scaling property is derived in [@Hu2005 Theorem 1]).

The function $f$ is continuously differentiable for $\gamma \geq 2$. Calculating the partial derivatives of $f$, we find that for $1 \leq m \leq d$, $1 \leq n \leq k$, $$\begin{gathered}
 \label{eq:graphenergygradient}
\frac{\partial f(x_1, \ldots, x_k)}{\partial x_{m \, n}} =
- \frac{\alpha \, \gamma}{2} \sum_{\{j {\ |\ }n \in e_j\}} \frac{c_j}{|e_j|} \sum_{i \in e_j} \|x_i - z_j\|^{\gamma - 2} \, (x_{m \, i} - z_{m \, j}) \\
+ \frac{\alpha \, \gamma}{2} \sum_{\{j {\ |\ }n \in e_j\}} c_j \, \|x_n - z_j\|^{\gamma - 2} \, (x_{m \, n} - z_{m \, j})
+ \underbrace{\beta \, \delta \sum_{\substack{i = 1 \\ i \neq n}}^k \frac{w_i \, w_n}{\|x_i - x_n\|^{\delta + 2}} \, (x_{m \, i} - x_{m \, n})}_{\text{repelling charges}}.\end{gathered}$$ Note that simplifies considerably if we pick $\gamma = 2$, as $\sum_{i \in e_j} \|x_i - z_j\|^{2 - 2} \, (x_{m \, i} - z_{m \, j}) = (\sum_{i \in e_j} x_{m \, i}) - |e_j| \, z_{m \, j} = 0$, which makes the first term disappear, and ensures that $f \in C^{\infty}(U, [0, \infty[)$ is smooth. This motivates us to pick $\gamma = 2$.

Calculating the repelling-charges part of for 