Changing spherical coordinates in a Lagrangian

  • #1
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In order to compute de lagrangian in spherical coordinates, one usually writes the following expression for the kinetic energy: $$T = \dfrac{1}{2} m ( \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2 \theta \dot{\phi}^2 )\ ,$$ where ##\theta## is the colatitud or polar angle and ##\phi## is the azimuthal angle.

My question is what happens if I want to give a different meaning to ##\theta##, for example the latitude itself or even the opposite angle from the South Pole, ##\pi - \theta##.

Let ##\alpha## be such an angle ##\alpha = \pi - \theta##. Since ##\dot{\alpha}=\dot{\theta}##, is it valid to just perform the change of the kinetic energy as $$T = \dfrac{1}{2} m ( \dot{r}^2 + r^2 \dot{\alpha}^2 + r^2 \sin^2 (\pi-\alpha) \dot{\phi}^2 )\ ,$$ or do I need to rebuild the expression using the new variable from the beginning? Or maybe the same original expression still holds true?.
 

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  • #2
kuruman
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Well, the trig identity says ##\sin(\pi-\alpha)=\sin\pi\cos\alpha-\cos\pi\sin\alpha=\sin\alpha##. So your kinetic energy would be ##T = \dfrac{1}{2} m ( \dot{r}^2 + r^2 \dot{\alpha}^2 + r^2 \sin^2 (\alpha) \dot{\phi}^2 )##. What's the difference? Answer: Just a name change.
 
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