Usually, Hamilton's equations of motion are given by: $$ (1)\;\; \frac{dp}{dt} = -\frac{\partial H}{\partial q} \;\;\; \text{ and } \;\;\;(2)\;\; \frac{dq}{dt} =\frac{\partial H}{\partial p}.$$ Using Poisson's bracket one can generalize Hamilton's equation i.e. the time derivative of anything is given by the Poisson bracket of that thing with the Hamiltonian: $$\frac{dF}{dt} = \{F,H\} $$ where $F$ is a function of $p$ and $q$, $H$ is the Hamiltonian, and Poisson bracket is defined as $$ \{A,B\} = \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q}. $$ See page 108 of 'No-nonsense Classical Mechanics' by Jakob Schwichtenberg and page 172 of 'The Theoretical Minimum' by Susskind & Hrabovsky for details. However, in Wikipedia, they are saying that the most general form of the Hamilton's equation of motion is $$\frac{dF}{dt} = \{F,H\} + \frac{\partial F}{\partial t}.$$ From where is the extra $\frac{\partial F}{\partial t}$ coming from? Which one is correct?

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    $\begingroup$ In general $F=F(q,p,t)$. Then you can define a function $\mathcal F(t) := F(q(t),p(t),t)$. Calculating its derivative (and employing the equations of motion) yields your last equation. Long story short: Observables can, in principle, depend on time explicitly and not only through $q$ and $p$. $\endgroup$ Commented May 6, 2022 at 21:09
  • $\begingroup$ Okay. Just for clarification, does that mean in Susskind/Jakob's text, they assumed that F doesn't depend explicitly on time, right? $\endgroup$ Commented May 6, 2022 at 21:18
  • $\begingroup$ I don't know the context (i.e. I haven't read the specific parts of the books), but it seems very likely to me. $\endgroup$ Commented May 6, 2022 at 21:20

1 Answer 1


Most generally the observable $F$ depends not only on $q(t)$ and $p(t)$, but also depends on $t$ explicitly, i.e. $$F=F(q,p,t)$$ Differentiating this with respect to time $t$ you get $$\begin{align} \frac{dF}{dt} &=\frac{\partial F}{\partial q}\frac{dq}{dt} +\frac{\partial F}{\partial p}\frac{dp}{dt} +\frac{\partial F}{\partial t} \\ &=\frac{\partial F}{\partial q}\frac{\partial H}{dp} -\frac{\partial F}{\partial p}\frac{\partial H}{dq} +\frac{\partial F}{\partial t} \\ &=\{F,H\} +\frac{\partial F}{\partial t} \end{align}$$


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