Hi Jason,
Yes. I already considered all three. My specific expression is 553*(0.9566-0.9572)^2 + 553*(0.9566-0.9572)^2 + 553*(1.5138-1.5136)^2, where 553 is the force constant for both OW-HW and HW-HW, 0.9572 and 1.5136 are the equilibrium distances for OW-HW and HW-HW in parm10, 0.9566 and 1.5138 are the O-H and H-H distances in my output geometry. The result is 0.00042. Since my geometry is very close to the equilibrium geometry, I couldn't figure out why AMBER gives 9.9065.
Thanks,
Sixue
-----Original Message-----
From: Jason Swails [mailto:jason.swails.gmail.com]
Sent: 2015年2月24日 13:58
To: AMBER Mailing List
Subject: Re: [AMBER] Bond Energy Term for a TIP3P Water
On Tue, Feb 24, 2015 at 2:37 PM, Zhang, Sixue <szhang74.illinois.edu> wrote:
> 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?
>
There are 3 bonds in a TIP3P water: O-H1, O-H2, and H1-H2. Make sure you add in all 3 contributions.
HTH,
Jason
--
Jason M. Swails
BioMaPS,
Rutgers University
Postdoctoral Researcher
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Received on Tue Feb 24 2015 - 13:30:04 PST
