General Relativity – Transformation Matrix from Cartesian to Polar of a Covariant Vector

Question

Suppose that there exists some contravariant 2-vector $V^{\alpha}$ . Now, let $V^{\alpha}$ be an element of the Cartesian coordinate system (x,y) such that it can be transformed into polar coordinates $(r, \theta)$. The transformation matrix $\Lambda^{\alpha'}_{\beta}$ can we written as $$\Lambda^{\alpha'}_{\beta}=\begin{align}\left(\begin{matrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} \end{matrix}\right)\end{align}$$ therefore the inverse transformation would just be the inverse of the transformation coefficients listed inside of the transformation matrix i.e this would be denoted $\Lambda^{\beta}_{\alpha'}$. Now suppose that we have a covariant covector $V_{\alpha}$ and we want to transform from cartesian to polar. How would we go about doing this process? Would its transformation matrix look like?

Answer 1

Let $x^\alpha = (x,y)$ and $x'^\alpha = (r,\theta)$. Then, $$ V'_\alpha(x') = \frac{\partial x^\beta}{\partial x'^\alpha} V_\beta(x) $$ Explicitly $$ V_r = \frac{\partial x}{\partial r} V_x + \frac{\partial y}{\partial r} V_y , \qquad V_\theta =\frac{\partial x}{\partial \theta} V_x + \frac{\partial y}{\partial \theta} V_y $$ Then, using the fact that $x=r\cos\theta$ and $y=r\sin\theta$, we find $$ V_r = \cos\theta V_x + \sin\theta V_y , \qquad V_\theta = - r \sin\theta V_x + r \cos\theta V_y $$ The inverse transformation is obtained by simply exchanging $x \leftrightarrow x'$, so $$ V_\alpha(x) = \frac{\partial x'^\beta}{\partial x^\alpha} V'_\beta(x'). $$ The transformation law for a contra-variant vector is $$ V'^\alpha(x') = \frac{\partial x'^\alpha}{\partial x^\beta} V^\beta(x) $$ Again, the inverse transformation is obtained by exchanging $x \leftrightarrow x'$.