From: Jérôme Hénin (jerome.henin_at_ibpc.fr)
Date: Thu Aug 31 2017 - 10:43:12 CDT
It's certainly not formally proper to enable LangevinHydrogen with
rigidBonds, because as you said the number of degrees of freedom is not
However, when you do that, the part of the Langevin forces that get applied
to the X-H atoms is canceled by the constraint (along the bond direction),
so I wouldn't be surprised if in the end you magically arrived at exactly
the right force for the actual number of degrees of freedom in the system.
On 31 August 2017 at 17:31, Benjamin Rousseau <benjamin.rousseau_at_duke.edu>
> Hello all,
> I have seen this topic pop up a few times before, with mixed answers. I
> would like to understand the proper combination of the langevinHydrogen and
> the rigidBonds settings.
> I have seen recommendations to set langevinHydrogen to off when using
> rigidBonds (http://www.ks.uiuc.edu/Research/namd/mailing_list/
> as is done in many NAMD configuration files.
> However, to my understanding, this removes the requirement that the
> average kinetic energy associated with each rotational degree of freedom of
> hydrogen must be ~ kT/2.
> On the other hand, setting rigidBonds to all and langevinHydrogen to on
> imposes this requirement of KE=kT/2 for each degree of freedom of hydrogen.
> However, as one of the degrees of freedom is missing due to a rigid bond to
> a heavy atom, this results in a kinetic energy of ~ kT rather than ~ 3/2kT
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