[gmx-users] constraint force

[Kay Gottschalk]

Hi all,

I wanted to calculate the constraint force between two molecules 
separated at a certain distance. The molecules should be free to rotate 
around their com, but not to change the com-com distance. I thought, 
the pull code could do the trick. Somehow, the manual is not so clear 
to me, therefore I tried different setups to figure out what is 
happening. After installing 3.2.1, I finally got some pull-code version 
running without crashing, but only if I use the following setup:

verbose = no
runtype = constraint
group_1 = chA
reference_group = chB
reftype = com
constraint_direction = 1.0 1.0 1.0
pulldim = Y Y Y.

The force is fluctuating (+/- 1000) strongly, but I found in the list 
that this is expected (due to some collisions with solvent??). If I use 
pulldim = N N N, no force is recorded (that makes sense, I think, since 
this probably means that no pulling force is applied, i.e. that the 
system is free to move). What puzzles me is the meaning of 
constraint_direction. If I use 'constraint_direction =  0 0 0' (which 
as far as I understand should mean that the distance between chA and 
chB should be restrained to the value found in the input structure), 
the run crashes immediately with first 'Warning: pressure scaling more 
than 1%', (I use berendse coupling, ref_p = 5), then 'relative 
constraint deviation after LINCS:', while if I use 
'constraint_direction = 1.0 1.0 1.0 (which as far as I understand means 
that the vector between the two com is somehow restrained??), 
everything seems to run fine. I am not sure I understand the meaning of 
constraint_direction and the difference between 0 0 0 and 1 1 1. IF 
what I understand is correct, something like 0 0 1 would allow chA to 
move freely in x and y, but be restrained in z. 1 1 1 would therefore 
mean to be restrained in xyz, which should keep the distance identical 
and therefore should be similar to 0 0 0. But somehow that doesn't seem 
to be the case. I'd be thankful for insight and enlightenment.
Thanks,
Kay.