun 29, 2006

### StatusX

Ok, so now I see that $(\vec A \cdot \nabla) f$ for a scalar f is just equal to $\vec A \cdot \nabla (f)$. I'm not sure why the former notation would ever be used. But the application of this operator to vector fields is still mysterious to me.

3. Jun 29, 2006

### Hurkyl

Staff Emeritus
(1) Because operator algebra is extremely useful.
(2) Would you really rather write:

$$\vec{A} \cdot \nabla( \vec{A} \cdot \nabla( \vec{A} \cdot \nabla(f)))$$

$$(\vec{A} \cdot \nabla)^3 f$$

?

(such things do arise. For example, in multivariable Taylor series)

4. Jul 3, 2006

### boarie

May i know what if f is a vector? $(\vec A \cdot \nabla) \vec F$ is this equivalent to $\vec A (\nabla \cdot (vec F))$.

5. Jul 3, 2006

### Hurkyl

Staff Emeritus
Nope. Everything in $(\vec{A} \cdot \nabla) \vec{F}$ is as written. You take the dot product, then apply the operator to $\vec{F}$.

If you want to associate it, then you have to take a tour through tensors -- $\nabla \vec{F}$ is a bivector, and then you contract it with $\vec{A}$.

Last edited: Jul 3, 2006
6. Jul 3, 2006

### boarie

hi Hurkyl

Thx for your advice.. thus may i know, if we are to take the del operator with A and then apply to F, the resultant will be: $(\vec{A1} \frac{\partial \vec{F1}}{\partial x}......$?

7. Jul 3, 2006

### Hurkyl

Staff Emeritus
Well, since you didn't write down what you think it is, I can only guess. I suspect you did not do what I said, and instead simply component-wise multipled all three parts.

8. Jul 4, 2006

### boarie

hi Hurkyl

My apology for not being clear.. is it wrong to state the result as:

$(\vec{A_{x}} \frac{\partial \vec{F_{x}}}{\partial x} + \vec{A_{y}} \frac{\partial \vec{F_{y}}}{\partial y} + \vec{A_{z}} \frac{\partial \vec{F_{z}}}{\partial z})$

Last edited: Jul 4, 2006
9. Jul 4, 2006

### arildno

Yes, it is wrong.
You should have a vector as the result, not a scalar.

10. Jul 4, 2006

### boarie

Roger on that...

Thus I shld perfom A dot del:

$(\vec{A_{x}} \frac{\partial }{\partial x} + \vec{A_{y}} \frac{\partial }{\partial y} + \vec{A_{z}} \frac{\partial }{\partial z})$

which gives me a scalar.

And multiply with vector F?

$((\vec{A_{x}} \frac{\partial }{\partial x} + \vec{A_{y}} \frac{\partial }{\partial y} + \vec{A_{z}} \frac{\partial }{\partial z})) (\vec F)$

Resulting in:

$[\vec{A_{x}} \frac{\partial \vec{F_{x}}}{\partial x} + \vec{A_{y}} \frac{\partial \vec{F_{x}}}{\partial y} + \vec{A_{z}} \frac{\partial \vec{F_{x}}}{\partial z}, \vec{A_{x}} \frac{\partial \vec{F_{y}}}{\partial x} + \vec{A_{y}} \frac{\partial \vec{F_{y}}}{\partial y} + \vec{A_{z}} \frac{\partial \vec{F_{y}}}{\partial z}, \vec{A_{x}} \frac{\partial \vec{F_{z}}}{\partial x} + \vec{A_{y}} \frac{\partial \vec{F_{z}}}{\partial y} + \vec{A_{z}} \frac{\partial \vec{F_{z}}}{\partial z}, ]$

Is this correct?

11. Jul 5, 2006

### StatusX

Here's how I think it works. When you have a vector function $\vec A(\vec r)[/tex], then for a given coordinate system, you can associate to it three scalar functions. For example, in cartesian coordinates, these are Ax, Ay, and Az. Now, when you have an operator involving del, you can classify it as behaving as a vector or a scalar. For example, del itself acts like a vector, and so can be dotted or crossed with a vector function to give a scalar or vector respectively. The laplacian operator (del squared) acts like a scalar, so can act on scalars or vectors, also to give a scalar or vector respectively. The operator that is the subject of this thread falls into the scalar operator category. Now, these operators are all linear, so: $$(\vec A \cdot \nabla) (\vec B + \vec C) = (\vec A \cdot \nabla) \vec B + (\vec A \cdot \nabla) \vec C$$ $$(\vec A \cdot \nabla) (b+c) = (\vec A \cdot \nabla) b + (\vec A \cdot \nabla) c$$ There are also analoques of the product rule, which I don't feel like completely working out, but these would include expressions for [itex] (\vec A \cdot \nabla) (b \vec C)$, $(\vec A \cdot \nabla) (b c)$, $(\vec A \cdot \nabla) (\vec B \cdot \vec C)$, and $(\vec A \cdot \nabla) (\vec B \times \vec C)$. The first one is the important one, and I believe the rule is:

$$(\vec A \cdot \nabla) (b \vec C) = \vec C (\nabla(b) \cdot \vec A) + b (\vec A \cdot \nabla) \vec C$$

Now you can combine these two rule to find the result of applying this operator to a vector function by expressing it as a linear combination of the basis vectors in your particular coordinate system. So if the basis vectors are $\vec u$,$\vec v$,$\vec w$, then:

$$(\vec A \cdot \nabla) \vec B = (\vec A \cdot \nabla) (B_u \vec u + B_v \vec v + B_w \vec w ) = B_u (\vec A \cdot \nabla) \vec u + \vec u (\vec A \cdot \nabla(B_u) ) +B_v (\vec A \cdot \nabla) \vec v + \vec v (\vec A \cdot \nabla(B_v) ) +B_w (\vec A \cdot \nabla) \vec w + \vec w (\vec A \cdot \nabla(B_w) )$$

This is the equation relating the result of this operator applied to vectors to the result of applying it componentwise to the associated scalar fields described in the first paragraph. In cartesian coordinates, $(\vec A \cdot \nabla) \hat x=(\vec A \cdot \nabla) \hat y= (\vec A \cdot \nabla) \hat z=0$, so the above reduces to the simple relation:

$$(\vec A \cdot \nabla) \vec B = \hat x(\vec A \cdot \nabla(B_x) ) + \hat y (\vec A \cdot \nabla(B_y) ) + \hat z (\vec A \cdot \nabla(B_z) )$$

In other coordinate systems, this does not happen (eg, in cylindrical coordinates $(\vec A \cdot \nabla) \hat r \neq 0$), so there will be extra terms involving the derivatives of the basis vectors. This was the main source of my confusion, but I think I'm satisfied now. I'm just a little confused why these important points are never (at least in my experience) explicitly talked about.

Last edited: Jul 5, 2006