[Pw_forum] about example 03 (Al relaxation)
baroni at sissa.it
Fri Feb 4 09:07:35 CET 2005
Dear Leonardo: all what has been said is correct. Let me add a general
theoretical consideration which may possibly help clarify the matter.
In any finite system, vanishing forces on individual atoms are a
NECESSARY and SUFFICIENT condition for mechanical equilibrium (whether
the equilibrium is stable or not depends on the second derivatives of
the energy: squared phonon frequencies in our case).
In an infinite system (such as a surface), instead, fanisihing forces
are a NECESSARY, but NOT SUFFICIENT, condition for mechanical
equilibrium. As strange as this fact might seem, it is indeed rather
intuitive. Think of a textbook linear harmonic chain with
first-neighbor interactions. If all the nearest-neighbor distances are
equal, then the forces acting on individual "atoms" will vanish.
Nevertheless, the energy per atom of the system will obviously depend
on such distances, and in general will be minimal for a well determined
value of them, only. If you cut the linear chain (you create a
"surface"), then the degrees of freedom perpendicular to the surface
(the "z" axis) will relax. In a 3D system, the degrees of freedom
parallel to the surface may or may not relax, depending on structure of
the surface. If the surface unit cell has only one atom (or else if its
symmetry is sufficiently high), then the parallel degrees of freedom
will not relax, even though the surface is not in mechanical
What is missing, to define the SUFFICIENT conditions for mechanical
equilibrium in an infinite system is the STRESS (whose concept is a
generalization of that of PRESSURE). This, however, wil be the subject
of a different thread ...
On Feb 3, 2005, at 7:04 PM, Cyrille Barreteau wrote:
> If you take a less symmetric surface such as a vicinal
> surface or a reconstructed (missing row reconstruction for
> example) then you will find x,y,z relaxation...
> cyrille barreteau
>> Aritz Leonardo wrote:
>>> I have a short question. I am trying to relax Magnesium surfaces in
>>> the same way as example03 for aluminium.
>>> When I run the example what I don't understand is that only
>>> Z-direction is relaxed. How come that atoms do not relax moving in X
>>> and Y direction? As far as I understand from the input, there are not
>>> fixed directions of atom positions so in principle they should all
>>> move in 3 directions.
>> symmetry (which is found by the code and preserved during relaxation)
>> is the reason
>> stefano deGironcoli
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>> Pw_forum at pwscf.org
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> Pw_forum at pwscf.org
Stefano Baroni --- SISSA & DEMOCRITOS National Simulation Center
via Beirut 2-4 34014 Trieste Grignano / [+39] 040 3787 406 (tel) -528
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