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Homework Help: Archived Find the vortices of a square after a transformation given by a tensor

  1. Nov 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Given a square and the respective distension tensor, ε, find the position on his vortices after the transformation.

    ε = 0.1.....0.25

    2. Relevant equations

    3. The attempt at a solution

    I got kind of lost in this question. I started thinking that maybe a vortic at the coordinates (a,b) would later be at the position (a',b') given by:

    (a',b') = (a,b)ε

    This got me some weird results tho, which led me to believe it was wrong.

    I later tried to solve it using each component of the tensor alone. I know that the components of the diagonal give me the elongation, therefore I used them to find the positions of the vortices after the elongation. The problem came when I had to deal with the distortion. How can I find the position of the vortices there? I managed to find the position of the vortices that were at either the x axis or y axis, using trigonometry

    sin(εxy)=b'/L , where L is the length of the square after the elongation.

    My problem now rests with finding the position of the vortix at (L,L).
    Using trigonometry I found that the size of the diagonal,D, after the distortion was:
    D = Lcos(π/4-εxy)
    Since the vortix will still have an y coordinate equal to the x coordinate after the distortion I can say that:

    2A^2=D^2 , where A will be the position of the vortix after the distortion.

    I think my way of solving the problem is correct, however I can't help but think there's a better way... If someone could throw me some light I'd appreciate.
  2. jcsd
  3. Mar 15, 2016 #2


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    Science Advisor

    I presume you mean "vertices". "Vortices" are completely different things!
    Is it (a, b)ε or ε(a, b)? What convention are you using?

    What were the vertices of your square to begin with? Is your square centered at (0, 0)? Are you taking into account rotation and dilation?

    You say that one vertex is at (L, L) so are you assuming a square with vertices at (0, 0), (L, 0), (0, L), and (L, L)? If so applying [itex]\epsilon[/itex] to each vertex gives
    [tex]\begin{bmatrix}0.1 & 0.25 \\ 0.25 & 0.1\end{bmatrix} \begin{bmatrix}0 \\ 0 \end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}[/tex]
    [tex]\begin{bmatrix}0.1 & 0.25 \\ 0.25 & 0.1\end{bmatrix} \begin{bmatrix}L \\ 0 \end{bmatrix}= \begin{bmatrix}.25L \\ .1L \end{bmatrix}[/tex]
    [tex]\begin{bmatrix}0.1 & 0.25 \\ 0.25 & 0.1\end{bmatrix} \begin{bmatrix}0 \\ L \end{bmatrix}= \begin{bmatrix} .1L\\ .25L \end{bmatrix}[/tex]
    [tex]\begin{bmatrix}0.1 & 0.25 \\ 0.25 & 0.1\end{bmatrix} \begin{bmatrix}L \\ L \end{bmatrix}= \begin{bmatrix}.35L \\ .35L \end{bmatrix}[/tex]
    Last edited by a moderator: Mar 17, 2016
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