Doubt on the difference between a rotational coordinate system and spherical coordinate system and the calculation of the Christoffel sysmbols I know basic differential geometry for general Relativity and classical mechanics. But an interesting fact was revealed in my calculations, namely, that I discovered that I didn't realize the difference between the spherical coordinate system and a rotational system. 
The problem arose when I tried to calculate the "generalized" force as
$$m\frac{d^{2}x^{i}}{dt^{2}} = - m\Gamma^{i}_{ij}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt} \tag{1}$$
in spherical coordinates $(r,\theta,\phi)$, to calculate the fictitious forces, namely, the centrifugal, Coriolis and (I think) the Euler forces. But in fact you reach those fictitious forces just in rotational coordinates like:
$$\begin{cases} x' = x cos\theta - y sin\theta \\ y' = x sin\theta + y cos \theta \end{cases} \tag{2}$$
I am now confused because, when you are spinning a ball in a circular motion, you use polar coordinates to describe the physical fact that something is under rotation where the polar coordinates are associated with a non-inertial frame. 
My doubt is why, using spherical coordinates metric tensor, I didn't get fictitious forces but in the rotational coordinates I did?
 A: you can calculate the the fictitious force vector $\vec{f}_c$ with this equation:
$$\vec{f}_c=m\,\frac{\partial (J\,\vec{\dot{q}})}{\partial \vec{q}}\,\vec{\dot{q}}$$
where:


*

*$\vec{q}=$ vector of the generalized coordinates

*$J=\frac{\partial{\vec(r}(\vec{q}))}{\partial \vec{q}}$ the Jacobi matrix

*$\vec{r}$ position vector


and the generalized fictitious force vector is:
$$\vec{f}_g=G^{-1}\,J^T\,\vec{f}_c\equiv  - m\Gamma^{i}_{ij}\frac{dx^{i}}{dt}\frac{dx^{j}}{dt} $$
where $G$ is the metric $G=J^T\,J$ 
Example
Sphere coordinates:
position vector :
$$\vec{r}= \left[ \begin {array}{c} \rho\,\cos \left( \vartheta  \right) \sin
 \left( \varphi  \right) \\ \rho\,\sin \left( 
\vartheta  \right) \sin \left( \varphi  \right) \\ 
\rho\,\cos \left( \varphi  \right) \end {array} \right] 
$$
generalized coordinates:
$$\vec{q}=\left[ \begin {array}{c} \rho\\ \varphi 
\\\vartheta \end {array} \right]  
=\begin{bmatrix}
  x^1 \\
  x^2 \\
  x^3 \\
\end{bmatrix}$$
$\Rightarrow$
$$G= \left[ \begin {array}{ccc} 1&0&0\\ 0&{\rho}^{2}&0
\\ 0&0&{\rho}^{2} \left( \sin \left( \varphi 
 \right)  \right) ^{2}\end {array} \right] 
$$
and the generalized fictitious force vector is:
$$\vec{f}_g= m\,\left[ \begin {array}{c}  \left( -{\dot{\varphi}}^{2}-{\dot{\vartheta}}^{2}+
 \left( \cos \left( \varphi  \right)  \right) ^{2}{\dot{\vartheta}}^{2}
 \right) \rho\\-{\frac {\cos \left( \varphi
 \right) \rho\,{\dot{\vartheta}}^{2}\sin \left( \varphi  \right) -2\,\dot{\rho}\,\dot{\varphi}}{\rho}}\\2\,{\frac {\dot{\vartheta}\,
 \left( \sin \left( \varphi  \right) \dot{\rho}+\rho\,\cos \left( \varphi
 \right) \dot{\varphi} \right) }{\rho\,\sin \left( \varphi  \right) }}
\end {array} \right]
$$
"My doubt is why, using spherical coordinates metric tensor, I didn't get fictitious forces but in the rotational coordinates I did?" this is not correct as you can see from  this example
