Lei,
I am also confused about the details of dgdl (or dHdl) calculation in
GROMACS. For example, if the lambda is fixed at certain value, how did the
program determine dHdl?
See below.
My initial understanding was kind of similiar to that of Mauricio that dHdl
equals the difference in total potential energy between two time frame
(snapshots).
That's never even approximately true unless you are changing lambda as
a function of time, and even then, it is not really true, as dH/dl
isn't evaluated using that approach, and it also contains an extra
factor of 1/dl relative to the difference in total potential energy,
plus there are dynamics going on...
However, this seems does not make sense when lambda is fixed.
Could you give more details about the calculation of the derivative dHdl at
each snapshot? Thank you.
Try reading the manual, for example, equation 4.116 and 4.120 in the
Gromacs 3.3 manual. If you read that, and my previous e-mail, where I
talked about taking the derivative, and you're still confused, write
back and try to be more specific about the nature of your confusion.
You're talking about evaluating dH/dl using a finite-difference
scheme, which is not how it's done -- it's done analytically, since
H(x,p,l) is known.
Take this example: Suppose f(l)=sin(l), and you want to know df/dl.
You ask, how do we evaluate this while keeping l fixed? Simple: Take
the derivative of sin(l) with respect to l. You could *also* do it
with a finite difference scheme (f(l+dl)-f(l))/dl or some such), but
there's no reason to do that, since the derivative approach is
simpler.
Isn't this clear from the manual? As much as I think the manual is
unclear sometimes, this isn't one of those times...
David
Best regards,
Lei Zhou
On 11/6/06, David Mobley <[EMAIL PROTECTED]> wrote:
> 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
> confused about.
>
> Best wishes,
> David Mobley
> UCSF
>
>
> On 11/5/06, Mauricio Sica <[EMAIL PROTECTED] > 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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