I am trying to find and solve the geodesics equation for polar coordinates. If I start by the definition of Christoffel symbols with the following expressions :(adsbygoogle = window.adsbygoogle || []).push({});

$$de_{i}=w_{i}^{j}\,de_{j}=\Gamma_{ik}^{j}du^{k}\,de_{j}$$

with $$u^{k}$$ is the k-th component of polar coordinates ($$1$$ is for $$r$$ and $$2$$ is for $$\theta$$).

Now, if I take :

$$de_{r} = d\theta e_{\theta}$$

$$de_{\theta} = -d\theta e_{r}$$

So : $$\Gamma_{12}^{2} = 1$$ and $$\Gamma_{22}^{1} = -1$$

All others Christoffel symbols seem to be zero.

Now, I can write the geodesics equation with :

$$\dfrac{d^{2}u^{i}}{ds^{2}} + \Gamma_{jk}^{i}\dfrac{du^{j}}{ds}\dfrac{du^{k}}{ds}$$

I get :

$$\dfrac{d^{2}r}{ds^{2}} = \dfrac{d\theta}{ds}\dfrac{d\theta}{ds}\,\,\,(1)$$

$$\dfrac{d^{2}\theta}{ds^{2}} = -\dfrac{dr}{ds}\dfrac{d\theta}{ds}\,\,\,(2)$$

By make appearing the logarithmic derivate : $$\dfrac{d\,ln(u)}{ds}=\dfrac{u'}{u}$$, I have :

for (2) : $$\dfrac{\theta'}{\theta} = - \dfrac{dr}{ds} ; \dfrac{d\theta}{ds} = e^{-r}$$

Finally, I have for (2): $$\theta(s)=s\,e^{-r}$$

for (1), taking $$\theta(s)=s\,e^{-r}$$, I have :

$$\dfrac{d^{2}r}{ds^{2}} = \bigg(\dfrac{d\theta}{ds}\bigg)^{2} = e^{-2r}$$

Finally, we get for (1) : $$r(s)=\dfrac{s^{2}}{2}e^{-2r} = \dfrac{1}{2}\theta^{2}$$

By using results from (1) and (2), I could write :

$$r=\theta^{2}/2$$

I don't understand this result knowing $$s$$ may be choose as a linear or curvilinear parameter (like the length on the geodesics).

If I set $$r$$ fixed, I expect to find $$s=r\theta$$. It doesn't seem clear for me. What should I find as final result ?

Surely I have done a mistake in my above calculus.

If someone could see what's wrong, this would be great.

Thanks in advance.

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# I Example of computing geodesics with 2D polar coordinates

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