Why Does the Curl Product Rule Seem Confusing?

[Saladsamurai]

Curl Product Rule confusion?

Homework Statement

In Griffith's Introduction to Electrodynamics, he gives the rule:

$$\nabla\times(\bold{A}\times\bold{B})=(\bold{B}\cdot\nabla)\bold{A}-(\bold{A}\cdot\nabla)\bold{B}+\bold{A}(\nabla\cdot\bold{B})-\bold{B}(\nabla\cdot\bold{A})$$

Now I know I am missing something stupid here, but what is the difference between $(\bold{A}\cdot\nabla)\bold{B}$ and $\bold{B}(\nabla\cdot\bold{A})$ ?


The dot product commutes doesn't it? And then we are left with a scalar times a vector

If $(\bold{A}\cdot\nabla)=(\nabla\cdot\bold{A})=k$

then what is the difference between kB and Bk ?


I know I am doing something wrong, but what?

Casey

 

[Unco]

Hi Saladsamurai,

The dot product on a vector space commutes because the scalars commute. The dot product in your formula isn't quite the same. Notice that, for example, the first terms of $$\mathbf{A}\cdot \nabla$$ and $$\nabla\cdot \mathbf{A}$$ are $$A_1\frac{\partial}{\partial x}$$ and $$\frac{\partial}{\partial x}A_1$$, respectively, which are not the same.

 

[Saladsamurai]

I still don't see it, why did you add a prime symbol?

If A is some vector with components $<A_x, A_y, A_z>$ and the operator $\nabla =<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}>$

oh nevermind... this is something I don't really need to understand.

I think my confusion stems from the inherently weird definition of $\nabla$.

(keep in mind I am an engineer )

 

[Phrak]

I still don't see it, why did you add a prime symbol?

That's a comma. Without the symbols inline, it does look like a prime.

If A is some vector with components $<A_x, A_y, A_z>$ and the operator $\nabla =<\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}>$

oh nevermind... this is something I don't really need to understand.

I think my confusion stems from the inherently weird definition of $\nabla$.

(keep in mind I am an engineer )

A and the parital derivative don't commute.

$$\nabla \cdot A$$
is a scalar that will act on the vector B.

$$A \cdot \nabla$$
is a derivative operator, scaled by A, that can act on the vector B.

It's goofy notation, but it's what we have.

 

[Defennder]

See this:
http://mathworld.wolfram.com/ConvectiveOperator.html

The del-dot operator isn't commutative.

 

[Saladsamurai]

That's a comma. Without the symbols inline, it does look like a prime.


A and the parital derivative don't commute.

$$\nabla \cdot A$$
is a scalar that will act on the vector B.

$$A \cdot \nabla$$
is a derivative operator, scaled by A, that can act on the vector B.

It's goofy notation, but it's what we have.

This makes more sense now. Thanks

See this:
http://mathworld.wolfram.com/ConvectiveOperator.html

The del-dot operator isn't commutative.

I will read in the morning; after that Divergence thread, I realize that I am toast at this point

thanks for the link