As I see it, the reason why newtons third law works is not because coulombs formula tells us, since that is an empirical formula that of course matches our observations: forces between 2 charges are equal and opposite. The reason why it is how it is, is because Noethers theorem, because of symmetry.

@user929304 please refer to my comment on SaudiBombsYemen's answer.

This "answer" doesnt explain why Newtons third law is true. You just ate 150 years of physics history.

@SaudiBombsYemen Thanks for your comment. I am not not acknowledging that Noether's theorem tells us that translational symmetry implies conserved momentum. Indeed, it can be used to explain why in homogeneous space we have a conservation of momentum, and hence Newton's laws of motion. However, it is not an equivalent statement to Newton's third law, for Newton's third law does not imply Noether's theorem. (Edit: I just realised that you have suggested Newton's third law implies Noether's theorem in one of your comments. My I ask where in Noether's theorem's derivation do we use N3L?)

@John I haven't suggested Newton's third law implies Noether's theorem, but the reciprocal. And as I said 100 times symmetry is also required for third law to work.

@SaudiBombsYemen Ah, we finally agree on something! Consider now a particle moving in a well. Roughly speaking, this is just SHM. Clearly there is no conservation of momentum. But N3L still holds. (Edit: Why am I stating this example? Because it shows that without symmetry, Newton's laws still hold)

@John there is conservation of momentum if you took into account the particles creating the potential. I'm curious to know where do you see a single system particle in simple harmonic motion obeys third law. What particle is suffering the reaction? And who is even causing the action? If you assume momentum is not conserved then $$\frac{dp_t}{dt}=F_t = F_{action}+F_{reaction}\neq 0 \implies F_{action}\neq F_{reaction}$$ No third law.

@SaudiBombsYemen I have extended my answer, which includes a clean example that should demonstrate Newton's third law does not require symmetry. I think that much of our disagreement arises because you like to think of a potential as being generated by some particle, but if you are thinking of Noether's theorem, you are really using the Lagrangian formalism which can just about contain any potential term. (Of course, whether or not this is physical is another issue; but I am more of a theoretical guy.)

I dont see how 3rd law holds in a particle in a well, where reaction being applied?

I think what reconcile our ideas is the notion of an infinitesimal translation

which will generate a symmetry so long as we take the translation to be infinitesimally small

(unless the potential term contains an infinity)

the particle is attracted by gravity, and the well responds to it but providing a normal upward force

by providing*

But newton's third law speaks about forces between particles, here you are just considering more abstract constructions like fields

And also I would like to say that a certain amount of symmetry is required, for instance, if potential depended on position rather than relative distance particles at different positions would exert different forces on each other, resulting in action not equaling reaction

Do you agree with my example though? The particle sitting (still) in a well.

(well there is I suppose, the gravitational field. But that is also in the realm of Newtonian dynamics?)

I don't agree where you say the field responds with another force, it is like directly taking third law as an axiom

Yes but newtonian dynamics speaks about particles exerting forces on each other, you require 2 particles to have an action and a reaction

In which case momentum will always conserve

I guess I was assuming Newton's laws hold. But that's because I wanted to give an example where Newton's laws can be applied but there's no symmetry.

What do you think about the case I stated with no symmetry?

which exact case?

The case where potential or force depends on some sort of absolute position (which implies no symmetry), case in which third law doesnt hold

I said to imply some degree of translational symmetry is needed for third law

Precisely. That's why I think our viewpoints are reconciled if we think of infinitesimal symmetries instead.

So for example if we have two particles moving in a 1d harmonic potential V(x) = x^2, and they may occasionally collide, then we can probably imagine applying N3L as they collide.

But that's because where they collide is a point

so if you 'zoom in' enough, it's basically translationally invariant.

as soon as they finish colliding, though, they will immediately be under the influence of the potential again, and overall momentum is not conserved. But in that instantaneous collision, we may think of momentum as being conserved.

Does this make sense to you?

Why wouldn't momentum be conserved in the collision?

But really, I think the example of a particle sitting still on a table requires the third law to explain. This is in turn due to the electrostatic repulsions, in this example momentum really plays no role.

Momentum will in that collision. But immediately after, since the particles are under the influence of a potential, they will accelerate/decelerate according to the potential, so momentum will vary with time.

Ok, now I see what you mean, I still disagree with this vision of forces being generated from nowhere, you are considering just part of the system, where of course momentum, energy and pretty much nothing is conserved.

Yeah, fine. But that's why the lagrangian is used extensively in theoretical physics - you can just add any potential term into your theory without having to introduce another particle to your system. I agree that it is non-physical, but unfortunately I study theoretical physics...

As you can imagine, if you're doing theoretical physics, you really don't want to add like a field generating machine to your system...

I agree with you that physically this doesn't happen. But I hope the other example of a particle sitting still is sufficient to explain why the third law doesn't need symmetries?

I also think it wouldn't be that easy to prove momentum wouldnt be conserved in the case you stated since momentum lost between the 2 particles could be compensated with the varying force (momentum change)

But in general you do need some sort of symmetry

Not for trivial cases but for more elaborate ones you do

Yeah I agree, since otherwise momentum wouldn't be conserved.

but the key quantity is infinitesimal symmetries, not global symmetries.

but certainly if something fails for trivial cases then it isn't true in general

infinitesimal symmetries imply global symmetries though

nope...that's not true.

Cant you construct global symmetries by infinitesimal changes?

not if the quantity in question changes with position...

more precisely we don't even have infinitesimal symmetries

we only have a symmetry when the translation goes to zero

but if we do?

if we do then as long as the shift that we do isn't position dependent then I think so

Yeah, see that's the thing with Newtonian mechanics

we assume pointlike interactions

which is why the third law will hold, 'coz obviously across a point we have translational symmetry

it's not even an infinitesimal translation

it's infinitely smaller than that

our discussion has again generated new insights!

I have got to go now. I realised we had a very long discussion in the comments, which isn't very appropriate according to the forum rules. I think what I'll do is delete my comments since they're probably not very helpful for other readers. Maybe you can do the same too. However, do feel free to chat about this further or leave fresh comments! Nice talking physics with you :)

I agree, nice discussion!