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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?

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  • $\begingroup$ The inverse transformation will be given by the matrix inverse of your $\Lambda$, not by the reciprocals of the matrix elements. $\endgroup$
    – J. Murray
    Commented Apr 7, 2022 at 18:42

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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 $x \leftrightarrow x'$.

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  • $\begingroup$ Would the case if $X_{\alpha}=(x,y)$ and $x_{\alpha'}=({r,\theta})$ be different? $\endgroup$
    – aygx
    Commented Apr 7, 2022 at 18:59
  • $\begingroup$ I do not understand the question. BTW, you can't lower the index on the coordinate. $x^\mu$ only makes sense with an upper index. $\endgroup$
    – Prahar
    Commented Apr 7, 2022 at 19:00
  • $\begingroup$ My mistake with typos. To begin, we let $x^{\alpha}=(x,y)$ and $x'^{\alpha}=(r,\theta)$ and obtained the transformation law for a rank (0,1) tensor in a primed frame. Would the case $x_{\alpha}=(x,y)$ and $x'_{\alpha}=(r,\theta)$ be any different? $\endgroup$
    – aygx
    Commented Apr 7, 2022 at 19:11
  • $\begingroup$ "the case $x_\alpha=$...be any different" I just don't understand this sentence. I have edited my answer - maybe it will help. Let me again clarify that you CANNOT lower the index on $x$. It MUST have an upper index. $\endgroup$
    – Prahar
    Commented Apr 7, 2022 at 19:12
  • $\begingroup$ Since thats the case then the transformation matrix for that covector $V_{\alpha}$ would just be the inverse matrix of our matrix for $V^{\alpha}$? $\endgroup$
    – aygx
    Commented Apr 7, 2022 at 19:23

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