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Questions about Index Notation

  1. Nov 11, 2006 #1
    I am playing around with learning index notation for tensors, and I came across the following where C is a 0th order tensor:

    [tex]E_{ijk} \partial_j \partial_k C = 0[/tex]

    I believe this equates to [tex]\nabla \times \nabla C[/tex]. I don't understand why this comes out to 0. Any ideas?

    Also, I am trying to understand in index notation how to represent the grad of a vector. The reason I am confused is that it seems that, taking C as a 0th order tensor, V as a 1st order tensor and T as a 2nd order tensor:

    [tex]div T = \nabla \cdot T = \partial_iT_{ij}[/tex]
    [tex]div V = \nabla \cdot V = \partial_iV_i[/tex]
    And of course, div C does not make sense as it would be a -1st order tensor.

    But, since:
    [tex]grad C = \nabla C = \partial_iC[/tex]
    I don't follow how to represent [tex]grad V = \nabla V[/tex] or [tex]grad T = \nabla T[/tex] in index notation; from what I have it seems there would be no difference in notation between grad and div! Any help would be greatly appreciated. Thanks!
     
    Last edited: Nov 11, 2006
  2. jcsd
  3. Nov 13, 2006 #2
    The first thing to note is that your notation here is a bit non-standard. If you assume that [itex]E_{ijk}[/itex] is just some arbitrary (0,3) tensor then [itex]\sum_{j,k}E_{ijk}\partial_j\partial_kC\ne 0[/itex] in general. However, if you take [itex]E_{ijk}=\epsilon_{ijk}[/itex], the totally antisymmetric or permutation tensor, then [itex]\sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC=0[/itex] is trivially satisfied. To see why this is so, note that since partial derivatives commute we can write

    [tex]\begin{equation*}
    \begin{split}
    \sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC
    &= \sum_{j,k}\frac{1}{2}\epsilon_{ijk}(\partial_j\partial_kC + \partial_k\partial_jC) \\
    &= \sum_{j,k}\frac{1}{2}(\epsilon_{ijk}\partial_j\partial_kC + \epsilon_{ijk}\partial_k\partial_jC) \\
    &= \sum_{j,k}\frac{1}{2}(\epsilon_{ijk}\partial_j\partial_kC + \epsilon_{ikj}\partial_j\partial_kC) \\
    &= \sum_{j,k}\frac{1}{2}(\epsilon_{ijk} + \epsilon_{ikj})\partial_j\partial_kC.
    \end{split}
    \end{equation*}[/tex]

    However, since one has [itex]\epsilon_{ijk}=-\epsilon_{ikj}[/itex] by definition, one can then write

    [tex]\sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC = \sum_{j,k}\frac{1}{2}(\epsilon_{ijk} - \epsilon_{ijk})\partial_j\partial_kC = 0.
    [/tex]

    You are then correct to say that [itex]\nabla\times\nabla C=\sum_{j,k}\epsilon_{ijk}\partial_j\partial_kC=0[/itex].

    Correct.

    You may or may not find this post helpful.
     
  4. Nov 15, 2006 #3
    Not sure if this is the source of your confusion, but notice that the 0 is a component of a tensor with rank(0,1) since only the j and k indices are repeated. i.e.

    [tex]A_{i} = E_{ijk} \partial_j \partial_k C = 0[/tex]
     
  5. Nov 15, 2006 #4
    coalquay,

    You are correct that I intended [tex]\epsilon_{ijk}[/tex]. Sorry for the bad notation. But I greatly appreciate the help, I did not think of that approach. Thanks!
     
  6. Nov 15, 2006 #5

    mathwonk

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    sillyme, i hoped this was about the index of an elliptic operator, but its the same old same old.
     
  7. Nov 16, 2006 #6
    A question about index theorems would be nice, wouldn't it? I suspect, however, we'll be waiting a while before we see one...
     
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