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Tensor notation

  1. Jun 15, 2009 #1
    I can't seem to wrap my mind around it. I understand the concept of it, but I can't figure out how to translate that concept into actually extracting the individual equations from tensor notation.

    For example,

    [tex]
    a^i \: b^j \: c^k \: \epsilon_{jqs} \: \epsilon_{krt} \: \tau_i^{qr} = 0_{3 \times 3}
    [/tex]

    note that [tex]a,b,c[/tex] are [tex]3 \times 1[/tex] and [tex]\tau[/tex] is [tex]3\times 3 \times 3[/tex].

    This represents 9 equations. I understand how to calculate the value of the http://mathworld.wolfram.com/PermutationSymbol.html" [Broken], but this is complicated by having both superscripts and subscripts, and I'm also not sure if the subscripts of [tex]\epsilon[/tex] count in Einstein summation. My biggest problem is that I don't understand the "method" that can be used to extract the actual equations out of this!

    If someone could show me how to extract just 1 of the equations that would help a lot
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 15, 2009 #2

    tiny-tim

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    Hi junglebeast! :smile:

    (btw, you needn't say [tex]a,b,c[/tex] are [tex]3 \times 1[/tex] and [tex]\tau[/tex] is [tex]3\times 3 \times 3[/tex] … it's obvious from the number of indices :wink:)

    Yes, it's 9 equations, for each of the 3 values of s and t.

    And yes, all subscripts and superscripts count, even in deltas and epsilons.

    Extracting one of the 9 equations simply involves fixing say s = 2, t = 3, and summing over all the rest … doesn't that .pdf (which I haven't looked at) give any examples?
     
  4. Jun 15, 2009 #3
    Hi tim!

    Let me start with something simpler...I'll try to show my process

    [tex]
    a^i b^j = R^{ij}
    [/tex]

    I understand how I can convert the above into linear algebra,

    [tex]
    \mathbf{a} \mathbf{b} ^\mathsf{T} = \mathbf{R}
    [/tex]

    ...it's just the outer product. I could write this out elementwise as

    [tex]
    \mathbf{R}(i,j) = a(i) b(j)
    [/tex]

    (where parenthesis are used to indicate the indices)

    Ok, so now I try to do this with 3 vectors. Written in tensor notation, it is

    [tex]
    a^i b^j c^k = R^{ijk}
    [/tex]

    If I'm not mistaken, this is a 3x3x3 cube which can be written out elementwise as

    [tex]
    \mathbf{R}(i,j,k) = a(i) b(j) c(k)
    [/tex]

    Now, [tex]\tau[/tex] is also a cube...and these somehow combine together to make a 3x3 matrix (ignoring the [tex]\epsilon[/tex] which just control the sign or cancellation).

    To extract one of the equations it makes more sense for me to think about holding q and r constant. So let's say I choose q = r = 1. Ignoring the $\epsilon$ factors, I will try to sum over the rest...

    [tex]
    \sum_i \sum_j \sum_k \sum_s \sum_t a(i) \: b(j) \: c(k) \: \tau(i,q,r) = 0
    [/tex]

    Well that seems to contradict what I was doing earlier which did not have summations
     
  5. Jun 15, 2009 #4

    tiny-tim

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    Hi junglebeast! :smile:
    No, you've lost me here. :confused:

    epsilon doesn't "just control the sign or cancellation" … it's an integral part of the summation, and all three of its indices have to be summed over.

    Yes, Rijk is a 3x3x3 cube, and so are Rijk and so on …

    in Cartesian coordinates, the "cubes" have the same entries except that some entries are multiplied by minus-one (though in other coordinate systems, it's more complicated).

    (and I haven't followed your final question)
     
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