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I Example of computing geodesics with 2D polar coordinates

  1. Aug 6, 2016 #1
    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.
     
  2. jcsd
  3. Aug 6, 2016 #2

    Orodruin

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    This is not correct. You are using the normalised basis vectors of polar coordinates here where you should be using the tangent vector basis ##E_i = \partial \vec x/\partial y^i##. These are not the Christoffel symbols of polar coordinates. Note that you are working in a torsion free space and therefore the Christoffel symbols must be symmetric in the lower indices.

    Also note that you should not use double $ for any math you want to write, it should be used for equations that need to stand alone only. If you use it for every small symbol, it will be very disruptive for people reading your text, use double # instead. This produces inline math mode such as the one used in my previous paragraph.
     
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