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

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

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.,

$\mathbf{r}_{ji} = \mathbf{r}_i - \mathbf{r}_j$

$r_{ji} = \sqrt{(r_i^x - r_j^x)^2 + (r_i^y - r_j^y)^2 + (r_i^z - r_j^z)^2}$

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 $k$ 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 $k$ is either $i$ or $j$, 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 $V_{kj}$ 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

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

which is non-zero when $k$ is either $i$ or $j$, 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 $\rho_{kj}$ 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

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

Since $i$ 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

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

If there is only type of atoms in the system, $\rho_{jk} = \rho_{kj}$, 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 $\rho_{kj}$ and $\rho_{ij}$ is not used during the derivation.