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For conceptual simplicity, let's restrict the discussion to systems with a two-dimensional phase space $\mathcal P$ with generalized coordinates $(q,p)$.

Hamiltonian is a function that maps a pair consisting of a point $(q,p)$ in phase space and a point $t$ in time, to a real number $H(q,p,t)$. When we say that we are taking the partial time derivative of $H$, we mean that we are taking a derivative with respect to its last argument (in my notation). When we say that we are taking a total time derivative, we have in mind evaluating the phase space arguments of the Hamiltonian on a parameterized path $(q(t), p(t))$ in phase space, then then taking the derivative with respect to $t$ of the resulting expression, like this; \begin{align} \frac{d}{dt}\Big(H(q(t), p(t), t)\Big) \end{align} If we use the chain rule, we find that this total time derivative can be related to the partial time derivative of $H$ as follows: \begin{align} \frac{d}{dt}\Big(H(q(t), p(t), t)\Big) = \frac{\partial H}{\partial q}(q(t), p(t), t) \dot q(t) + \frac{\partial H}{\partial p}(q(t), p(t), t) \dot p(t) + \frac{\partial H}{\partial t}(q(t), p(t), t) \end{align}

So if we say the Hamiltonian is time-independent, it automatically also means by definition that $\frac{\partial H}{\partial t}(q(t), p(t), t) = 0$, and not only $\frac{d}{dt}\Big(H(q(t), p(t), t)\Big)=0$ right?

Qmechanic
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Alex Santeri
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1 Answers1

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In fact we only mean that the partial derivative $\frac{\partial H}{\partial t}(p,q,t) = 0$. Note that $p$ and $q$ are independent arguments here, they are not the components of a curve parameterized by $t$.

When this holds, and $(p(t), q(t))$ is a parameterization of the curve that satisfies the Hamilton equation, then this implies that $\frac{dH}{dt}(q(t), p(t), t)$ vanishes on this curve.

doetoe
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