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On page 188 of Abraham and Marsden Foundations of classical mechanics, how "by construction" does \begin{equation} i_{X_{H}}dq^i=\frac{\partial H}{\partial p_i}\ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ i_{X_{H}}dp_i=-\frac{\partial H}{\partial q^i} \end{equation} I don't know if this is an obvious fact of the mathematics (Im a chemist not a mathematician or physicist) but it seems that without this fundamental piece of information then saying $i_{X_{H}}\omega=\omega (X_{H},\cdot )=dH$ doesn't make sense on its own. Thank you for your help !

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First, $i_{X_H} dq^i = dq^i(X_H)$. (This is the definition of what the operation $i_{X_H}$ means).

Next, $dq^i(X_H)$ is the component of the vector field $X_H$ in the direction of the basis vector $\partial/\partial q^i$. (This is the definition of what the one-form $dq^i$ does when you feed it a vector field.)

And since they start the proof on the previous page by saying "Let $X_H$ be defined by the formula (3)", we can just look at that formula (which is given in terms of canonical coordinates): $$ X_H = \left( \frac{\partial H}{\partial p_i},-\frac{\partial H}{\partial q^i} \right) = J \cdot dH ,\tag{3} $$ or, if we write out exactly what they mean by this, $$ X_H = \sum_{i=1}^n \frac{\partial H}{\partial p_i} \, \frac{\partial}{\partial q^i} + \sum_{i=1}^n \left(-\frac{\partial H}{\partial q^i}\right) \, \frac{\partial}{\partial p_i} . $$ Here we see that the $\partial/\partial q^i$ component of $X_H$ is indeed by construction $\partial H/\partial p_i$.

Similarly for the other one.

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  • $\begingroup$ Thank you so much, I really appreciate it! $\endgroup$ – AngusTheMan Mar 22 '15 at 21:34

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