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Homework Help: Vector identity

  1. Mar 19, 2006 #1
    Hi, I am having an immense amount of trouble trying to show that the following identity is true.

    Q. Let F be a C^2 vector field. Prove that

    [tex]\nabla \times \left( {\nabla \times \mathop F\limits^ \to } \right) = \nabla \left( {\nabla \bullet \mathop F\limits^ \to } \right) - \nabla ^2 \mathop F\limits^ \to [/tex]

    Firstly, I know that this can be done by using the definitions to find the ith component of the LHS. However, the use of this method requires something to be noted about the curl of a vector field and since I don't understand how one can just somehow notice that, I will not use this method.

    I attempted this question by using:

    \nabla \times \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \times \partial _i \mathop F\limits^ \to }

    \nabla \bullet \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \bullet \partial _i \mathop F\limits^ \to }

    \nabla ^2 \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\partial _i ^2 \mathop F\limits^ \to }

    where e_i is the usual basis vector, the d_i is the ith first order partial derivative.

    Using the above I proceeded as follows.

    \nabla \times \left( {\nabla \times \mathop F\limits^ \to } \right) = \nabla \times \left( {\sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \times \partial _i \mathop F\limits^ \to } } \right)

    = \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \times \left( {\mathop {e_j }\limits^ \to \times \partial _j \partial _i \mathop F\limits^ \to } \right)} \right)} }

    = \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \partial _j \partial _i \mathop F\limits^ \to } \right)} } \mathop {e_i }\limits^ \to - \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \mathop {e_i }\limits^ \to } \right)} } \partial _j \partial _i \mathop F\limits^ \to

    Where I have used the vector triple product.

    = \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \partial _j \partial _i \mathop F\limits^ \to } \right)} } \mathop {e_i }\limits^ \to - \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\partial _j ^2 \mathop F\limits^ \to } }
    [/tex] since the second sum kind of disappears when i and j are not equal and when i and j are equal the dot product is equal to one.

    = \sum\limits_{j = 1}^3 {\sum\limits_{i = 1}^3 {\left( {\mathop {e_j }\limits^ \to \bullet \partial _j \partial _i \mathop F\limits^ \to } \right)} } \mathop {e_i }\limits^ \to - \nabla ^2 \mathop F\limits^ \to

    At this point I'm having trouble trying to get a div in there somewhere. Is there anything I can do with the stuff inside the remaining double sum? Any help would be good thanks.
    Last edited: Mar 19, 2006
  2. jcsd
  3. Mar 19, 2006 #2
    notice that the LHS equals to:

    [ F'yz - F'zy , F'xz - F'zx , F'xy - F'yx ]
    in differential math, F'ab = F'ba (where a,b = x,y,z)
    so you get [0 , 0 , 0] on the LHS.
    hope this helps...
  4. Mar 19, 2006 #3


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    Benny, your notation is not familiar to me. (What is the first term in your summation for the curl?) What would your notation for the gradient of scalar field be? Are you familiar with the alternating epsilon notation?

    If the LHS were identically zero, then (for example) the usual derivation to obtain the wave equation from the Maxwell Equations would fail.

    Note that F'ab = F'ba (which appears to be the equality of mixed partials) is true when the object whose mixed-partial-derivatives are being taken is a scalar field.
  5. Mar 20, 2006 #4
    \nabla \times \mathop F\limits^ \to = \sum\limits_{i = 1}^3 {\mathop {e_i }\limits^ \to \times } \partial _i \mathop F\limits^ \to

    So if i = 1 then e_i = e_1 = (1,0,0) and [itex]\partial _i \mathop F\limits^ \to = \partial _1 \mathop F\limits^ \to = \frac{{\partial \mathop F\limits^ \to }}{{\partial x_i }}[/itex].

    The gradient of a scalar field would be [itex]\nabla f = \sum\limits_{i = 1}^n {\mathop {e_i }\limits^ \to } \partial _i f[/itex].

    In any case I think I've got it now but it involved moving the squiggly d's around a bit.:bugeye:
  6. Oct 25, 2007 #5
    Allowing summation convention,
    [tex]\nabla \times \left( {\nabla \times \mathop F\limits^ \to } \right) [/tex]
    [tex]= \frac{\partial}{\partial x^{i}} \mathop {e_i }\limits^ \to \times \left( {\frac{\partial}{\partial x^{j}} \mathop {e_j }\limits^ \to \times \mathop F\limits^ \to } \right) [/tex]
    [tex]= \frac{\partial}{\partial x^{i}} \frac{\partial}{\partial x^{j}} \left( { \mathop {e_i }\limits^ \to \times \left( { \mathop {e_j }\limits^ \to \times \mathop F\limits^ \to } \right) } \right) [/tex]
    [tex]= \frac{\partial}{\partial x^{i}} \frac{\partial}{\partial x^{j}} \left( { \mathop {e_j }\limits^ \to \left( { \mathop F\limits^ \to \bullet \mathop {e_i }\limits^ \to } \right) - \mathop F\limits^ \to \left( { \mathop {e_i }\limits^ \to \bullet \mathop {e_j }\limits^ \to } \right) } \right) [/tex]
    [tex]= \mathop {e_j }\limits^ \to \frac{\partial^{2} F_i}{\partial x^{i} \partial x^{j}} - \frac{\partial^{2} \mathop F\limits^ \to}{\partial x^{i} \partial x^{j}} \delta_{ij} [/tex]
    [tex]= \mathop {e_j }\limits^ \to \frac{\partial}{\partial x^{j}} \left( { \frac{\partial F_i}{\partial x^{i}} } \right) - \frac{\partial^{2} \mathop F\limits^ \to}{\partial x^{i} \partial x^{i}} [/tex]
    [tex]= \nabla \left( { \nabla \bullet \mathop F\limits^ \to } \right) - \nabla \bullet \left( { \nabla \otimes \mathop F\limits^ \to } \right) [/tex]
    [tex] \nabla \bullet \left( { \nabla \otimes \mathop F\limits^ \to } \right) = \nabla^{2} \mathop F\limits^ \to [/tex]
    Last edited: Oct 25, 2007
  7. Oct 25, 2007 #6
    I can't believe this thread was revived after well over a year. It brings back memories though. This was one of the problems I was struggling on while working through my problem booklet.

    Since then I've learned to use the summtation convention and I can get the answer out in a few lines without going to double sums! Although I think the notation I use is slightly different to yours.
    Last edited: Oct 25, 2007
  8. Oct 25, 2007 #7
    I happened to find this thread while searching for 'vector identity'.
    I would appreciate if you show the slight difference for my study in your spare time.
    Thank you in advance.
  9. Oct 26, 2007 #8
    There isn't much difference, it's just that I've been using cartesian tensor methods and an identity involving the epsilon and delta tensors to do problems with with cross products. I won't have much free time in the next few weeks so if you're interested in different notations my suggestion would be to just look them up in books.
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