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Tensors, matrices and vectors

  1. Mar 30, 2009 #1
    Show that a tensor T can be written as

    [tex]T_{ij}=\lambda \delta_{ij} + F_{ij} +\epsilon_{ijk} v_{k}[/tex]

    for the tensor
    [itex]\[ \left( \begin{array}{ccc}
    1 & 2 & 3 \\
    4 & 5 & 6 \\
    7 & 8 & 9 \end{array} \right)\] [/itex]

    find [itex] \lambda, F_{ij}, v_k[/itex]

    i can't get anywhere whatsoever with this question???
     
    Last edited by a moderator: May 25, 2014
  2. jcsd
  3. Mar 31, 2009 #2
    Re: Tensors

    If i and j are the rows/columns, then what is k? Is there something missing in the problem description?
     
  4. Mar 31, 2009 #3

    lanedance

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    Re: Tensors

    i'm not sure about the problem itself, as i'm not sure of the context or meaning of the symbols (what are vk's & F?), but k is a contracted sum, so i think its ok
     
  5. Mar 31, 2009 #4
    Re: Tensors

    hi. it's straight out a past exam q so should be ok.

    not sure if this is what you mean but i noted that because of the levi civita,

    if k is 3 then you can have either i=1,j=2 or j=1,i=2

    i.e. we get [itex]v_3-v_3[/itex] but wouldnt that just make this last term trivial as all the terms would just cancel out similarly for k=1 and k=2???
     
  6. Mar 31, 2009 #5

    Ben Niehoff

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    Re: Tensors

    I think the tensor F should be symmetric. Basically what the question is asking is to separate the tensor into three components:

    1. Its trace.

    2. A symmetric, traceless part.

    3. An antisymmetric part.
     
  7. Mar 31, 2009 #6
    Re: Tensors

    yes but [itex]\lambda \delta_{ij}[/itex] puts a [itex]\lambda[/itex] on each diagonal element so in order to get 1,5 and 9 on the diagonals, at least one of the other two terms must have non zero diagonal components. clearly the third (antisymmetric) one cannot. so its down to the [itex]F_{ij}[/itex] and it must be symmetric and traceless.

    incidentally - this is the only term i dont understand the origin of - why must it be symmetric and traceless?

    then how do i solve for F when all i know is that

    [itex]F_{11}=1-\lambda.F_{22}=5-\lambda,F_{33}=9-\lambda,F_{11}+F_{22}+F_{33}=0[/itex]
    this gives [itex]\lambda=5[/itex]
    so then [itex]F_{11}=-4.F_{22}=0,F_{33}=4[/itex]

    how do i find the rest of F using v???
     
  8. Mar 31, 2009 #7

    Ben Niehoff

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    Re: Tensors

    Try first writing

    [tex]\epsilon_{ijk}v_k = A_{ij}[/tex]

    where A is an antisymmetric tensor. Now relate the components of A, F, and T. Also use the symmetry of F and the antisymmetry of A. You should get enough equations to obtain the rest of the elements.
     
  9. Mar 31, 2009 #8

    Ben Niehoff

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    Re: Tensors

    The purpose of this exercise, as far as I can tell, is for you to separate a general tensor T into parts that transform under different representations of the rotation group.

    1. The trace is a scalar, and is invariant under rotation: a spin-0 object.

    2. The antisymmetric part has 3 independent components, and transforms like a vector under rotations: a spin-1 object.

    3. The symmetric, traceless part has 5 independent components, and transforms like a tensor under rotations: a spin-2 object.
     
  10. Mar 31, 2009 #9
    Re: Tensors

    hang on,

    first of all how did you know to write [itex]\epsilon_{ijk}v_k=A_{ij}[/itex]

    secondly i get equaitons like

    [itex]A_{12} + F_{12}=2, A_{21}+F_{21}=4 \Rightarrow F_{12} + F_{21} = 6[/itex]

    but none of the other equaitons refer to either the 12 or 21 components so im struggling to get an exact value here
     
  11. Mar 31, 2009 #10

    tiny-tim

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    Hi latentcorpse! :smile:

    Hint: what are Tij + Tji,

    and Tij - Tji ? :wink:
     
  12. Mar 31, 2009 #11
    Re: Tensors

    do you want me to apply that to F or A?

    for the symmetric F the add one will give twice the original and the subtract one will give 0. vice versa for the antisymmetric tensor A.

    how does that help us though?
     
