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Homework Help: Coordinate transformation of a tensor in 2 dimensions

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Given a symmetric tensor [itex]T_{\mu\nu}[/itex] on the flat Euclidean plane ([itex]g_{\mu\nu}=\delta_{\mu\nu}[/itex]), we want to change to complex coordinates [itex]z=x+iy, \,\overline{z}=x-iy[/itex].
    Show, that the components of the tensor in this basis are given by:
    [itex]T_{zz}=\frac{1}{4}(T_{00}-2iT_{10}-T_{11}),\, T_{\overline{z}\overline{z}}=\frac{1}{4}(T_{00}+2iT_{10}-T_{11}),\, T_{\overline{z}z}=T_{z\overline{z}}=\frac{1}{4}(T_{00}+T_{11})[/itex]

    2. Relevant equations
    [itex]T'_{\alpha\beta} = U_{\alpha}{}^{\mu}U_{\beta}{}^{\nu}T_{\mu\nu}[/itex]

    3. The attempt at a solution
    In priciple, this should be an easy linear algebra problem. I just don't get the right result.
    I use [itex]T'[/itex]for the tensor in the new coordinates. Then, there should be a transformation matrix [itex]U[/itex], s.t.
    [itex]T'_{\alpha\beta} = U_{\alpha}{}^{\mu}U_{\beta}{}^{\nu}T_{\mu\nu}[/itex]
    But what is [itex]U[/itex]? Neiter with the transformation matrix from the new to the old coordinates [itex]\begin{pmatrix}1 & 1 \\ i & -i\end{pmatrix}[/itex] nor with the transformation matrix from the old to the new coordinates [itex]\begin{pmatrix}\frac{1}{2} & \frac{-i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix}[/itex] it is working. What am I doing wrong?

  2. jcsd
  3. Jun 4, 2012 #2
    It's difficult to say without seeing the calculation. You do remember that there are two transformation matrices, right? (one for each index)
  4. Jun 5, 2012 #3
    Thanks, but I still get a wrong result.

    The matrix [itex]U = \begin{pmatrix}\frac{1}{2} & \frac{-i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix}[/itex] transforms from the old basis [itex]x,y[/itex] to the new basis [itex]z=x+iy, \overline{z}=x-iy[/itex].
    The original metrix is [itex]g_{\mu\nu}=\delta_{\mu\nu}[/itex]. The transformed metric is [itex]g'_{\mu\nu}=\begin{pmatrix} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{pmatrix}[/itex]
    Therefore, the tensor in the new basis is given by
    [itex]T'^{\alpha}{}_\beta = U^{\alpha}{}_\mu T^{\mu}{}_\nu (U^{-1})^\nu{}_\beta = \begin{pmatrix}\frac{1}{2}T^0{}_0+\frac{1}{2}T^1{}_1 & \frac{1}{2}T^0{}_0-\frac{i}{2}T^0{}_1- {\frac{i}{2}}T^1{}_0-T^1{}_1 \\ \frac{1}{2}T^0{}_0+ \frac{i}{2}T^0{}_1+ \frac{i}{2}T^1{}_0-T^1{}_1 & \frac{1}{2}T^0{}_0+\frac{1}{2}T^1{}_1 \end{pmatrix}[/itex]

    Now, I use the metric to lower the indices. Since the original metric ist Euclidean one can simply lower the indices of [itex]T[/itex]without further factors, but not for [itex]T'[/itex].
    [itex] T'_{00}=g'_{00}T'^0{}_0+g'_{01}T'^1{}_0 = \frac{1}{2}T'^1{}_0 = \frac{1}{2}(\frac{1}{2}T_{00}+ \frac{i}{2}T_{01}+ \frac{i}{2}T_{10}-T_{11}) = \frac{1}{4}(T_{00}+2iT_{10}-T_{11}) [/itex]
    I have used the symmetry of [itex]T_{\mu\nu}[/itex].
    The plus sign in the middle of the expression is noch correct. Does someone see my mistake?

    Last edited: Jun 5, 2012
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