Thanks for the help. I resolved this issue. Some parameters are wrong in my topology file. After I corrected them, it gives the energy of 0.00021, which is half of the energy I calculated by hand. I guess the program adds 1/2 in front of the kb(r-r0)^2 when it gives the final bond energy. Am I right?
Sixue
-----Original Message-----
From: Zhang, Sixue [mailto:szhang74.illinois.edu]
Sent: Tuesday, February 24, 2015 1:37 PM
To: AMBER Mailing List
Subject: Re: [AMBER] Bond Energy Term for a TIP3P Water
There is typo in my previous email, it's 9.9065 instead of 9,9065. Sorry about that.
-----Original Message-----
From: Zhang, Sixue [mailto:szhang74.illinois.edu]
Sent: 2015年2月24日 13:31
To: AMBER Mailing List
Subject: [AMBER] Bond Energy Term for a TIP3P Water
Hello,
Recently, I'm doing tests on AMBER Forcefield. I found one thing that I can't explain. After I ran one step of minimization of a TIP3P water, the output bond energy for this water is 9.9065 kcal/mol. However, the two OH distances are 0.9566 Å and the HH distance is 1.5136 Å, which are very close to the equilibrated distances in the forcefield parameter file (parm10.dat). If I put these values into the bond energy expression stated in the AMBER manual, SUM(kb(r-r0)2), using the force constants and equilibrium distances in the parameter file, I got a value of 0.00042 for the bond energy. Does anyone know why the program give 9,9065 for bond energy in the output file?
Thanks a lot,
Sixue
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Received on Tue Feb 24 2015 - 17:00:02 PST
