Simple question about change of coordinates

Suppose we have two coordinate systems (Cartesian and spherical)

$$x^{\mu} = (t,x,y,z)$$

$$x'^{\mu'} = (t',r,\theta,\phi)$$

where $$r= \sqrt{x^2 + y^2 + z^2} , \theta = \cos^{-1}(z/r), \phi = \tan^{-1} (y/x)$$. My question is, in general, what are the components of a vector $$A_{\mu} = (A_t,A_x,A_y,A_z)_{\mu}$$ in the primed coordinates? From GR, I believe the answer is $$A'_{\mu'} = (A_{t'},A_{r},A_{\theta},A_{\phi})_{\mu'} = \frac{\partial x^{\mu}}{\partial x'^{\mu'}} A_{\mu}$$, with the inverse matrix used for upper-index vectors.

If this is the case, in particular it should work for position vectors. That is, $$x'^{\mu'} = \frac{\partial x'^{\mu'}}{\partial x^{\mu}} x^{\mu}$$. However, applying this transformation gives $$x'^{\mu'} = (t',r,0,0)$$, not $$(t',r,\theta,\phi)$$. Am I doing something wrong?

Edit: The second paragraph incorrectly applies the formula I've cited, as pointed out by mike stone.

As for the first question, since we have $$x'_r = \sqrt{x_1^2 +x_2^2 + x_3^2}, x'_{\theta} =\cos^{-1}(x_3/x'_r)$$,$$x'_{\phi} = \tan^{-1}(x_2/ x_1)$$, does it follow for any vector $$A'_{\mu}$$ (for instance, the EM gauge field) that $$A'_r = \sqrt{A_1^2 + A_2^2 + A_3^2}$$, $$A'_{\theta} = \cos^{-1}(A_3/ A'_r)$$, and $$A'_{\phi} = \tan^{-1}(A_2/A_1)$$?

• You don't need to list an edit history. This is automatically done. Just make the appropriate edits to your equations. Sep 21 '18 at 17:32

\begin{align*} &\text{The position vector for a sphere is: } \\ &\vec{R_s}= \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \left[ \begin {array}{c} r\cos \left( \vartheta \right) \cos \left( \varphi \right) \\ r\cos \left( \vartheta \right) \sin \left( \varphi \right) \\ r\sin \left( \vartheta \right) \end {array} \right]&(1) \\\\ &\text{we can now calculate the transformation matrix R:}\\\\ &R=J\,H^{-1}\\ &\text{J is the Jakobi matrix }\quad\,, J=\frac{\partial\vec{R_s}}{\partial\vec{q}}\quad \text{with:}\\ &\vec{q}=\begin{bmatrix} r \\ \varphi \\ \vartheta \\ \end{bmatrix}\quad, H=\sqrt{G_{ii}}\,,H_{ij}=0\quad\text{and } G=J^{T}\,J\quad\text{the metric.}\\\\ &\Rightarrow\\\\ &R=\left[ \begin {array}{ccc} \cos \left( \varphi \right) \cos \left( \vartheta \right) &-\sin \left( \varphi \right) &-\cos \left( \varphi \right) \sin \left( \vartheta \right) \\ \sin \left( \varphi \right) \cos \left( \vartheta \right) &\cos \left( \varphi \right) &-\sin \left( \varphi \right) \sin \left( \vartheta \right) \\ \sin \left( \vartheta \right) &0&\cos \left( \vartheta \right) \end {array} \right]&(2) \end{align*} \begin{align*} &\text{We can solve equation (1) for r\,,\varphi and \vartheta}\\\\ &r=\sqrt{x^2+y^2+z^2}\\ &\varphi=\arctan\left(\frac{y}{x}\right)\\ &\vartheta=\arctan\left(\frac{z}{\sqrt{x^2+y^2}}\right)\\\\ &\text{and with equation (2):}\\\\ &R= \left[ \begin {array}{ccc} {\frac {xz}{\sqrt {{y}^{2}+{x}^{2}}r}}&-{ \frac {y}{\sqrt {{y}^{2}+{x}^{2}}}}&-{\frac {x}{r}} \\ {\frac {yz}{\sqrt {{y}^{2}+{x}^{2}}r}}&{\frac {x} {\sqrt {{y}^{2}+{x}^{2}}}}&-{\frac {y}{r}}\\ {\frac {\sqrt {{y}^{2}+{x}^{2}}}{r}}&0&{\frac {z}{r}}\end {array} \right] &(3)\\\\ &\text{The components of a vector can transformed either with equation (2) or with equation (3) } \end{align*}
The GR vector transformation you cite applies to elements of the tangent space at a point. Positions are not vectors in any tangent space, so coordinates $$x^\mu$$ do not transform as vectors.