# Hamiltonian time-independent, partial derivative always zero?

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?

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.
• @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. – Michael Seifert Jan 29 '19 at 13:57
• @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. – doetoe Jan 29 '19 at 14:12