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I am trying to apply a time integration technique for a system of discrete particles using Hamiltonian symplectic mechanics.

In his relatively well-known paper (CERN,PDF) on symplectic integrators, Ruth gives a few integration methods of various order's accuracy. They all apply the rule for the coordinates that $$x = x_i + p t$$ or some variant of that where momentum is integrated directly as a velocity. I notice that his kinetic energy term is actually one mass unit too many with $p^2/2$ but this should compound the problem rather than balance it out.

Is there some implicit assumption about unit mass here? When implementing these higher order methods in my own code I would just do the usual $$x=x_i +tp/m.$$ Is that still correct?

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    $\begingroup$ Likely just setting $m=1$ so that they didn't have to carry around a parameter they didn't need in the equations. $\endgroup$
    – Andrew
    Commented Dec 25, 2022 at 17:41
  • $\begingroup$ This question is similar to Peskin and Schroeder, where is the mass in the denominator of the simple harmonic oscillator Hamiltonian?. It is just notation, but it might be worth an answer explaining how operators can be redefined to make equations simple but their interpretation confusing for novice readers. For example, I was certainly confused the first time I saw the harmonic oscillator Hamiltonian written as $H=P^2+X^2$, not realizing that the author was using $X=\sqrt{m/2}\omega x$ and $P=p/\sqrt{2m}$. $\endgroup$ Commented Dec 25, 2022 at 19:32
  • $\begingroup$ yes but $x = x + pt$ strikes me as the undiscerning error of a undergrad student rather than the brilliant work of a particle physicist. I'd have thought the distinction between Hamiltonian and Lagrangian mechanics were to be emphasized not blurred. $\endgroup$ Commented Dec 25, 2022 at 20:21
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    $\begingroup$ Only slightly paraphrasing Humpty-Dumpty “When a theorist uses a symbol, it means what they choose it to mean – neither more nor less.” I certainly don't disagree with your irritation. I have felt it many times myself when trying to figure out a paper's notation. $\endgroup$ Commented Dec 25, 2022 at 20:47
  • $\begingroup$ Yes it certainly doesn’t make a difference to the paper‘s purpose but it does feel like an abuse of something. Perhaps one could show that such a momentum variable could be a Legendre transform of the velocity. $\endgroup$ Commented Dec 26, 2022 at 23:51

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As Andrew states in a comment, the author is simply setting $m=1$ to avoid excess notational clutter.

yes but $x=x+pt$ strikes me as the undiscerning error of a undergrad student rather than the brilliant work of a particle physicist.

A more advanced physicist tends to be aware of when a particular symbol is crucial to the context in which it appears and when it is not. In this case, the presence of $m$ is irrelevant to the algorithm; if you wish to include it, it can be easily replaced via dimensional analysis.


Alternatively, note that the equations of motion for a particle with Hamiltonian $H:= p^2/2m + V(x,t)$ are $$\matrix{\dot x = p/m \\\dot p = -\partial V/\partial x}$$ If we define a new time variable $\tau \equiv t/m$ and a scaled potential $\tilde V \equiv m V$ and denote differentiation with respect to $\tau$ with a prime rather than a dot, the equations of motion become $$\matrix{x' = p\\ p' = -\partial \tilde V / \partial x}$$

Once we've used the algorithm to obtain $x(\tau)$, we have all of the information we need to reconstruct the trajectory as a function of $t$ (e.g. the position at time $t$ is $x(t/m)$).


While it may not appear this way to you at the moment, extensive experience with dimensional analysis makes the equivalence between these approaches about as trivial as the equivalence between $6/8$ and $3/4$. The presence of $m$ is irrelevant for understanding how the algorithm works, so it can be omitted without fear of ambiguity; though it may be not so obvious, it can be replaced with no difficulty either by dimensional analysis (replace $pt$ with $pt/m$ while writing your code) or by implicitly scaling the variables.

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  • $\begingroup$ thanks. I guess this works for multiple particle masses too except we would be differentiating with different time scales so could be tricky to get interactions right $\endgroup$ Commented Dec 26, 2022 at 13:15
  • $\begingroup$ @Notachance Multiple particle masses would be problematic because dimensional analysis could not distinguish between them, so in that case you could set at most one mass equal to $1$. You'd need to also include the relative mass $m/M$ of the other in the Hamiltonian/equations of motion. $\endgroup$
    – J. Murray
    Commented Dec 26, 2022 at 14:32

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