[lammps-users] langevin - temperature decay

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1

Hi,

I am cooling my sytem from 3000K to 0K, using the following commands (metal units):

fix 1 coreatoms langevin 3000 0.0 0.06 48279
fix 2 coreatoms nve

and run for about 20 ps (0.1 fs timestep). I would need 2 clarifications about this procedure:

  1. By the end of my run, the temperature is not exactly 0K, but around 10K. Is it just a matter of increasing the number of steps, or modifying the damping parameter? (I am currently investigating this possibility)

  2. Also, I observed that the temperature is decreasing linearly during the run. I would have rather expected an exponential decay, driven by some tau. Is the linear decay always used with this fix command? And/or is there a way to modify the way this is done?

Thanks a lot

2

If there are non-zero forces at the end of your run, then
the T will be non-zero. Langevin won't really take out the
last bit of energy.

The target T at any time is a value linearly interpolated
from Tstart to Tstop. If you want to choose a target T
that depends on (Tstart, Tstop, time) in some other way,
e.g. exponential, then you'd need to modify the ocde or
write your own fix.

Steve

3

I see. I am working on a simulation of annealing, so I will test if this remaining energy isn’t too much of a problem.

Thanks (one more time) for your time and your answer.

2008/8/21 Steve Plimpton <[email protected]>