Re: [AMBER] help in diffusion coefficient analysis

From: Jason Swails <>
Date: Fri, 23 Jan 2015 04:36:55 -0500

> On Jan 23, 2015, at 3:02 AM, Robin Jain <> wrote:
> Dear all,
> As a test run, we are trying to compute diffusion constant of water.
> We have taken 64 water molecules. I am confused about the "center" and
> "image" command. Using the following command, we are getting the value
> 2.918 and without using the two commands we are getting a different
> value (2.026). kindly suggest.

Try the commands in isolation and look at the resulting trajectories -- this should *show* you what the commands do.

> trajin X.mdcrd 1 last 1
> unwrap :1-64

This “unwraps” a trajectory, so that trajectories trace the paths of specific water molecules (rather than the paths of all molecules within a particular unit cell, which will track *different* molecules as they enter and leave that unit cell).

> center :1-64 mass origin
> image origin center familiar

This wraps all of the water molecules *back* into the same unit cell -- i.e., it reverses the “unwrap” command above.

> diffusion 0.25 diff.out :1-64 average

An imaged trajectory will always give diffusion constants larger than (or equal to if you are lucky) unimaged (or unwrapped) trajectories. The reason for this is that when wrapping occurs, a single water molecule appears to *jump* across the periodic box. Obviously this discontinuous jump introduces a fictitiously large “movement” (which is about the size of the periodic box), which makes particles *seem* like they are moving farther than they really are.


Jason M. Swails
Rutgers University
Postdoctoral Researcher
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Received on Fri Jan 23 2015 - 02:00:03 PST
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