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Del calculus

  1. Jan 8, 2006 #1
    Hi. I have this exam in vector calculus tomorrow, but i'm having trouble sorting the following formula out. Could someone help me on the track or show me why this is an equality. Feels meaningsless to merely memorize the formula.
    \nabla \times (\bar{u} \times \bar{v}) = (\bar{v} \cdot \nabla) \bar{u} - \bar{v} (\nabla \cdot \bar{u}) + \bar{u} (\nabla \cdot \bar{v}) - (\bar{u} \cdot \nabla) \bar{v}
  2. jcsd
  3. Jan 8, 2006 #2

    matt grime

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    expand each side in components (preferably use the einstein summation convention to make it easier) to verify if it is true.
  4. Jan 8, 2006 #3
    i was more thinking about some rules i can use when i'm sitting at my exam, can't remember the formula and need to derive it... i'm sure it's pretty easily proved but that's not really what i need.
  5. Jan 8, 2006 #4


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    The way I think of it is in terms of the bac-cab rule you use for triple cross products:

    [tex]\vec a \times (\vec b \times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)[/tex]

    The first and last terms of the identity you displayed fit the bac-cab rule. Obviously, the [itex]\nabla[/itex] has to operate on something so the dot products get placed before the vector corresponding the grad operator. You can think of the middle two terms as compensating for the fact that [itex]\nabla \vec a[/itex] and [itex]\nabla \vec b[/itex] are tensors and remove terms that don't belong in the bac-cab expansion.
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