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I have some cumbersome with derivative problems

  1. Aug 17, 2013 #1
    I am confronting with how to realize the difference between two formulae belows
    [itex]\left( {\vec a \times \nabla } \right) \times \vec b[/itex], and [itex]\left( {\nabla \times \vec a} \right) \times \vec b[/itex], in here
    [itex]\nabla [/itex] is Del (also known as nabla)
    Thanks
     
  2. jcsd
  3. Aug 17, 2013 #2
    Why these results are so different in formulae
     
  4. Aug 17, 2013 #3

    lurflurf

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    Homework Helper

    Ah yes recall that
    $$(\vec{a}\cdot \nabla)\vec{b}+\vec{a} \times (\nabla \times \vec{b})=\vec{a}(\nabla \cdot \vec{b})+(\vec{a} \times \nabla) \times \vec{b}$$

    By convention the nabla operator acts to the right. For this reason we must be careful. In your formula
    $$(\vec{a} \times \nabla )\times \vec{b}$$
    Since nabla acts to the right it acts on b and not a
    The left acting nabla would act on a and not b
    $$(\vec{a} \times \overset{\leftarrow}{\nabla} )\times \vec{b}$$
    This is nonstandard notation, but we can write it as
    $$-(\nabla \times \vec{a} ) \times \vec{b}=\vec{b} \times (\nabla \times \vec{a} )$$
    which is your second formula (with a minus sign)
    The sum of the left and right forms is the bidirectional form
    $$(\vec{a} \times \overset{\leftrightarrow}{\nabla} )\times \vec{b}=(\vec{a} \times \nabla )\times \vec{b}+\vec{b} \times (\nabla \times \vec{a} )$$
    This is useful when you get to integrals because you will have
    $$\int_{\partial V} (\vec{a} \times n )\times \vec{b} \, \mathrm{d}S=\int_V ((\vec{a} \times \nabla )\times \vec{b}+\vec{b} \times (\nabla \times \vec{a} )) \, \mathrm{d}V$$
    Where we change a surface integral into a volume integral.

    The short version if any of that was confusing is that since nabla acts to the right it acts on a and not b in one formula and b and not a in the other, they are very different.
     
  5. Aug 17, 2013 #4
    I am new and unfamiliar with these operation, so what is the name of the first identity you wrote, then I could look up online for reference
    thank you
     
  6. Aug 17, 2013 #5

    lurflurf

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    Homework Helper

    I don't know if that identity has a name. I brought it up to illustrate that
    $$( {\vec a \times \nabla } ) \times \vec b \text{ and } ( {\nabla \times \vec a} ) \times \vec b$$
    have different vectors being operated on while
    $$( {\vec a \times \nabla } ) \times \vec b \text{ and } ( {\nabla \times \vec b} ) \times \vec a$$
    have the same vector being operated on
    These two forms arise in the gradient of dot product identity
    $$\nabla_\vec{b} (\vec{a} \cdot \vec{b})=(\vec{a}\cdot \nabla)\vec{b}+\vec{a} \times (\nabla \times \vec{b})=\vec{a}(\nabla \cdot \vec{b})+(\vec{a} \times \nabla) \times \vec{b}$$
    The nabla subscript b means that a is not operated on.

    In your original question the key point was that one expression had a being operated on while the other had b being operated on. This is similar to in single variable calculus if we have
    (D indicates the derivative)
    D(uv)=uDv+vDu
    uDv and vDu are not equal because
    in uDv
    v is being operated on while u is not
    and
    in vDu
    u is being operated on while v is not

    http://en.wikipedia.org/wiki/Vector_calculus_identities
    Does not include the identity I mention, but it includes others that may be of interest.
     
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