An interresting "method" that allows you to know the acceleration vector with respect to any coordinate system is just a matter of recognize some key formulas.

1) Given the metric of a particlar line elemente of a particular coordinate system. Find the metric tensor:

$$ds^{2} = g_{\gamma \beta} dx^{\gamma}dx^{\beta}$$

2) Given the "definition" of Christoffel symbols. Compute all the Christoffel symbols:

$$ \displaystyle \Gamma ^{\alpha} _{\mu \nu} = g^{\delta \alpha} \Big\{ g_{\mu \delta,\nu} + g_{\nu \delta,\mu} - g_{\mu \nu,\delta} \Big\} $$

3) Calculate the "generalized force" components:

$$ \vec{F} = m\vec{a} = m \Big(a^{\alpha}\frac{\partial}{\partial x^{\alpha}} \Big) = \Big(m a^{\alpha} \Big) \frac{\partial}{\partial x^{\alpha}} = f^{\alpha} \frac{\partial}{\partial x^{\alpha}} $$

$$ f^{\alpha} = \Big(m a^{\alpha}\Big) = m \Big( \frac{d^{2}x^{\alpha}}{dt^{2}}+\displaystyle \Gamma ^{\alpha} _{\mu \nu} \frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt} \Big) $$

4) Calculate The "physical components" of the generalized force (contravariant) vector and unitary basis vectors for spherical coordinate system:

According to Sochi,pages (19-20) [https://arxiv.org/pdf/1610.04347.pdf] :

For a contravariant vector we have:

$$ \vec{A} = A^{\mu}\frac{\partial}{\partial x^{\mu}}$$

Now, the Physical Representation of basis (covariant) vectors (unitary basis vectos) are:

$$\hat{e}_{\mu}= \frac{\partial}{\partial x^{\mu}} \frac{1}{\sqrt{g_{\mu \mu}}} \equiv \frac{\partial}{\partial x^{\mu}} \frac{1}{h_{\mu}}$$ (No Sum in $\mu$)

And the Physical Representation for (Contravariant) componentes are:

$$A^{\mu}_{physical} = \sqrt{g_{\mu \mu}} A^{\mu} \equiv h_{\mu}A^{\mu}$$

Then, for the force we have:

$$ \vec{F}_{physical} = m \vec{a}_{physical} = \Big[\sqrt{g_{\alpha \alpha}}f^{\alpha}\Big] \Big[ \frac{1}{\sqrt{g_{\alpha \alpha}}} \frac{\partial}{\partial x^{\alpha}} \Big]$$

5) Setting $m=1$, then you have the accerelation (physical) vector :

$$ \vec{F}_{physical} = 1 \vec{a}_{physical}$$

$$ \vec{a}_{physical} = \Big[\sqrt{g_{\alpha \alpha}}a^{\alpha}\Big] \Big[ \frac{1}{\sqrt{g_{\alpha \alpha}}} \frac{\partial}{\partial x^{\alpha}} \Big]$$

For instance:

1) We know the metric tensor from line element of spherical coordinates:

$$ds^{2} = dr^{2} + r^{2}d\theta^{2} + r^{2}sin^{2}(\theta) d\phi^{2} \implies $$

$$ g^{spherical}_{\gamma \beta} = Diag (1, r^{2}, r^{2}sin^{2}(\theta))$$

2) The non zero Christoffel symbols are:

$$\Gamma ^{r}_{\theta \theta} = -r \\ \Gamma ^{r}_{\phi \phi} = -rsin^{2}(\theta) \\ \Gamma ^{\theta}_{\phi \phi} = -sin(\theta) cos(\theta) \\ \Gamma ^{\theta}_{r\theta} = \Gamma ^{\theta}_{\theta r} = \frac{1}{r}\\ \Gamma ^{\phi}_{r\phi} = \Gamma ^{\phi}_{\phi r} = \frac{1}{r} \\ \Gamma ^{\phi}_{\theta \phi} = \Gamma ^{\phi}_{\phi \theta} = cotg(\theta)$$

3) The generalized force components are:

$$ f^{r} = m\Big( \ddot{r} - r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big)$$

$$f^{\theta} = m\Big(\ddot{\theta} +2\frac{1}{r}\dot{r}\dot{\theta} - sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big)$$

$$f^{\phi} = m\Big(\ddot{\phi} +2\frac{1}{r}\dot{r}\dot{\theta} + 2cotg(\theta) \dot{\phi} \dot{\theta} \Big)$$

