Example of computing geodesics with 2D polar coordinates

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SUMMARY

This discussion focuses on computing geodesics in 2D polar coordinates, specifically addressing the application of Christoffel symbols. The user initially defines the Christoffel symbols and derives equations for geodesics, ultimately arriving at the relationships $$r=\theta^{2}/2$$ and $$\theta(s)=s\,e^{-r}$$. However, a critical error is identified regarding the use of normalized basis vectors instead of the tangent vector basis for polar coordinates, which invalidates the derived Christoffel symbols. The discussion emphasizes the importance of using correct mathematical notation for clarity.

PREREQUISITES
  • Understanding of geodesics and their equations in differential geometry.
  • Familiarity with Christoffel symbols and their role in curved spaces.
  • Knowledge of polar coordinates and their mathematical representation.
  • Proficiency in mathematical notation, particularly in LaTeX formatting.
NEXT STEPS
  • Study the derivation and properties of Christoffel symbols in polar coordinates.
  • Learn about tangent vector bases and their application in differential geometry.
  • Explore the concept of geodesics in various coordinate systems, including Cartesian and polar.
  • Review mathematical notation best practices, particularly for LaTeX typesetting.
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Mathematicians, physicists, and students studying differential geometry, particularly those interested in geodesics and polar coordinate systems.

fab13
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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.
 
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fab13 said:
Now, if I take :

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

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

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

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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