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Math Help - tensor and einstein summation notation

  1. #1
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    tensor and einstein summation notation

    Hello,

    I have some basic problems with tensors and Einstein summation notation. As I understand, if a suffix appears twice in a tensor then summation over repeated indices is assumed, i.e.
    S_{ii} = S_{11} + S_{22} + S_{33},
    where S is a second order tensor and i is the repeated suffix. Is S_{ii} indirectly a zeroth order tensor, i.e. a scalar?

    Einstein summation convention also states that a suffix cannot appear more than twice in an expression. I came across an expression for fourth order tensor T_{iijj}, where it is mentioned that T_{iijj} is a scalar. Why is T_{iijj} a zeroth order tensor, i.e. a scalar? I would not expect that since two different suffices are used.
    Last edited by Opalg; May 19th 2010 at 09:02 AM. Reason: (Sorry – edited in error)
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  2. #2
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    Quote Originally Posted by amrasa81 View Post
    Hello,

    I have some basic problems with tensors and Einstein summation notation. As I understand, if a suffix appears twice in a tensor then summation over repeated indices is assumed, i.e.
    S_{ii} = S_{11} + S_{22} + S_{33},
    where S is a second order tensor and i is the repeated suffix. Is S_{ii} indirectly a zeroth order tensor, i.e. a scalar?

    Einstein summation convention also states that a suffix cannot appear more than twice in an expression. I came across an expression for fourth order tensor T_{iijj}, where it is mentioned that T_{iijj} is a scalar. Why is T_{iijj} a zeroth order tensor, i.e. a scalar? I would not expect that since two different suffices are used.
    When you sum over a repeated suffix, it reduces the order of the tensor by 2. For the fourth order tensor T_{iijj}, when you sum over i the order goes down to 2, then when you sum over j it goes down to 0, so you are left with a scalar.

    For a simple example, a square matrix with entries A_{ij} is a second-order tensor. The sum A_{ii} is the trace of the matrix, which is a scalar.
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