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?


1 Answer 1


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.

  • $\begingroup$ Alright, but are you implying that there are special cases when the partial derivative with respect to time is zero but the total derivative is not? Isnt the parametrization (p(t),q(t)) of the curve a requirement to satisfy the Hamilton equation? Or just that it follows everytime $\endgroup$ Jan 29, 2019 at 13:46
  • 1
    $\begingroup$ @Alex Santeri: (1) The reparametrization issue is a bit of a red herring. If we use a different parameter $\tau(t)$, we have$$\frac{d}{d\tau} H(q, p,\tau) = \frac{dt}{d\tau} \frac{d}{dt} H(q, p, t(\tau)) = 0,$$as before. (2) The statement is "if $\partial H/\partial t = 0$, and if $p(t)$ and $q(t)$ obey Hamilton's equations, then $dH/dt = 0$." This means that for a curve in phase space that doesn't follow the Hamiltonian flow, you will generally not have $H = \text{const.}$ along that curve. $\endgroup$ Jan 29, 2019 at 13:57
  • $\begingroup$ @AlexSanteri Note that the partial derivative is defined on the whole of $\mathcal P\times \mathbb R$, while the total derivative is defined on the composition of $H$ with a curve parameterized by $t\mapsto (q(t), p(t), t)$, which is a function of $t$ alone. If the partial derivative is 0 and this curve satisfies the Hamilton equations, the total derivative always is 0. $\endgroup$
    – doetoe
    Jan 29, 2019 at 14:12
  • $\begingroup$ Thank you for your explanations doetoe and Professor Seifert. Could you perhaps suggest me further reading on the subject? I am currently studying Landaus mechanics with a bachelor degree in applied physics. $\endgroup$ Jan 29, 2019 at 14:53
  • $\begingroup$ @AlexSanteri I think Landau and Lifschitz is an excellent book. A standard text that is very good as well is Goldstein's. As a mathematician, my favorite is Arnold's Mathematical Methods of Classical Mechanics $\endgroup$
    – doetoe
    Jan 29, 2019 at 17:02

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