Soldering Spinors in cylindrical coordinate

Question

For my computation, I need to have Solderings in cylindrical coordinate(I think). I know what Solderings are in cartesian coordinate: $\sigma^{\mu}_{A\dot{A}}=(\sigma^{0}_{A\dot{A}},\sigma^{1}_{A\dot{A}},\sigma^{2}_{A\dot{A}},\sigma^{3}_{A\dot{A}})$ and we know that $\sigma^{0}_{A\dot{A}}=I,\quad \sigma^{x}_{A\dot{A}}=\sigma_x \quad \sigma^{y}_{A\dot{A}}=\sigma_y \quad \sigma^{z}_{A\dot{A}}=\sigma_z$ and the spinorial indices give their components. For having its cylindrical form I use changing of coordinate in this way:$$\sigma^{\rho}_{A\dot{A}}=\frac{\partial\rho}{\partial x}\sigma^{x}_{A\dot{A}}+\frac{\partial\rho}{\partial y}\sigma^{y}_{A\dot{A}}+\frac{\partial\rho}{\partial z}\sigma^{z}_{A\dot{A}}$$ $$\sigma^{\phi}_{A\dot{A}}=\frac{\partial\phi}{\partial x}\sigma^{x}_{A\dot{A}}+\frac{\partial\phi}{\partial y}\sigma^{y}_{A\dot{A}}+\frac{\partial\phi}{\partial z}\sigma^{z}_{A\dot{A}}$$ Knowing that $\sigma^{t}_{A\dot{A}}$ and $\sigma^{z}_{A\dot{A}}$ do not change.Could anybody say that what I have done is meaningful? I am confused. To be more precise, is the index of $\mu$ is a real space-time index?

Answer 1

It seems that the answer is yes and we can write the soldering in deferent coordinate system.