# EAM potential

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

E = \frac{1}{2}\sum_i\sum_{j \atop j\neq i} V_{ij}(r_{ij}) + \sum_i F(\bar{\rho}_i)

where $ is the pair potential, $ is the embedding potential, and $ is the host electron density, i.e.,

\bar{\rho}_i = \sum_{j \atop j \neq i} \rho_{ij}(r_{ij})

where $ is the local electron density contributed by atom $ at site $.

Let $ be the vector from atom $ to atom $ with norm $, i.e.,

\mathbf{r}_{ji} = \mathbf{r}_i - \mathbf{r}_j
r_{ji} = \sqrt{(r_i^x - r_j^x)^2 + (r_i^y - r_j^y)^2 + (r_i^z - r_j^z)^2}

where

\mathbf{r}_j = r_j^x\mathbf{e}^x + r_j^y\mathbf{e}^y + r_j^z\mathbf{e}^z

Now, let's prove an important identity,

\frac{\partial r_{ji}}{\partial \mathbf{r}_j} = \frac{\partial r_{ji}}{\partial r_j^x} \mathbf{e}^x + \frac{\partial r_{ji}}{\partial r_j^y} \mathbf{e}^y + \frac{\partial r_{ji}}{\partial r_j^z} \mathbf{e}^z = - \frac{r_{ji}^x}{r_{ji}} \mathbf{e}^x - \frac{r_{ji}^y}{r_{ji}} \mathbf{e}^y - \frac{r_{ji}^z}{r_{ji}} \mathbf{e}^z = -\frac{\mathbf{r}_{ji}}{r_{ji}}

which will be used in the force formulation derivation later.

The force on atom $ is

\mathbf{f}_k = -\frac{\partial E}{\partial \mathbf{r}_k} = -\frac{1}{2} \frac{\partial \sum_i \sum_{j \atop j \neq i}V_{ij}(r_{ij})}{\partial \mathbf{r}_k}-\frac{\partial \sum_i F(\bar{\rho}_i)}{\partial \mathbf{r}_k}

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

-\frac{1}{2} \left[\frac{\partial \sum_{j \atop j \neq k} V_{kj}(r_{kj})}{\partial \mathbf{r}_k}+\frac{\partial \sum_{i \atop k \neq i}V_{ik}(r_{ik})}{\partial \mathbf{r}_k}\right] = -\frac{1}{2} \left[\frac{\partial \sum_{j \atop j \neq k} V_{kj}(r_{kj})}{\partial r_{kj}}\frac{\partial r_{kj}}{\partial \mathbf{r}_k} - \frac{\partial \sum_{i \atop k \neq i}V_{ik}(r_{ik})}{\partial r_{ik}}\frac{\partial r_{ik}}{\partial \mathbf{r}_k}\right]

With the help of the identity, the term becomes

\frac{1}{2} \left[\frac{\partial \sum_{j \atop j \neq k} V_{kj}(r_{kj})}{\partial r_{kj}}\frac{\mathbf{r}_{kj}}{r_{kj}} - \frac{\partial \sum_{i \atop k \neq i}V_{ik}(r_{ik})}{\partial r_{ik}}\frac{\mathbf{r}_{ik}}{r_{ik}}\right]

where $ and $ are the pair potentials for the atomic pairs $ and $, respectively, while $ and $. Since $ is atom type-specific, $ and $ are likely not the same unless atom $ and $ are of the same type. Thus, if there are two types of atoms in the system, there will be three $, 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

-\sum_i\frac{\partial F(\bar{\rho}_i)}{\partial \mathbf{r}_k} = -\sum_i\frac{\partial F(\bar{\rho}_i)}{\partial \bar{\rho}_i}\frac{\partial \bar{\rho}_i}{\partial \mathbf{r}_k} = -\sum_i\frac{\partial F(\bar{\rho}_i)}{\partial \bar{\rho}_i}\sum_{j \atop j \neq i}\frac{\partial \rho_{ij}(r_{ij})}{\partial \mathbf{r}_k} = -\sum_i\frac{\partial F(\bar{\rho}_i)}{\partial \bar{\rho}_i}\sum_{j \atop j \neq i}\frac{\partial \rho_{ij}(r_{ij})}{\partial r_{ij}}\frac{\partial r_{ij}}{\partial \mathbf{r}_k}

