From hinsen@ibs.ibs.fr Wed Nov 6 04:11:34 1996 Received: from ibs.ibs.fr for hinsen@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@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@felix.anorg.chemie.tu-muenchen.de Cc: chemistry@www.ccl.net In-Reply-To: <9611051739.ZM23541@felix> (rochus@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@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 ------------------------------------------------------------------------------- From hinsen@ibs.ibs.fr Wed Nov 6 05:11:33 1996 Received: from ibs.ibs.fr for hinsen@ibs.ibs.fr by www.ccl.net (8.8.2/950822.1) id EAA07801; Wed, 6 Nov 1996 04:27:44 -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 KAA05400; Wed, 6 Nov 1996 10:28:17 +0100 Message-Id: <199611060928.KAA05400@ibs.ibs.fr> Received: by lmshp2.ibs.fr (1.37.109.4/16.2) id AA08228; Wed, 6 Nov 96 10:28:17 +0100 Date: Wed, 6 Nov 96 10:28:17 +0100 From: Konrad Hinsen To: echamot@xnet.com Cc: rochus@felix.anorg.chemie.tu-muenchen.de, chemistry@www.ccl.net In-Reply-To: Subject: Re: CCL:MD/MM combination > Now I have a question. What "temperature" is this type of MD going to > simulate? Temperature normally just reflects the average kinetic energy of > the system, so it can be calculated from the velocities. But if part of the > system is frozen, should those atoms be included in the "average"? Strictly speaking, no. Equipartition of energy assigns an equal amount of kinetic energy to each degree of freedom in the system, but fixed coordinates are not degrees of freedom. > Obviously, if all atoms are included, it will add a bunch of zero > velocities, so the calculated temperature will be lower than what is > reflected by the movement of molecules actually allowed to move. But would If the fixed atoms ever constitute a significant part of your system, then you are too far away from the thermodynamic limit anyway. So it shouldn't make a difference in practice how you calculate the temperature. -- ------------------------------------------------------------------------------- Konrad Hinsen | E-Mail: hinsen@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 ------------------------------------------------------------------------------- From hrusak@ims.ac.jp Wed Nov 6 09:11:36 1996 Received: from ims1.ims.ac.jp for hrusak@ims.ac.jp by www.ccl.net (8.8.2/950822.1) id IAA09518; Wed, 6 Nov 1996 08:15:48 -0500 (EST) Received: from ikarus.ims.ac.jp by ims1.ims.ac.jp (8.6.8+2.4Wb2/TISN-1.3M/R2) id WAA20526; Wed, 6 Nov 1996 22:14:34 +0900 Received: by ikarus.ims.ac.jp with Microsoft Mail id <01BBCBCC.5E363B80@ikarus.ims.ac.jp>; Wed, 6 Nov 1996 10:22:09 +-900 Message-ID: <01BBCBCC.5E363B80@ikarus.ims.ac.jp> From: Jan Hrusak To: "'CHEMISTRY@www.ccl.net'" Subject: Help with G94 needed Date: Wed, 6 Nov 1996 10:22:07 +-900 Return-Receipt-To: Encoding: 24 TEXT Hi Netters, Doing geometry optimization with Gaussian 94 on a triatomic system I meet the following problem: After few gradients (analytical) the program stops with the message Error permuting atoms in Fill, LPerm : 0 0 0 Symmetry turned off : Internal error in symmetry package. Error termination via Lnk1e ....... Could somebody tell me what is going on and how to avoid this? The result does not change when using correlated, SCF or DFT method. Thanks Jan Hrusak hrusak@ims.ac.jp hrusak@jh-inst.cas.cz Web: http://zeus1.ims.ac.jp/Iwata/Hrusak/index.html From noy@einstein.sc.mahidol.ac.th Wed Nov 6 09:19:37 1996 Received: from einstein.sc.mahidol.ac.th for noy@einstein.sc.mahidol.ac.th by www.ccl.net (8.8.2/950822.1) id JAA09818; Wed, 6 Nov 1996 09:05:57 -0500 (EST) Received: (from noy@localhost) by einstein.sc.mahidol.ac.th (8.7.4/8.7.3) id UAA32384 for chemistry@www.ccl.net; Wed, 6 Nov 1996 20:52:52 -0700 From: "Dr. Teerakiat Kerdcharoen" Message-Id: <199611070352.UAA32384@einstein.sc.mahidol.ac.th> Subject: Re: CCL:MD/MM combination To: chemistry@www.ccl.net Date: Wed, 6 Nov 1996 20:52:51 -0700 (GMT+7) In-Reply-To: from "Ernest Chamot" at