# EAM potential

As mentioned earlier, the EAM formulation for the potential energy is

where V is the pair potential, F is the embedding potential, and \bar{\rho} is the host electron density, i.e.,

where \rho_{ij} is the local electron density contributed by atom j at site i.

Let \mathbf{r}_{ji} be the vector from atom j to atom i with norm r_{ji} (= r_{ij}), i.e.,

where

Now, let's prove an important identity,

which will be used in the force formulation derivation later.

The force on atom k is

The first term in the force formulation is non-zero only when k is either i or j, thus it becomes

With the help of the identity, the term becomes

where V_{kj} and V_{ik} are the pair potentials for the atomic pairs kj and ik, respectively, while V_{kj} = V_{jk} and V_{ik} = V_{ki}. Since V is atom type-specific, V_{kj} and V_{ik} are likely not the same unless atom i and j are of the same type. Thus, if there are two types of atoms in the system, there will be three V, between type 1 and type 1, between type 2 and type 2, and between type 1 and type 2.

The second term in the force formulation can be written as

which is non-zero when k is either i or j, i.e., the term becomes

Again, with the help of the identify, the term becomes

Note that \rho_{kj} is the local electron density contributed by atom j at site k. In general, \rho_{kj} \neq \rho_{jk}. This is different from the pair potential V, for which generally V_{kj} = V_{jk}. Also, generally \rho_{kj} \neq \rho_{ij} unless atom k and atom i are of the same type.

In classical EAM, \rho_{kj} = \rho_{ij} even when atom k and atom i are of different type. If there are two types of atoms in the system, there are only two \rho, for the contribution from type 1 atom and for that from type 2 atom, regardless of which type of atomic site it contributes to. This is different from the pair potential V, which would have three expressions in this case. Extensions of \rho to distinguish contributions at different types of atomic sites have been proposed, e.g., in the Finnis-Sinclair potential.

Adding the two terms in the force formulation together yields

Since i and j are just dummy indices, it is safe to replace all i with j. After that, with \mathbf{r}_{jk} = -\mathbf{r}_{kj}, r_{jk} = r_{kj}, V_{jk} = V_{kj}, and \rho_{jk} \neq \rho_{kj}, the force on atom k becomes

If there is only type of atoms in the system, \rho_{jk} = \rho_{kj}, and the force formulation is simplified to

which is Equation 15 of Xu et al. Note that the last two equations hold for both classical EAM and Finnis-Sinclair potentials, because the relation between \rho_{kj} and \rho_{ij} is not used during the derivation.