Then,the force vector is:

$$\vec{F} = f^{r}\frac{\partial}{\partial x^{r}} + f^{\theta}\frac{\partial}{\partial x^{\theta}}+f^{\phi}\frac{\partial}{\partial x^{\phi}} \implies $$

$$\vec{F} =\Big[ m\Big( \ddot{r} - r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big) \Big] \frac{\partial}{\partial x^{r}}\\+ \Big[ m\Big(\ddot{\theta} +2\frac{1}{r}\dot{r}\dot{\theta} - sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \Big]\frac{\partial}{\partial x^{\theta}}\\+\Big[ m\Big(\ddot{\phi} +2\frac{1}{r}\dot{r}\dot{\theta} + 2cotg(\theta) \dot{\phi} \dot{\theta} \Big) \Big]\frac{\partial}{\partial x^{\phi}} $$

4) The physical components and vector are then:

$$\vec{F}_{physical} = \Big[\sqrt{g_{\alpha \alpha}}f^{\alpha}\Big] \Big[ \frac{1}{\sqrt{g_{\alpha \alpha}}} \frac{\partial}{\partial x^{\alpha}} \Big] = \sqrt{g_{rr}}f^{r} \frac{1}{\sqrt{g_{rr}}} \frac{\partial}{\partial x^{r}} + \sqrt{g_{\theta \theta}}f^{\theta} \frac{1}{\sqrt{g_{\theta \theta}}} \frac{\partial}{\partial x^{\theta}}+ \sqrt{g_{\phi \phi}}f^{\phi} \frac{1}{\sqrt{g_{\phi \phi}}} \frac{\partial}{\partial x^{\phi}} \implies $$

$$\vec{F}_{physical}= \Big[ m\Big( \sqrt{g_{rr}}\ddot{r} -\sqrt{g_{rr}} r\dot{\theta}^2 - \sqrt{g_{rr}}rsin^{2}(\theta) \dot{\phi}^{2} \Big) \Big]\frac{1}{\sqrt{g_{rr}}} \frac{\partial}{\partial x^{r}}\\+ \Big[ m\Big( \sqrt{g_{\theta \theta}}\ddot{\theta} +\sqrt{g_{\theta \theta}}2\frac{1}{r}\dot{r}\dot{\theta} -\sqrt{g_{\theta \theta}} sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \Big] \frac{1}{\sqrt{g_{\theta \theta}}} \frac{\partial}{\partial x^{\theta}}\\+\Big[ m\Big(\sqrt{g_{\phi \phi}}\ddot{\phi} +\sqrt{g_{\phi \phi}}2\frac{1}{r}\dot{r}\dot{\theta} +\sqrt{g_{\phi \phi}} 2cotg(\theta) \dot{\phi} \dot{\theta} \Big) \Big] \frac{1}{\sqrt{g_{\phi \phi}}}\frac{\partial}{\partial x^{\phi}} \implies$$

$$\vec{F}_{physical}=\Big[ m\Big( 1 \ddot{r} -1 r\dot{\theta}^2 - 1 rsin^{2}(\theta) \dot{\phi}^{2} \Big) \Big] \frac{1}{1}\frac{\partial}{\partial x^{r}}\\+ \Big[ m\Big( r \ddot{\theta} +r2\frac{1}{r}\dot{r}\dot{\theta} -r sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \Big] \frac{1}{r}\frac{\partial}{\partial x^{\theta}}\\+\Big[ m\Big( \sqrt{r^{2}sin^{2}(\theta)}\ddot{\phi} +\sqrt{r^{2}sin^{2}(\theta)}2\frac{1}{r}\dot{r}\dot{\theta} +\sqrt{r^{2}sin^{2}(\theta)} 2cotg(\theta) \dot{\phi} \dot{\theta} \Big) \Big] \frac{1}{rsin(\theta)}\frac{\partial}{\partial x^{\phi}} \implies$$

