derivatives transformation I'm currently doing a calculation for the connection coefficients using the standard space-time coordinates, namely $x_0,x_1,x_2,x_3$. The setup is a spherically symmetric problem.
In my expressions, I have derivatives with respect to these functions (i.e. $d/dx_0,d/dx_1$ etc). Now I am trying to rewrite my results in terms of spherical coordinates $t,r$ (no $\theta$ and $\phi$ because of spherical symmetry), with $r=\sqrt{x_1^2+x_2^2+x_3^2}$ and need to replace the $x$'s derivatives with the spherical ones.
That is where my problem is. The usual transformation from cartesian to spherical derivatives involves coefficients that depend on $\theta$ and $\phi$. But I don't have those to start with!
So, my question is, how do I find the derivatives w.r.t. $x$'s in terms of spherical coords for the spherically symmetric case?
Many thanks in advance!
 A: You are trying to rewrite derivatives $\frac{\partial f}{\partial x_i}$ in terms of spherical coordinates $r, \theta, \phi$. They are gonna be not as nice as you might think based on spherical symmetry. These derivatives rather reflect how $f$ looks in cartesian coordinates, and in general they will depend on all of $r,\theta$ and $\phi$ when transformed to spherical coords.
You might want to see instead how del ($\nabla$) or Laplacian ($\nabla^2$) look in spherical coordinates. Expressions for them acting on spherically-symmetric function will be then indeed independent of $\theta$ and $\phi$.
A: When $\theta=0$ at the equator you have
$$r=\sqrt{x_1^2+x_2^2+x_3^2},\quad \phi={\rm arg}(x_1,x_2),\quad \theta={\rm arg}\bigl(\sqrt{x_1^2+x_2^2},x_3\bigr)\ .$$
For a function $f:{\mathbb R}^3\to{\mathbb R}$ that does not depend on $\phi$ or $\theta$ we therefore obtain using  the chain rule:
$${\partial f\over\partial x_i}={\partial f\over\partial r}{\partial r\over\partial x_i}+ {\partial f\over\partial \phi}{\partial \phi\over\partial x_i}+{\partial f\over\partial \theta}{\partial \theta\over\partial x_i} ={\partial f\over\partial r}{x_i\over r}+0+0\ .\tag{1}$$
A coordinate-free version of $(1)$ is the following: Dentote by $r\mapsto \tilde f(r)$ the underlying function of $r$ alone. For a given point $p\in{\mathbb R}^3$ and a "displacement vector" $X\in T_p$ one then has
$$df(p).X=\nabla f(p)\cdot X=\tilde f'(|p|){X\cdot p\over|p|}\ .$$