  13. Mar 31, 2009 #12

    tiny-tim

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    No … to T :rolleyes:
     
  14. Mar 31, 2009 #13
    Re: Tensors

    [itex]T_{ij}+T_{ji}=2\lambda \delta_{ij} + 2 F_{ij}, T_{ij}-T_{ji}=2 \epsilon_{ijk} v_k[/itex]

    can you explain why [itex]\epsilon_{ijk} v_k = A_{ij}[/itex] i can see that it must be anti symmetric because of the levi civita but how does the k fit into it all?

    and why is F symmetric and traceless - how do we know that?
     
  15. Mar 31, 2009 #14

    tiny-tim

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    Yes, half the sum (this is for anything, not just tensors) is the symmetric part, and half the difference is the anti-symmetric part. :smile:
    k is a dummy index …

    just draw the matrix, and you'll see where the v coordinates fit in :wink:
    We don't … we choose F and lambda so that trace(F) = 0, by subtracting trace(Tij + Tji) from Tij + Tji :smile:
     
  16. Apr 1, 2009 #15
    Re: Tensors

    so the sums involving T_ij and T_ji give me the extra equations i was needing to find all the constants but im still having trouble seeing whats going on:

    i agree that k is a dummy index but i cant really see how it works here:
    if i pick k=1 that restricts i and j to 1 or 2
    so the matrix has a 1 in the (12) component and a -1 in the (21) component - is this along the right lines?

    and i still dont follow how/why we choose F and lambda so that trace(F)=0 or why we'd even want trace(F) to be 0 in the first place???
     
  17. Apr 1, 2009 #16

    lanedance

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    Re: Tensors

    so i think it means
    [itex]\epsilon_{ijk} v_{k} =
    \[ \left( \begin{array}{ccc}\epsilon_{11k} v_{k} & \epsilon_{12k} v_{k} & \epsilon_{13} v_{k} \\\epsilon_{21k} v_{k} & .. & .. \\.. & .. & .. \end{array} \right)\] [/itex]

    if i remember correctly about the levi civita
    [itex]\epsilon_{ijk} = 0 [/itex], if [itex] i=j [/itex] or [itex]j=k [/itex] or [itex]k=i [/itex]
    [itex]\epsilon_{ijk} = 1 [/itex], for even permutation of i,j,k
    [itex]\epsilon_{ijk} = -1 [/itex], for odd permutation of i,j,k

    which gives
    [itex]\epsilon_{ijk} v_{k} =
    \[ \left( \begin{array}{ccc} 0 & \epsilon_{123} v_{3} & \epsilon_{132} v_{2} \\\epsilon_{213} v_{3} & .. & .. \\.. & .. & .. \end{array} \right)\] [/itex]

    which gives
    [itex]\epsilon_{ijk} v_{k} =
    \[ \left( \begin{array}{ccc} 0 & +v_{3} & -v_{2} \\ -v_{3} & .. & .. \\.. & .. & .. \end{array} \right)\] [/itex]

    which is antisymmetric with zero trace
     
    Last edited: Apr 1, 2009
  18. Apr 1, 2009 #17

    tiny-tim

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    No, you're missing the point (and it's not 1, it's 3 :wink:) …

    the point of the εijk is to pick up bits of v and plonk them down into a matrix …

    if you pick k=3 that restricts i and j to 1 or 2
    so the matrix has v3 in the (12) component and -v3 in the (21) component …

    see lanedance's :smile: last matrix!
     
  19. Apr 1, 2009 #18
    Re: Tensors

    kl. can i ask a quick unrelated question while your at it:

    let [itex]V=span(1,1)^T[/itex]

    that is the set of all linear combinations, i.e. [itex]V=\{\lambda \vec{i} + \mu \vec{j} \}[/itex] if we're in [itex]\mathbb{R^2}[/itex]. so would it be correct to plot this in [itex]\mathbb{R^2}[/itex] as a series of infinite lines parallel to the line y=x. or is it just the line y=x?
     
  20. Apr 1, 2009 #19

    tiny-tim

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    uhh? :confused: [itex]\{\lambda \vec{i} + \mu \vec{j} \}[/itex] is R2
     
  21. Apr 1, 2009 #20
    Re: Tensors

    so what does the span of (1,1)^T look like then is it just the line y=x?
     
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