I'm having problems with strange BAC-CAB inconsistencies which makes me loose faith in my ability of solving these types of problems. I need someone to check if this is correct or incorrect.

del = d/dx,d/dy,d/dz
a = constant vector=(a,a,a)
(fi) = scalar field
del^2(fi)=0

Question:

My attempt at solution:
Using einsteins way with eidi of solving, this would take too long to show in plain text( I dont know LaTeX yet ) so for simplification I use classic vector identities to proove it.

rot(a x grad(fi) = del x (a x grad(fi)) ={BAC-CAB}= a(del del(fi)) - del(fi)(del a)={First part is 0 according to previous definition}= -del(fi)(del * a).

-del(fi)(a*del)

Discussion:
The difference between (del*a) and (a*del) is huge, esp. as a is a constant vector and will be zero for (del*a). I can not wrap my head around the answer of this problem as all I really have to go on is the acual BAC-CAB rule and nothing else. Please explain to me what might be wrong.

This is what I got (both with vector identities and actually expanding it out)

$\displaystyle \nabla \times \left(\bf{a} \times \nabla \phi\right) = \bf{a} \nabla \cdot \nabla \phi - \left( \bf{a} \cdot \nabla\right) \nabla \phi = - \left( \bf{a} \cdot \nabla\right) \nabla \phi$.

Originally Posted by Danny
This is what I got (both with vector identities and actually expanding it out)

$\displaystyle \nabla \times \left(\bf{a} \times \nabla \phi\right) = \bf{a} \nabla \cdot \nabla \phi - \left( \nabla \cdot \bf{a} \right) \nabla \phi = - \left( \bf{a} \cdot \nabla\right) \nabla \phi$.
Okay, so we know that what i'm doing is wrong. Here is what I know about the BAC-CAB rule;

$\displaystyle \bf{A} \times \left(\bf{B} \times \bf{C} \right) = \bf{B} \left( \bf{A} \cdot \bf{C} \right) - \bf{C} \left( \bf{A} \cdot\bf{B} \right)$

Directly applying this to the original question will yeild my first answer and not the one you provided.

$\displaystyle \nabla \times \left(\bf{a} \times \nabla \phi\right) = \bf{a} \nabla \cdot \nabla \phi - \left(\nabla \cdot \bf{a} \right) \nabla \phi = - \left( \nabla \cdot \bf{a} \right) \nabla \phi$

Please tell me why direct computation of the BAC-CAB rule wont work. As I see it it's now the BAC-CBA rule. I've had this problem many times before, and I think I never noticed because I was working around the answer rather than the theory. Please help me understand why I get it wrong pretty much every time. Isn't the rule what is says it is?

EDIT: Lol the LaTeX code worked! First time I used it!

Well, let me ask you this question. Suppose that

$\displaystyle A = x$ and $\displaystyle B = y$

What's the difference between $\displaystyle AB$ and $\displaystyle BA$?

Now let $\displaystyle A = x$ and $\displaystyle B = \frac{d}{dx}$

and ask the same question. What's the difference between $\displaystyle AB$ and $\displaystyle BA$?

Originally Posted by Danny
Well, let me ask you this question. Suppose that

$\displaystyle A = x$ and $\displaystyle B = y$

What's the difference between $\displaystyle AB$ and $\displaystyle BA$?

Now let $\displaystyle A = x$ and $\displaystyle B = \frac{d}{dx}$

and ask the same question. What's the difference between $\displaystyle AB$ and $\displaystyle BA$?
$\displaystyle A = x$ and $\displaystyle B = y$
$\displaystyle AB = xy$ & $\displaystyle BA = yx$
So there is no difference really other than order of apperance

$\displaystyle A = x$ and $\displaystyle B = \frac{d}{dx}$
$\displaystyle AB = x \frac{d}{dx} = x \frac{d}{dx} = 1$ & $\displaystyle BA = \frac{d}{dx} x = 1$

Indeed you are correct, as $\displaystyle A \cdot B = B \cdot A$, so both the books answer and mine is correct? Is the answer simply cosmetic?

Actually no.

$\displaystyle AB = x \dfrac{d}{dx}$ - this is an operator
$\displaystyle BA = \frac{d}{dx} x = 1$ - this is a scalar.

The point is $\displaystyle AB \ne BA$ (these are not the same).

You said this in your original post. The difference is $\displaystyle \bf{ huge}!$