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Metric orthonormal basis

  1. Dec 19, 2013 #1
    1. The problem statement, all variables and given/known data

    For the orthonormal coordinate system (X,Y) the metric is

    \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

    Calculate G' in 2 ways.

    1) G'= M[itex]^{T}[/itex]*G*M
    2) g[itex]\acute{}[/itex][itex]_{ij}[/itex] = [itex]\overline{a}\acute{}_{i}[/itex] . [itex]\overline{a}\acute{}_{j}[/itex]

    2. Relevant equations

    \begin{pmatrix} \overline{a}\acute{}_{1} \\ \overline{a}\acute{}_{2} \end{pmatrix}
    = \begin{pmatrix} -cos(\phi).\overline{a}\acute{}_{1} -\overline{a}\acute{}_{2} \\

    M= \begin{pmatrix} -cos(\phi) & 0 \\ -1 & cos(\phi) \end{pmatrix}
    M[itex]^{T}[/itex] = \begin{pmatrix} -cos(\phi) & -1 \\ 0 & cos(\phi) \end{pmatrix}
    G=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
    => G' = \begin{pmatrix} cos²(\phi) +1 & -cos(\phi) \\ -cos(\phi) & cos²(\phi) \end{pmatrix}

    3. The attempt at a solution

    So I'm having problems with the 2nd method i.e. g[itex]\acute{}[/itex][itex]_{ij}[/itex] = [itex]\overline{a}\acute{}_{i}[/itex] . [itex]\overline{a}\acute{}_{j}[/itex]

    g[itex]\acute{}[/itex][itex]_{11}[/itex] = (-cos[itex](\phi)[/itex] . [itex]\overline{a}_{1}[/itex] -[itex]\overline{a}_{2}[/itex]) . (-cos[itex](\phi)[/itex] . [itex]\overline{a}_{1}[/itex] -[itex]\overline{a}_{2}[/itex]) = ?

    What are the values for [itex]\overline{a}_{1}[/itex] and [itex]\overline{a}_{2}[/itex] ?
    I think they're both 1 because they're both unit vectors of length 1 but I'm not sure.
    Also, this is the first time ever I have used LaTeX so sorry if it's a bit sloppy.
  2. jcsd
  3. Dec 19, 2013 #2
    Also, I can't fall asleep because of this. I REALLY want to know the answer :p
  4. Dec 19, 2013 #3
    If there's something confusing, just ask :)
  5. Dec 19, 2013 #4


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    Homework Helper

    I don't know what problem you are trying solve. I know what "orthogonal coordinates" means and I know what "orthonormal vector fields" are. I not sure I know what "orthonormal coordinates" are. I can guess but could you explain that? Might be best to start with the definitions before you plow into trying to prove things. That might make it a lot easier than you think. And what IS this G' you are trying to calculate anyway?
    Last edited: Dec 19, 2013
  6. Dec 20, 2013 #5

    I'm sorry, I meant orthogonal coordinate system :)
    Basically I'm trying to find G' which is the metric of the coordinate system (X', Y').
    There are 2 ways to find this :
    1) G' = M^T * G * M (M=inverse transformationformula // M^T = the transposed matrix of M // G = metric of the coordinate system (X,Y) )

    2) Via the multiplication of the primed basic vectors a1 and a2 (see first matrix) and that's where I'm stuck. I don't know if I'm supposed to fill in those basis vectors with the value 1 or just leave them like that BUT that would mean that the 2nd method has a different result from the 1st method which can't be possible. I think I have to fill it in.

    Sorry for all the confusion!
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