# Canonical equations of motion

The Hamiltonian is obtained as the Legendre transform of the Lagrangian: $$$$H(q,p,t)=\sum_i p_i \dot{q_i} - L(q,\dot{q},t)\tag{1}$$$$ If the Hamiltonian is expressed in canonical coordinates, the equations of motion would have the form $$$$\begin{split} & \frac{\partial H}{\partial q}&=-\dot{p},\newline & \frac{\partial H}{\partial p}&=\dot{q}. \end{split}\tag{2}$$$$ This can be seen by taking the differential of $$H$$: $$$$\begin{split} dH&= \frac{\partial H}{\partial q} dq + \frac{\partial H}{\partial p} dp + \frac{\partial H}{\partial t} dt\\ &= -\dot{p}dq + \dot{q} dp -\frac{\partial L}{\partial t} dt \end{split}\tag{3}$$$$ and equating the coefficients. But if the coordinates are not canonical, the form of the equations of motion would be different. However, nothing was said about the nature of the coordinates in which $$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?

• What do you mean, Poisson brackets, in the context of Lagrangian mechanics, where momentum is yet to be defined and we just have coördinates and their associated velocities? Jul 17, 2023 at 9:27
• It's not a question about Poisson brackets in the Lagrangian formalism. It's about how the Hamiltonian is constructed from the Lagrangian which does not have any restrictions on its coordinates.
– EM_1
Jul 17, 2023 at 9:42
• Your question statement explicitly mentions it right there at the crucial part, namely where the question mark appears. Also, we tend to try to use good coördinates, rather than explicitly seek for weird ones; the whole point of the standard prescription, to write down the Lagrangian by guessing with simple, tabularised set of known good stuff to combine, and then Legendre transform, is precisely because it is almost always guaranteed to give you canonical results. The Lagrangian formalism only has DoF in coördinates, and so it is much more difficult to get incompatibility between $q$ vs $p$ Jul 17, 2023 at 9:48

2. 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}
3. 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.
• I don't see how that addresses the question. We still didn't make any assumptions about the $q$'s to begin with.