Electromagnetic linear momentum for a system of two moving charges

angrystudent
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Homework Statement
I want to obtain the "total" EM linear momentum for a system of two moving charges
Relevant Equations
Lienard-Wiechert potentials, Lorentz force law
When you write out the equations of motion for a system of two isolated charges, you can add both of the equations and get the increase in the particles linear momentum on one side. On the other side, you get the sum of all the forces between the particles. I understand that this sum of forces could be written as the negative time derivative of the system's total electromagnetic momentum. But since the forces are quite complicated, I just can't seem to deduce the expression of this EM momentum

I guess that you could write the ##\mathbf{E}## and ##\mathbf{B}## fields due to both charges and integrate the momentum density ##\mathbf{D}\times\mathbf{B}## over all space to get the desired momentum, but I'm not too skilled with the integration of retarded functions. I have also tried to write the forces without deriving the potentials explicitly,
since
$$\mathbf{F}_{12} = -q_2\nabla_2V_{12} -q_2\partial_t(V_{12} \mathbf{v}_1/c^2) + q_2\mathbf{v}_2 \times \nabla_2 \times(V_{12} \mathbf{v}_1/c^2)$$
and
$$\mathbf{F}_{21} = -q_1\nabla_1V_{21} -q_1\partial_t(V_{21} \mathbf{v}_2/c^2) + q_1\mathbf{v}_1 \times \nabla_1\times(V_{21} \mathbf{v}_2/c^2)$$
but that doesn't seem to give an inspiring result

tl;dr I'm trying to write the sum of ##\mathbf{F}_{12}+\mathbf{F}_{21}## as a total time derivative. ##V_{12}## and ##V_{21}## refer to the Lienard-Wiechert potential
 
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I just realized that electromagnetic radiation contradicts my statement. As soon as the charges get accelerated, then the sum of linear momentum plus electromagnetic momentum is not conserved anymore because the momentum flux is non-zero at infinity. My initial approach doesn't make sense due to radiation phenomena This means that the Griffiths book chapter on conservation laws gets contradicted by the chapter on radiation.
 
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