Are fields ignored in conservation of 4-momentum problems?

[SiennaTheGr8]

When we encounter particle-collision problems that call for invoking the conservation of four-momentum, are we tacitly assuming a field-free idealization (or at least negligible potential energy)?

For example, say particles 1 and 2 collide elastically. Then the conservation of four-momentum says:
$$\mathbf{P}_{1,i} + \mathbf{P}_{2,i} = \mathbf{P}_{1,f}+ \mathbf{P}_{2,f}$$ (where $i$ means initial and $f$ means final).

But in reality, there's potential energy associated with the (changing) relative positions of the particles, isn't there? So to express the full picture, would we add $\mathbf{P}_{\textrm{field},i}$ to the left side and $\mathbf{P}_{\textrm{field},f}$ to the right side?

 

[robphy]

We do something similar in nonrelativistic physics.

 

[ChrisVer]

But in reality, there's potential energy associated with the (changing) relative positions of the particles, isn't there?

Well, you'd better ask yourself what would happen if you consider the momentum of the particles 1,2,3,4 pretty "far-away" , that is final corresponding to $t \rightarrow \infty$ and initial to $t \rightarrow - \infty$ (or you can see infinity as 'very large').
As long as no new particles as asymptotic states are produced by the interaction of 1,2 to 3,4 the momenta of initial and final should be equal by conservation of energy/momentum... no matter what happened inbetween, since anything that happens inbetween is going to conserve the momentum..

 

[Khashishi]

You do need to include the energy+momentum of the fields. A bound positron and electron (positronium) has less energy than a free positron and electron. That extra energy has to come from somewhere!

 

[ChrisVer]

You do need to include the energy+momentum of the fields. A bound positron and electron (positronium) has less energy than a free positron and electron. That extra energy has to come from somewhere!

the bound state of electron and positron [positronium] is again giving you some photons... and the result is again to take: Let's say you have this process:
$e^- e^+ \rightarrow P(^1S_0) \rightarrow \gamma \gamma$
again you can use $p_{e-} + p_{e+} = p_{\gamma} + p_{\gamma}$... as if you forget what happened at the intermediate step.