Hello A
I am scared....that means that the computed work is dependent on the
velocity applied
which sounds strange to me....there should be a way to factor out the
work done by
the biasing force.
I ran various tests and it turns out that the higher the rate of motion
of the force,
the higher the work I get, . . . Is that correct?.
The free energies I am computing should be flat at the end, but instead
it looks as
if the biasing work has to be subtracted.
puzzled E.
Adrian Roitberg wrote:
>
>
> E.M. wrote:
>> Hello,
>>
>> I am doing some pulling experiments using Jarzinsky's relationship,
>> the output file using AMBER9 is something like
>>
>> x xo whatever work
>>
>> My question is, do I have to substract the biasing potential after I
>> get the output
>> or that is taking care of by the program?. I just want to make sure
>> I understand
>> what is going on here.
>>
>>
>> Regards
>>
>> E.M.
>
> Hi,
>
> The 'wathever' in the list is the crucial factor, in fact it is the
> only one one computes !
>
> It is the external force applied by the external spring (as defined in
> the input file). Basically, it is k*(x-x0)
>
> Work is the integral of that force over the run, so there is no need
> to subtract the biasing potential. The subtraction would be needed
> only if work was computed from the spring energy, which is not.
>
> a.
>
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Received on Wed May 07 2008 - 06:07:23 PDT
