# [gmx-users] (dH/dl) calculation

David Mobley dmobley at gmail.com
Mon Nov 6 19:19:05 CET 2006

```Mauricio,

I'm somewhat confused by your question and notation. However, I think
the basic answer is something like this: In molecular dynamics, you
know the Hamiltonian from which you are sampling; call it H(x,p, l),
where x denotes all of the positions, p the momentums, and l lambda.
This, of course, is closely linked to the potential energy. Anyway, at
any snapshot, you can simply take the derivative dH(x,p,l)/dl, and you
have dH/dl at that snapshot. This is usually straightforward since you
know the dependence of all of the terms in your Hamiltonian on lambda,
so you actually have the functional form for dH/dl as well -- so it
just involves taking the appropriate combination of positions,
momentums, etc. This is of course all handled internally by the code.
<dH/dl>, then, is just the time-average of dH/dl, which can be
evaluated every step by the code.

I am not sure if that's helpful at all, as I'm not entirely sure what
problem you're having. After all, whenever you do TI calculations in
GROMACS, the code gives you back dG/dl (or dH/dl, or dA/dl) for every
snapshot in an xvg output file. Are you just confused about how the
code gets this (I think I just answered that above), or are you trying
to figure out how to use it? If you're confused about how to use it,
try to ask a question that relates to the specific issue you're

Best wishes,
David Mobley
UCSF

On 11/5/06, Mauricio Sica <msica at unq.edu.ar> wrote:
> Dear experts
>
> I am doing FEP (thermodynamic integration method) simulations.
> I have a questions about <dH/dl> calculation in GROMACS.
> Take in mind equation 3.77 from the GROMACS 3.3 manual.
> There, dA/dl is calculated as
>
> dA/dl = SS{ (dH/dl) exp()dp dq } / SS{ exp()dp dq = <dA/dl>NVT;l }
>
> where SS are doble integrals (sorry for the notation).
>
> My question is: how is (dH/dl) (in the middel-term of the equation)
> calculated?
> My idea is that the difference V(L=1)-V(L=0) is calculated for every time
> step (irrespective of the lambda value of the simulation) and <dG/dl> is
> the time average of that difference.
>
> <dG/dl> = < V(L=1)(i)-V(L=0)(i)/1 >
>
> Is this correct?
>
>
> Thanks
>
>
>
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