The Hamiltonian of a simple pendulum is given by \(H = \frac{p_\theta^2}{2mL^2} + mgL(1 - \cos \theta) \ \). What is the corresponding Lagrangian L?
1
\(\frac{1}{2}mL^2\dot{\theta}^2 - mgL(1 - \cos \theta)\ \)
2
\(\frac{1}{2}mL^2\dot{\theta}^2 + mgL(1 - \cos \theta) \ \)
3
\( -\frac{1}{2}mL^2\dot{\theta}^2 - mgL(1 - \cos \theta) \ \)
4
\( -\frac{1}{2}mL^2\dot{\theta}^2 + mgL(1 - \cos \theta)\ \)