Dynamic group with atoms such that Ek > Eo

Dear all,

I have a system where I would like only to follow the evolution of atoms that have a kinetic energy > Eo. Would it be possible to create a dynamic group (called “interesting” eg) only containing these atoms?
I would like to have a dynamic group since atoms can gain or loose energy by interactions and can thus have sometimes Ek > Eo and sometimes Ek < Eo.

Then, to follow the evolution of atoms in this group I would use a fix id interesting nve, and would ignore the evolution of atoms that are not in this group.

Is it possible to create such dynamic group with the condition on the kinetic energy of atoms?

Many thanks in advance and best regards,
Christophe

Dear all,

I have a system where I would like only to follow the evolution of atoms
that have a kinetic energy > Eo. Would it be possible to create a dynamic
group (called "interesting" eg) only containing these atoms?

​yes. you can build dynamic groups based on an atom style variable and such
variables can access or compute all kinds of per-atom​ properties.

I would like to have a dynamic group since atoms can gain or loose energy
by interactions and can thus have sometimes Ek > Eo and sometimes Ek < Eo.

Then, to follow the evolution of atoms in this group I would use a fix id
interesting nve, and would ignore the evolution of atoms that are not in
this group.

​now, this part doesn't make any sense to me. what would be the benefit of
this? and how would that be physically meaningful?

axel.

Great! Thanks!

I knew you would say that :-D! I have some idea I want to test…

Christophe

I did it. It seems to work. I can see the number of atoms belonging to the group evolves with time.
This when I do fix all nve. See below…

Little problem, that I actually expected.
When I do
fix 1 interesting nve

it means that atoms that are not in the dynamic group ‘interesting’ are not included in the evolution.
Hence, these atoms do not interact with ‘interesting’ atoms, they do not receive any energy from them. I observe thus that the number of ‘interesting’ atoms stays constant. It does not vary with time, which is not physical.

Is it posible to solve this?
In other words, though ‘non-interesting’ atoms are not included in the fix nve, is it possible to make they interact with atoms from ‘interesting’ group, and receive some energy from them such that they can be added to the dynamic group ‘interesting’?

With best regards,
Christpohe

Dear all,

I have a system where I would like only to follow the evolution of atoms
that have a kinetic energy > Eo. Would it be possible to create a dynamic
group (called "interesting" eg) only containing these atoms?

​yes. you can build dynamic groups based on an atom style variable and
such variables can access or compute all kinds of per-atom​ properties.

I did it. It seems to work. I can see the number of atoms belonging to the
group evolves with time.
This when I do *fix all nve*. See below...

I would like to have a dynamic group since atoms can gain or loose
energy by interactions and can thus have sometimes Ek > Eo and sometimes Ek
< Eo.

Then, to follow the evolution of atoms in this group I would use a fix
id interesting nve, and would ignore the evolution of atoms that are not in
this group.

​now, this part doesn't make any sense to me. what would be the benefit
of this? and how would that be physically meaningful?

Little problem, that I actually expected.
When I do
*fix 1 interesting nve*

it means that atoms that are not in the dynamic group 'interesting' are
not included in the evolution.
Hence, these atoms do not interact with 'interesting' atoms, they do not
receive any energy from them. I observe thus that the number of
'interesting' atoms stays constant. It does not vary with time, which is
not physical.

Is it posible to solve this?

​no. what you propose is unphysical and makes no sense. your explanations
are incorrect, too.

In other words, though 'non-interesting' atoms are not included in the fix
nve, is it possible to make they interact with atoms from 'interesting'
group, and receive some energy from them such that they can be added to the
dynamic group 'interesting'?

​this is​ why i said what you were suggesting makes no sense. if you time
integrate only select atoms, you create an unphysical situation. the
kinetic energy of immobile atoms cannot change; it is the time integration,
that changes it. and more importantly, whether you time integrate atoms or
not has no impact *at all* on the computation of the forces.

axel.

Does that mean that if a selected atom comes close to an unselected one, the force between them will be computed? The unselected atom will feel the force (may the force be with it :-D) but its kinetic energy will not change because it is not in the group of the time integration.
Do I understand correctly?

Christophe

​[...]

Does that mean that if a selected atom comes close to an unselected one,
the force between them will be computed? The unselected atom will feel the
force (may the force be with it :-D) but its kinetic energy will not change
because it is not in the group of the time integration.
Do I understand correctly?

do i understand correctly, that you have never looked into a textbook on
MD? or if you did, that you have forgotten everything?

axel.

