# Hamiltonian dependence of variables

How can one say that in Hamilton mechanics the $$q$$'s are independent of the $$p$$'s while if I have the Lagrangian $$L = \frac{1}{2}\dot{x}^2 + \frac{1}{2}x^2\dot{y}^2$$ then $$p_y = \frac{\partial{L}}{\partial\dot{y}} = x^2\dot{y}$$ and the $$p_y$$ depends on $$x$$?

• I think there should be a duplicate in PSE somewhere, but I cannot find one at the moment. For the time being, think about what you are assuming: that an equals sign indicates dependence. This is certainly not always true. For example, Newton's second law tells us $F=ma$. Does this mean the net force acting on an object depends on its mass? – Aaron Stevens Feb 11 at 16:34
• This question is not a duplicate question, but I think the accepted answer will help you out a bit :) – Aaron Stevens Feb 11 at 16:37
• If $\dfrac{\partial L}{\partial \dot q _{i}}\neq f\left( q_{i}\right)$ then $\dot p_¡$ is independent of $q_¡$ – Eli Feb 12 at 15:15