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 $A'_{\phi} = \tan^{-1}(A_2/A_1)$?

 A: Your transformation matrix:
I will ignore the "t" coordinate 
\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*}
A: 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.
