1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Interpretation of a tensor

  1. Sep 20, 2015 #1
    1. The problem statement, all variables and given/known data

    [itex]M= \begin{pmatrix}
    2 & -1 & 0\\
    -1 & 2 & -1\\
    0 & -1 & 2\\

    Compute [itex]\frac{1}{6}\epsilon_{ijk}\epsilon_{lmn} M_{il} M_{jm} M_{kn}[/itex] .

    3. The attempt at a solution

    I computed the result which is 4, by realizing that there are 36 non-zero levi-civita containing components to sum. Within this group, for each fixed {i,l} there are 4 possible sets of {in},{km}. As the order of the M_{ab}'s inside the product doesn't matter they can be rearranged to form sums of 2 each at a time. The positive contribution to the sum arises from the (now) three occurences of the product of the diagonal elements [itex]M_{ii} M_{jj} M_{kk} = 2 \cdot 2 \cdot 2 = 8[/itex] (not meant to be read as sums). The negative contributions arise from products of specific non-diagonal components multipied by a specific diagonal component. Because of interchangeability inside the product (or alternatively, in this case, symmetry) all get bundled in pairs of two. Each non-zero contribution (i.e. non-{1,3} permutation) then equals [itex]2 \cdot (-1 \cdot -1 \cdot 2) = 4[/itex], and there are three of them. So the final result turns out to be

    [itex]2 (3 \cdot 8 - 3 \cdot 4) = 24[/itex], which divided by 6 leaves us with 4.

    However, it feels like I'm missing something here (I'm not using the symmetry in any kind of crucial way). Is this the way you would solve this problem? Would it be possible to "see" the result by simply looking at the expression (i.e. interpreting it before doing the math)? Or for example by expressing the epsilons in deltas?

    ( Edit: I now see it's the determinant. But still, it isn't super apparent from just doing the math mindlessly (which is my type of thing until I feel I grasp the basics.)
    Last edited: Sep 20, 2015
  2. jcsd
  3. Sep 25, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Sep 25, 2015 #3
    Apologies for not having Latex here. As you have discovered, your expression is just the determinant of the matrix. This is evident by expanding
    eps(i,j,k) eps(l,m,n) M(i,l) M(j,m) M(k,n) as six lines, one for each sequence of possible i,j,k that result in non zero epsilon components:

    = eps (1,2,3) eps(l,m,n) M(1,l) M(2,m) M(3,n)
    + eps (2,3,1) eps(l,m,n) M(2,l) M(3,m) M(1,n)
    + eps (3,1,2) eps(l,m,n) M(3,l) M(1,m) M(2,n)
    + three other lines, you get the idea.

    But each line is equal to the determinant which is defined as eps(l,m,n) M(1,l) M(2,m) M(3,n). To see this you have to reorder the l,m,n indices in each line except the first.

    Your fundamental question, why can't this be done by looking at the symmetries of the M matrix, is equivalent to asking: can the determinant of a symmetric matrix be quickly evaluated by looking at symmetries? Not as far as I know.
  5. Sep 25, 2015 #4
    Thanks, that's a very clear and nice explanation.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Interpretation of a tensor