[gmx-users] Question about COM-Pulling

Christian Mücksch muecksch at rhrk.uni-kl.de
Mon Dec 13 22:17:10 CET 2010


Dear Chris,

thx a lot for your help!

Unfortunately, the information you provided is nowhere to be found in the manual.

With best wishes,
Christian


--------------------------------------------------------------
for your setup options, the spring potential is something like:

U=0.5*k*[(z_ref - z_pull) - (initial_dist + time*0.01)]^2

(please refer to the manual for more details, I don't do AFM and didn't check the exact functional form).

So the spring is not on the displacement but on the difference between the actual displacement and the desired displacement, where your desired displacement is changing over time.

Chris.

-- original message --


>/  >     Here you have time, the z position of your reference group and the
/>/  >     difference along z between your pull group and your reference group.
/>/  >     There is no "dummy particle" as the spring extends between the
/>/  >     "reference" group at z=1.1675 and the "pull" group, which in the first
/>/  >     frame is located at z=1.1675 + 3.47701.
/>/  >
/>/  >     I have used the pull code since v3.3.1 and I do not recall there ever
/>/  >     being a dummy particle when using:
/>/  >
/>/  >     pull = umbrella
/>/  >     pull_geometry = distance
/>/  >
/>/  >     It is the distance of the restraint that changes over time in your setup.
/Sorry to ask again but I'm a bit confused. I thought that a spring (harmonic potential)
is connected to the pull-group (by "dummy-particle" I meant the top of the spring=zspring)
and then moved at a certain pull_rate in the desired direction.

If the spring extends from the pull-group to the reference group and the reference group stays in place while the pull-group is moved away,
then the spring would extend and extend and therefore the force rise and rise. What obvious fact am I missing?

Is there a difference between AFM pulling and Umbrella pulling?

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