- #1
Rick16
- 174
- 51
- TL;DR
- find the Lagrangian of a somewhat complex system
This problem has been addressed before under https://www.physicsforums.com/threads/trouble-understanding-coordinates-for-the-lagrangian.1006528/ I also copied the following problem statement with Landau's very sketchy solution from this old post, because I don't have the English edition of the text.
The old thread only addressed the question of how to determine the coordinates of the masses. My problem is how to get the kinetic energy term from the displacement term. For one mass ##m_1## Landau writes the (square of its) displacement as ##dl_1^2=a^2d\theta^2+a^2sin^2\theta d\phi^2##. His kinetic energy term for this mass is ##\frac{1}{2}m_1a^2(\dot{\theta}^2+\dot{\phi}^2sin^2\theta)##. I don't know how he gets to this expression from his displacement term.
It seems easier not to lump together the displacement term into one expression right away, but to write the horizontal and vertical coordinates separately. I could then write (for ##m_1##): ##dy=ad\theta## and ##dx=asin\theta d\phi##. Then I take the derivatives ##\dot{y}=a\dot{\theta}## and ##\dot{x}=a\dot{\theta}cos\theta d\phi + a sin\theta \dot{\phi}##. If I now neglect the term ##a\dot{\theta}cos\theta d\phi## because of the infinitesimally small angle ##d\phi## I get Landau's result. Is it legitimate to do it like this? In any case, I don't understand why Landau immediately wrote the square of the displacement and what he did next. Is there a direct and easy way to get from the square of the displacement to the kinetic energy term?
The old thread only addressed the question of how to determine the coordinates of the masses. My problem is how to get the kinetic energy term from the displacement term. For one mass ##m_1## Landau writes the (square of its) displacement as ##dl_1^2=a^2d\theta^2+a^2sin^2\theta d\phi^2##. His kinetic energy term for this mass is ##\frac{1}{2}m_1a^2(\dot{\theta}^2+\dot{\phi}^2sin^2\theta)##. I don't know how he gets to this expression from his displacement term.
It seems easier not to lump together the displacement term into one expression right away, but to write the horizontal and vertical coordinates separately. I could then write (for ##m_1##): ##dy=ad\theta## and ##dx=asin\theta d\phi##. Then I take the derivatives ##\dot{y}=a\dot{\theta}## and ##\dot{x}=a\dot{\theta}cos\theta d\phi + a sin\theta \dot{\phi}##. If I now neglect the term ##a\dot{\theta}cos\theta d\phi## because of the infinitesimally small angle ##d\phi## I get Landau's result. Is it legitimate to do it like this? In any case, I don't understand why Landau immediately wrote the square of the displacement and what he did next. Is there a direct and easy way to get from the square of the displacement to the kinetic energy term?
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