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    I'm asked to prove, using suffix notation, the following identity:

    \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B})

    I know that [(\mathbf{A} \times \mathbf{B})]_i = \epsilon_{ijk} A_j B_k,  (\nabla \times \mathbf{A})_i = \epsilon_{ijk} \dfrac{\partial A_k}{\partial x_j}, and so on... I also know that \nabla \cdot \mathbf{F}= \dfrac{\partial F_i}{\partial x_i} where F is some vector field and that \mathbf{B} \cdot \mathbf{F} = B_i F_i and so on.

    I'm struggling to link the definitions above, however. For example, for \nabla \cdot (\mathbf{A} \times \mathbf{B}) I'm struggling to show this in suffix notation. I tried writing it as \dfrac{\partial}{\partial x_l} (\epsilon_{ijk} A_j B_k)_l . Is it wrong to sum over a different index l or can I sum it over, say, i as well?

    I also tried writing out the other terms in suffix notation, and got:

    \mathbf{B} \cdot (\nabla \times \mathbf{A})_i = B_l (\epsilon_{ijk} \dfrac{\partial B_k}{\partial x_j})_l

    again, is it wrong to sum over l and should I sum over i/j/k instead?

    Thanks for any clarification
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    (Original post by trm90)
    \mathbf{B} \cdot (\nabla \times \mathbf{A})_i = B_l (\epsilon_{ijk} \dfrac{\partial B_k}{\partial x_j})_l
    You can't write that. The quantity inside the bracket is already a scalar, so you can't have a further index on it. It's \mathbf{B} \cdot (\nabla \times \mathbf{A}) = B_i \epsilon_{ijk} \partial_j A_k.
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    (Original post by Zhen Lin)
    You can't write that. The quantity inside the bracket is already a scalar, so you can't have a further index on it. It's \mathbf{B} \cdot (\nabla \times \mathbf{A}) = B_i \epsilon_{ijk} \partial_j A_k.
    Silly of me to forget that (especially how the whole point of suffix notation is to make vectors look cleaner :P), thanks very much!
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    Sorry to bump this thread, just one more question

    I have to use suffix notation again to show:

    \nabla \times \nabla \times \mathbf{A} = -(\nabla \cdot \nabla) \mathbf{A} + \nabla (\nabla \cdot \mathbf{A})

    I've already been told that:

    [(\nabla \cdot \nabla) \mathbf{A}]_i = (\dfrac{\partial^2 A_i}{\partial x_{1}^2} + \dfrac{\partial^2 A_i}{\partial x_{2}^2} + \dfrac{\partial^2 A_i}{\partial x_{3}^2})

    I next evaluated the last term in the equation:

    \nabla \cdot \mathbf{A} = \dfrac{\partial A_i}{\partial x_i}

    thus \nabla (\nabla \cdot \mathbf{A}) = \dfrac{\partial}{\partial x_i} (\dfrac{\partial A_i}{\partial x_i}) = \dfrac{\partial^2 A_i}{\partial x_{i}^2}

    Combining the two terms together (taking into account the negative sign in nabla^2 A) I should get

    -(\nabla \cdot \nabla) \mathbf{A} + \nabla (\nabla \cdot \mathbf{A}) = -(\dfrac{\partial^2 A_2}{\partial x_{1}^2} +\dfrac{\partial^2 A_3}{\partial x_{1}^2}) - (\dfrac{\partial^2 A_1}{\partial x_{2}^2} + \dfrac{\partial^2 A_3}{\partial x_{2}^2}) - (\dfrac{\partial^2 A_2}{\partial x_{3}^2} + \dfrac{\partial^2 A_1}{\partial x_{3}^2})

    Next I figured I'd go about evaluating the left hand side in hopes that its final expression will lead to a summation similar to above.

    \nabla \times \mathbf{A} = \epsilon_{ijk} \dfrac{\partial A_k}{\partial x_j}

    I think I might have got a little lost here though...

    \nabla \times (\nabla \times \mathbf{A}) = \epsilon_{ilm} \dfrac{\partial}{\partial x_l} (\epsilon_{ijk} \dfrac{\partial A_k}{\partial x_j})_m = \epsilon_{ilm} \epsilon_{ijk} \dfrac{\partial}{\partial x_l} (\dfrac{\partial A_k}{\partial x_j})_m = (\delta_{jl} \delta_{km} - \delta_{lm} \delta_{kl}) \dfrac{\partial}{\partial x_l} (\dfrac{\partial A_k}{\partial x_j})_m

    I can kind of visualise the answer from here as if j = k = l = m then the bracket containing the delta comes to 1-1 and in my RHS there are no expressions which involve dA_i / dx_i in it. I am, however, struggling to do the last line or two which proves that my expression above works? I can't really seem to understand what it means to have an 'm' suffix next to my differentiation term... I've also toyed around with different combinations of j, k, l, m and can't seem to get much.

    any help much appreciated.
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    (Original post by trm90)
    \nabla \times (\nabla \times \mathbf{A}) = \epsilon_{ilm} \dfrac{\partial}{\partial x_l} (\epsilon_{ijk} \dfrac{\partial A_k}{\partial x_j})_m = \epsilon_{ilm} \epsilon_{ijk} \dfrac{\partial}{\partial x_l} (\dfrac{\partial A_k}{\partial x_j})_m = (\delta_{jl} \delta_{km} - \delta_{lm} \delta_{kl}) \dfrac{\partial}{\partial x_l} (\dfrac{\partial A_k}{\partial x_j})_m
    It's the same mistake again - you can't have a subscript on a scalar! It's \left[\nabla \times (\nabla \times \mathbf{A})\right]_i = \epsilon_{ijk} \partial_j \epsilon_{klm} \partial_l A_m.
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    (Original post by trm90)
    thus \nabla (\nabla \cdot \mathbf{A}) = \dfrac{\partial}{\partial x_i} (\dfrac{\partial A_i}{\partial x_i}) = \dfrac{\partial^2 A_i}{\partial x_{i}^2}
    \nabla \times \mathbf{A} = \epsilon_{ijk} \dfrac{\partial A_k}{\partial x_j}
    Be careful when you write something like this - you mean to say the i-th component of the vector on the LHS is equal to the RHS.
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    (Original post by Zhen Lin)
    It's the same mistake again - you can't have a subscript on a scalar! It's \left[\nabla \times (\nabla \times \mathbf{A})\right]_i = \epsilon_{ijk} \partial_j \epsilon_{klm} \partial_l A_m
    Hmm, I'm definitely not understanding it now then; where in my expression was I summing a scalar?

    Also, I'm looking at the new expression and it'sstarting to make sense now So that means the product of the epsilon terms will be

    \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl} yes?
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    (Original post by trm90)
    Hmm, I'm definitely not understanding it now then; where in my expression was I summing a scalar?
    (\epsilon_{ijk} \dfrac{\partial A_k}{\partial x_j})_m — the expression inside the brackets is a scalar. (To be precise, it's a family of scalars indexed by i.) It doesn't have any components to be indexed by m.

    On the other hand, \epsilon_{ijk} \mathbf{e}_j \times \mathbf{e}_k is a family of vectors indexed by i, so you can have a further subscript m to indicate its components: in particular, \left[ \epsilon_{ijk} \mathbf{e}_j \times \mathbf{e}_k \right]_m = \delta_{im}. However, mixing vectors and suffix notation like this is considered bad form and should be avoided.
 
 
 
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