Dear all,
I want to calculate Mie Gruneisen parameter of a system i.e. V(P/U).
I have computed U i.e lammps total energy in Kcal/mol unit, P in atm, V in meter ^3 .
To get the total energy, I have divided the energy which I get from thermo_style etot by 6.023*10^23.
But my Mie Gruneisen parameter value differs a lot from the original value. Is there any other way to compute Mie Gruneisen parameter from lammps.
thanks in advance.
This sounds like a units issue, not a LAMMPS issue.
It sounds like you are computing VP/U using
V in m^3, P in atm, and U in Kcal/atom (after dividing
by 6E23). They seem like 3 inconsistent units to
use in computing one new value.
Steve
Dear Steve,
I get energy in Kcal/mol for real unit, does it mean that to get E in Kcal unit I have to multiply by mass(all)/N_a ; N_a = Avogadro no ?
I don’t know - I’m just saying that using
Kcal, atm, and meters in one expression
are not consistent units. Consistent units
would be (for example): joules, newton/m^2,
and meter^3.
Steve
Dear Steve,
I was trying to compute Mie Grunesian parameter of water at ambient condition from molecular dynamics i.e. by plugging the Lammps TD output variables. Here is the more details of the total calculation------ — –
For my system water (TIP4P), I got------------
The energy output from LAMMPS is for the whole system, not per atom. Divide by number of molecules to get 9.117 Kcal/mol. Then I think it should work out.
You can choose whether to output the potential energy per atom or not, but the default behaviour differs based on your units. So be careful with that. Also, some checking should have shown you that your PE per atom couldn’t possibly be correct!
Actually to compute Mie Grunsein parameter I need the total energy of the system, so I have divided the E value by Avogardo’s number to get per atom and have multiplied by number of atoms to get the total energy i.e.
E = 298745 Kcal/mol = ( 298745 * 10^3 * 4.1840 * 98304 ) / (6.023E23) Joul { total no of atom = 98304 }
Where I have done the wrong ?
I have computed thermo_style …etotal i.e. total energy (pe + ke) which is actually the total internal energy of my system, isn’t it ?
Dear Steve,
I was trying to compute Mie Grunesian parameter of water at ambient
condition from molecular dynamics i.e. by plugging the Lammps TD output
variables. Here is the more details of the total calculation------ --- --For my system water (TIP4P), I got------------
------------------------------------------------------------------------------------------------------------------------------------------------------------
P = 1 atm = 1.01325 * 10^5 N/m^2
V= 10^-24 m^3
E ( in lammps real unit )= 298745 Kcal/mol = ( 298745 * 10^3 * 4.1840 ) /
(6.023E23) Joul/atom = 2.07529 e-15 Joul ( Total no. of atom was 98304 )
this is not joule/atom, but plain joule. total energy is the total
energy of the entire system.
so VP/E = Mie Grunesian parameter = 0.0000488245
based on your numbers, i get thus: 0.00488244
--------------------------------------------------------------------------------------------------------------------------------------------------------------
But as per the existing literature is concerned it is nearly 0.2 ~ 0.3.
Please help me out.
what is the literature value for? experiment? simulation under the
same conditions?
please note that most simple water models can be quite far off from
experimental thermodynamic properties. like for compressibility,
diffusivity, melting point, boiling point. and without manybody
contributions you also want get the density anomaly near the melting
point.
axel.
That’s not how it works. The energy here 298745 kcal/mol is already for the whole system. When you divide this by the number of molecules you end up with about 9.117 kcal/mol per molecule. -9.117 kcal/mol = -38.12 kJ/mol is at least in the ballpark for the configurational energy of liquid water. So that’s why I thought that was the mistake you made.
But now that I try to work it all out, I get the same answer you got. System PE with your data is 9.117 kcal/mol/molecule x 4.184e3 J/kcal * (98304/3) molecules / 6.022e23 /mol = 2.08 e-15 J, as you have. So… I guess that’s not the problem. Sorry.
Just where are you getting this literature value? Is this parameter even used for liquids like water? As far as I can tell it’s more of a parameter used in solid state physics.
Apart from all the difficulties with units, it should also be noted that the Gruneisen parameter is not V(P/U), rather gamma = V(dP/dU). The fact that the original poster did not know this is a little troubling. Calculating gamma from MD is by no means straightforward. I suggest starting with something easier but requiring similar skills, such as bulk modulus = V(dP/dV).