From hinsen: at :ibs.ibs.fr Wed Nov 6 04:11:34 1996 Received: from ibs.ibs.fr for hinsen#* at *#ibs.ibs.fr by www.ccl.net (8.8.2/950822.1) id EAA07660; Wed, 6 Nov 1996 04:07:31 -0500 (EST) Received: from lmshp2.ibs.fr (lmshp2.ibs.fr [192.134.36.228]) by ibs.ibs.fr (8.6.12/8.6.12) with SMTP id KAA05266; Wed, 6 Nov 1996 10:08:38 +0100 Message-Id: <199611060908.KAA05266-!at!-ibs.ibs.fr> Received: by lmshp2.ibs.fr (1.37.109.4/16.2) id AA08126; Wed, 6 Nov 96 10:08:38 +0100 Date: Wed, 6 Nov 96 10:08:38 +0100 From: Konrad Hinsen To: rochus #at# felix.anorg.chemie.tu-muenchen.de Cc: chemistry-!at!-www.ccl.net In-Reply-To: <9611051739.ZM23541%!at!%felix> (rochus[ AT ]felix.anorg.chemie.tu-muenchen.de) Subject: Re: CCL:MD/MM combination > Let's consider the following example: > We have a rather flexible molecule which has functional groups allowing the > formation of hydrogen bonds. Now we put it in a box of water molecules and > calculate the potential energy of the system with a forcefield. If we do an MD > calculation with a fixed geometry of the molecule (just the solvent is moving) > we'll get an ensemble and we can average the potential energy. Sure, but I don't see what this "averaged potential energy" is good for. It is not a well-defined thermodynamic quantity. > Question 2: > If I calculate all internal forces on the atoms of the molecule and average all > forces excerted by the solvent on the molecule, I'll get the gradient on that > potential energy surface mentioned above. (Right?) Not quite. What you described above is the calculation of some energy for isolated configurations, not a surface. The average force is (by definition) the gradient of the potential of mean force, which is a well-defined thermodynamic quantity. > Question 3: > As far as I understood the Thermodynamic Integration method and related > methods, the free energy difference can be calculated by defining a reaction > coordinate and integrating over the averaged forces along that reaction > coordinate. Right, and that will give you the profile of the potential of mean force along the reaction coordinate. > I'm not quite clear, where this deviates from the aproach above. If I asume the The difference is that you have to calculate a whole path between two configurations, whereas your original idea was to calculate something for only the two end points. A separate calculation of the two end points means that you have an unknown additive constant for *each* of them, and hence no information about the difference in free energy. By calculating the path you end up with only *one* unknown additive constant, and you can evaluate the energy. -- ------------------------------------------------------------------------------- Konrad Hinsen | E-Mail: hinsen(+ at +)ibs.ibs.fr Laboratoire de Dynamique Moleculaire | Tel.: +33-76.88.99.28 Institut de Biologie Structurale | Fax: +33-76.88.54.94 41, av. des Martyrs | Deutsch/Esperanto/English/ 38027 Grenoble Cedex 1, France | Nederlands/Francais -------------------------------------------------------------------------------