Nov 5, 96 06:25:37 pm Content-Type: text > Now I have a question. What "temperature" is this type of MD going to > simulate? Temperature normally just reflects the average kinetic energy of > the system, so it can be calculated from the velocities. But if part of the > system is frozen, should those atoms be included in the "average"? > Obviously, if all atoms are included, it will add a bunch of zero > velocities, so the calculated temperature will be lower than what is > reflected by the movement of molecules actually allowed to move. But would > the "temperature" calculated only by averaging the moving molecules have any > more meaning? I suppose that either way this is just an artificial > situation, but if there is something that corresponds to temperature, I hope > any answer to Rochus's question can also answer my question. Thanks. Hi! This point has confined me in a cycle of confusion for quite a long time. Traditionally, we calculate "temperature" from the ensemble average of kinetic energy. The kinetic energy is unambiguously defined as a square of velocity. Well, we assign thermodynamic temperature with a factor of boltzmann constant and the total degree of freedom of the system. Things are clear if we consider system as a system of N atoms moving only translationally and so there are 3N degrees of freedom. Things are more ambiguous when people try to devide motions into 3 modes, translation (of the centre of mass), rotation (about the C.O.M.) and vibration. Then they introduce the equipartition theorem to cope with this. They calculate translational and rotational energies separately and they have translational and rotational temperature that are believed to be equal due to the equipartition theorem (frequently they are much different in some systems that two modes of motion are not coupled then the macroscopic temperature is calculated by averaging translational and rotational degrees of freedom). In a system of rigid molecules, vibration is absent and then the degrees of freedom from vibration are not employed to calculate temperature. In a constraint method, system of rigid molecules can be consider as a system of flexible molecules with a set of constraint. For this, only translation motions of N atoms are involved but the degree of freedom is not 3N, but substracted by an amount of constraint. Your problem may be consider like that, the solute is fixed with constraint and so not used to calculate temperature. take care, Teerakiat ---------------------------------------------------------------------------- Teerakiat Kerdcharoen, Ph.D. Profession: Lecturer and Information Technology Consultant Address: Department of Physics, Mahidol University, Bangkok 10400 Phone: 2461381 FAX 2461381 Cellular: 01-4906089 E-mail: noy@einstein.sc.mahidol.ac.th, noy@atc.atccu.chula.ac.th Homepage: http://www.sc.mahidol.ac.th/noy/ Research: Computer Aided Molecular Design (CAMD) ----------------------------------------------------------------------------- From hinsen@ibs.ibs.fr Wed Nov 6 12:11:39 1996 Received: from ibs.ibs.fr for hinsen@ibs.ibs.fr by www.ccl.net (8.8.2/950822.1) id LAA10874; Wed, 6 Nov 1996 11:13:34 -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 RAA09057; Wed, 6 Nov 1996 17:14:31 +0100 Message-Id: <199611061614.RAA09057@ibs.ibs.fr> Received: by lmshp2.ibs.fr (1.37.109.4/16.2) id AA11295; Wed, 6 Nov 96 17:14:32 +0100 Date: Wed, 6 Nov 96 17:14:32 +0100 From: Konrad Hinsen To: noy@einstein.sc.mahidol.ac.th Cc: chemistry@www.ccl.net In-Reply-To: <199611070352.UAA32384@einstein.sc.mahidol.ac.th> (noy@einstein.sc.mahidol.ac.th) Subject: Re: CCL:MD/MM combination > Things are more ambiguous when people try to devide motions > into 3 modes, translation (of the centre of mass), rotation (about the > C.O.M.) and vibration. Then they introduce the equipartition theorem > to cope with this. They calculate translational and rotational energies And this division must be done very carefully. The equipartition theorem is valid *only* in a set of canonical coordinates and momenta. An important practical case of non-canonical variables are angular velocities (there are even no corresponding coordinates). It is therefore *not* allowed to simply assign equal kinetic energy to rotation about each of the principal axes of a molecule. However, you can use the equipartition theorem for rotation if you use some canonical cooridinate set to describe orientation, such as three Euler angles per molecule, plus the associated momenta. -- ------------------------------------------------------------------------------- Konrad Hinsen | E-Mail: hinsen@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 ------------------------------------------------------------------------------- From Jennie.Weston@org.chemie.uni-giessen.de Wed Nov 6 12:32:24 1996 Received: from hermes.hrz.uni-giessen.de for Jennie.Weston@org.chemie.uni-giessen.de by www.ccl.net (8.8.2/950822.1) id LAA11226; Wed, 6 Nov 1996 11:50:39 -0500 (EST) Received: from maier.org.chemie.uni-giessen.de by hermes.hrz.uni-giessen.de; Wed, 6 Nov 96 17:49:35 +0100 Received: by maier.org.chemie.uni-giessen.de (AIX 3.2/UCB 5.64/4.03) id AA13440; Wed, 6 Nov 1996 17:36:49 +0100 From: Jennie.Weston@org.chemie.uni-giessen.de (Jennie Weston) Message-Id: <9611061636.AA13440@maier.org.chemie.uni-giessen.de> Subject: help with nbo analysis in g94 To: CHEMISTRY@www.ccl.net Date: Wed, 6 Nov 1996 17:36:48 +0100 (CET) X-Mailer: ELM [version 2.4 PL23] Mime-Version: 1.0 Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit Hello everybody, I have a little problem with the nbo implementation in gaussian94. According to the NBO manual, in order to do matrix element deletions, i need to use a non-standard gaussian input stream. Well, i have an nbo manual version 3.0, which describes the implementation in gaussian88. The code has changed a little since then........ I managed to get the manual's example working for a normal nbo analysis, but when it enters the link to calculate the matrix deletion, my job dies wanting a g94-5262.nbo file. Does any Guru out there know how to induce the program to write this file? Or is there a simpler way to do this? A working example file would be AWFULLY nice....... Thanks.... J. Weston. Reply to Jennie.Weston@org.chemie.uni-giessen.de to save bandwidth Here's the error msg: Leave Link 303 at Wed Nov 6 17:29:24 1996, MaxMem= 6000000 cpu: 0.6 (Enter /home/u3/tmp/g94/l401.exe) Projected INDO Guess. open(/work/ge91/ge91.5257/g94-5262.nbo): No such file or directory apparent state: unit 48 named lately reading direct unformatted external IO *** Execution Terminated (2) *** --------- Here's my version of the manual's input example (See manual pg 98 for orig): it dies with 6/40=2/7(1); it works through 6/7=2,8=2,9=2,10=2,19=1,28=1,40=1/1,7; ---- $ RunGauss #P NONSTD 1/38=1/1; 2/12=2,17=6,18=5/2; 3/5=5,11=1,25=1,30=1/1,2,3; 4/7=1/1; 5/5=2,32=1,38=4/2; 6/7=2,8=2,9=2,10=2,19=1,28=1,40=1/1,7; 6/40=2/7(1); 99/5=1,9=1/99; 5/7=1,13=1/2; 6/40=3/7(-3); methylamin 0 1 c n 1 cn h 1 ch 2 tet h 1 ch 2 tet 3 120. 0 h 1 ch 2 tet 3 240. 0 h 2 nh 1 tet 3 60. 0 h 2 nh 1 tet 3 300. 0 cn 1.47 ch 1.09 nh 1.01 tet 109.4712 $NBO $END $DEL NOSTAR $END -------------------- From jasna@cucbs.chem.columbia.edu Wed Nov 6 13:11:39 1996 Received: from cucbs.chem.columbia.edu for jasna@cucbs.chem.columbia.edu by www.ccl.net (8.8.2/950822.1) id MAA11568; Wed, 6 Nov 1996 12:32:27 -0500 (EST) Received: from localhost (jasna@localhost) by cucbs.chem.columbia.edu (8.7.5/8.7.3) with SMTP id MAA15342 for ; Wed, 6 Nov 1996 12:32:26 -0500 (EST) Date: Wed, 6 Nov 1996 12:32:25 -0500 (EST) From: Jasna Klicic Reply-To: Jasna Klicic To: CHEMISTRY@www.ccl.net Subject: CCL: Gaussian94 versus Gaussian92 timings In-Reply-To: <01BBCBCC.5E363B80@ikarus.ims.ac.jp> Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Dear netters, I would like to know what is the speed-up of Gaussian94 versus the older version, Gaussian92 on IBM or SGI plarforms. I would mostly be