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Double curl identity

  1. May 9, 2012 #1
    This isn't a homework problem, but it won't let me post on the other page.

    A well known vector identity is that rot(rot(E)) = grad(div(E)) - div(grad(E)).
    I've actually used this before without encountering any problems, so I don't know if I'm just having a brain fart or something, but shouldn't grad(div(E)) be equal to a vector and div(grad(E)) be equal to a scalar? How can you add or subtract them? It doesn't make any sense.
     
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  3. May 9, 2012 #2

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    Hi ShamelessGit! :smile:

    In that identity E is a vector and grad(E) is a component wise gradient yielding a matrix.
    The divergence is taken of the gradient of each of the components of E.
     
  4. May 9, 2012 #3
    Isn't it ∇ × ∇ × E = ∇ (∇ · E) - ∇ ^2 E, where ∇ ^2 is the vector Laplacian (as in, ∇ ^2 E instead of ∇ · (∇ E) or the scalar Laplacian)? That way they are both vectors. At least that's what I've been told.
     
    Last edited: May 9, 2012
  5. May 9, 2012 #4

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    The Laplacian is defined as the divergence of the gradient.
    It's just a shorthand notation.
     
  6. May 9, 2012 #5
    (I'm sure you're correct, don't get me wrong. I just want to understand this myself. I don't actually know that much about the maths behind this, I've only had to use nablas in physics and haven't even encountered partial derivatives in actual mathematics classes yet.)

    Yeah, I know that using ∇ is just a shorter notation, but isn't ∇ ^2 A for a vector field A defined as the vector Laplacian

    [itex]\nabla^2 \overline{A} = \frac{\partial^2 A_{x}}{\partial^2 x}\hat{i} + \frac{\partial^2 A_{y}}{\partial^2 y}\hat{j} + \frac{\partial^2 A_{z}}{\partial^2 z}\hat{k} [/itex]

    and ∇ ^2 A for a scalar field A separately defined as the scalar Laplacian

    [itex]\nabla^2 {A} = \nabla \cdot (\nabla A) = \frac{\partial^2 A_{x}}{\partial^2 x} + \frac{\partial^2 A_{y}}{\partial^2 y} + \frac{\partial^2 A_{z}}{\partial^2 z} [/itex]

    Or have I just understood something wrong?
     
    Last edited: May 9, 2012
  7. May 9, 2012 #6

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    Yes, I'm afraid you've misunderstood.

    [tex]\Delta \mathbf{A} = \nabla^2 \mathbf{A} = (\frac{\partial^2 A_{x}}{\partial^2 x} + \frac{\partial^2 A_{x}}{\partial^2 y} + \frac{\partial^2 A_{x}}{\partial^2 z})\hat{i} + (\frac{\partial^2 A_{y}}{\partial^2 x} + \frac{\partial^2 A_{y}}{\partial^2 y} + \frac{\partial^2 A_{y}}{\partial^2 z})\hat{j} + (\frac{\partial^2 A_{z}}{\partial^2 x} + \frac{\partial^2 A_{z}}{\partial^2 y} + \frac{\partial^2 A_{z}}{\partial^2 z})\hat{k} [/tex]

    The vector laplacian is the same as the scalar laplacian, but it is applied component-wise on the vector.
     
    Last edited: May 9, 2012
  8. May 9, 2012 #7
    Ok, thanks a lot for the clarification! :smile:

    IIRC my lecturer actually used the word 'vector Laplacian' (not in English, but still), when deriving the wave equation for electromagnetic waves. I might just remember wrong, though, or it might've been a careless mistake by him. Now I know better, so thanks!
     
  9. May 9, 2012 #8

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  10. May 9, 2012 #9
    Yeah, that's what I figured from your previous post.

    I thought he did, in the very same "curl of the curl"-sense, when taking the curl of Maxwell's third and fourth equations. But I tried to search our lecture notes for the word and couldn't find it, so now I'm starting to think I just remember wrong or something. Nevertheless, sorry for the trouble I caused (apologies to OP as well!) & thanks for explaining this to me.
     
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