Christophe

​--​

Dr. Axel Kohlmeyer [email protected] http://goo.gl/1wk0
College of Science & Technology, Temple University, Philadelphia PA, USA
International Centre for Theoretical Physics, Trieste. Italy.

Does that mean that if a selected atom comes close to an unselected one,
the force between them will be computed? The unselected atom will feel the
force (may the force be with it :-D) but its kinetic energy will not change
because it is not in the group of the time integration.
Do I understand correctly?

That's exactly right. The unselected atoms will act like immobile
obstacles. If you want to omit them from force computations you would have
to change their type and make the interesting and uninteresting type not
interact, or somehow exclude them from neighbour list builds. fix nve is
just doing time integration (i.e., changing position and velocity) _given_
the forces. It does not calculate the forces itself.

I am just trying to understand how you made LAMMPS. You are the one who knows all its secrets. What is sure, is that sarcasms do not need to be learnt in text books…and often you could keep them for you. When one writes to this mailing list, he expects some answers or guidance, and not sarcasms about his knowledge.

Christophe

do i understand correctly, that you have never looked into a textbook on
MD? or if you did, that you have forgotten everything?

I am just trying to understand how you made LAMMPS. You are the one who
knows all its secrets. What is sure, is that sarcasms do not need to be
learnt in text books...and often you could keep them for you. When one
writes to this mailing list, he expects some answers or guidance, and not
sarcasms about his knowledge.

​this has *nothing* to do with LAMMPS. this is about how MD works at the
very fundamental level. look at how a velocity verlet integrator works (and
that should be shown in any good text book) and you should see the point i
was making. there is no big secret here.

you were simply just repeating what i had explained to you as a question.
that is a waste of your and particularly my time.
just read what you get told more carefully. i have told you multiple times,
that what you are proposing doesn't make sense. that is a strong hint, that
you are about to do something that is not very smart.

when i am withholding details, it is often for a very good reason: this is
something that you should think about and need to understand all the way
through. not to mention. you have heard the saying: give a man a fish and
he is sated for the day, teach him how to fish, and he is sated for the
rest of his life.
well, something similar holds true for science as well. there are topics,
where it is better to make you think and figure it out by yourself. it
sometimes hurts at the beginning, but it pays off big time in the long run.
if it to work, you just have to pay more attention to details and take a
hint here or there. if you expected well formed and ready to consume answer
for even the most fundamental issues of the research you are doing, then
you have to re-think how you learn.

also, you have to realize that this mailing list is a voluntary effort and
you are in no way *entitled* to get any particular level of service here.
when people feel they are being abused, you have to expect some kind of
less helpful reaction. often this is done with the best intentions, as some
people seem to need to feel some level of shame or pain to have their
brains jolted to start thinking in the right way again.

if you ask well formed and meaningful questions, you will get carefully
worded and competent answers. if you ask superfluous questions, you will
either get no answer or something that tries to point out that you have
overstepped the bounds of what is worth asking.

in summary, what you are dealing with here is a quite common phenomenon in
open source software and there are essays like this one
http://www.catb.org/esr/faqs/smart-questions.html that try to explain it.

so get over it, go back and revisit and rethink what you were advised on,
and try to ask smarter questions the next time.

axel.

ok - further discussion on this thread should go off-line.

Chistophe is right, there’s no need for sarcasm on mail list replies.

Axel has given lots of useful feedback already on this thread and Qs.

Axel - when your patience runs out, please send a private email, not a public

posting.

And Stefan’s answer is the answer to the original Q - the integration fixes just update velocities and positions, the atoms not integrated are just immobile obstacles which exert

force but do nothing with the force on themselves.

Steve

ok - further discussion on this thread should go off-line.

Chistophe is right, there’s no need for sarcasm on mail list replies.

Axel has given lots of useful feedback already on this thread and Qs.

Axel - when your patience runs out, please send a private email, not a public

posting.

And Stefan’s answer is the answer to the original Q - the integration fixes just update velocities and positions, the atoms not integrated are just immobile obstacles which exert

force but do nothing with the force on themselves.

Thanks Steve. This leads to my following question. If the force between pairs is calculated (although not integrated atoms will not have their positions and velocities updated) is it possible to have access to the force value that is exerted on “not integrated atoms” ? Would compute pair/local force give this value?

Christophe

Just try dumping forces on all atoms (including the

non-integrated ones). I think they should be non-zero.

Steve