[gmx-users] polarizable ff and free energy calculation
Mark.Abraham at anu.edu.au
Thu Apr 19 05:09:44 CEST 2012
On 19/04/2012 2:18 AM, Mark Abraham wrote:
> On 18/04/2012 10:32 PM, Tom Kirchner wrote:
>> Of my first mail, only the attachment was printed in the mailing
>> list. I am sorry for the inconvenience. The attachment can be sent by
> They came through.
>> Hi all,
>> Recently, I posted a problem concerning the usage of polarizable ff
>> in conjunction with free energy calculation. Sadly I got no answer,
>> but after some research, I hope I have isolated the problem.
> Thanks for the clear report.
>> Using a polarizable ff without free energy makes no problems. As soon
>> as free energy is turned on, Gromacs gives the error:
>> /Program mdrun_d, VERSION 4.5.5
>> Source code file: [...]/src/gromacs-4.5.5/src/mdlib/shellfc.c, line: 365
>> Fatal error:
>> polarize can not be used with qA != qB/
>> This error occurs, even if the atom and the shell particle have the
>> same charge or are unaltered by the free energy option, e.g.
>> lambda=0.0 and multiplication with (1-lambda). Because of this, I
>> guess there is a problem with different charge variables being single
>> and double precision due to multiplication with lambda.
> ... well, likely some kind of floating-point-algebra failure, anyway.
Hmm, no actually the issue is that qA is -1 and qB is 0 for the second
atom. I have no idea how that arises.
>> The error occurs in this segment of shellfc.c:
>> case F_POLARIZATION:
>> if (qS != atom[aS].qB)
>> gmx_fatal(FARGS,"polarize can not be used with qA != qB");
>> If anybody knows, how to correct this, I would be very grateful.
> I've uploaded a draft fix to https://gerrit.gromacs.org/#/c/746/
>> Moreover I'm wondering about the physical meaning of the if-statement.
> No idea.
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