CCL: Amber free energy equation

From: Daquan Gao <>
Date: Tue, 17 Jun 2003 12:33:39 -0400

Dear CCLers or AMBER users:

 Having a question about the equation with free energy perpurbation:

The equation is , among other variations:

  Delta G = -kT ln <exp(-(H(lambda + d lambda) - H (lambda))/kT) >

Now in gibbs, I used nstlim = 1, nrun = 100, almdel =0.01, isldyn = -3.

The reason I used nstlim =1 is I wanted to test the equation.

The isldym = -3 means I have a starting state at lambda = 1.0, and a
final state of lambda =0.0 which is when the ligand disappears.

Now my questions:

After the first nrun, there is still only one set of energies, (either
Etot or EPtot or something other terms.). There is already a free energy
for "reverse" given. How was this achieved? There has not been any
energy differences yey.

At nrun = 2, I have lambda = 0.99, the previous state is lambda = 1.00.
Now I have two sets of energies, I would be able to get a delta H =
H(0.99)- H(1.00). This is the ansemble average, since there is only one
such delta H! But the result is not that number. I don't understand the
number given in gibbs in that stage.

What exactly is the equation that is being used in gibbs?

I have looked at the references, all says the above equation. Only
Amber7 mentioned a reference by Brooks in JCP. It looked a little
different, but only added DS and DE terms. Don't you get DG direcly from
the above eqution?

If any one knows the answer, please help. It may help me to improve my
results of binding free energy calculations.

I have posted once today, it did not appear, so again, sorry if you get
this twice.
TIA and have a good day.
Daquan Gao
Received on Tue Jun 17 2003 - 17:53:02 PDT
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