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Simple(?) index notation

  1. Oct 22, 2015 #1
    Hello all, long time lurker, first time poster. I don't know if I am posting this in the proper section, but I would like to ask the following:
    In index notation the term [itex]σ_{ik}x_{j}n_{k}[/itex] is [itex]\bf{σx}\cdot\bf{n}[/itex] or [itex]\bf{xσ}\cdot\bf{n}[/itex], where ##σ## is a second order tensor and ##x,n## are vectors.

    On the same note, is ##\frac{\partialσ_{ik}}{\partial x_{k}}x_{j}## equivalent to ##\nabla\cdot(\bf{xσ})## or ##\nabla\cdot(\bf{σx})## ? For some reason there is an index notation rule that eludes me.

    Pardon me for the fundamendality or even stupidity of my questions!
     
    Last edited: Oct 22, 2015
  2. jcsd
  3. Oct 22, 2015 #2

    Fredrik

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    I'm not familiar with how people use that dot product notation when there's more than one index on a tensor component. It looks really ambiguous to me, but maybe there's some convention that removes the ambiguity. If your book explains the notation, maybe you can tell us how.

    Since ##\partial_i f_i=\nabla\cdot f##, I guess ##\frac{\partial\sigma_{ik}}{\partial x_k}x_j## would have to correspond to something like ##(\nabla\cdot\sigma)x## or ##x(\nabla\cdot\sigma)## in that notation.
     
  4. Oct 22, 2015 #3
    I totally agree, thats why I am confused! But what about my first question?
     
  5. Oct 22, 2015 #4

    Geofleur

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    I agree with what Fredrik said, except that I would write the same thing as ## (\nabla \cdot \mathbf{\sigma}) \otimes \mathbf{x}##. ## \sigma_{ik} x_j n_k ## also looks to me like a dyadic product, so I would write it as something like ## (\mathbf{\sigma} \cdot \mathbf{n}) \otimes \mathbf{x} ##. As a simpler example, if we had ## x_j y_i ##, that would be the ## ji ## component of the dyadic ## \mathbf{x}\otimes\mathbf{y} ##.
     
  6. Oct 23, 2015 #5
    Fredrik's final interpretation is exactly correct for cartesian coordinates. Otherwise, ##\partial σ_{ik}/\partial x_k## does not represent the components of ##\vec{∇}\centerdot \vec{σ}##.

    Chet
     
  7. Oct 23, 2015 #6

    Geofleur

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    Very true!
     
  8. Oct 23, 2015 #7

    dextercioby

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    The dyadic notation is tricky to learn to learn and use, but it's undeniably correct, because it's a statement of a relationship between tensors, not about components in a particular base.

    ## \vec{v} ## is a vector, ## \overleftrightarrow{\sigma} ## is a dyad, typically a 2nd rank tensor. We use ## \otimes## for the dyadic (tensor) product and ## . ## for the scalar (contracted tensor) product. And one puts an arrow -> over nabla, too. So:

    $$ \frac{\partial \sigma_{ik}}{\partial x_k} x_j \mapsto \left(\vec{\nabla} \bullet \overleftrightarrow{\sigma}\right) \otimes \vec{x} $$.
     
  9. Oct 23, 2015 #8
    In what way does this differ materially from the sum total of what the other responders said?
     
  10. Oct 23, 2015 #9

    dextercioby

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    I don't quite get your question. "The other responders" seem to agree to a posting which starts with "I'm not familiar with" and contains "it looks really ambiguous". I just thought to write something that leaves no room to debate/uncertainty.

    As a further note, for a divergence of a(n Euclidean) tensor, you usually contract by the first slot of the tensor. Older books I came against called that 'left divergence'. You can also contract by the 2nd (last to the right) slot and you'll have the 'right divergence'. Dyadic notation is really old-fashioned. Even engineering schools (should) teach tensors nowadays.
     
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