How is Newton's third law not valid in special relativity while momentum is still conserved? I'm assuming questions about the third law being invalid in relativity have already been asked on this site, but I'm specifically asking about how momentum is still conserved inspite of it.
Consider only two particles in the universe. In some reference frame, they're moving with velocities $v_1(t)$ and $v_2(t)$ at time $t$.
The relativistic momentum is:
$$\gamma (v_1(t))m_1v_1(t)+\gamma (v_2(t))m_2v_2(t)$$
The relativistic momentum after a small time $dt$ is:
$$\gamma (v_1(t+dt))m_1v_1(t+dt)+\gamma (v_2(t+dt))m_2v_2(t+dt)$$
If both these quantities are same, we can set them equal. After setting them equal, we get:
$$\gamma (v_1(t+dt))m_1v_1(t+dt)-\gamma (v_1(t))m_1v_1(t)=-(\gamma (v_2(t+dt))m_2v_2(t+dt)-\gamma (v_2(t))m_2v_2(t))$$
Dividing both sides by $dt$ and letting $dt\rightarrow 0$, we get:
$$\frac{d(\gamma(v_1(t))m_1v_1(t))}{dt}=-\frac{d(\gamma(v_2(t))m_2v_2(t))}{dt}$$
$$F_{12}=-F_{21}$$
So what's wrong with the above?
Edit- That other question is about General Relativity and the answers do not address the computations I've provided here. I want to know what's wrong with this specifically
Edit- After reading the other links, what I got is that the momentum conservation, as I've stated stated it in this post, is incorrect. This is because I did not account for the momentum of the field. So that means that Newton's third law is also correct if we extend the notion of force from particle-particle interactions to particle-field interactions. Is this correct? And in what sense do fields carry momentum? What is the mass and velocity of fields?
 A: Good question, and the answer is that Newton's third law remains valid in Special Relativity as long as one applies it the right way, and that means it has to be applied locally at each event where forces are acting, not non-locally by comparing a force
at some location $A$ with another force at some other location $B$. The third law applies to forces acting in a pair at any one location, say $A$.
When a force acts at the boundary between solid objects, this is straightforward. Each object pushes on the other.
When a force acts throughout a solid, one can analyse it the same way; for the details you need to invoke the concept of pressure and/or tension and stress. This is done in full via the stress-energy tensor.
The case where people say the third law breaks down is for example when charged objects attract or repel one another at a distance. It is true that in such cases the force on one object is not necessarily of equal size and opposite direction to the force on the other object. But one should ask: how is the force arising? It arises by an interaction between the charge on any given body and the electromagnetic field right there at the body. If the force causes the body to accelerate, for example, then, by conservation of momentum, one must find that momentum is moving out of the field and into the accelerating body. Force is, by definition, rate of change of momentum. One concludes that there is a pair of forces: one acting on the charged body, and the other, equal and opposite, acting on the electromagnetic field. These forces cause momentum to go into the charged body, and an equal and opposite momentum to go into the electromagnetic field. They are both present at the same location. They are equal and opposite.
It might seem odd to think of a force acting on a field, but the stress energy tensor makes no distinction between matter and field. Anything that can carry momentum can in principle have a force act on it.
P.S. The calculation offered in the original question is ok if the two particles are colliding at a single event, but if they are at different places and the momenta are changing with time (owing to the force from a field, for example) then you can't assume each particle's momentum will change by the same amount during some small time. In this case the sum of the two particles' momenta is not constant, because they are interacting with a third party, namely the field.
