Let $(M,\omega)$ with $M$ a $2n$-dimensional manifold and $\omega$ a $2$-form on $M$, furthermore let $(M,\omega)$ be a symplectic manifold (smooth, differentiable, continuous, whatever is needed for the physics). Then we can define the phase space of a classical system as $M = T^\star Q$ for some configuration manifold $Q$. We can define the Hamiltonian as a map $H: T^\star Q \rightarrow \mathbb{R}$ with the induced vector field $X_H$ on $T^\star Q$. Let's have a look at one of the Hamilton equations:

$$\frac{dq^i}{dt} = \frac{\partial H}{\partial p_i}.$$

I understand that the right hand side of the equation is properly read as (for a chart $(U,\chi)$ in the atlas of $T^\star Q$ with $\chi: U \rightarrow \mathbb{R}^{2n}$ describing the Darboux coordinates $q^1,...,q^n,p_1,...,p_n$ (each of these is a mapping from $U$ to $\mathbb{R}$))

$$\frac{\partial H}{\partial p_i} = \Big(\frac{\partial}{\partial p_i}\Big)_m H = \partial_i (H \circ \chi^{-1})(\chi(m)),$$

for some $m \in U$.

The left hand side is not clear for me. I know that $X_H$ induces an integral curve $\gamma: I \subseteq \mathbb{R} \rightarrow T^\star Q$. But my question is the following: Is it true that if the Hamilton equations are satisfied that we set $I = \mathbb{R}$ (i.e. complete integral curve) and how to interpret $\frac{d q^i}{dt}$, is this just $\frac{d}{dt} (q^i \circ \gamma)$ evaluated at $t=0$ (with $\gamma(0)=m$)?

  • $\begingroup$ I don't understand your question. Are you wondering about whether or not the solutions to Hamilton's equations are valid for all times? If so, this is not a question about physics, but about ordinary differential equations. $\endgroup$ Jan 9, 2020 at 21:01
  • $\begingroup$ I am asking about the mathematically rigorous meaning of $dq^i/dt$. $q^i$'s are just coordinate maps as I described. Maybe this question should be asked at math exchange I agree. $\endgroup$ Jan 9, 2020 at 21:14
  • 2
    $\begingroup$ Well, they are just the derivatives of a curve parametrized by $t$. It doesn't matter where it lives. Write the solutions to Hamilton's equations as $\gamma(t,q_0,p_0)$, and take the derivative wrt $t$. $\endgroup$ Jan 9, 2020 at 21:19
  • 1
    $\begingroup$ $\gamma$ cannot depend on 3 variables, it maps from $\mathbb{R}$. That does not make sense $\endgroup$ Jan 9, 2020 at 21:39
  • 3
    $\begingroup$ The solution to Hamilton's equations is an isotopy. It is a family of curves with respect to $t$, each of them changing with respect to the initial point: $\gamma: T^*M \times \mathbb{R} \to T^*M$ $\endgroup$ Jan 9, 2020 at 21:49

1 Answer 1


Your interpretation in the last paragraph is almost right, but the point of evaluation, and some logical quantifiers seem incorrect (or atleast it's phrased a little awkwardly). Also your index for interpreting the RHS should be $\partial_{n+i}$. Here's a general rule in physics: is you have a quantity which should be a function of a quantity "$\ddot{\smile}$", but somehow miraculously becomes a function another quantity "$:)$", then what it really means is that there is an extra composition which is going on, which the author has not made explicit.

Hamilton's Equations:

Let me just spell out in full gory detail how (I sometimes like) to "read" Hamilton's equations. Start with a $2n$-dimensional smooth symplectic manifold $(M, \omega)$, and a certain smooth $H: M \to \Bbb{R}$. Then, we get an induced vector field on $M$ via $X_H = \omega^{\sharp}(dH)$ (the musical isomorphism). Then, the fully geometric way of stating Hamilton's equations is that a smooth curve $\gamma: I \subset \Bbb{R} \to M$ ($I$ an open interval which doesn't have to be all of $\Bbb{R}$) is said to satisfy Hamilton's equations (with respect to $H$) iff $\gamma$ is an integral curve of the vector field $X_H$.

