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I How to prove the curl curl of a vector?

  1. Mar 25, 2016 #1
    I've got ∇×(∇×R)=∇(∇.R)-∇2R [call it eq.1]

    However I have the identity ∇×(A×B)=A(∇.B)-B(∇⋅A)+ (B⋅∇)A-(A⋅∇)B [call it eq.2]

    Substituting in A=∇ and R=B into eq.2 we get ∇×(∇×R)=∇(∇.R)-R(∇⋅∇)+ (R⋅∇)∇-(∇⋅∇)R

    which i work out to be ∇×(∇×R)=∇(∇.R)-R(∇⋅∇)+ (R⋅∇)∇-∇2R

    Basically I don't understand what happens to the two terms -R(∇⋅∇)+ (R⋅∇)∇ from eq.2 when we get to eq.1, why do they disappear?
     
  2. jcsd
  3. Mar 25, 2016 #2

    BvU

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    Can't be done:$$ \vec A \times \vec B = - \vec B \times \vec A $$ but $$ \vec \nabla \times \vec A \ne - \vec A \times \vec \nabla $$the left hand side is a vector, the righthand side an operation
     
  4. Mar 25, 2016 #3

    DrDu

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    The disappearing terms are differential operators. As long as they don't act on anything, they will be zero.
     
  5. Mar 25, 2016 #4

    Why is this though?
     
  6. Mar 28, 2016 #5

    BvU

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    basically then you might put a factor 1 behind the expression and ##\nabla 1=0##, but as I indicated in #2, the equals sign doesn't hold, just like ##{\delta f\over \delta x }\ne f{\delta 1\over x} = 0##
     
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