Griffiths' argument that Newton's third law is invalid in special relativity

Question

I believe that Newton's third law is valid in special relativity. But I read an argument in Griffiths' textbook Introduction to Electrodynamics (3rd edition), where he argued on page 517, that Newton's third law is violated:

Indeed, if the two objects in question are separated in space, the third law is incompatible with the relativity of simultaneity. For suppose the force of $\mathbf{A}$ on $\mathbf{B}$ at some instant t is $\mathbf{F}(t)$, and the force of $\mathbf{B}$ on $\mathbf{A}$ at the same instant is $-\mathbf{F}(t)$; then the third law applies, in this reference frame. But a moving observer will report that these equal and opposite forces occurred at different times; in his system, therefore, the third law is violated.

Currently I could not find a argument against his. I found several posts here in Physics SE arguing that Newton's third law is equivalent to conservation of momentum, which is correct in special relativity. However, their reasons do not seem to invalidate Griffiths' argument. I would like to know where did Griffths' went wrong in his argument.

---

Update: I realize that there are several similar posts:

Is Einstein's Special Relativity completely inclusive of Newton's 3 laws of motion?

Violation of Newton's 3rd law and momentum conservation

But both of them centered on the momentum conservation, whereas what I want to ask is about the action-reaction view of Newton's third law.

Answer 1

Griffith's argument is correct, when we try to conceive one particle acting instantaneously on another. This is idea pretty explicitly disallowed in special relativity -- if we can't even have a force acting over a distance, then surely we can't have Newton's third law in the same way!

But, as some have noted, momentum is conserved, and this should be equivalent to the opposition of force in some way. The best context to understand this is in electromagnetism. If you've seen any relativistic electromagnetism, you know that when the charges are moving, the electric field changes with time as well, going out radially at the speed of light. Only when that wave (lit. photons) gets to the other particles does it do anything. Well, once we start to ask why all the charged particles are moving, we'll indeed find that some of them move in ways that appear to contradict the conversation of momentum. But it turns out, those ripples of electric field also contain momentum themselves.

If particle $A$ starts to move at time $t=1$, (from whatever other forces: other EM waves, or ropes, or whatever), the electromagnetic wave that gives off sort of contains all the "correction" momentum. If some particle $B$ is being accelerated at $t=2$ by the field $A$ produced at $t=0$, it's "out of date": $B$ is getting accelerated the "wrong way". But this is okay, because all that missing momentum is in the wave. Then the wave will reach $B$, impart some of the momentum on it (the rest will continue flying out to infinity), and most of the difference will be made up. The "equal and opposite" force terms come from $B$ was acting on $A$.

Obviously when the particles move much slower than the speed of light, the fields stay "correct" and we get normal Newtonian mechanics. But at relativistic rates, the momentum that is radiated transiently through the field -- and that is radiated out to infinity -- can be pretty significant.

So, how do we state the "equal and opposite force" idea, if we can't have simultaneous forces at different locations? Well, there is an equal and opposite force, but it's the particle acting on the EM field, and the EM field acting on the particle. As the particle moves around, it gives the EM field at that point momentum -- which is exactly what applying a force is! Similarly, the EM field will act on a particle, giving it momentum. In this way, the interaction is completely local, and self-consistent. The quantum perspective twist on things is to now consider the EM field being built of photons, and each interaction with a charged particle is an emission or absorption of a photon. These interactions are instantaneous, so the "force" is infinite and over 0 time, but the idea still holds that there is an equal and opposite instantaneous transference of momentum.

Answer 2

From the question,

I found several posts here in Phy.SE arguing that Newton's Third Law is equivalent to Conservation of Momentum, which is correct in Special Relativity.

Newton's third law and conservation of momentum imply one another in a Newtonian setting. That Newtonian setting includes forces only arising in pairs (but not triples, etc.), forces being communicated instantaneously ( regardless of distance), and universality of simultaneity. The latter two are problematic in a relativistic setting. I Conservation of momentum plus the assumption that forces always come in pairs implies Newton's third law in a Newtonian context.

Newton's third law and conservation of momentum do not imply Newton third law in a non-Newtonian context. Forces are not communicated instantaneously, regardless of distance, and simultaneity is relative.

From a comment,

But equation (12.60) of the same book says: $\mathbf F= \frac{d\mathbf P} {dt}$. If the relativistic momentum is conserved, then an application of (12.60) on a closed system tells that the total force is zero, hence the third law of Newton.

No, it doesn't. You are implicitly assuming a number of things here. In addition to universality of simultaneity and instantaneous propagation, you are assuming that $\mathbf F = \gamma m \mathbf a$. That is not the case. Since force is not transmitted instantaneously, a field (e.g., electromagnetic field) must itself contain momentum (and it does).