# Hamilton equations and phase space

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 $$\gamma(0)=m$$)?

• 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. – QuantumBrick Jan 9 at 21:01
• 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. – Mathphys meister Jan 9 at 21:14
• 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 $t$. – QuantumBrick Jan 9 at 21:19
• $\gamma$ cannot depend on 3 variables, it maps from $\mathbb{R}$. That does not make sense – Mathphys meister Jan 9 at 21:39
• 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: $\gamma: T^*M \times \mathbb{R} \to T^*M$ – QuantumBrick Jan 9 at 21:49

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.