Thank you very much!
Dividing by 0.2082 is correct.
Best regards,
Xiao
________________________________
From: Daniel Roe <daniel.r.roe.gmail.com>
Sent: Thursday, January 3, 2019 2:52:13 AM
To: AMBER Mailing List
Subject: Re: [AMBER] Calculated dipole moment in AMBER
Hi,
> I read one post, which says the unit for dipole moment calculated from AMBER cpptraj is not Debye(D). And, the code for calculating dipole moment may be out-dated.
CPPTRAJ (like the rest of Amber) uses the AKMA unit system, so charge
is in electrons and length is in Angstroms, meaning the dipole is in
units of e * Ang; you'll have to convert to get to Debye (I believe
the conversion is dividing by ~ 0.2082 but you should double check).
The output format of 'dipole' command is outdated - the 'vector
dipole' code is fine.
-Dan
On Wed, Jan 2, 2019 at 10:22 AM Xiaocong Wang <wangxiaocong.outlook.com> wrote:
>
> Dear all,
> I used cpptraj to calculate the dipole moment for a sugar. But, the value is very different from that calculated in QM.
> Codes are as follows:
>
> trajin sugar.rst7
> vector v1 :1-2 dipole out sugar_dipole.txt
> go
>
> I read one post, which says the unit for dipole moment calculated from AMBER cpptraj is not Debye(D). And, the code for calculating dipole moment may be out-dated.
> Can I get some help on this issue?
> Thank you very much!
> Best,
> Xiao
>
> _______________________________________________
> AMBER mailing list
> AMBER.ambermd.org
> http://lists.ambermd.org/mailman/listinfo/amber
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Received on Wed Jan 02 2019 - 18:00:03 PST
