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Relativistic angular momentum and cyclic coordinates

  1. Jan 10, 2012 #1
    I'm getting myself confused here. If my relativistic Lagrangian for a particle in a central potentai is

    [tex]L = \frac{-m_0 c^2}{\gamma} - V(r) [/tex]

    should

    [tex] \frac{d L}{d \dot{\theta}} [/tex]

    not give me the angular momentum (which is conserved)? Instead I get

    [tex] \frac{d L}{d \dot{\theta}} = -4 m v r^2 \dot{\theta}\gamma [/tex]
     
  2. jcsd
  3. Jan 11, 2012 #2
  4. Jan 11, 2012 #3
    [tex]L = - {m_0}{c^2}\sqrt {1 - \frac{{{{\dot r}^2} + {r^2}{{\dot \theta }^2}}}{{{c^2}}}} - V\left( r \right)[/tex]

    so

    [tex]\frac{{\partial L}}{{\partial \dot \theta }} = \gamma {m_0}{r^2}\dot \theta [/tex]

    What's the problem?
     
  5. Jan 11, 2012 #4
    Absolutely nothing apparently. I just did it again this morning and got the right answer. Sorry for the time waste.
     
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