Relativistic angular momentum and cyclic coordinates

[maverick_starstrider]

I'm getting myself confused here. If my relativistic Lagrangian for a particle in a central potentai is

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

should

\frac{d L}{d \dot{\theta}}

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

\frac{d L}{d \dot{\theta}} = -4 m v r^2 \dot{\theta}\gamma

 

[maverick_starstrider]

Anyone?

 

[netheril96]

L = - {m_0}{c^2}\sqrt {1 - \frac{{{{\dot r}^2} + {r^2}{{\dot \theta }^2}}}{{{c^2}}}} - V\left( r \right)

so

\frac{{\partial L}}{{\partial \dot \theta }} = \gamma {m_0}{r^2}\dot \theta

What's the problem?

 

[maverick_starstrider]

Absolutely nothing apparently. I just did it again this morning and got the right answer. Sorry for the time waste.