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Given a dynamical system with canonical variables $q$ and $p$. The equations of motion are given by Hamilton's equations

\begin{equation} \dot{q}=\frac{\partial H}{\partial p} \\ \dot{p}=-\frac{\partial H}{\partial q} \end{equation}

where a dot represents a derivative respect to time and $H=H(q,p)$ is the hamiltonian of the system. I'm going to use this as an example to show what my question is, but my concern is more general. I understand that the derivatives on the rhs of the equations are not taken along the trajectories. If they were, any system with constant energy would give identically zero, since $H=$ constant. However, the lhs are evaluated on the trajectories since the canonical variables $q$ and $p$ have time dependence. This means that the right hand side is supposed to be evaluated on the phase space trajectory $\big(q(t),p(t)\big)$after we apply the derivative. Why is this not explained in any book (Goldstein, Lanczos, etc.) that I have encountered when studying Classical Mechanics? Is it that obvious?

This same problems appears when dealing with the Poisson brackets. When we calculate

\begin{equation} \{q,p\}=\frac{\partial q}{\partial q}\frac{\partial p}{\partial p}-\frac{\partial q}{\partial p}\frac{\partial p}{\partial q}=1-0=1 \end{equation}

I assume that we are using $q$ and $p$ as canonical variables, without any time dependence. In other words, the variables are not evaluated along the solutions of Hamilton's equations. However, when we write Hamilton's equations using Poisson brackets

\begin{equation} \dot{q}=\{q,H\}=\frac{\partial q}{\partial q}\frac{\partial H}{\partial p }- \frac{\partial q}{\partial p }\frac{\partial H}{\partial q}= \frac{\partial H}{\partial p} \end{equation}

The Poisson bracket is clearly evaluated at the trajectories after we apply the derivatives, just like Hamilton's equations.

However, I have also encountered terms like $\frac{\partial q}{\partial p}$ that were actually evaluated at the trajectory before taking the derivative. Specially when dealing with changes of coordinates and Canonical transformations. So the question is: Is there a way of indicating when derivatives are applied before evaluating on the trajectory and when they are applied after that?

  • The rule here isn't complicated. Whenever you take a total time derivative (i.e. whenever you see a dot that appear), you're talking about a trajectory. That's it. – knzhou May 08 '20 at 02:51
  • Things are more annoying in Lagrangian mechanics because there, $\dot{q}$ can either stand for a genuine time derivative along a trajectory, or some quantity formally independent of $q$ (e.g. see here). But it's not really an issue in Hamiltonian mechanics. – knzhou May 08 '20 at 02:54
  • Encountered where? Which reference? Which page? – Qmechanic May 08 '20 at 07:28
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    Yes, it is that obvious upon consideration of the paraboloid of the oscillator hamiltonian in phase space. The phase-space velocity, hence motion is specified so that your trajectory lies on a constant energy section. – Cosmas Zachos May 08 '20 at 20:12

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