# Relativistic angular momentum and cyclic coordinates

1. Jan 10, 2012

### 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

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

2. Jan 11, 2012

### maverick_starstrider

Anyone?

3. Jan 11, 2012

### 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?

4. Jan 11, 2012

### maverick_starstrider

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