Role of immobile rigid atoms in pressure calculations

"From what I understand, in your system, you initially had a non-rigid
immobile sheet, i.e. the sheet was fixed in space but the atoms
constituting the sheet were not immobile"

No. atoms were immobile as they were not time integrated. e.g. let us
say sheet is made up of 2 type of atoms type 3 and type 4. Then
following commands were given

group nonMobile type 3 4
group mobile subtract all nonMobile
fix 1 mobile npt temp 300 300 100 z 1.0 1.0 1.0 dilate mobile

This gives pressure up to 6000 atm. Just tried delete_bonds, that was
also of no help. After delete bonds, system does expand a little bit
but still far from required density/pressure.


group nonMobile type 3 4
group mobile subtract all nonMobile
fix sheet nonMobile rigid single
fix 1 mobile npt temp 300 300 100 z 1.0 1.0 1.0 dilate mobile

results in rapid expansion of system, and reaches 'closest' to what
should be actual pressure. Only few hundred atm off!

so my main worry is the physics of my simulation. fix rigid discards
internal degree of freedom for the sheet, but as sheet is immobile and
only interaction sheet has with system is pairwise lj and coulombic
interaction, should it matter so much?
Now i want to relax my system by increasing volume and taking to
desired pressure,should i do it with just non time integrated sheet,
fix nvt mobile temp 300 300 100
or do it with fix rigid?
fix sheet nonMobile rigid single
fix nvt mobile temp 300 300 100

Also I am not sure why internal degree of freedom of a rigid body
shall influence any external particle? As per my understanding, motion
of each particle is governed by the force it feels.
What difference it should make if a immobile particle is having
internal strain? So if my sheet is not moving, rigid or not, results
shall be similar (whether they are right or wrong is separate), should
they not be?

Not following all the details of these long emails.

For rigid bodies, you should typically turn off

pairwise interactions within the body, so that they

don’t contribute to the pressure. The dynamics of

the bodies are the same regardless. You can turn

off internal interactions via the neigh_modify exclude



Sorry for the delayed reply.
I am running lammps on gpu, neigh_modify exclude gives error that it
cannot be run on gpu. (LAMMPS version August 2016, ICMS branch).
My labmate achieved equivalent by setting force coefficients of pair
wise interaction between atoms of rigid body to zero. I tried same by
giving sheet as fix rigid, which seem to ignore internal interactions
then managing pressure manually by fix deform + fix nvt and p-v curve
of system.

My main confusion was that in Hamiltonian for constant pressure
simulation we only have terms regarding momentum and temperature
(which is calculated from kinetic energy), then why internal energy of
a rigid body shall have an affect on final pressure. But i think it is
more of a question of MD rather than LAMMPS. :confused:

Thank you for your time.

If you have low internal energy (non-overlapped atoms),
then I think the virial term from interactions within
the body will cancel with the rigid body constraint virial,
So that the pressure is no different whether you include
or exclude the internal interactions.

But for highly overlapped, the internal energy and virial
may be huge and round-off errors from adding many
huge (nearly canceling) numbers may lead to inaccuracies.