The Hamiltonian is obtained as the Legendre transform of the Lagrangian: \begin{equation} H(q,p,t)=\sum_i p_i \dot{q_i} - L(q,\dot{q},t)\tag{1} \end{equation} If the Hamiltonian is expressed in canonical coordinates, the equations of motion would have the form \begin{equation} \begin{split} & \frac{\partial H}{\partial q}&=-\dot{p},\newline & \frac{\partial H}{\partial p}&=\dot{q}. \end{split}\tag{2} \end{equation} This can be seen by taking the differential of $H$: \begin{equation} \begin{split} dH&= \frac{\partial H}{\partial q} dq + \frac{\partial H}{\partial p} dp + \frac{\partial H}{\partial t} dt\\ &= -\dot{p}dq + \dot{q} dp -\frac{\partial L}{\partial t} dt \end{split}\tag{3} \end{equation} and equating the coefficients. But if the coordinates are not canonical, the form of the equations of motion would be different. However, nothing was said about the nature of the coordinates in which $L$ is expressed, or in the derivation, and hence, we don't know the kind of coordinates in which $H$ is written. Without assuming anything, the equations of motion came out in canonical form anyway. How does that happen? When did the restriction get placed on the coordinates?
1 Answers
Starting from a regular Lagrangian formulation, the Legendre transformation ensures that we can consistently define a Hamiltonian formulation with a canonical Poisson structure in phase space. For details see e.g. this Phys.SE post.
To derive OP's eq. (3) from eq. (1) it is implicitly used that: $$\begin{align}p~=~&\frac{\partial L}{\partial \dot{q}}\quad(\text{Lagr. def. of momentum})\cr \dot{p}~=~&\frac{\partial L}{\partial q}\quad(\text{EL eqs})\cr 0~=~&\frac{\partial H}{\partial \dot{q}}\quad(\text{standard assumption}). \end{align}$$
It should perhaps be stressed that it is possible to consider a Hamiltonian formulation $$ \dot{z}^I~=~\{z^I,H(z,t)\}$$ in non-canonical coordinates $z^I$, $I\in\{1,\ldots,2n\}$, but then the form of OP's eq. (1) is modified, cf. e.g. this Phys.SE post.
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