Homework Help: Coordinate transformation of a tensor in 2 dimensions

1. Jun 4, 2012

physicus

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

2. Relevant equations
$T_{\mu\nu}=T_{\nu\mu}$
$T'_{\alpha\beta} = U_{\alpha}{}^{\mu}U_{\beta}{}^{\nu}T_{\mu\nu}$

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 $T'$for the tensor in the new coordinates. Then, there should be a transformation matrix $U$, s.t.
$T'_{\alpha\beta} = U_{\alpha}{}^{\mu}U_{\beta}{}^{\nu}T_{\mu\nu}$
But what is $U$? Neiter with the transformation matrix from the new to the old coordinates $\begin{pmatrix}1 & 1 \\ i & -i\end{pmatrix}$ nor with the transformation matrix from the old to the new coordinates $\begin{pmatrix}\frac{1}{2} & \frac{-i}{2} \\ \frac{1}{2} & \frac{i}{2} \end{pmatrix}$ it is working. What am I doing wrong?

physicus

2. Jun 4, 2012

clamtrox

It's difficult to say without seeing the calculation. You do remember that there are two transformation matrices, right? (one for each index)

3. Jun 5, 2012

physicus

Thanks, but I still get a wrong result.

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

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

physicus

Last edited: Jun 5, 2012