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Vector differential identity proof

  1. Oct 23, 2011 #1
    Hi,
    I am a engineering student and I am currently upgrading my maths level on my own to follow physics courses. While reading a book, I came across a vector differential identity that I don't manage to prove using index notation.
    The identity is the following:
    [tex]
    \nabla(\vec{A}\cdot\vec{B}) =
    \vec{A} \times (\nabla \times \vec{B}) + (\vec{A} \cdot \nabla)\vec{B}
    + \vec{B} \times (\nabla \times \vec{A}) + (\vec{B} \cdot \nabla)\vec{A}
    [/tex]
    Could you please give me a hint on how to prove this ?
    Thank you for your time.
     
  2. jcsd
  3. Oct 24, 2011 #2

    lurflurf

    User Avatar
    Homework Helper

    Why do you want to use index notation? In any case what makes this formula a bit strange is that it can be cast into many forms and that is in some sense not the natural one.

    For example the identity

    [tex]\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}[/tex]

    shows we can express each of those four terms in terms of the others
    also we can chose to use or not use dyatic terms

    two links
    https://www.physicsforums.com/showthread.php?t=297027
    https://www.physicsforums.com/showthread.php?t=273630
     
  4. Oct 24, 2011 #3
    Thank you, I wanted to use index notation because it proved really usefull for developping vector identities. But if this identity's proof does not lend itself to this notation, I don't see the necessity to use it.

    In one of your links, you posted this:

    [tex]
    \nabla (\vec{A} \cdot \vec{B}) = \nabla_{\vec{A}}(\vec{A} \cdot \vec{B}) + \nabla_{\vec{B}}(\vec{A} \cdot \vec{B})
    [/tex]
    [tex]
    \nabla_{\vec{A}}(\vec{A} \cdot \vec{B}) = B \times (\nabla \times \vec{A}) + (\vec{B} \cdot \nabla)\vec{A}
    [/tex]
    [tex]
    \nabla_{\vec{B}}(\vec{A} \cdot \vec{B}) = A \times (\nabla \times \vec{B}) + (\vec{A} \cdot \nabla)\vec{B}
    [/tex]
    [tex]
    \nabla(\vec{A}\cdot\vec{B}) =
    \vec{A} \times (\nabla \times \vec{B}) + (\vec{A} \cdot \nabla)\vec{B} + \vec{B} \times (\nabla \times \vec{A}) + (\vec{B} \cdot \nabla)\vec{A}
    [/tex]

    But how do you develop the 2nd and 3rd lines from the partial gradient ?
     
  5. Oct 26, 2011 #4

    lurflurf

    User Avatar
    Homework Helper

    [tex]\mathbf{\nabla_b (a\cdot b)=(a\times\nabla)\times b+a(\nabla\cdot b)=a\times(\nabla\times b)+(a\cdot\nabla)b}[/tex]

    follows from the vector algebra identity

    [tex]\mathbf{c(a\cdot b)=(a\times c)\times b+a(c\cdot b)=a\times(c\times b)+(a\cdot c)b}[/tex]

    or in index form the identity (not guaranteed typo free)

    δilδjkmjlεimkijδklijmεmlkikδjl

    can be applied to

    δilδjkajbk,l

    Care must be taken when using algebra identities to deduce calculus identities, mainly because by normal convention differential operators are right acting so we must not change multiplication order. We would not like to make an error like

    [tex]\nabla\cdot b=b\cdot\nabla[/tex]

    It is also possible and convenient to change to bidirectional differential operators and back so that manipulations can be unrestricted.
     
  6. Oct 26, 2011 #5
    Thanks a lot ! That was really helpful !
     
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