Dear forum members,
I want to find the minimal enthalpy under application of (not necessarily hydrostatic) stress.
In the description of the input variable "strtarget" I find:
"The optimization ... will target the stress tensor defined by by strtarget,...
...Presently, this required target stress is not taken into account for the determination of the symmetries...
...Also, presently, the thermodynamical potential to be used in this situation (the free energy) does not replace the total energy, so that, for exemple, ionmov=3 cannot be used, since this algorithm is taking into account the total energy."
From the first statement I conclude that either the target functional contains the stress or the total energy is target und minimized under constraint.
From the second statement I conclude that I should better calculate with nsym=1 ?
The third statement confuses me: the thermodynamic potential used is called "the free energy" (Helmholtz free energy ? Where is stress then? And what is meant with: does not replace the total energy ?
Who knows what is actually implemented in abinit ?
- target functional = Etot and minimization under constraint or
- target functional enthalpy and minimization without constraint
- can I impose symmetry or do I have to set nsym=1 to get reliable results ?
Alfred Kersch
University of Applied Sciences,
Munich, Germany
relaxation under stress with abinit
Moderator: bguster
Re: relaxation under stress with abinit
Hello,
the correct Free energy functional in this case contains a stress*strain term, to account for fixed pressure instead of fixed volume.
You don't need to restrict to nsym 1, but if (e.g.) you start with a cubic system, and impose a non-isotropic stress, the initial symmetries found will be the cubic ones instead of the correct ones (e.g. tetragonal). You should break the symmetries in the way you want, from the beginning.
Matthieu
the correct Free energy functional in this case contains a stress*strain term, to account for fixed pressure instead of fixed volume.
You don't need to restrict to nsym 1, but if (e.g.) you start with a cubic system, and impose a non-isotropic stress, the initial symmetries found will be the cubic ones instead of the correct ones (e.g. tetragonal). You should break the symmetries in the way you want, from the beginning.
Matthieu
Matthieu Verstraete
University of Liege, Belgium
University of Liege, Belgium