forked from lijiext/lammps
129 lines
4.5 KiB
HTML
129 lines
4.5 KiB
HTML
<HTML>
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<CENTER><A HREF = "http://lammps.sandia.gov">LAMMPS WWW Site</A> - <A HREF = "Manual.html">LAMMPS Documentation</A> - <A HREF = "Section_commands.html#comm">LAMMPS Commands</A>
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<HR>
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<H3>fix uniaxial command
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</H3>
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<P><B>Syntax:</B>
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</P>
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<PRE>fix ID group-ID uniaxial N keyword dim amount
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</PRE>
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<UL><LI>ID, group-ID are documented in <A HREF = "fix.html">fix</A> command
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<LI>uniaxial = style name of this fix command
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<LI>N = perform uniaxial rescaling every this many timesteps
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<LI>dim = <I>x</I> or <I>y</I> or <I>z</I>
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<LI>strain = uniaxial strain in dim (2.0 = 2x larger)
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</UL>
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<P><B>Examples:</B>
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</P>
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<PRE>fix 1 all uniaxial 100 x 2.0
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</PRE>
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<P><B>Description:</B>
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</P>
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<P>Enable a uniaxial dilation/contraction of the simulation box during a
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simulation. For example if the direction is X and the strain is 2,
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then the final box size is 2L, L/sqrt(2), L/sqrt(2), where L**3 is a
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cube with the same volume as the initial box, which need not be cubic.
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</P>
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<P>The chosen direction is ramped linearly during the course of the run
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to its final value. The <A HREF = "run.html">run</A> command documents how to make
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the ramping take place across multiple runs.
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</P>
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<P>If the two perpendicular box sizes are equal then the deformation
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pathway is uniaxial at each timestep. If the two perpendicular box
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length sizes differ, then their aspect ratio will be linearily ramped
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down to 1. Irregardless of the initial box shape the total volume is
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constant during the deformation. Additional details provided by
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Carsten Svaneborg (Max Planck Institute for Complex Systems, Dresden,
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Germany) who authored this fix, are at the bottom of this page.
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</P>
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<P>The initial simulation box boundaries at the beginning of a run are
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specified by the <A HREF = "create_box.html">create_box</A> or
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<A HREF = "read_data.html">read_data</A> or <A HREF = "read_restart.html">read_restart</A> command
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used to setup the simulation, or they are the values at the end of the
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previous run. Every Nth timestep during the run, the various
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dimensions are expanded or contracted. The coordinates of all atoms
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in the group are also scaled to the new box size.
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</P>
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<P><B>Restrictions:</B>
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</P>
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<P>To use this fix, all dimensions of the system must be periodic.
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</P>
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<P><B>Related commands:</B> none
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</P>
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<P><B>Default:</B> none
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</P>
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<P><B>Extra Notes:</B>
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</P>
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<P>The uniaxial deformation is performed as follows:
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</P>
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<P>For notational simplicity the deformation is assumed to be in
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the X direction with final strain lambda. alpha denotes an
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arbitrary Cartesian direction.
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</P>
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<P>The initial strain is obtained from box dimensions:
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</P>
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<PRE>lambdai_alpha = Box(alpha)/power(Box(0)*Box(1)*Box(2),1/3)
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</PRE>
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<P>The final strain lambda in dir is specified:
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</P>
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<PRE>lambdaf_x = lambda
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lambda_y = lambda_z = 1/sqrt(lambda)
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</PRE>
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<P>Volume conservation implies lambda_x(t)*lambda_y(t)lambda_z(t) = 1.0
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for all times. Rather than time, let delta is denotes reduced time in
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the interval from 0 to 1.
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</P>
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<P>We want a linear ramp in the specified strain component, such that MD
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time steps and uniaxial strain are linearly related:
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</P>
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<PRE>lambda_x(delta) = lambdai_x (1-delta) + lambdaf_x
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</PRE>
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<P>The problem that remains is to choose a deformation pathway for
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lambda_y(delta) and lambda_z(delta) that agrees with the initial and
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final strains, and at all times conserves volume. Secondly the pathway
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should be symmetric if the box has y<->z symmetry.
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</P>
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<P>In the case where the initial box is symmetric in yz, this follows
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from volume conservation:
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</P>
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<PRE>lambda_y(d) = lambda_z(d) = 1/sqrt(lambda_x(d))
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</PRE>
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<P>However, in general the initial box dimensions in the y and z directions
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need not be the same so assume a relation:
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</P>
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<PRE>lambda_y(d) = alpha(d)lambda_z(d)
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</PRE>
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<P>From volume conservation it follows that
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</P>
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<PRE>lambda_y(d) = sqrt(alpha(d)/lambda_x(d))
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lambda_z(d) = 1/sqrt(alpha(d)*lambda_x(d))
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</PRE>
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<P>The asymmetry parameter has to fulfill the following boundary
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conditions:
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</P>
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<PRE>initial alpha(d=0) = lambdai_y/lambdai_z
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final alpha(d=1) = 1
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</PRE>
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<P>Any interpolation that does this will by define a continuous volume
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conserving deformation from the initial to the desired final state.
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The freedom of choice here is e.g. to relax the asymetry of the box
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very fast, and then slowly elongate along x, or to do this more
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slowly.
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</P>
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<P>The choice used in the code is:
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</P>
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<PRE>alpha(d) = lambdai_y/lambdai_z (1-d) + d
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</PRE>
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<P>Note in some cases like strain <1 2 0.5> -> strain <2 0.707107
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0.707107> the perpendicular strains do not follow a monotonic curve.
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</P>
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