lammps-users Digest, Vol 133, Issue 109


From: Lamm Gro <lammgro@…24…>

Subject: Re: [lammps-users] Free energy of water

Date: 27 June 2017 at 12:07:22 GMT+2

To: Axel Kohlmeyer <[email protected]>

Cc: lammps-users <[email protected]>

Hi .

I have some questions about fep examples in LAMMPS and I will be grateful if you can answer them :

1 - in equilibration section of ( lammps/examples/USER/fep/CH4hyd/fep01 ) why we set the charges of methane atoms to zero ?!

fix TPSTAT all npt temp {TK} {TK} 100 iso {PBAR} {PBAR} 1000

set type 1*2 charge 0.0

The charges are initially set to zero so that the solute does not interact with the solvent, both through the real-space and the k-space parts of the Coulomb potential. (For the real-space part using the lambda would suffice)

run 20000

2 - fix adapt/fep and compute fep commands ( lammps/examples/USER/fep/CH4hyd/fep01 ) :

fix ADAPT all adapt/fep 10000 &
pair lj/cut/tip4p/long/soft lambda 12 34 v_lambda &
atom charge 1 v_q1 &
atom charge 2 v_q2 &
after yes

compute FEP all fep ${TK} &
pair lj/cut/tip4p/long/soft lambda 12 34 v_dlambda &
atom charge 1 v_dq1 &
atom charge 2 v_dq2

What is the reason of using these atom types separately ? 12 : Methane atoms ----- 34 water atoms
If we want to set a parameter on methane atoms why we need to mention water molecules ( 3*4 ) in these two commands ?

What you modify during the free energy calculation are the I-J interaction parameters (cross or unlike interactions).

3 - Can you please let me know why we need v_dlambda with v_lambda ?
Actually i can’t understand the reason of using this command : variable dlambda equal 0.05

lambda is the value of the activation parameter (0: initial system, 1: final)
dlambda is the step in lambda between each stage of activation.

You should study the fundamentals of free energy calculations.

Best regards,
Agilio Padua