5) Setting $m=1$, we have the well known acceleration vector in Spherical Coordinares:

$$ \vec{a}_{physical} =\Big( \ddot{r} -r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big)\frac{1}{1}\frac{\partial}{\partial x^{r}}\\+\Big( r \ddot{\theta} +2\dot{r}\dot{\theta} -r sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \frac{1}{r}\frac{\partial}{\partial x^{\theta}}\\+\Big( rsin(\theta) + 2sin(\theta)\dot{r}\dot{\phi}+2rcos(\theta)\dot{\phi}\dot{\theta} \Big)\frac{1}{rsin(\theta)}\frac{\partial}{\partial x^{\phi}} \implies$$

$$ \vec{a}_{physical} =\Big( \ddot{r} -r\dot{\theta}^2 - rsin^{2}(\theta) \dot{\phi}^{2} \Big)\hat{e}_{r}\\+\Big( r \ddot{\theta} +2\dot{r}\dot{\theta} -r sin(\theta)\cos(\theta)\ddot{\phi}^{2} \Big) \hat{e}_{\theta}\\+\Big( rsin(\theta) + 2sin(\theta)\dot{r}\dot{\phi}+2rcos(\theta)\dot{\phi}\dot{\theta} \Big)\hat{e}_{\phi}$$

Now, my question is: how can I compute the velocity and position vectors?

The reason for these awful calculations is simple, you just need to know a few formulas and how to calculate partial derivatives properly. Then you earn a powerful method to compute the equations of motion in any coordinate system (also in a flat and curved space).

  • $\begingroup$ Minor comment to the post (v3): In the future please link to abstract pages rather than pdf files. $\endgroup$
    – Qmechanic
    Commented Oct 14, 2018 at 6:00

1 Answer 1


There is a quicker way to derive the equations of motion via an action principle using the line element. Consider the variational principle:

$$ \mathcal A = \int \mathrm{d}\lambda \left[g_{ab}(x)\frac{\mathrm dx^a}{\mathrm d\lambda}\frac{\mathrm dx^b}{\mathrm d\lambda} -V(x)\right] $$

where $\lambda$ is an affine parameter (time in the case of classical mechanics). The resultant Euler-Lagrange equations which give you the equations of motion are:

$$ \frac{\mathrm d}{\mathrm d\lambda}\left(g_{ab}\frac{\mathrm dx^a}{\mathrm d\lambda}\right) = \frac{1}{2}(\partial_b g_{ij})\frac{\mathrm dx^i}{\mathrm d\lambda}\frac{\mathrm dx^j}{\mathrm d\lambda} - \partial_b V$$

So consider $\mathrm ds^2 = \mathrm dr^2 + r^2\mathrm d\theta^2 + r^2\sin^2\theta\,\mathrm d\phi^2 \implies g_{ab} = \text{diag}(1,r^2,r^2\sin^2\theta)$:

$$ \mathcal A = \int \mathrm d\lambda \left[\dot r^2 + r^2\dot \theta^2 + r^2\sin^2\theta\,\dot\phi^2 \right], \quad \dot x^a = \frac{\mathrm dx^a}{\mathrm d\lambda}$$

And derive the equations of motion component-wise by reading off the Euler-Lagrange equations:

$$ \ddot r = r\dot\theta^2 + r\sin^2\theta\,\dot\phi^2 + a_r, \quad [a_b =- \partial_b V]$$

$$ \left(\frac{\mathrm d}{\mathrm d\lambda}(r^2\dot\theta) = r^2\ddot\theta + 2r\dot r\dot\theta\right) = r^2\sin\theta\cos\theta\,\dot\phi^2 + a_{\theta}$$

$$ \left(\frac{\mathrm d}{\mathrm d\lambda}(r^2\sin^2\theta\,\dot\phi) = r^2\sin^2\theta \,\ddot\phi +\dot\phi[2r\dot r\sin^2\theta + r^2\dot\theta\sin\theta\cos\theta] \right) = a_\phi$$

And choose your basis (either coordinate or non-coordinate; this changes the definition of $a_b$, for example, in your case: $a_b = \sqrt{-g_{bb}}\partial_b V$), then read off your Christoffel symbols without having to do needless computations that end up in zeros.

To answer your main question about finding velocity and position vectors, you need to specify what external forces are acting on your system (by specifying the potential $V(x)$) and solve the resultant differential equations, then integrate the components of your acceleration vector. Unfortunately, there's no general method for finding the solutions to ordinary differential equations (coupled in this case) as far as I know. So there's no general algorithmic procedure (except numerically) for determining the solutions to a classical system without specifying the cases.


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