1
$\begingroup$

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?

$\endgroup$
0
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.