interested in timings for HF/6-31g** type jobs, but other results will be welcome too. Thanks in advance for your help. -- Jasna Jasna Klicic Columbia University jasna@cucbs.chem.columbia.edu From walt@panix.com Wed Nov 6 14:11:39 1996 Received: from panix3.panix.com for walt@panix.com by www.ccl.net (8.8.2/950822.1) id NAA12067; Wed, 6 Nov 1996 13:40:02 -0500 (EST) Received: from localhost (walt@localhost) by panix3.panix.com (8.7.6/8.7/PanixU1.3) with SMTP id NAA14188; Wed, 6 Nov 1996 13:38:55 -0500 (EST) Date: Wed, 6 Nov 1996 13:38:51 -0500 (EST) From: Walter Polkosnik To: Konrad Hinsen cc: echamot@xnet.com, rochus@felix.anorg.chemie.tu-muenchen.de, chemistry@www.ccl.net Subject: Re: CCL:MD/MM combination In-Reply-To: <199611060928.KAA05400@ibs.ibs.fr> Message-ID: Errors-To: walt@panix.com MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On Wed, 6 Nov 1996, Konrad Hinsen wrote: > > Now I have a question. What "temperature" is this type of MD going to > > simulate? Temperature normally just reflects the average kinetic energy of > > the system, so it can be calculated from the velocities. But if part of the > > system is frozen, should those atoms be included in the "average"? > > Strictly speaking, no. Equipartition of energy assigns an equal > amount of kinetic energy to each degree of freedom in the system, > but fixed coordinates are not degrees of freedom. > There seems to be some confusion here with understanding of the equipartition theorm. The classical equipartition theorem assigns a total energy (note not just kinetic energy) of kT/2 per each term in the system's Hamiltonian (momentum or position) that is quadratic. Only for ideal gases is the total energy kinetic, for other molecules this is not true. Also, non quadratic terms will contibute to the energy of the system but not necessarily in the simple kT/2 way. For empirical potential energy functions used in MD how many quadratic terms you have depends on the choice of force field. Another complication are the VDW and Coulomb terms in the FF, which are not quadratic, so the classical equipartition theorem can't be used. This is always something that has bothered me about MD, the thermodynamics does not seem incorporated into simulations well. The statistical temperature is defined as 1/T= dS/dE, where S is the system's entropy. This really gives and unambiguous definition of temperature to your system, but I'm not sure if the current state MD/MM simulations allow S or dS/dE to be calculated easily. -- Walter Polkosnik walt@panix.com Physics Department http://www.physics.qc.edu/~walt Queens College, City University of New York (718)-997-3364 voice Credo quia absurdum est. (718)-997-3349 fax From walt@panix.com Wed Nov 6 15:11:44 1996 Received: from panix3.panix.com for walt@panix.com by www.ccl.net (8.8.2/950822.1) id OAA12478; Wed, 6 Nov 1996 14:33:46 -0500 (EST) Received: from localhost (walt@localhost) by panix3.panix.com (8.7.6/8.7/PanixU1.3) with SMTP id OAA16907; Wed, 6 Nov 1996 14:33:31 -0500 (EST) Date: Wed, 6 Nov 1996 14:33:30 -0500 (EST) From: Walter Polkosnik To: Konrad Hinsen cc: echamot@xnet.com, rochus@felix.anorg.chemie.tu-muenchen.de, chemistry@www.ccl.net Subject: Re: CCL:MD/MM combination In-Reply-To: <199611061913.UAA10578@ibs.ibs.fr> Message-ID: Errors-To: walt@panix.com MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII On Wed, 6 Nov 1996, Konrad Hinsen wrote: > > There seems to be some confusion here with understanding of the > > equipartition theorm. The classical equipartition theorem assigns a total > > energy (note not just kinetic energy) of kT/2 per each term in the system's > > Hamiltonian (momentum or position) that is quadratic. Only for ideal gases > > is the total energy kinetic, for other molecules this is not true. Also, > > non quadratic terms will contibute to the energy of the system but not > > necessarily in the simple kT/2 way. For empirical potential energy > > I don't see the problem. The equipartition theorem says that