# 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?

## 1 Answer

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