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Homework Help: Prove nabla(a dot b)

  1. Jul 2, 2010 #1
    please prove :
    nabla(a dot b)=(b dot nabla)a+(a dot nabla)b+b cross nabla cross a+a cross nabla cross b;
    (a and b are all vectors)
    when I was proving it,I found it impossible to go from right side to left side. I don't know whether '(b dot nabla)a = b (nabla a)' is right or not , when proving ,I found only when it is right can I prove the equation.
    Can you help me to prove '(b dot nabla)a = b (nabla a)'? Thank you very much!
    (I am a freshman)
     
  2. jcsd
  3. Jul 2, 2010 #2

    Dick

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    (b dot nabla)a is a vector. b (nabla a) isn't really a vector, is it? b is a vector. (nabla a) is some sort of two index object. That makes b (nabla a) a three index object. Put indexes on stuff and work it out concretely.
     
  4. Jul 2, 2010 #3
    sorry,what I want to prove is '(b dot nabla) a=b (nabla dot a)'.It is clearly that they are not in a same direction , but when I was using Einstein summation convention , they have to be the same,which confused me a lot .I would show you my procedure of proving.Please wait.
     
    Last edited: Jul 3, 2010
  5. Jul 2, 2010 #4

    Dick

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    Take a=(x,0,0) and b=(0,y,0). The left side is y*d/dy(a)=(0,0,0). The right side is b*1=(0,y,0). They aren't the same.
     
  6. Jul 2, 2010 #5
    left=[tex]\nabla(\vec{a} \cdot \vec{b})[/tex]
    =[tex]\partial_{i}\vec{e}_{i}(a_{j}b_{j})[/tex]
    =[tex]\vec{e}_{i}(b_{j}\partial_{i}a_{j}+a_{j}\partial_{i}b_{j})[/tex]
    =[tex]\vec{e}_{i}b_{j}\partial_{i}a_{j}+\vec{e}_{i}a_{j}\partial_{i}b_{j}[/tex]
    right=[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{b}\times\nabla\times\vec{a}+\vec{a}\times\nabla\times\vec{b}[/tex]
    =[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+(\vec{e}_{i}\epsilon_{ijk}b_{j}\partial_{k})\times\vec{a}+\vec{a}\times\nabla\times\vec{b}[/tex]
    =[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{a}\times\nabla\times\vec{b}+(\vec{e}_{i}\epsilon_{ijk}b_{j}\partial_{k}) a_{m}\vec{e}_{m}[/tex]
    =[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{a}\times\nabla\times\vec{b}+\epsilon_{ijk}\vec{b}_{j}(\partial_{k}a}_{m})\vec{e}_{n}\epsilon_{nim}[/tex]
    =[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{a}\times\nabla\times\vec{b}+b_{j}(\partial_{k}a_{m})\vec{e}_{n}(\delta_{jm}\delta_{kn}-\delta_{jn}\delta_{km})[/tex]
    =[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{a}\times\nabla\times\vec{b}+b_{m}(\partial_{k}a_{m})\vec{e}_{k}-b_{n}(\partial_{k}a_{k})\vec{e}_{n}[/tex]
    =[tex](\vec{b}\cdot\nabla)\vec{a}+(\vec{a}\cdot\nabla)\vec{b}+\vec{a}\times\nabla\times\vec{b}+(b_{m}(\partial_{k}a_{m})\vec{e}_{k}-b_{n}(\partial_{k}a_{k})\vec{e}_{n})[/tex]

    then, if [tex](\vec{b}\cdot\nabla)\vec{a}+(b_{m}(\partial_{k}a_{m})\vec{e}_{k}-b_{n}(\partial_{k}a_{k})\vec{e}_{n})=b_{m}(\partial_{k}a_{m})\vec{e}_{k}[/tex]

    we can easily prove the equation,
    however only when
    [tex](\vec{b}\cdot\nabla)\vec{a}[/tex]
    =[tex](b_{i}\vec{e}_{i}\partial_{j}\vec{e}_{j})a_{k}\vec{e}_{k}[/tex]
    =[tex]b_{i}\partial_{j}a_{k}\vec{e}_{i}\vec{e}_{j}\vec{e}_{k}[/tex]
    =[tex]b_{i}(\partial_{j}a_{j})e_{j}[/tex]
    can satisfy the requirments,but this equation is obviously wrong.

    so,I am confused now.
     
    Last edited: Jul 3, 2010
  7. Jul 3, 2010 #6
    well,the editing drives me crazy,the 6th and 7th line are not printed properly,but it seems not very vital,the main process starts from the 8th line.

    thank you everyone!!!
    I was puzzelled by this problem for almost 2 days.
     
  8. Jul 3, 2010 #7

    Dick

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    It's really difficult to comment on sort of stuff just because it's really hard to read. But it doesn't seem like you are really done. You are doing the right kind of things by changing the epsilon products into deltas. Change everything into index form and cancel equal things. If you do it right, it will work. There's no magic trick.
     
    Last edited: Jul 3, 2010
  9. Jul 3, 2010 #8
    ok ,I will try again
     
  10. Jul 4, 2010 #9
    I made it!
    Another form of this equation is:
    [tex]\nabla ( \vec{A} \cdot \vec{B} ) = (\vec{B} \cdot \nabla ) \vec{A} + ( \vec{A} \cdot \nabla) \vec{B} + \vec{A} \times ( \nabla \times \vec{B}) + \vec{B} \times ( \nabla \times \vec{A})[/tex]

    So, if I prove this form of it ,the proving process will be very easy.
    In the right side , the 1st and 4th will make [tex]\vec{e}_{i} B_{j} (\partial_{i} A_{j})[/tex],
    and the 2nd and 3rd will make [tex]\vec{e}_{i} A_{j} (\partial_{i} B_{j})[/tex]
    after being added together,they are equal to the left side, that's what I want.
    Hah hah.
    Thank you Dick!
    (What makes me crazy is that I found this form in a Japan website ! I wish our forum would set a special zone and give some equations' right form. Although the initial form above is right , but for a freshman , it is really hard to read and even prove it. )
    P.S.:
    Can someone tell me some appliances of this equation ? It's my year vacation in college and I really haven't got enough knowledge and experience to see the appliance of the equation .
     
    Last edited: Jul 4, 2010
  11. Jul 4, 2010 #10

    Dick

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    That's the same form you gave in post #5. It's also in the definitive vector identity guide which is the cover notes to Jackson's "Classical Electrodynamics". So it's likely you can find an application where A and B are vector potentials for electromagnetic fields. Good job on solving it! It really just takes patience.
     
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