# Re: AMBER: Steered MD & Jarzynski's equality

Date: Wed, 02 May 2007 09:26:30 +0200

James W wrote:
> Dear all ,
> #the reference : (http://en.wikipedia.org/wiki/Jarzynski_equality)
> I confused the Jarzynski's equality . The equality is shown:
> exp (-F/KT) = < exp (-W/KT) > .
> &
> F = < W > + ......
> I used SMD module of CHARMM and obtained the reaction coordinates &
> work ,like this :
> ______________________________________________________
> 57.00000 57.73576 -1471.51842 0.00000
> 57.00093 57.34648 -691.10171 -1.00562
> (RC) (WORK)
> *RC : reaction coordinates
> I wanted to obtain the " < W > " form my data . My method is :
> 1. S = sum { exp (-W(i) /kT )} , i = 1 ....N
> 2. p(i)= exp (-W(i) /kT ) / S
> 3. <W(i)> = W(i)*p(i) / S
>
> Could you tell me that my method is right ?

James,
I am not sure I fully understand your question, because you do not
mention what the subindex i is in your formulas.
Now, the formula from wikipedia is basically right, except for a couple

First, it must be clear that F is really \Delta F. This is designed for
free energy differences and NOT absolute free energies. Second, the
exponential average of work values is over a set of runs, ALL starting
from an equilibrated ensemble at a certain value of the coordinates (s)
to be changed. The average then involves MANY runs.

The second line F = < W > + ...... is a cumulant expansion of
exponential average. I do not recommend using using this unless you
really know what you are doing.

Your formulas on how to implement this are not quite right.
If you are thinking about the index i meaning time steps on one
simulation, then it is not right.
The subindex i should refer to the same distance and DIFFERENT simulations.

Once that is done:

1. S = ( sum { exp (-W(i) /kT )}/N) , i = 1 ....N
2. \Delta F = - kT ln(S)

I hope this makes things a bit clearer.

```--
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Quantum Theory Project and Department of Chemistry
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