$$\textbf{F}(\textbf{r}_1,\textbf{r}_2)=\textbf{F}(\textbf{r}_2,\textbf{r}_1)$$
The simplest possibilities are
$$r=|\textbf{r}_1-\textbf{r}_2|$$$$\textbf{r}=\textbf{r}_1+\textbf{r}_2$$
Alireza Jamali's Blog
$$\textbf{F}(\textbf{r}_1,\textbf{r}_2)=\textbf{F}(\textbf{r}_2,\textbf{r}_1)$$
The simplest possibilities are
$$r=|\textbf{r}_1-\textbf{r}_2|$$$$\textbf{r}=\textbf{r}_1+\textbf{r}_2$$
If we have macroscopic quantum phenomena, why not low-speed relativistic phenomena?!
In fact, the first example already exists: Graphene! which has
$$E=pv_F,$$
where \(v_F\) is the Fermi velocity, and \(p=\hbar k\). This suggests that \(v_F\) plays the role of \(c\)
$$\gamma_F=\frac{1}{\sqrt{1-(\frac{v}{v_F})^2}}.$$