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Math Help - Strange BAC-CAB inconsistencies, please help check my answer

  1. #1
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    Strange BAC-CAB inconsistencies, please help check my answer

    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:
    Simplify rot(a x grad(fi)

    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).

    The answer in my book:
    -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.
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

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

    \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.
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

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

    \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;

    \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.

    \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!
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

    Well, let me ask you this question. Suppose that

    A = x and B = y

    What's the difference between AB and BA?

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

    and ask the same question. What's the difference between AB and BA?
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

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

    A = x and B = y

    What's the difference between AB and BA?

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

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

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

    Indeed you are correct, as A \cdot B = B \cdot A , so both the books answer and mine is correct? Is the answer simply cosmetic?
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

    Actually no.

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

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

    You said this in your original post. The difference is \bf{ huge}!
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

    Yes I told you I start to grasp anything that makes sense with the answer rather than the acual theory. Now, please tell me why BAC-CAB does not apply for this case. Why even have a BAC-CAB rule if it does not apply for every case of Ax(BxC)? It makes no sence and it's not like math definitions at all to stray from case to case.
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

    Your BAC - CAB works as long as you have vectors with real (or complex) components. The problem arises when you allow the components to be operators. You lose commutativity.
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    Re: Strange BAC-CAB inconsistencies, please help check my answer

    Indeed, that must be the case. That is also why we use einsteins eidi method which I at first thought was stupid as we alredy had an arsenal of laws to prove, as I belived, anything. Thank you, I've got to try and find more on this eidi method so I really understand it all.
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