Coordinate transformation problems

Question

Two questions here, both related to one another.

SOLVED Question 1. I am looking to transform a dual vector $p_a$ from spherical polar coordinates to Cartesian coordinates. My dual vector is given generally by $p_a=(p_r,p_\theta,p_\phi)$. Transforming this would usually be done using the transformation matrix \begin{equation} \Lambda^a_{\;a'}= \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} & \frac{\partial z}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \theta} \\ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \end{pmatrix}, \end{equation} but the result of $\Lambda^a_{\;a'}p_a$ is: \begin{equation} \Lambda^a_{\;a'}p_a= \begin{pmatrix} p_\theta r \cos\theta \cos\phi + p_r \cos\phi \sin\theta - p_\phi r \sin\theta\sin\phi \\ p_\phi r \cos\phi \sin\theta + p_\theta r \cos\theta \sin\phi + p_r \sin\theta \sin\phi\\ p_r \cos\theta- p_\theta r \sin\theta \end{pmatrix}. \end{equation} This would equal $(p_x,p_y,p_z)$ if not for the factors of $r$ beside $p_\theta$ and $p_\phi$. I'm clearly missing something in this transformation, can someone point me in the right direction?

Question 2. I am also working on a transformation of a rank 3 tensor by the same means: $$T^{a'b'c'}=\Lambda^{a'}_{\;a}\Lambda^{b'}_{\;b}\Lambda^{c'}_{\;c}T^{abc}.$$ The $\Lambda$'s here should include the flipped derivatives of above (i.e. $\frac{\partial x}{\partial r}\rightarrow \frac{\partial r}{\partial x}$). When I do the final multiplication of $T^{a'b'c'}p_{b'}p_{c'}$ (where $p_{c'}$ is $(p_x,p_x,p_z)$) I get a result which seems to include factors such as $(p_x^2+p_y^2+z^2p_z^2)$, which obviously has some unit problems. Should the components of the $\Lambda$'s include any prefactors such as $r$ or $r\sin\theta$, i.e.

$\begin{pmatrix} \frac{\partial r}{\partial x} & \frac{\partial \theta}{\partial x} & \frac{\partial \phi}{\partial x} \\ \frac{\partial r}{\partial y} & \frac{\partial \theta}{\partial y} & \frac{\partial \phi}{\partial y} \\ \frac{\partial r}{\partial \phi} & \frac{\partial \theta}{\partial z} & \frac{\partial \phi}{\partial z} \end{pmatrix}$ or $\begin{pmatrix} \frac{\partial r}{\partial x} & r\frac{\partial \theta}{\partial x} & r \sin\theta\frac{\partial \phi}{\partial x} \\ \frac{\partial r}{\partial y} & r\frac{\partial \theta}{\partial y} & r\sin\theta\frac{\partial \phi}{\partial y} \\ \frac{\partial r}{\partial \phi} & r\frac{\partial \theta}{\partial z} & r\sin\theta\frac{\partial \phi}{\partial z} \end{pmatrix}$

Answer 1

If you transform a dual vector $p_a$ (covariant object) from spherical coordinates labelled $a$ to cartesian ones $a'$ the transformation law is:
$p_{a'} = \Lambda^a_{a'} p_a$
where $\Lambda^a_{a'}$ are the partial derivatives of the spherical coordinates (old coordinates) vs. the cartesian ones (new coordinates), that is $\frac{\partial r}{\partial x}$, etc.
The matrix in question 1. shows the opposite.
As for question 2. the comment is similar, remembering that now you transform a contravariant object, hence the $\Lambda^{a'}_a$ require the partial derivatives of the cartesian coordinates (new coordinates) vs. the spherical ones (old coordinates), that is $\frac{\partial x}{\partial r}$, etc.