there > will be kT/2 of (kinetic) energy for each degree of freedom whose > momentum enters quadratically into the Hamiltonian, and kT/2 of > (potential) energy for each degree of freedom whose coordinate enters > quadratically into the Hamiltonian. The first part remains true even > if the second part can't be applied. As far as I know, the equipartition > theorem is used in MD only for the momenta, not for the coordinates. > The form of the potential energy function is therefore irrelevant. > What I was trying to point out (perhaps in a very roundabout way) is that defining 'temperature' in this way for an MD situation is a problem (and it seems like all MD simply uses ideal gas results). This 'temperature' is not simply related to the real physical temperature. -- Walter Polkosnik walt@panix.com Physics Department http://www.physics.qc.edu/~walt Queens College, City University of New York (718)-997-3364 voice Credo quia absurdum est. (718)-997-3349 fax From hinsen@ibs.ibs.fr Wed Nov 6 15:22:28 1996 Received: from ibs.ibs.fr for hinsen@ibs.ibs.fr by www.ccl.net (8.8.2/950822.1) id OAA12339; Wed, 6 Nov 1996 14:12:42 -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 UAA10578; Wed, 6 Nov 1996 20:13:26 +0100 Message-Id: <199611061913.UAA10578@ibs.ibs.fr> Received: by lmshp2.ibs.fr (1.37.109.4/16.2) id AA12939; Wed, 6 Nov 96 20:13:26 +0100 Date: Wed, 6 Nov 96 20:13:26 +0100 From: Konrad Hinsen To: walt@panix.com Cc: echamot@xnet.com, rochus@felix.anorg.chemie.tu-muenchen.de, chemistry@www.ccl.net In-Reply-To: (message from Walter Polkosnik on Wed, 6 Nov 1996 13:38:51 -0500 (EST)) Subject: Re: CCL:MD/MM combination > There seems to be some confusion here with understanding of the > equipartition theorm. The classical equipartition theorem assigns a total > energy (note not just kinetic energy) of kT/2 per each term in the system's > Hamiltonian (momentum or position) that is quadratic. Only for ideal gases > is the total energy kinetic, for other molecules this is not true. Also, > non quadratic terms will contibute to the energy of the system but not > necessarily in the simple kT/2 way. For empirical potential energy I don't see the problem. The equipartition theorem says that there will be kT/2 of (kinetic) energy for each degree of freedom whose momentum enters quadratically into the Hamiltonian, and kT/2 of (potential) energy for each degree of freedom whose coordinate enters quadratically into the Hamiltonian. The first part remains true even if the second part can't be applied. As far as I know, the equipartition theorem is used in MD only for the momenta, not for the coordinates. The form of the potential energy function is therefore irrelevant. For those who want it precisely, here's the equipartition theorem for the kinetic energy in its exact form: / d H \ kT < p ------- > = ---- \ i d p / 2 i (Of course it's a partial derivative, but ASCII didn't provide for this!) -- ------------------------------------------------------------------------------- Konrad Hinsen | E-Mail: hinsen@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 ------------------------------------------------------------------------------- From xiannong@pauling.chem.uga.edu Wed Nov 6 18:11:41 1996 Received: from charon.chem.uga.edu for xiannong@pauling.chem.uga.edu by www.ccl.net (8.8.2/950822.1) id RAA13771; Wed, 6 Nov 1996 17:28:03 -0500 (EST) Received: from pauling.chem.uga.edu ( [128.192.5.50]) by charon.chem.uga.edu (8.7.4/8.6.9) with SMTP id RAA15865 for ; Wed, 6 Nov 1996 17:30:46 -0500 Date: Wed, 6 Nov 1996 17:27:48 -0500 (EST) From: Xiannong Chen To: chemistry@www.ccl.net Subject: Help: chemical inventary control Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Hi, Dear Netters: I work for a small pharmaceutical company. We want to use barcode to keep track of our compounds. We use IsisBase to manage structural as well as other information. Does anyone have experience with using barcode system integrated with IsisBase or other database, spreadsheet