Dear all

I have a question about the calculation of pair potential of MEAM in LAMMPS

The implementation of MEAM is based on the paper of I. Huang, MSMSE, 3:615(1995), which gives the formulation about pair potential calculation as following

Phi_ij=[2*Eu-F_i(rho_j)-F_j(rho_i)]/Z_ij

But in lammps, it gives the pair potential like

Phi_ij=[2*Eu-F_i(rho_i)-F_j(rho_j)]/Z_ij

I firstly changed this and do the example for comparison. The SiC example showed some difference

And if looking up other literature, e.g., Baskes, MSMSE, 2:505(1994), the pair potential for like atoms is then given by

Phi_ii=2*[E_u-F_i(rho0_i/Z_i)]/Z_i

Then I tried to implement this in. The Ni example gave me same result compared to the original one but the error of losing atoms appeared in this case for SiC example.

Also, there are other pair potential calculations in some papers, i.e. Baskes, 75, 054106(2007)…

Is there some easy way to add the pair potential calculation in or actually they actually are same since it gives same result for some problem?

Thanks for your advice in advance

Best

Yi

I'll see if Greg Wagner wants to answer this.

Steve

I chipped in these responses on the MEAM part, but neglected to do the “rely-all” thing. Comments on the LAMMPS implementation are certainly in order. Steve Valone

Yi,

I have a question about the calculation of pair potential of MEAM in LAMMPS

The implementation of MEAM is based on the paper of I. Huang, MSMSE, 3:615(1995), which gives the formulation about pair potential calculation as following

Phi_ij=[2*Eu-F_i(rho_j)-F_j(rho_i)]/Z_ij

The particulars of pair potentials in MEAM (Baskes PRB 1992 formulation) for compounds and alloys depend on the crystal structure implied by “Eu”. Eu corresponds to the reference state to which the compound or alloy model is being calibrated. This formula is specific to whatever crystal Huang was talking about. If you look at the canonical MEAM 1992 PRB paper by Baskes you will see this form and see discussion for specific crystal structures. You should infer from that discussion that if your reference crystal structures do not conform to those structures then, the formulae need to be adjusted.

For the specific case of stoichiometric SiC, most of the crystal structures that you might use (or were used) as a reference state conform to this form.

But in lammps, it gives the pair potential like

Phi_ij=[2*Eu-F_i(rho_i)-F_j(rho_j)]/Z_ij

I don’t know so much about the LAMMPS implementation. Your first Eq. should be the correct one.

And if looking up other literature, e.g., Baskes, MSMSE, 2:505(1994), the pair potential for like atoms is then given by

Phi_ii=2*[E_u-F_i(rho0_i/Z_i)]/Z_i

Is there some easy way to add the pair potential calculation in or actually they actually are same since it gives same result for some problem?

In DYNAMO we treat like and unlike pairs as separate cases. For unlike pairs we have a catalog of possibilities depending on the reference crystal structure.

As you deduced, both of your first 2 Eqs. reduce to the last when i and j are the same type and with rho_i = rho0_i/Z_i. You can’t distinguish which form of the unlike pair is correct form the single component test.

Hope that this helps.

Steve Valone

Yi,

Thanks a lot for your elaborate explanations.

Based on my understanding, in alloy case, most cubic structure B1, B2, or dimer structure can use my first Eq. to calculate the pair potentials. And normally, people define E0_ij, alpha_ij and R0_ij to calculate Eu_ij in this case.

You are right for L12. I can give you the code section for L12 for DYNAMO but it would not be much help. I think that it would be more helpful to look at the J of Metals paper from 2003 by Baskes, Stan and some others of us, describing the Pu-Ga system. We use the L12 structure of Pu_3Ga as the reference state.

But for L12 or HCP, I have to use other formulation to get pair potentials. Am I right?

HCP is slightly different again because the 3rd-order partial density contributions entire the background densities in your first equation. When an HCP alloy is used as a reference state, then other components may also entire. One has to think through the symmetries, which I confess I have not done.

For MEAM in lammps, it doesn’t treat likely and unlikely atoms separately because like what you mentioned, when i=j, the Eq. 1 is reduced to Eq3. The error I got when I tried to add Eq3 into lammps was because of the different way of calculating the background density of the reference structure in lammps and the literatures. I think I should not add 1/Z_i here in Eq3 when adding it into lammps.

Yes, one does have to be careful that the normalization convention is consistent between the two formulas. There is a second matter regarding the relative scaling for \rho_0 for the different elements. That is another potential source of discrepancy, but usually it does not come up in the single element case.

Sorry that I did not get this posted to the message board. I guess I didn’t do it right. What was I suppose to do? Reply to all?

Steve

I’ll see if Greg Wagner wants to answer this.

Steve

Dear all

I have a question about the calculation of pair potential of MEAM in LAMMPS

The implementation of MEAM is based on the paper of I. Huang, MSMSE,

3:615(1995), which gives the formulation about pair potential calculation as

following

Phi_ij=[2*Eu-F_i(rho_j)-F_j(rho_i)]/Z_ij

But in lammps, it gives the pair potential like

Phi_ij=[2*Eu-F_i(rho_i)-F_j(rho_j)]/Z_ij

I firstly changed this and do the example for comparison. The SiC example

showed some difference

And if looking up other literature, e.g., Baskes, MSMSE, 2:505(1994), the

pair potential for like atoms is then given by

Phi_ii=2*[E_u-F_i(rho0_i/Z_i)]/Z_i

Then I tried to implement this in. The Ni example gave me same result

compared to the original one but the error of losing atoms appeared in this

case for SiC example.

Also, there are other pair potential calculations in some papers, i.e.

Baskes, 75, 054106(2007)…

Is there some easy way to add the pair potential calculation in or actually

they actually are same since it gives same result for some problem?

Thanks for your advice in advance

Best

Yi

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Steven M. Valone

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Los Alamos National Laboratory

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