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This might be a stupid question, but I am having trouble understanding the derivation of Standard map by integrating Hamilton's equation of motion over one period. I am going through this dissertation (Daniel Adam Steck, “Quantum Chaos, Transport, and Decoherence in Atom Optics,” Ph.D. dissertation, The University of Texas at Austin (November, 2001)), and am not able to wrap my head around the equations (4.15) and (4.16). I might be missing something trivial but If $p$ is a function of time then how can we reduce it's integral over time to as mentioned in equation (4.16)?

That is, how is the following equation correct?

$$\int_{t_n-\epsilon}^{t_{n+1}-\epsilon}p\ dt = \epsilon p (t_n-\epsilon) + (1-\epsilon) p (t_{n+1}-\epsilon) $$

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  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. Also consider to write the mentioned eqs. to make the question self-contained. $\endgroup$
    – Qmechanic
    Commented Jan 11, 2019 at 7:14
  • $\begingroup$ Sudheesh, I've implemented Qmechanic suggestions for you in this question, please remember to follow them in future posts. $\endgroup$
    – stafusa
    Commented Jan 12, 2019 at 2:55

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You're forgetting that $p$ is constant between kicks.

Together with the fact that the (normalized) time between kicks is $t_{n+1}-t_n=1$, the figure bellow should make things clearer (I've used $p_n \equiv p(t_{n}), p_{n+1} \equiv p(t_{n+1})$, etc.).

enter image description here

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  • $\begingroup$ Thank you for amending the question, would keep the points in mind. It makes sense now thanks for explaining it! $\endgroup$
    – Sudheesh
    Commented Jan 12, 2019 at 6:14
  • $\begingroup$ Awesome! I had a trouble understanding exactly the same dissertaion with these equations. Is that we assume that we can normalize the time difference because of the fact that we are looking at the map instead of a continous equation? So that in map we only are interested in steps not in a real time? $\endgroup$ Commented Jul 6, 2023 at 15:02
  • $\begingroup$ @AndrisErglis It's just a convenient choice of units. In page 103 you find the explicit explanation: "Before proceeding, we will transform to scaled units to simplify our discussion". $\endgroup$
    – stafusa
    Commented Jul 7, 2023 at 7:36

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