On Thu, Apr 2, 2015 at 3:06 PM, Brian Radak <brian.radak.accts.gmail.com>
wrote:
> Can you actually do that or are you off by the 1 time step as usual due to
> leapfrog? This appears to still be the case with amdlog files too.
>
> I would likely accept the argument that the difference is below statistical
> error, but the imin = 5 solution is guaranteed to be correct and not
> particularly expensive, especially if your print step is 10^3-10^4 times
> larger than your time step.
>
​True, the trajectories are 1 timestep off (but not because of leapfrog --
it's because the snapshots are printed after they are propagated; this
could easily be fixed by moving the trajectory write *before* the
propagation step). If you need the energies to correspond to the snapshots
in the trajectory (rather than just a time series of energies), the imin=5
solution is what you need to do.
Thanks for the catch,
Jason
--
Jason M. Swails
BioMaPS,
Rutgers University
Postdoctoral Researcher
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Received on Thu Apr 02 2015 - 13:30:03 PDT
