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In Verlet integration, why is $a*dt*dt$ considered "force vector"?

Intuitively, since Verlet integration is:

$$x_{i+1}=x_i+(x_i - x_{i-1}) + a * dt * dt$$

then this doesn't have other "force" terms than $a * dt *dt$. However I'm unsure as to what the $dt*dt$ does here.

mavavilj
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    Use a velocity Verlet or leapfrog Verlet variant to reduce the method to difference order one steps. This might also be more stable numerically for small $dt$. – Lutz Lehmann Aug 24 '19 at 08:27

1 Answers1

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It takes the centered difference approximation to the second derivative of $x$, namely $\frac{x_{i+1}-2x_i+x_{i-1}}{\Delta t^2}$, and sets it equal to $a(x_i)$ (implicitly assuming a force that is dependent only on position), and then solves for $x_{i+1}$.

Ian
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  • Okay, but this doesn't explain why force is $adtdt$ rather than just $a$? Also if "force is dependent on position", then why aren't the $x$ terms considered "force"? – mavavilj Aug 24 '19 at 08:46
  • @mavavilj The contribution of force to the change in the position after just one time step of that force being applied should be proportional to $\Delta t^2$. The interesting thing is actually that Verlet integration is using $a \Delta t^2$ instead of $\frac{1}{2} a \Delta t^2$. This is an improvement (for nice $x(t)$) because of the error in extrapolating $x_i-x_{i-1}$ as the approximation for $x'(t_i)$. – Ian Aug 24 '19 at 13:30