Curl identities symbol question

[bobfei]

Hello,

I would like to ask a question on curl.

The wikipedia page http://en.wikipedia.org/wiki/Vector_calculus_identities" gives formulas of various operations, among which:

[tex]\nabla \times (A \times B) = A(\nabla \cdot B) - B(\nabla \cdot A) + (\underbrace {B \cdot \nabla }_{{\rm{meaning?}}})A - (A \cdot \nabla )B[/tex]

What does [tex]\underbrace {B \cdot \nabla }_{{\rm{meaning?}}}[/tex] mean in this formula? It appears to me should not be the same as divergence, but besides that what can it stand for?

Can anyone explain this, and preferably give a reference to this?


Bob

 

[0ddbio]

Just use:
[tex]\nabla=\frac{\partial}{\partial x}\hat{i}+\frac{\partial}{\partial y}\hat{j}+\frac{\partial}{\partial z}\hat{k}[/tex]

Then [itex]\vec{B}\cdot\nabla[/itex]
is just:
[tex]B_{x}\frac{\partial}{\partial x}+B_{y}\frac{\partial}{\partial y}+B_{z}\frac{\partial}{\partial z}[/tex]

I don't think it really has any sort of interpretation. It's easy to use mathematically but don't try to read too much into it's properties, just treat it as a mathematical operator.

So it looks similar to divergence as you said, but it is not anything like it. So then when this ([itex]\vec{B}\cdot\nabla[/itex]) is multiplied by some function the function is just distributed to the partial derivatives.

The main difference coming from the fact that when the A is then distributed you are not taking the derivatives of the components, but the entire vector.. Because it then reads:
[tex]B_{x}\frac{\partial}{\partial x}\vec{A}+B_{y}\frac{\partial}{\partial y}\vec{A}+B_{z}\frac{\partial}{\partial z}\vec{A}[/tex]
The B's are the components of B, but the A is the entire vector A.

 

[bobfei]

Dear Oddbio,

I have checked and your answer is correct, they fit into the expansion of [tex]\nabla \times (A \times B)[/tex].

In driving this, I actually find what really takes is some re-combination, which finally yields the [tex]B_{x}\frac{\partial}{\partial x}+B_{y}\frac{\partial}{\partial y}+B_{z}\frac{\partial}{\partial z}[/tex] term which exist in each of the vector components. What is your opinion here?


Bob

 

[0ddbio]

Have you given this much thought as to what [itex]\nabla[/itex] itself is?

It's a vector operator, it really doesn't mean much on its own. That is the same with what you have here
[tex]B_{x}\frac{\partial}{\partial x}+B_{y}\frac{\partial}{\partial y}+B_{z}\frac{\partial}{\partial z}[/tex]
There are partial derivatives but they aren't taking the derivative of anything.
So we call it an operator. It really only has meaning when you multiply it by something.

You will never just have that as an answer, it will always be multiplied by something. Because of that, it is difficult to say what meaning it really has.

What you are doing is really like looking at something like:
[tex]\frac{dF}{dx}[/tex]
and asking what is the meaning of
[tex]\frac{d}{dx}[/tex]

So you could say that the real equation is:
[tex]B_{x}\frac{\partial}{\partial x}\vec{A}+B_{y}\frac{\partial}{\partial y}\vec{A}+B_{z}\frac{\partial}{\partial z}\vec{A}[/tex]
But just to simplify it we define what [itex]\nabla[/itex] is, and we can rewrite it as:
[tex]\left(\vec{B}\cdot\nabla\right)\vec{A}[/tex]
It's just a nicer way of writing it. But you have to look at the whole thing, not just the dot product part.

 

[bobfei]

Dear Oddbio,

This Wikipedia page http://en.wikipedia.org/wiki/Nabla_symbol" contains the information that William Hamilton first invented it. I actually have used these symbols for pretty some days, and from them, although I know little of Hamilton's other achievements, my humble view is that he was indeed, as its inventor, a first class mathematician in history as now recognized.

Symbols are not arbitrarily chosen. Although the underlying operation is really independent of the [tex]\nabla [/tex] symbol or whatever else, by using this triangle, lots of calculations involving cross product can be greatly simplified. From this perspective, triangular symbols ([tex]\nabla [/tex] or [tex]\Delta [/tex]) are uniquely suited for these operations.

For operations other than cross product, http://en.wikipedia.org/wiki/Del" also has information on how matrix form can be used instead of lengthy derivative calculations.

The initial post was due to my unfamiliarity with the [tex]A \cdot \nabla [/tex] operation. Besides that I am moderately familiar with them.Bob

 

[dextercioby]

[tex] \left(\vec{B}\cdot\nabla\right){A} [/tex]

(A is a scalar, B is a vector)

is the projection of the gradient of A along the vector B. (=Contracted tensor product between the gradient of a 0-tensor A and a 1-tensor B).

[tex] \left(\vec{B}\cdot\nabla\right)\vec{A} [/tex]

(A is a vector, B is a vector)

is the projection of the gradient of A along the vector B. (=Contracted tensor product between the gradient of a 1-tensor A and a 1-tensor B).

 

[bobfei]

This succinct word "projection" is everything.

Thanks,
Bob