software? Any suggestion about chemical inventary control with any software system is appreciated. THank you very much. I am not on the list. Please sned your response to me at xiannong@ccmsd.chem.uga.edu Xiannong From rjm@theory.chem.ubc.ca Wed Nov 6 18:15:25 1996 Received: from mail-relay.ubc.ca for rjm@theory.chem.ubc.ca by www.ccl.net (8.8.2/950822.1) id SAA13948; Wed, 6 Nov 1996 18:04:38 -0500 (EST) Received: from seymour.chem.ubc.ca (root@seymour.chem.ubc.ca [137.82.31.13]) by mail-relay.ubc.ca (8.7.6/1.14) with SMTP id PAA04644 for ; Wed, 6 Nov 1996 15:04:39 -0800 (PST) Received: from crown.chem.ubc.ca by seymour.chem.ubc.ca (AIX 3.2/UCB 5.64/1.14) id AA21677; Wed, 6 Nov 1996 15:04:38 -0800 Sender: rjm@theory.chem.ubc.ca Message-Id: <32811984.41C6@theory.chem.ubc.ca> Date: Wed, 06 Nov 1996 15:04:36 -0800 From: Richard Moss Organization: Dept Chemistry, University of British Columbia X-Mailer: Mozilla 3.0 (X11; I; AIX 2) Mime-Version: 1.0 To: chemistry@www.ccl.net Subject: MD pressure calculation problem/question Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit Hi ya, I'm trying to calculate the pressure tensor in a MD simulation for different regions within the system (ie divide my "box" up into sub-regions and look at the tensor separately for each of them). But I have several problems when trying to do this: I can't use an addition to the force routine to sum the components of the tensor, because the force routine loops over i=1,N-1 j=i+1,N, which though fine for calculating the tensor over the entire system, _doesn't_ work for separate regions, as you miss certain interactions. (ie the tensor is bigger in some regions, smaller in other, over what it should be). The second problem is if I try and use a simple loop over i, calculating the x/y/z components of R(i).F(i). This doesn't work when you use periodic boundaries (ref Allen&Tildesly) - only the R(ij).F(ij) form can be used. So, as I see it the only way I can get the tensor properly is to do an "extended" force calculation type routine, and do a double sum, both over N (opposed to the normal force where it is an i,j>i sum). Is this correct? Are there any other methods for calculating the pressure tensor (or at least the diagonal components). thanks, Richard From hrusak@ims.ac.jp Wed Nov 6 19:11:41 1996 Received: from ims1.ims.ac.jp for hrusak@ims.ac.jp by www.ccl.net (8.8.2/950822.1) id TAA14191; Wed, 6 Nov 1996 19:06:38 -0500 (EST) Received: from ikarus.ims.ac.jp by ims1.ims.ac.jp (8.6.8+2.4Wb2/TISN-1.3M/R2) id JAA28681; Thu, 7 Nov 1996 09:05:16 +0900 Received: by ikarus.ims.ac.jp with Microsoft Mail id <01BBCC27.4A2EEA00@ikarus.ims.ac.jp>; Wed, 6 Nov 1996 21:13:00 +-900 Message-ID: <01BBCC27.4A2EEA00@ikarus.ims.ac.jp> From: Jan Hrusak To: "'CHEMISTRY@www.ccl.net'" Subject: G94 help Date: Wed, 6 Nov 1996 21:12:58 +-900 Return-Receipt-To: Encoding: 31 TEXT Thanx to all the people who responded to my posting from yesterday evening. The fact that few hours later my problem seems to be solved, once again and unambiguously demonstrates that the CCL list is very very useful. Back to the problem: I optimized the triatomic (doublet state) starting from a bend geometry and met a problem which forced the G94 program to stop. > Error permuting atoms in Fill, LPerm : > 0 0 0 > Symmetry turned off : > Internal error in symmetry package. > Error termination via Lnk1e ....... > I am not going to list all the answers I got (they are too many), but the suggestions I obtained are basically two : 1) To solve the problem using the NOSYMM keyword 2) To optimize the system with restriction to linearity In fact both ways work, eventhought, the first one seems not to be very "clean", since symmetry breaking may lead to artificial lowering of the energy. The reason why I did not consider the linear structure from the beginning, was simply because the program stopped quite far from linearity. Jan Hrusak hrusak@ims.ac.jp hrusak@jh-inst.cas.cz http://zeus1.ims.ac.jp/Iwata/Hrusak/index.html