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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 $x$?

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    $\begingroup$ 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? $\endgroup$ – Aaron Stevens Feb 11 at 16:34
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    $\begingroup$ This question is not a duplicate question, but I think the accepted answer will help you out a bit :) $\endgroup$ – Aaron Stevens Feb 11 at 16:37
  • $\begingroup$ If $\dfrac{\partial L}{\partial \dot q _{i}}\neq f\left( q_{i}\right) $ then $\dot p_¡$ is independent of $q_¡$ $\endgroup$ – Eli Feb 12 at 15:15

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