momentum - The number of independent variables in the Lagrangian and Hamiltonian methods in Classical Mechanics

It's told in Landau - Classical Mechanics, that in the Hamiltonian method, generalized coordinates $q_j$ and generalized momenta $p_j$ are independent variables of a mechanical system. Anyway, in the case of Lagrangian method only generalized coordinates $q_j$ are independent. In this case generalized velocities are not independent, as they are the derivatives of coordinates.

So, as I understood, in the first method, there are twice more independent variables, than in the second. This fact is used during the variation of action and finding the equations of motion.

My question is, can the number of independent variables of the same system be different in these cases? Besides that, how can the momenta be independent from coordinates, if we have this equation

$$p=\frac{\partial L}{\partial \dot{q}}.$$

Answer

$q_j$ and $\dot q_j$ are independent. I think it's more straightforward (at first) to think of this in terms of Newton's equations of motion, where the force determines the accelerations of the various particles, than in terms of the more abstract Hamiltonian methods. Because the forces determine the accelerations, not the velocities, both the initial positions and the initial velocities have to be given to determine the trajectories, which is just to say that the $q_j$ and the $\dot q_j$ independently determine the trajectories.

Note that the Lagrangian function is written as a function of both $q_j$ and $\dot q_j$, $L(q_j,\dot q_j)$, which makes sense of the equation for the momenta that you cite, $p_j=\frac{\partial L(q_j,\dot q_j)}{\partial \dot q_j}$.

So, there are the same numbers of independent variables in the Lagrangian and in the Hamiltonian formalisms.