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

-\frac{\partial F(\bar{\rho}_k)}{\partial \bar{\rho}_k}\sum_{j \atop j \neq k}\frac{\partial \rho_{kj}(r_{kj})}{\partial r_{kj}}\frac{\partial r_{kj}}{\partial \mathbf{r}_k}-\sum_{i \atop i \neq k}\frac{\partial F(\bar{\rho}_i)}{\partial \bar{\rho}_i}\frac{\partial \rho_{ik}(r_{ik})}{\partial r_{ik}}\frac{\partial r_{ik}}{\partial \mathbf{r}_k}

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

\frac{\partial F(\bar{\rho}_k)}{\partial \bar{\rho}_k}\sum_{j \atop j \neq k}\frac{\partial \rho_{kj}(r_{kj})}{\partial r_{kj}}\frac{\mathbf{r}_{kj}}{r_{kj}}-\sum_{i \atop i \neq k}\frac{\partial F(\bar{\rho}_i)}{\partial \bar{\rho}_i}\frac{\partial \rho_{ik}(r_{ik})}{\partial r_{ik}}\frac{\mathbf{r}_{ik}}{r_{ik}}

Note that $ is the local electron density contributed by atom $ at site $. In general, $. This is different from the pair potential $, for which generally $. Also, generally $ unless atom $ and atom $ are of the same type.

In classical EAM, $ even when atom $ and atom $ are of different type. If there are two types of atoms in the system, there are only two $, 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 $, which would have three expressions in this case. Extensions of $ 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

\mathbf{f}_k = \frac{1}{2} \left[\frac{\partial \sum_{j \atop j \neq k} V_{kj}(r_{kj})}{\partial r_{kj}}\frac{\mathbf{r}_{kj}}{r_{kj}}-\frac{\partial \sum_{i \atop k \neq i}V_{ik}(r_{ik})}{\partial r_{ik}}\frac{\mathbf{r}_{ik}}{r_{ik}}\right] + \frac{\partial F(\bar{\rho}_k)}{\partial \bar{\rho}_k}\sum_{j \atop j \neq k}\frac{\partial \rho_{kj}(r_{kj})}{\partial r_{kj}}\frac{\mathbf{r}_{kj}}{r_{kj}}-\sum_{i \atop i \neq k}\frac{\partial F(\bar{\rho}_i)}{\partial \bar{\rho}_i}\frac{\partial \rho_{ik}(r_{ik})}{\partial r_{ik}}\frac{\mathbf{r}_{ik}}{r_{ik}}

Since $ and $ are just dummy indices, it is safe to replace all $ with $. After that, with $, $, $, and $, the force on atom $ becomes

\mathbf{f}_k = \sum_{j \atop j \neq k}\left[\frac{\partial V_{kj}(r_{kj})}{\partial r_{kj}}+\frac{\partial F(\bar{\rho}_k)}{\partial \bar{\rho}_k}\frac{\partial \rho_{kj}(r_{kj})}{\partial r_{kj}}+\frac{\partial F(\bar{\rho}_j)}{\partial \bar{\rho}_j}\frac{\partial \rho_{jk}(r_{kj})}{\partial r_{kj}}\right]\frac{\mathbf{r}_{kj}}{r_{kj}}

If there is only type of atoms in the system, $, and the force formulation is simplified to

\mathbf{f}_k = \sum_{j \atop j \neq k}\left[\frac{\partial V_{kj}(r_{kj})}{\partial r_{kj}}+\left(\frac{\partial F(\bar{\rho}_k)}{\partial \bar{\rho}_k}+\frac{\partial F(\bar{\rho}_j)}{\partial \bar{\rho}_j}\right)\frac{\partial \rho_{kj}(r_{kj})}{\partial r_{kj}}\right]\frac{\mathbf{r}_{kj}}{r_{kj}}

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 $ and $ is not used during the derivation.