- #1
fab13
- 318
- 6
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 :
$$de_{i}=w_{i}^{j}\,de_{j}=\Gamma_{ik}^{j}du^{k}\,de_{j}$$
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.
$$de_{i}=w_{i}^{j}\,de_{j}=\Gamma_{ik}^{j}du^{k}\,de_{j}$$
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.