- #1

Jufa

- 101

- 15

Let us consider a sphere of a unit radius . Therefore, by choosing the canonical spherical coordinates ##\theta## and ##\phi## we have, for the differential length element:

$$dl = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2} $$

In order to find the geodesic we need to extremize the following:

$$ \int_{\lambda_0}^{\lambda_f} {\sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2}} d\lambda $$

We can do it so by imposing the Euler-Lagrange equations for the lagrangian ## L = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2} ## or equivalently, for the lagrangian ## L' = L^2 ##. These equations for L' look like:

$$ \ddot{\theta} +sin(\theta)cos(\theta)\dot{\phi}^2 = 0 $$

$$ \ddot{\phi}+ cot(\theta) \dot{\phi}\dot{\theta}=0 $$

I am pretty sure that I am right until here.

What does not make sense to me is the following:

Suppose you choose a curve in which ## \theta = \pi/2## i.e. both its first and second derivative vanish.

Then we get the following condition:

$$ \ddot{\phi} = 0 $$

But why does that happen? Once we have fixed the angle ## \theta## we have also fixed the curve (geodesic) and the only condition on ##\phi(\lambda)## should be that it is continuous and injective (i.e., it does not make ##\phi## go back and forth).

For example, the parametrization ##\phi(\lambda) = \frac{\lambda^2}{2\pi}## (which does not have a null second derivative) from ##\lambda_0 = 0## to ##\lambda_f =2 \pi## should be as valid as the parametrization ##\phi(\lambda) = \lambda## from ##\lambda_0 = 0## to ## \lambda_f = 2\pi ##.Thanks in advance.

$$dl = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2} $$

In order to find the geodesic we need to extremize the following:

$$ \int_{\lambda_0}^{\lambda_f} {\sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2}} d\lambda $$

We can do it so by imposing the Euler-Lagrange equations for the lagrangian ## L = \sqrt{\dot{\theta}^2+sin^2(\theta)\dot{\phi}^2} ## or equivalently, for the lagrangian ## L' = L^2 ##. These equations for L' look like:

$$ \ddot{\theta} +sin(\theta)cos(\theta)\dot{\phi}^2 = 0 $$

$$ \ddot{\phi}+ cot(\theta) \dot{\phi}\dot{\theta}=0 $$

I am pretty sure that I am right until here.

What does not make sense to me is the following:

Suppose you choose a curve in which ## \theta = \pi/2## i.e. both its first and second derivative vanish.

Then we get the following condition:

$$ \ddot{\phi} = 0 $$

But why does that happen? Once we have fixed the angle ## \theta## we have also fixed the curve (geodesic) and the only condition on ##\phi(\lambda)## should be that it is continuous and injective (i.e., it does not make ##\phi## go back and forth).

For example, the parametrization ##\phi(\lambda) = \frac{\lambda^2}{2\pi}## (which does not have a null second derivative) from ##\lambda_0 = 0## to ##\lambda_f =2 \pi## should be as valid as the parametrization ##\phi(\lambda) = \lambda## from ##\lambda_0 = 0## to ## \lambda_f = 2\pi ##.Thanks in advance.

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