In terms of a Darboux chart $(U, \chi)$, where the coordinate functions are labelled as $\chi(\cdot) = \left(q^1(\cdot), \dots, q^n(\cdot), p_1(\cdot), \dots, p_n(\cdot) \right)$, the condition of $\gamma$ being an integral curve of $X_H$ can be equivalently written as: for all $i \in \{1, \dots, n\}$, and all $t \in I \cap \gamma^{-1}[U]$, \begin{align} \begin{cases} (q^i \circ \gamma)'(t) = \dfrac{\partial H}{\partial p_i}\bigg|_{\gamma(t)} &\equiv \partial_{n+i}(H \circ \chi^{-1})_{\chi(\gamma(t))} \\\\ (p_i \circ \gamma)'(t) = -\dfrac{\partial H}{\partial q^i}\bigg|_{\gamma(t)} &\equiv -\partial_{i}(H \circ \chi^{-1})_{\chi(\gamma(t))} \end{cases} \end{align} where $\equiv$ above means "same thing written more explicitly/in different notation".

Euler-Lagrange Equations:

Although you didn't ask about this, let me give a second illustration of my "general rule" above. The Euler-Lagrange equations are typically written as \begin{align} \dfrac{d}{dt} \dfrac{\partial L}{\partial \dot{q}^i} &= \dfrac{\partial L}{\partial q^i} \end{align}

This is particularly infamous for causing a lot of confusion on its meaning, especially with regards to "how can $\dot{q}^i$ and $q^i$ be treated as independent variables" and so on.

Anyway, here, the setup is that we have a smooth $n$-dimensional manifold $Q$ (almost always we assume we have a Riemannian/Lorentizian metric $g$ depending on what purposes we have in mind... but for now we don't need it). Next, we have a smooth function $L: TQ \to \Bbb{R}$. In this case, since everything is a function on manifolds, how can we even take time derivatives of $\dfrac{\partial L}{\partial \dot{q}^i}$?

Well, what we actually mean is that given a chart $(U, \chi)$ on $TQ$, whose coordinate functions we label as $\chi(\cdot) = \left( q^1(\cdot), \dots, q^n(\cdot), \dot{q}^1(\cdot), \dots, \dot{q}^n(\cdot)\right)$ a smooth curve $\gamma : I \to Q$ satisfies the Euler-Lagrange equations on the chart $(U, \chi)$ if the tangent lift $\gamma' : I \to TQ$ is such that for all $t \in I \cap (\gamma')^{-1}[U]$ and all $i \in \{1, \dots, n\}$,

\begin{align} \dfrac{d}{ds}\bigg|_{s=t} \left(s \mapsto \dfrac{\partial L}{\partial \dot{q}^i}\bigg|_{\chi(\gamma'(s))}\right) &= \dfrac{\partial L}{\partial q^i} \bigg|_{\chi(\gamma'(t))} \end{align} which being even more explicit means that \begin{align} \dfrac{d}{ds}\bigg|_{s=t} \left( s \mapsto \partial_{n+i}\left( L \circ \chi^{-1}\right)_{\chi(\gamma'(s))} \right) &= \partial_{i}\left( L \circ \chi^{-1}\right)_{\chi(\gamma'(t))}. \end{align}

BTW, here $\gamma': I \to TQ$ obviously doesn't mean the limit of a difference quotient, since the target space of $\gamma$ is a manifold $Q$. In this context, for each $t \in I$, $\gamma'(t) := T\gamma_t(1_t) \in T_{\gamma(t)}Q$ is the tangent mapping/push-forward mapping of $\gamma$ at $t$, namely $T\gamma_t : T_tI = T_t\Bbb{R} \to T_{\gamma(t)}Q$, applied to the unit tangent vector $1_t \in T_t \Bbb{R}$.

As you can see, the notation gets pretty cumbersome very easily, and it requires a LOT of words to define all the objects before hand. This is partly why people often use the coordinate functions $q^i, \dot{q}^i, p_i$ with two meanings: sometimes they really mean it as functions from an open subset of a manifold into $\Bbb{R}$, but sometimes they mean it as a "function of time" by composing with a curve which maps from an interval into a manifold.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.