special relativity - Relativistic 2-body problem with instantaneous Colomb force

Setup

Assume that we have two (isolated) massive and electrically charged particles, moving in the vacuum space, that interact through the instantaneous Coulomb force. I am an inertial observer and I want to build the equations of motion for each particle. Let ${{\mathbf{r}}_{i}}\left( t \right)$ be trajectory* of the particle with rest mass ${{m}_{i}}$ and electric charge ${{q}_{i}}$, and ${{\mathbf{F}}_{i}}\left( t \right)$ the force applied to that particle. Then, in my reference frame, the equations of motion for each particle would be: $$\tfrac{d}{dt}{{\mathbf{p}}_{i}}\left( t \right)={{\mathbf{F}}_{i}}\left( t \right)$$ where ${{\mathbf{p}}_{i}}\left( t \right)\equiv {{m}_{i}}{{\gamma }_{i}}\left( t \right){{\mathbf{\dot{r}}}_{i}}\left( t \right)$, ${{\gamma }_{i}}\left( t \right)\equiv \gamma \left( {{v}_{i}}\left( t \right) \right)$ and ${{v}_{i}}\left( t \right)\equiv \left\| {{{\mathbf{\dot{r}}}}_{i}}\left( t \right) \right\|$.

${*}$ as it is observed in my reference frame, where $t$ is the reading of my clock

The question

Let, $\mathbf{r}\left( t \right)\equiv {{\mathbf{r}}_{1}}\left( t \right)-{{\mathbf{r}}_{2}}\left( t \right)$ and $K\equiv {{{q}_{1}}{{q}_{2}}}/{\left( 4\pi {{\varepsilon }_{0}} \right)}\;$. Which expression should I use for ${{\mathbf{F}}_{i}}\left( t \right)$? Possible candidates are:

1. the Coulomb force between two particles that are ${r}\left( t \right)$ far apart from each other, i.e.: $${{\mathbf{F}}_{1}}\left( t \right)=K\frac{1}{{{r}^{2}}\left( t \right)}\mathbf{\hat{r}}\left( t \right)=-{{\mathbf{F}}_{2}}\left( t \right)$$


2. the Coulomb force that takes into account the phenomenon of length contraction, i.e.: $${{\mathbf{F}}_{1}}\left( t \right)=K\frac{1}{{{\left( r\left( t \right){{\gamma }_{1}}\left( t \right) \right)}^{2}}}\mathbf{\hat{r}}\left( t \right)\quad ,\quad {{\mathbf{F}}_{2}}\left( t \right)=-K\frac{1}{{{\left( r\left( t \right){{\gamma }_{2}}\left( t \right) \right)}^{2}}}\mathbf{\hat{r}}\left( t \right)$$

If non of the above is true, which would be the correct expression? Let it be noted that we disregard any retardation effects.

My intuition

I would say with certainty that it is case 2., unless I had noticed the following: $${{\mathbf{F}}_{1}}\left( t \right)+{{\mathbf{F}}_{2}}\left( t \right)=K\frac{1}{{{r}^{2}}\left( t \right)}\left( \frac{1}{\gamma _{1}^{2}\left( t \right)}-\frac{1}{\gamma _{2}^{2}\left( t \right)} \right)\mathbf{\hat{r}}\left( t \right)=K\frac{1}{{{c}^{2}}{{r}^{2}}\left( t \right)}\left( v_{2}^{2}\left( t \right)-v_{1}^{2}\left( t \right) \right)\mathbf{\hat{r}}\left( t \right)$$ which, in general, implies that $\tfrac{d}{dt}\left( {{\mathbf{p}}_{1}}\left( t \right)+{{\mathbf{p}}_{2}}\left( t \right) \right)\ne 0$, unless $v_{1}^{2}\left( t \right)=v_{2}^{2}\left( t \right)$ for any $t\in \mathbb{R}$.

More questions

If my intuition is correct, then why, in general, is the total relativistic momentum not conserved? Is it because we neglected the retardation effects, and treated the problem semi-relativistically. Or is there a logical fallacy in the above considerations?